UNIVERSITY OF CALIFORNIA GIFT OF Col. Glen F. PE.PRRTr«QE.1MT DOCUMENT NO. ZO35 THEORY AND DESIGN OF RECOIL SYSTEMS AND GUN CARRIAGES Prepared in the Office of the Chief of Ordnance SEPTEMBER, REPRODUCTION PUANT WA5H\NQTON BARRRCKS O.C. -rinf Lilrary U.F ORDNANCE DEPARTMENT Document No. 2035 Off ic« of the Chief of Ordnance WAR DEPARTMENT, Washington, October 1921. The following publication entitled "Theory and Design of Recoil Systems and Gun Carriages" is pub- lished for the information and guidance of all students of the Ordnance training schools, and other similar educational organizations. The contents should not be republished without Authority. By order of the Secretary of C. C. WILLIAMS, MAJOR GENERAL, CHIEF OF ORDNANCE. Jjlraiy FOREWORD This edition is published in its present form with lioeral margins and spacing so that corrections or additions may be freely made. In a document of this kind it is almost inevitable that ambiguities, errors and misstatenents will appear, and it is only in extended and repeated use that these are fully exposed. It will therefore be appreciated if all those to whom this volume comes and who use it critically will forward criticisms, corrections or necessary additions to the Artillery Division, Ordnance Office, Washington, D. C., so that these nay be incorporated in the master volume. After such changes have been received for a suitable period, it is expected to have the text printed in usual book form. PREFACE Although strictly artillery design may be considered a highly specialized branch of machine design, there are so many features that differentiate this work from ordinary machine design, it has been felt that a volume covering the specialized points is of fundamental importance in order that our designing engineers may have in a readily accessible from reference data covering the general subject and in particular those features of modern development not now covered in published works. Such is the purpose of this volume. Artillery design may be subdivided into the design of cannon, and the design of gun carriages and recoil systems. During the late war the ex- tensive introduction of self propelled gun mounts, such as caterpillar vehicles, has introduced automotive problems in the design of these types of gun mounts in addition to the ordinary consideration affecting design of gun mounts. Further in the design of artillery we have three important aspects, - (l) the technical and theoretical considerations of a design, (2) the fabrication, standardization and production features, and (3) the service and field require- ments to be fulfilled. All three aspects are equally important and a successful design results only from a balanced consideration of the three. This discussion has been written under the auspices of Colonel G. F. Jenks, Chief of the Artillery Division, Ordnance Department, U. S. A. and of Colonel J. B. Rose, Chief of the Mobile Gun Carriage Section of that Division. Effort has been made to arrange systematically in a form for reference tne great quantity of engineering data in the files of the office. In order to develop and analyze this data, it has been necessary to introduce a considerable number of original discussions and deductions. The work is an attempt to cover only the *» technical aspect of the design of gun carriages and recoil systems. The fabrication and field service phases, though of course inherently coordinate in. a design are subjects of such com- plexity and broadness that they require for their full appreciation a separate treatment. These aspects have therefore necessarily been entirely onitted, except in so far as they are directly connected with the technical features involved. Acknowledgement and thanks are especially due to Colonel J. B. Rose, who has proof read the complete work in the view of bringing the data into conformity with the practice and standards of the Ordnance Department. It should be stated, however, that this has been done only to the degree which was found possible without destroying original con- clusions and discussions or without alteration of the system of nomenclature used. The latter is in partial but not complete agreement with the most general practice. Further acknowledgement and thanks for suggestions on the various parts of the work are due to: - Mr. D. A. Gurney, Ordnance Engineer, Mobile Gun Carriage Section, Artillery Division. Prof. B. V. Huntington, Professor of Mathematics and Mechanics, Harvard University Professor C. E. Fuller, Professor of Ap- plied Theoretical Mechanics, Massachusetts Institute of Technology. Professor G. Lanza, Professor Emeritus in charge of Mechanical Engineering Department, Massachusetts Institute of Technology. Acknowledgement of assistance on the Computation work is due to Mr. E. V. B. Thomas, Mr. Kasargian and Mr. McVey of the Artillery Division, also to Messrs. Murray H. Resni Coff and 0. L. Garver for preparing this data for publication. RUPBN BKSERGIAN, Formerly Captain, Ordnance Dept.U.S.A, Chapter I Chapter II Chapter III Chapter IV Chapter V Chapter VI Chapter VII Chapter VIII Chapter IX Introduction and Elements - Types of Cannon and Carriages - Classifi- cation of Carriages and Recoil Systems. Dynamics of Interior Ballistics as Af- fecting Recoil Design - The Para- bolic Trajectory. External Reaction on Carriage during Recoil - Stability - Jump. Internal Reactions throughout a Carriage during Recoil and Counter Recoil. Gun Hydraulic Principles as Applied to Various Systems of Recoil and Counter Recoil - General Theories on Orifices and Flow of Oil. The Dynamics of Recoil and Counter Re- coil - Differential Equations of Resistance, Braking, etc. and of Velocity - General "Formulas for Recoil and Counter Recoil, Classification of Recoil Systems - Derivation of General Formulas for Design and Computation - General Linitations, etc. Hydro-pneumatic Recoil Systems. Hydro-pneumatic Recoil Systems (Continued) 8 Chapter X Railway Gun Carriages. Chapter XI Gun Lift Carriages. Chapter XII Double Recoil Systems. Chapter XIII Miscellaneous Problems — Discussions of Various Types of Carriages. CHAPTER I TYPES AND PRINCIPAL ELEMENTS OF CANNON AND CARRIAGES. The fundamental principles of gun carriage design are entirely the same as those of engine and machine design, and it is the object of this volume merely to bring out the specific ap- plication of these principles to the design of gun carriages . A gun carriage is a machine exercising primarily the following functions: (1) To provide a fixed firing platform which dissipates the energy given to the recoiling parts in reaction to the energy imparted to the projectile and powder gases. (2) To return the recoiling parts to their initial position for further firing. (3) To provide the mechanism for elevat- ing the gun for different ranges and angles of site, and for traversing the gun for changes in the direction of fire. The effect of allowing the gun and a part of the carriage to recoil is to reduce many times the stresses in the carriage and to maintain its equilibrium. A properly designed recoil system will give reactions consistent with the strength and stability of the carriage, and a smoothness of action which is essent- ial for long service and accuracy. The success of one design over another is due to perfection of many de- tails, which insures smooth action and long service and to a judicious compromise between many opposing conditions and requirements. To approach the study of carriage design, it is necessary to know the elements of interior ballistics and the characteristics of guns for meeting different 9 10 ballistic conditions in so far as these affect the form of carriage and determine the forces acting upon it. These subjects will, therefore, be briefly considered, but a complete discussion must be obtained from works treating them specifically. From one view point a cannon may be considered as a tube of proper thickness for strength, having a chamber in the rear of somewhat larg- er diameter which contains the powder charge. The powder charge is inserted by opening a breech block in the rear end of the cannon. This breech block necessarily must withstand the maximum powder pressure over its cross section and a power- ful locking device is therefore needed. The details of this mechanism are complicated, but need not be considered in carriage design, except in special cases where the breech mechanism is operated during counter recoil. The design of the rifling grooves and capacity of powder chamber will be considered later. The elements of a gun are shown in figure (1). "A" is the powder chamber, "B" the rifled portion of the bore, "C" the breech block, "D" the gun lug for the attachment of piston rods, which restrain the gun in recoil. — * — V- il \ . -L fV ^ The caliber of a gun is the diameter of the bore and is expressed usually in millimeters or inches. Speaking very roughly, small guns range from 37 m/m 11 to 75 m/m, and are suitable for mounting on aeroplanes or for use with infantry. Light field guns range from 75 m/m to 105 m/m. Ordinary medium artillery ranges from 105 m/m to 8 inches* Heavy artillery ranges from 8 inches upward. The above classification refers to mobile field materiel only. PQWTT7FRS AND flUNS Carriages are designed for either howitzers or guns. Howitzers are for high angle fire, the striking angle being generally above 25 de- grees. They have a medium or low muzzle velocity. A gun is designed for range and, there- fore, has a high muzzle velocity. The angle of elevation of howitzers is usually between 20"* and 70°and the muzzle velocities from 400 to 1800 feet per second. The angle of elevation of a gun is usually from minus 5 degrees to plus 45 degrees with muzzle velocities ranging from 1700 to 3000 feet per second. The angular velocity of the projectile is also considerably higher for a gun than for a howitzer. In modern practice the line of demarcation between guns, howitzers and mortars has become somewhat less distinct, and we may consider all of them as cannon which decrease in power in the order named and generally for use at elevations which increase ia the order named. Against aircraft, firing is at elevations from 0 to 80 degrees, hut the muzzle velocity is high; hence, the pieces used in such work are properly classified as "guns". The traversing limitations of a gun and howitzer may be the same or different but do not enter in the differentiation between a gun and howitzer. The muzzle velocity of howitzers being lower than that of guns, it is possible with the same total weight of materiel to fire a much heavier projectile. 12 BECOIU BG_PABT-g The recoiling parts consist of the gun together with the various parts attached to it and recoiling with it. We have two methods of arrangement of recoiling parts: (1) the piston rods with their pistons attached to the gun lug and recoiling with the gua. (2) the pistons and their rods held stationary. So far as the recoil mechanism is concerned we are only concerned with the relative motion between the rods and pistons and their cylinders. The greater part of our guns in the service translate in recoil directly along the axis of the bore, others as on certain Barbette mounts and double recoil systems have a translation in addition to that along the axis of the bore. Guns on Dis- appearing carriages and special other types have rotation in addition to translation. In ordinary recoil systems the center of gravity of the recoiling parts is usually located slightly below the axis of the bore. This insures a positive jump (muzzle up) during the powder pressure period. If the center of gravity of the recoiling parts is great- ly below the axis of the bore considerable stresses are brought upon the elevating rack and pinion, due to the fact that the powder pressure causes an ex- cessive turning effect about the trunnions the amount depending also upon the location of these. For this reason when the cylinders recoil with the gun, extra weight is very often introduced on the top of the gun. This, of course, raises the center of gravity of the recoiling parts nearer the axis of the bore. The recoiling parts are constrained to recoil parallel with the axis of the bore by gun clips en- gaging in guides in a fixed cradle or by the gun itself sliding in a fixed cylindrical sleeve. Due to the fact that the braking forces developed in the 13 cylinders are usually considerably below the axis of the bore during recoil, considerable pinching action takes place at the front and rear clip contact with the guides. This causes somewhat greater friction than would be obtained by mere sliding friction. The clips attached to the recoiling parts, or rather to the gun itself, which in turn engage in the guides of the cradle, are usually either continuous or three to four in number. In order to maintain a con- stant friction throughout the recoil, clips should be evenly spaced along the gun and the front clip should engage in the guides before the rear clip leaves the guides. When the gun recoils in a sleeve or cy- linder which is a part of the cradle, it is conetiroes possible to distribute the various pistons and cy- linders symmetrically about the axis of the bore. As we shall see, this decreases the friction during the recoil and counter recoil. Figure (2) shows the recoiling mass where the pistons and their rods recoil with the gun. Below in figure (3) is shown a recoiling mass consisting of the cylinders grouped together in a single forging in a so-called slide or sleigh, and rigidly attached to the gun. 14 . 3 nt£_CRADLE_ The cradle serves as a constrain- ing member for the sliding of tlie gun to the rear in recoil and as a support for elevating the gun. The cradle and the recoiling parts together are known as the tipping parts and turn about horizontal trun- nions fixed to the cradle and resting in bearings in the top carriage. To elevate the tipping parts, an elevating arc bolted to the cradle engages in a pinion fixed to the top carriage, or vice versa. The cradle and therefore the tipping parts are supported in the top carriage at two points: (1) at the trunnions and (2) at the tooth contact of the elevating arc and pinion. See figure (4). IS ng. 4- When the cylinders do not recoil they are in turn an integral part of the cradle, and therefore, the recuperator forgings and the cradle are one and the same. A sleigh may or may not he interposed between the gun and cradle. With guns, where the cylinders recoil with gun, the cradle merely serves the purpose of a constraining guide for the recoiling parts and rigid- ly attached to it are the piston rods and their pis- tons . TJPPIflG_FABT§ The term"tipping parts" applies to those parts of a carriage which move in the process of elevating the gun. In order to rapidly elevate the gun, it is considered very important that the tipping parts are nicely balanced about the trunnions. Thus the center of gravity of the tipping parts must be located at the trunnions. As the height of the trunnions and axis of the bore are governed by stability at horizontal elevation, clearance in traveling and accessibility for leading, the length of recoil at maximum elevation becomes limited. If a minimum elevation of about 20 degrees is allowed for a howitzer, we might raise the trunnions, thereby in- 16 crease the length of recoil, and thus maintain stability. When, however, a gun must fire at high ele- vation as in antiaircraft materiel, or when a carriage serves the double purpose of supporting a gun or a howitzer at high elevations, the maximum possible re- coil at maximum elevation becomes greatly limited. The recoil displacement at maximum elevation may be increased most satisfactorily by placing the trunnions to the rear and introducing a balancing gear for balancing tbe tipping parts about the trunnions. The balancing gear usually consists of an os- cillating spring or pneumatic cylinder, the trunnions of which rest in bearings in the top carriage, the end of the piston rod being attached to the cradle. Since it is difficult to obtain perfect balance by this method throughout the elevation, the maximum unbalanced moment in the process of elevation should be considered in the design of the elevating gear mechanism. A method by which exact balance can be maintained throughout the elevation is obtained by use of a cam and chain connecting tbe cradle with the spring or pneumatic cylinder. In this case the cam is fixed to the cradle and the spring cylinder to the top carriage. However, due to the variation in trunnion friction and other similar factors the former method is probably better since a very close approximation in balance throughout the elevation can be obtained. The reaction on the elevating arc and the trunnion reaction are modified by the introduction of tbe balancing gear, though ordinarily where the weight of tipping parts is relatively small as com- pared with the recoil reaction the effect of the balancing gear on the reactions may be neglected. When it is desired to use an independent line of sight, a rocker is introduced between the elevating pinion and cradle. The rocker, when moving, is a part of the tipping parts. In the process of 17 elevating the gun an elevating pinion rotates the rocker about the trunnions until the proper line of sight is obtained; the cradle is then brought into its proper position by gearing con- necting the rocker and cradle. TOP CARRIAGE The top carriage serves as an. intermediary piece connecting the tipping parts with the bottom car- riage, or in semi-fixed mounts, with the bottom platform. The top car- riage is supported at its bottom by a vertical pintle block and circular traversing clips. At the top it supports the tipping parts on its trun- nion bearings and elevating pinion bearing. The top carriage -together with the tipping parts are known as the traversing parts. To traverse the gun, the top carriage with the tipping parts are rotated in a horizontal plane about the pintle block by a circular traversing reck and pinion or worm gear. In certain types of field artillery the top carriage is an integral part of the trail, in which case traversing is obtained with respect to the wheels and axle by moving the trail along the axle and about the spade point as a pivot. Traverse by this method is naturally very limited as com- pared to traverse with a rotating top carriage. All stationary mounts or field platform mounts have a separate top carriage which serves this specific function of traversing about the vertical pintle support. In very large carriages the top carriage is supported by a circular ring of horizontal rollers, the pintle bearing merely serving as a constraining pivot. In certain types where the bottom carriage itself is traversed, the top carriage is used for translation only. It is then supported on rollers moving along an inclined or horizontal plane and the braking is affected by a recoil cylinder in the top carriage which 18 connects the top carriage with the bottom carriage through the piston rods. Fig. 5 Top carriages way be roughly classified into: (1) the ordinary type of side frames connected at the front or rear by cross beams or transoms which contain the pivot bearing. (2) pivot yoke type used on small mobile mounts and (3) trail carriages. The ordinary side frame type of top carriage is extensively used on the stationary mounts on mobile platform mounts and even on trail supported carriages. The pivot yoke type is especially useful when split trails are introduced, since it supports the equalizer bar for balancing the distribution of the load between the two trails. 19 TRAIL AND SPADE With mobile field artillery it is customary to use a trail and spade for the double purpose of preventing a backward motion of the carriage on firing of tbe gun, and of giving sufficient stability to the carriage in order that the wheels may not leave the ground. We have two classifications of trails, - (1) the single or box trail, (2) the split trail. With a single trail it is necessary to have a large U- shaped aperture or fork arrangement at the forward end in order to elevate, load and traverse the gun without interference. When split trails are used we have two separate single trails which may turn at the wheel ends about the axle. It is customary with the split trails to introduce an equalizing mechanism which connects the two trails and distributes the load between the trails on firing. The spade and float support tbe trail and are designed to take up the horizontal and vertical re- actions at the rear end. In tbe design of the spade and floats it is important that tbe unit bear- ing pressure be held to a low value. This should not be more than about 30 Ib. per sq. in. for the float and 40 Ib. per sq. in. for the spade. For wide traverse of the gun it is necessary to lift the spade from the ground and turn the carriage to the desired line of fire. For this reason the static load on the spades should not ex- ceed about 100 Ibs. for light carriages. Thus in a preliminary lay-out of tbe carriage, it is necessary to locate the center of gravity of tbe total system in battery very close to the axle in order that the static load under the float does not exceed the desired amount. This inherently makes the counter recoil stability in battery very small especially at horizontal recoil and requires con- siderable care in the design of a counter recoil 20 system. At horizontal elevation the carriage is usually designed with a very small margin of stability. Therefore, in firing the vertical load on the float practically equals the weight of the total system. The bending moment in the trail gradually increases from the spade toward the wheel axle. We have max- imum bending moment at the attachment of the trail to the top carriage or wheel axle. PLATFORM MOUNTS. With fixed mounts and heavier types of field artillery it is customary to support the travers- ing parts on a platform, that is, the top carriage rests upon a platform which serves as a bottom carriage. When a platform bottom carriage is used, it must be either bolted to a concrete foundation as in fixed mounts or else it must have a vertical projection similar to a spade on a field carriage to take up the horizontal reaction in firing. Further, the bearing surfaces of this platform must be suf- ficient to prevent overturning of the carriage firing at low angles of elevation or change in level in firing at any elevation. That is, the center of pressure of the reaction of the earth must be within the middle third of the length or diameter of the platform in the line of fire. Since platform mounts vary considerably in con- struction of detail no attempt will be made to catalogue the various types used. With fixed mounts the bottom carriage or plat- form is usually secured to a concrete foundation by a distribution of bolts along a circular flange; and since with fixed mounts all round traverse is possible, each bolt should be designed for maximum tension. 21 DC 22 CATERPILLAR MOUNTS. To increase mobility during the World War, cater- pillar mounts were developed extensively. A caterpillar mount consists of an ordinary gun mount including the tipping parts and top car- riage mounted on a bottom carriage which fits with- in the frame of the caterpillar. The caterpillar is propelled by its own engine, and traverse can be readily made by keeping one of the caterpillar tracks stationary and moving the other. For more delicate traversing the top carriage is provided with limited traverse about the bottom carriage. The essential features of the caterpillar proper are: (1) The frame which supports the bottom carriage and the principal bearings for the driving mechanism. The caterpillar frame in turn is generally supported on a series of roller trucks which travel on the caterpillar tracks. Between the roller trucks and caterpillar frame, spring supports are usually provided, and the roller trucks are built to have more or less up and down movement at their ends to conform with the contour of the ground. The frame may be either a casting or built up of structural steel. The structural steel frame is perhaps lighter but more subject to objectional de- flections . .,» The reactions on the frame consist of the various spring supports from the supporting roller trucks, the reactions of the bearings of the running gear and the reactions of the gun mount transmitted by the bottom carriage to the frame on firing. 23 The frame of a caterpillar is subjected to a complicated system of stresses. Due to various possible loading conditions during traveling such as the entire weight of the caterpillar being carried in the center or else at the ends, we have different types of loading reactions. Further a wrenching action with corresponding large transverse, stresses are induced by the supporting reactions being on either side at the further extremities of the track. This requires considerable lateral bracing. In fact outside of fabrication and con- struction considerations, the design of the cater- pillar frame should be based on a careful analysis of the various types of supporting reaction com- binations that may take place in the traveling of the caterpillar. It will be usually found that the traveling stresses are somewhat greater than the firing stresses and are often of an opposite character. The driving mechanism of the caterpillar con- sists of two tracks each consisting of a continuous track or belt of linked shoes. The caterpillar track is driven, by sprockets usually at the rear end. The drive shaft contains at one end the track sprocket, and at the other end the drive sprocket gear, which meshes by a suitable gearing to a clutch, the system of gearing &nd clutch being symmetrically the same for either track. The clutches are driven by bevel gears or other forms of reduction gearing through a gear box, and sometimes a master clutch, to the engine crank shaft. The traction gearing is straight forward and is very similar to other types of drive gear transmission. Mechanical steering is obtained by operating either the right or left track, holding it stationary or sometimes reversing the motion and running the track backward. 24 Electric drive caterpillar mounts are in two (2) units and possess certain advantages'; first, the transmission can be greatly reduced in either unit by the use of compact motors a_nd gearing'; second, the units can be made similar and the mobility thereby increased; third, a better design cf gun mount is possible due to less limitations on clearance and other corresponding factors. The electric drive consists of the gun mount unit and the power plant unit. The power plant unit sup- plies power for driving itself, as well as the gun mount; fourth, the caterpillar is braked in traveling by suitable band brakes in the trans- mission. When, however, the gun is fired, it is necessary to brake the caterpillar from running back. The braking and torque being usually in an opposite direction and necessarily of a large value as compared with the traveling braking1; it is usually customary to introduce a band brake on the final drive shaft and thus eliminate the stresses in the transmission during firing. The braking should be designed to produce a traction reaction equal to approximately 80 percent of the total caterpillar. Fifth, in a design of caterpillar mounts, stability is of prime im- portance due to the limited wheel base and necessity of maintaining as light a mount as possible. Stability may be increased by the use of outriggers attached to the caterpillar body. To decrease the overturning reaction of the recoil on firing and thus increase the stability, double recoil systems have been successfully introduced on larger caterpillar guns. A double recoil system consists of an ordinary recoil system between the gun and cradle of the top carriage and a lower recoil system between the top carriage and frame. The top carriage is designed to roll up an inclined plane of sufficient elevation to bring the recoiling masses into battery and the cate-rpillar lies in a r k tf 26 horizontal plane. This elevation is usually at from 6 to 7 degrees. By the use of double re- coil systems fhe stability is greatly enhanced, since the inertia resistance of the top carriage creates a stabilizing moment which is added to the inertia resistance of the upper recoiling parts. In the design of the double recoil system cater- pillar mount, it is highly desirable that the top carriage recoil as far as possible up the inclined plane. Due to less limitations and clearance, an electric drive of the two supporting units offers a very suitable gun mount and a long recoil of the lower recoil system is usually possible. Fig. 6 RAILWAY MOUNTS Railway mounts developed during the late war consist of three (3) systems: (1) these where the car mounted on suitable tracks, rolls back on firing;(2). those sliding back on a special track the tracks being disengaged, (3) platform or stationary railway mounts with suitable outriggers, the 27 trucks being entirely disengaged. In types (1) and (2) a very limited traverse is possible, whereas, in type (3) considerable amount of traverse is possible. Railway mounts of type (1), rest upon suitable girders, supported by the trucks at either end. The girder must be designed to carry the maximum firing load stresses at maximum elevation, as well as stresses due to the dead load weights. The trucks take the supporting reactions from the girders of the dead weight load as well as fhe firing load at maximum elevation. Great care is needed in dis- tributing the loading from the various axles by properly formed truck equalizers. In type (2) a special built-up track is necessary, the trucks being disengaged merely carrying the dead weight of the mount. The mount is designed to have a considerable bearing surface, and thereby the bearing pressures are greatly reduced In sliding railway types, recoil systems have in certain types been completely eliminated, the recoil being merely resisted by the friction of the track. Due, however, to the enormous stresses due to high caliber guns at maximum elevation, recoil systems should always be introduced. With stationary or platform mounts the question of stabilizers of corresponding outriggers become a fundamental feature in this type of design. Platform railway mounts have similar characteristics as ordinary field platform mounts in mobile artillery. 28 rig. 9 Fig. 10 CHAPTER II DYNAMICS OF INTERIOR BALLISTICS AS AFFECTING RECOIL DESIGN. The object of interior ballistics is partly to derive expressions for the acceleration and velocity of the projectile during the travel in the bore, and the corresponding pressures on the base of the shell and breech in terms of tne powder loading, the form of powder grain, the initial volume of powder chamber in the gun, and other variables upon which the velocity and pressure depend. In the design of the recoil mechanism as well as the carriage for its maximum stresses, it is very important to know the accelerations, velocities, and pressures in the gun to a considerable degree of accuracy throughout the time the powder gases act. In the study of interior ballistics, it is con- venient to divide the powder pressure interval into two periods: (1) The interior period while the shot travels up the bore to the muzzle. (2) The after effect period while the powder gases expand after' the shot has left the muzzle. During the interior period, we have considerable combustion of the charge and corresponding gas evolved in the powder chamber before the shot has left its initial position in the breech end of the bore, the temperature rising and the pressure reaching a value sufficient to force the projectile into the rifling groove and to overcome initial frictions, usually a considerable fraction of the powder pres- sure obtained. The projectile then moves up the bore followed by further combustion and expansion of the gases evolved from the combustion of the powder. The combustion exceeds the expansion up 29 30 to the tine of maximum powder pressure which is reached after a travel up the bore roughly from 1/6 to 1/3 the length of the bore depending greatly on the type of cannon, charge, etc. The energy of combustion is expended: (a) In Kinetic Energy of translation ? of the projectile. (b) In Kinetic Energy of translation of the recoiling mass (assuming the recoiling mass free). (c) In the Kinetic Energy of the charge itself. (d) In the work on the rifling and in friction. (e) In the angular energy given to the projectile. (f) In dissipated heat. The last three are very small as compared with (a), (b) and (c). Further (b) and (c) are small as compared with (a). "Ingalls" states that about 83* of the total energy of the work of expansion goes into the Kinetic Energy of translation of the shot, the re- mainder 17* going into the forms b, c, d, e and f. The rate of combustion depends upon the forn and size of the grain, it being an observed fact that powder burns in layers always parallel to the initial surface. Further the rate of combustion is a function of the actual pressure generated, vary- ing as some power of the pressure. The value used for this exponent is one of the most tentative features in the whole subject of interior ballistics. DYNAMIC RELATION- Let IT = the mass of the pro- jectile. * = the weight of the SHIPS IN INTERIOR BALLISTICS. projecti le . m = the mass of the charge 31 it = the weight of the charge. mr= the mass of the recoiling parts. wr= the weight of the recoiling parts. u = the travel up the bore. x = the absolute displacement of the shot in the bore . X = the corresponding displacement of the recoiling parts . v = the absolute velocity of the shot in the bore. vo= the muzzle velocity of* the shot. V = the free velocity of the recoiling parts (absolute ). Pxj= the total pressure on the breech. P = the total pressure on the base of the shot. Pk= the intensity of pressure on the breech (Ibs. per sq . in. ) . p = the intensity of pressure on. the base of the shot (sq. in.), f = the component of the rifling reaction parallel to the axis of the bore. Then, dv P - f = m -— , for the motion of the d t, projectile dV and P^ - f = mr — , for the motion of the recoiling mass in free recoil (2) and further assuming the charge to expand in parallel laminae with the successive laminae having velocities as a linear function of the end velocities, we have, p. _ p = 5 /dv dV . D 9 V - — ) f n\ 2 dt dt (3> where dv dV dt dt = the mean acceleration 32 of the powder. Combining (l), (2) and (3) I , dV ffi x dv <*<• * — > at = (n + T ? at (4) Integrating successively, (m * JL) V ••<• + 4-)v (5) (mr * -1-) x - (» + ~)x (6) c £ The absolute displacement of the shot in the bore is connected with the travel (u) up the bore by the following relation: x = u - X since the positive value of X is assumed opposite to x. Substituting in (6), we have, i . (m + — - ) u X - 2_ (7) mr+ m •*• m which gives the relation of free recoil to the travel of the shot up the bore. Obviously (5), (6) and (7) may be written immediately from the principle of "linear momentum" (that is, the total momentum of the system remains constant unless acted on by external forces) and the principle that tha center of gravity remains fixed unless acted upon by external forces. In free recoil the exterior forces are nil. The pressure on the breech exceeds that on the shot by the inertia resistance offered by the of the powder gases, 33 >nrdV mdv Neglecting ^ as small compared with mr, 2 pb - f (m P - f hence Since the rifling reaction expecially during the movement of the shot up the bore is roughly 2 per cent or less of the value of p, we may entire ly neglect the term f . ' in the above expression, which simplifies to ft m + — Pb = - 2P (8) • From a series of experiments conducted by the United States Navy the value E m + - - 2m - = 1.12 a constant, approx. • hence pb- = 1.12 F approx. (9) It is to be noted that the acceleration of the powder is very likely somewhat different from the assumption upon which (8) was de- rived, but nevertheless equations (8) and (9) give a good approximation of the increase of breech pressure over that at the base of the projectile. During the "forcing in of the rifling" before the commencement of motion of the shot, obviously Pb = P- 34 According to the previous assumptions the pres sure varies progressively, decreasing from its maximum value at the breech block to a slightly smaller value at the base of the projectile. Therefore, if we let p^ be the average or mean instantaneous pressure or rather the pressure in the -powder chamber and bore, we have, p + p • P = -— *» 2 In terms of the total pressure at the base of the projectile, m dV dv m p , . 7T . ,21 -3T, dv m * • 2^ dt but dv hence -•_ = p=p(i + ^_) = p(1+ _L_) (10) » 4 in Aw or in terras of the total breech pressure dV "rar dV mr dt + ^»_ ~tt Pm « 5 ~ (1 m " * 2 (11) ID * 1 2 35 EQUIVALENT The riflil1g grooves in MASS OF the gun come in contact with PROJECTILE the copper rifling band on the projectile and angular motion is transmitted to the projectile in addition to the translatory motion. The object of the angular motion is to give the projectile a gyroscopic effect maintaining, with a combination of the air reaction, the axis of the projectile parallel to the tangent of the trajectory and further making an oblong projectile possible with greater ballistic efficiency. Let P = the reaction of the powder on the base of the shell. m = the mass of the projectile, f = the total rifling reaction normal to the rifling groove, uf = the friction component of the rifling reaction tangent to the rifling groove. 6 = the angle of pitch of the rifling, (i. e. the angle the rifling makes with the axis of the bore ). p = the pitch of the rifling d = the diameter of the bore k = the radius of gyration of the projectile.. x = the displacement of the projectile up the bore from its initial position. £5 = the corresponding angular displacement twist of the projectile. Then we have, P - f(sin e 4 u cos 9 ) = m 4!*- dt \i£>) f(cos e - u sin 9 ) | = mk' £f (^} 36 Further since, the number of complete turns or revolutions of the projectile in its linear displacement x or its angular displacement ft, is JL or JL p 2* i»e have da# = 2* dax (14) ar*" ~p~ "31"*"" In terms of the angle of pitch of the rifling, - 16 = x tan e or n tan e = "" hence 2 d*x T tan 6 dF~ (is) Substituting (14) or (15) in equation (13) we have * dax d*x mk* 4 "P" dT» mkMtan --- — -------- = - - (cos 9 - u sin e)d (cos 9 - u sin e)da (16 ) which shows the reaction f is always proportional to the linear acceleration of the projectile. Therefore, the friction uf , is also proportional to the linear acceleration. Substituting (16) in (12), we have or in terms of the rifling angle, .sin e * u cos e \ 4k*tan@. d*x P - I 1 ^cos 6 - u 51B b > ~li - J ffi ip 37 which shows that the powder reaction P is also directly proportional to the linear acceleration of the projectile. Evidently the equivalent mass of the projectile, is .sin e «• i i cos 6 < cos e - i i sin 6 j .sin Q + i i cos 6 > ^cos e - ' j sin e ; dp 4 tan e ^a -, 0\ " Hence the rifling reaction and friction due to rifling are directly proportional to the powder re- action, that is the pressure on the rifling grooves always varies at any instant directly with the powder reaction. fbus we have the relationship that rifling. frlc.tj.on behaves exactly like, an additional mass: that jLaf it has an inertia^ef f ect since it is_prft- portional to the aeeeleratipn. The true equivalent mass due to the linear and angular inertia of the projectile alone, can be ob- tained by assuming the rifling friction zero, ( i.e., putting u = o) 4 n8k*v n' • (1 — - - ) m (i " ::" ° ) . (20) The true equivalent mass may be readily checked by a consideration of the total energy of the projectile,. that is, \ m'v2 « i mva + y In* 2nv 2v tan and w • - - 38 and I = mka where k = radius of gyration about its longitudinal axis. hence . » . * 4n k . ,, 4k* tan* 9 N B1 - ( 1 + = — ) m = (1 + — ) m D* d* EQUIVALENT MASS For a differential layer of OF the powder charge at the base of POWDER CHARGE the projectile, its velocity evidently is equal to that of the projectile while for a dif- ferential layer at the breech, the velocity is equal to that of the gun. For intermediate layers, we must assume some law of variation of velocities, between the two end limits. For simplicity and probably a fairly close approximation, we will assume for the various laminae, a linear variation of velocity be- tween the end limits. Further since the velocity of the gun is small as compared with that of the projectile, in virtue of the approximation of the whole analysis, we are entirely justified in assum- ing the recoil velocity entirely negligible. If, Velocity of projectile * v (ft. sec.) Distance between breech and base of projectile = x (ft.) Velocity of any inter - mediate lamina = v1 (ft/sec.) Distance from breech to lamina = x1 (ft.) Then v1 = — v (ft. sec.) u If we assume the density of the powder is uniform through the distance x, so that the weight of the lanina is x dx, then the kinetic 39 energy of the lanina is W v ' W V 2 o V 1 I " * I £• * • TT o dx or — . -7T * *X 2 g x» 2 g and the Kinetic energy for the total charge becomes, K. E. of w - f. _Ii rx x'2 dx1 x» 2g ^o ' J (i. ) T- (21) That is, the equivalent mass, when dealing with the energy equation, is 1/3 the mass of the total charge. It is important to note that when dealing with momentum, the momentum for the total charge becomes, on the same assumption w u x1 w JL / JLvdx'=.JLv (22) gx o x 2g that is the equivalent mass from the moment or aspect is 1/2 the mass of the total charge. EQUIVALENT MASS It is convenient in deriv- OF ing the energy equation, THE RECOILING PARTS to express the Kinetic Energy of the recoiling parts in terms of the velocity of the projectile. Neglecting m ds small as compared with mr 2 and, neglecting the recoil brake reaction as small, we have, by the principle of linear momentum, f ™ \ ( \ mr V = (m + - ) v ( approx.) 40 hence (m * ) Therecoil Energy, becomes, ? »r v* = i I va (23) and therefore the equivalent mass of the recoil ing parts, in terms of the velocity of the projectile, becomes, (m + JL )« 2 ENERGY EQUATION The mechanical work expended by the gases of the powder charge in the bore is equal to the external work ex- erted on the projectile and gun, plus the Kinetic Energy given to the gases themselves, plus the heat energy lost in radiation through the walls of the gun. If W = the Potential Energy of the Gases at any instant. P|,j= the total reaction exerted on the breech of the gun. P = the total reaction exerted on the base of the projectile. X = the displacement of the gun measured in the direction of its movement. x = the displacement of the projectile measured in the direction of its motion. E = the Kinetic Energy of the powder charge. Q = the loss of heat due to radiation. 41 J » the mechanical equivalent of heat = 778 *** lb* B. T . U , Then for the energy equation of the powder gases, ire have, -Pb dX -P dx * d(E + W) * JdO (1) hence -dW » Pb dX + P dx + dE + JdQ. (2) that is the loss of the potential energy of the gases, due to a differential expansion goes into mechanical work (PjjdX + P dx + dE) and radiation JdO. Further by (19), (23) and (21), P dx = d[y(m"va)) pbax AS = tlf (5 v» )] 0 so that Further, in terms, of a hypothetical mean pressure Pm(over the cross section of the bore) equation (3) may be expressed in terms of the travel up the bore u, (i.e. the relative displace- ment between the gun and projectile). where (6) (6) jec the bore of the gun, approximately since dv where v --— = the acceleration of the projectile up du 42 dv dx dv dv l~ dx v dV = df dx + dx = d? l~TdT> and dX is pared with dx, and m" = the equivalent mass of the projectile which takes care of its angular acceleration as well as the rifling friction, see equation (19). EXPANSION It will be assumed, that the OF expansion of the gases due to POWDER GASES the combustion of the powder charge obeys the law of a perfect gas. Hence, we have, PV = RwT where p = the Intensity of Pressure exerted by the gas Ibs/sq. ft. V = the volume of the gas. (cu. ft.) w = the weight of gas (Ib.) R = a coefficient (ft. Ibs. per Ib . gas.) T = absolute temperature reached. Further, with a perfect gas, the internal energy of the molecules of the gas is entirely in a Kinetic or Vibratory form, and therefore, is directly proportional to the temperature. Hence, we have, dQ dQ = cwd T and c = — where dQ = the heat required to raise the gas for a change of temperature dT. c = Specific heat or the heat required to raise one Ib . of gas one degree of temperature at the temperature considered. We are concerned especially with the expansion of a gas at constant pressure or at constant volume or a combination of the two. 43 pdV Hence, dQ. = wCp dT = Cp =-- — at constant pressure Cv dQ » wC dT = R at constant volume. If the volume and pressure vary together, then, we have the sum of the partial variations, above, that is, dQ »—- (C p dV + Cv V dp) Rw " but since, dT = — (p dV + V dp) R we have, dQ = CywdT + °P ' °v p dV and Q _£ a = wCv/dT + p R v — /p dv This relation can be interpreted, physically immediately, since the internal energy being entirely of a kinetic or vibratory form, must be proportional to the change in temperature at constant volume other- wise additional heat must be added for the external work. Hence wCy/dT measures the molecular kinetic energy. Considering an expansion at constant pressure, the total heat required is, Q = V 4-pdV •I where U * the internal energy = wCy dT. Since the heat is added at constant pressure, we a Iso have, pdV = wR dT But the heat added at constant pressure is, Q » wCp dT hence, substituting in the total heat equation, wCp dT = wCv dT — w R dT ^Dd R _R _ Cp - Cv ' T- or J - P - <-v 44 If now the specific heats Cv and Cp are assumed constant for the range of temperatures during the ex- pansion of the powder gases, we have, Cp - Cv Q = *CV(T - TJ + -^ W (7) where If is the external work performed. Tt is the initial temperature. Neglecting the loss of heat by radiation as small, we have practically an adiabatic expansion in the bore of a gun; that is, - C, wRT Since p = -=-, dividing by T, we have Cv " + <Cp - Cv) |L 0 and if we let Cp — = " **«" j X" ./v dV Cp <jx -i- = n, then — + (n-1)^- = 0 uv and , It ' V Therefore, T- . c^)"-1 Ta V ; Now for an adiabatic expansion, Q. = 0 and therefore, (eq.7) becomes, RwCv(Tt - T) T, » = -7; « — (Tt - TJ Cp - Cy n - hence 7 n-1 = wCv J Tt [1 - (•—-) 1 (8) This equation 1* in convenient form since it it in terms of the initial and final volume in the bore. The equivalent length of the powder chamber of the gun in terms of the area of the bore, becomes, V, Vi uo 0.786 d* so that V u0+ u In terns of the displacement up the bore the work of the adiabatic expansion of the gases, becomes, W » wCv J T,. f 1 - ( - — )" " l] (ft - Ibs) 1 (ft - Ibs.) ) ) (10) * i (- • ) n - uo * u Here w» weight of gases (Ibs.) W « external work performed during the (adiabatic) expansion (ft. Ibs.) Cv* specific heat for constant volume (B- T. U. per Ib. per deg. ) Cp= specific heat for constant pressure (B. T. U. per Ib. per deg.) T,= initial temperature "(degrees) Vt» initial volume (i.e. volume of powder chamber) V«. u0 = _± - where 0.7854d* • area of bore 0.7854d'« u = displacement of projectile up the bore J = mechanical equivalent of heat (• 778 ft. Ib. per B. T. IT.) 46 R» (C CV)J= the gas constant (ft. per degree) From this equation we may deduce the differential equations of velocity and powder pressure for the movement of the projectile up the bore, provided we know the manner of burning of the powder gases, etc. The energy equation therefore becomes:- 2 wRT* (i . .!£.. )n - l = ." ! 1 • v2 11 ' n - 1 u0* u ; mr 3 J (11) From which we may determine v in teras of u. The factor dg which represents the loss due to beat radiation must be determined by experiment. Based on the energy equation (or its derivation, the force equation of the motion of the projectile in terms of the displacement up the bore) various interior ballistic formulae have been derived differing in the method assumed as to the combustion and expansion of the charge. The formulae of Ingalls and Hugoniot have been used by our Ordnance from time to time especially in ballistic calculations. In recoil design, however, rough approximations are sufficient since the manner of combustion has small effect on the recoil. The formula of Leduc is sufficiently condensed with sufficient approximation to be admirably suited for recoil design. TOROUE REACTION It is important in OF the design of traversing THE PROJECTILE gear for guns shooting at high angles of elevation to compute the average torque reaction of the projectile upon the gun. Let w s ang. vel. of projectile at any point in the bore (rad/sec) 47 v = linear velocity in bore along X axis (ft/sec) r = radius of bore or of projectile (ft) Pjj = powder pressure at base of projectile (ibs) f = normal reaction of rifling groove (Ibs) T = torque on projectile (Ib. ft.) 6 = angle of rifling grooves with XX I = mk* = moment of Inertia of Projectile IS = angle turned by projectile Then u = v tan e d*0 = tan e dgx (21) dt* r dt2 hence , d 0 mk* „ d*x T = nk* = — tan 0 — - , ^. dt* r dt* (12) but Tt2" m (approx.) hence T = ~ tan 0 Pb (Ib ft.) (13) 2*r where tan 0 = p being the pitch of the rifling at the point considered . From equation (23) we see that the torque is proportional to the powder reaction on the projectile, and the "slope" of the rifling grooves, the steeper the grooves being the greater the torque reaction with a given powder pressure. Further, if the rifling pitch is made constant throughout the greater part of the bore, the torque 48 varies as the powder pressure curve along the bore and therefore is a maximum at the beginning of the travel of the projectile. For the average torque, we have, d»0 Tav s »k* ( - ~\v u t The moment of inertia of the projectile may be roughly evaluated by assuming a solid cylindrical projectile:- If C * mean or equivalent length D = density or weight per cu. ft. DC r DC - ' 2 n r dr = ~ DC but m * -— hence -E£-nr*k-* -— -S~ and k« «-^- ; g 2 k - 0.707 r Further, it is customary to designate the "rifling" as a "twist" of 1 turn in "g" calibers. Therefore, if w,e let, Twist: = 1 turn in "g" calibers Time of travel in bore (approx.) 3 t » 3/2 - v Radius of gyration of projectile » k * 0.7 r Number of rev. per sec. = n Then, n g 2 r = v ) an£ ) at the nuzzle d0 HV ) —— * 2nn = - dt gf d0 } Therefore, since T...t = mk(r — ) , dt m T -r— - mx 0.49 r* - 2v gr and rev " T » i.05---r (24) ug 49 which gives the average torque reaction on the gun due to the angular acceleration of the projectile. Since the slope of the rifling grooves is small, we may roughly assume, that, a fr = T and f = 1.05 - (35) ug which gives the mean pressure on the rifling band. It is of interest to compare the maximum torque reaction to the average torque reaction in an actual gun. Type of Gun: 240 m/m Howitzer Muzzle Velocity: 1700 ft/sec. Weight of Projectile: 356 Ibs. Max. Powder Pressure: 32000 Ibs/sq.in. Rifling = 1 turn in 20 cals. Travel up bore = 13.33 ft. Then, for the max. powder reaction, we have, Pbmax. = 32000 x 0.7854 * - = 2,242,000 Ibs. and for the rifling slope, hence, for the max. torque reaction, 9 45 T = _: — x 0.157 x 2,242,000 = 69,500 Ibs. ft. 48 where as for average torque, we have, u 1.05 x 356 x 1700 x 9.45 T „ „ = - — _ = 49600 Ib. ft. av 13.33 x 32.2 x 20 x 24 Therefore, the ratio of max. torque to tne average torque becomes, Tmax _ 69500 f av " 49600 = Due to the short time action of the travel up the bore, the effect on the traversing gear depends upon the average torque rather than the maximum. 50 MUZZLE BRAKE GENERAL The muzzle brake consists of curved DESCRIPTION vanes secured to the end of the muzzle upon which a portion of the powder gases are deflected in the second part of the powder period after the projectile has left the muzzle. The gases are deflected somewhat to the rear, and we have a forward reaction due to the change of momentum of the gases, which materially checks the recoil. The design and best arrangement of vanes requires a considerable experimental investigation and it is merely proposed here to outline certain general limitations based on an elemantary theory. ELEKANTARY If it were possible to calculate THEORY the mass of gas discharged through the vanes, as well as the mean extrance and exit velocities, the reaction on the vanes could be determined. But the •etbod is complicated, since the pcmder pressure after the shot has left the muzzle falls off accord- ing to a complicated function of the time, and, con- sidering the variable volume of gas, this makes it dif- ficult to approximate the mass of the gas as a function of the time. Further the amount discharged through the vanes depends upon the initial mean muzzle velocity of the gases, the caliber of the fcore and the entrance areas to the vanes, as well as the variation of muzzle velocity of the gases against time. We see therefore to approximate roughly the problem from a theoretical point of view would require an elaborate analysis com- bined with a long and eleborate experimental research. Let w » weight of projectile (Ibs) 5 * weight of total charge (Ibs) v » nuzzle velocity ft/sec. 51 Vo = velocity of recoiling parts when the projectile leaves the muzzle ft/sec. to = time for shot to reach muzzle. T = total time of powder period. Pjj = total pressure on breech due to powder gases (Ibs) Wr = weight of reeoiling parts vw = mean velocity of gases after free expansion (ft/sec. ) T / Hdt = impulsive reaction on vanes (Ibs) „ charge through vanes Cw = ratio of s — - — r5 total charge Then, without vanes, we have, pb dt = ° Pb dt +{pbdt Total o t0 reaction on the Gun during the (w + - ) v Powder period. Now / ° Pb dt = o g and since the powder charge has a mean velocity = v — — when the projectile leaves the bore, H ± W hence / pb dt = v + - vw o £ g therefore, WrVm = wv * w vw as we should expect from the principle of the con- servation of momentum. With the muzzle brake acting, the total reaction 52 during the second period (T -t0), on the gun becomes, /T Pb dt - ;T8 dt ' and therefore the momentum imparted to the gun be- wr (v V \ dt » - <vra - o ' comes,.? _ T wr (v V P dt - o If it were possible to deflect the total charge entirely backward and maintaining the same expansion, then, for the total reaction on the gun during the expansion period of the powder gases, we have, V since the change in velocity * vw + - ft/sec ft Therefore, the momentum given to the gun during the powder period becomes, wv - w v_. which gives the impulse imparted if we had muzzle brake, with the sane expansion backward as forward through the vanes. With the same expansion to the rear the maximum possible recoil energy that can be absorbed with a muzzle brake, becomes, Aab 2 w w v vw * »r and the maximum possible percentage of the total recoil energy absorbed by the ideal muzzle brake becomes, 53 Aab 4 w w v vm A (wv + w vw)* Since w vw is always less than wv, we see that even with an ideal muzzle brake and complete expansion, it is impossible to completely check the recoil energy, unless greater expansion is obtained to the rear than forward. The total gas reaction on the gun due to the com- bined expansion and deflection of the gases, is represented by the impulsive reaction, T T / R dt -/ Pjj dt in a forward direction (i.e. to to towards the muzzle) If we have complete expansion of the gases, before entrance into the muzzle vanes, then, ;T pb dt - ^ (v . y. ) «* '0 g ^ W U and if now the total gases are deflected entirely back, then T 2 w v.. r R dt = - • * and as before, If, however, the gases are accelerated to a mean velocity v1 before entrance into the vanes, then 1* V /T Pb dt - - (v- _ ' ) g and further expansion takes place through the vanes to the maximum value vw to the rear, we have / R dt = - (v« + v ) since v1 + vw is the change in velocity. As before the total impulsive reaction on the gun, becomes, £H at -£ pb at .j (,„ , J> Without the vanes, the reaction on the gun breech be- comes,- R v {I ?* « • ; K - I ) and with the vanes the reaction on the breech is pro- bably different and modified to, /T Pfc dt = * ( V' - J) to since some expansion probably takes place within the vanes, themselves. Now as to the actual reaction obtained, the ideal brake differs from actual conditions, essentially in the following points:- (1) Only a part of the total charge can be deflected through the vanes. (2) The entrance velocity can only be a component of the actual muzzle velocity of the gases. (3) Only a partial expansion of the gases can take place before entrance into the vanes . (4) The exit velocity can not, for practical considerations, be entirely to the rear, 30* from the rear, being like- ly the maximum angle that the gases can be deflected. (5) Only a very small expansion can take place through the vanes themselves; of the gases passing through the vanes the total expansion is small. In consideration of (1) (unless the vanes are ex- tended a considerable way out) the higher the muzzle 55 velocity the less the total charge passing through the vanes. It has been found experimentally that it is useless to add more than a given column of vanes, further addition of vanes having very little effect on the reaction. Further the first one or two vanes nearest the muzzle, are subjected to an intensity of pressure practically equal to that of the gases at the muzzle. Further development of the muzzle brake should be directed in obtaining greater expansion to the rear by a suitable combination of vanes, curvatures of same, etc. LEDUC'S FORMULA The empirical formula estab- lished by Leduc is especially service- able and sufficiently accurate for a predetermination of the reaction of the powder, during the powder period and its effect on the recoil. Leduc's formula, assumes that the velocity curve of the projectile during its travel up the bore follows that of an equilateral hyperbola, with parameters a and b, that is, If v = the velocity of the projectile at any point in the bore (ft/sec) u = the corresponding travel up the bore (ft) a and b being parameters of the hyperbola, then v = r-r: — (ft/sec) b + u where a and b must be determined by the elemantary principles of Interior Ballistics. Determination of the parameters a and b:- When u is made infinite, that is u = a and v - a - a then a 56 a is therefore determined by considering the expansion in a gun of an infinite length. c If n » the ratio of the heat capacities (—*- *1.4l) and for an adiabatic expansion pV = k, v Then the work of an expansion from, initial Volume Vt to final Volume V , becomes, i = / p d V, but p = — * where k and n are con- Now when Vf becomes infinite W * — — * t (ft. Its.) n - 1 Vt Since. 1 Ib. of water * 27.68 cu. in. for unit density, k 1 i - 1 27.68" Expansion at Unit Work for an Infinite Expansio Density. Weight of given volume of powder gas Weight of same volume of water and if 7C » the given volume of the chamber (cu. in.) Vt » the volume of 1 Ib . of gas (cu. in.) then ^ ** 27.68 A * » — - — per Ib. of powder gas. *a 27.68 hence the specific volume of the gas, becomes, 27.68 V*a ' "I" 57 Therefore, the work of expansion of 1 Ibs. of the gas to oc becomes, \f An~l W 1 27.68° Since the gas evolved is proportional to the weight of the charge w, and a = v for an infinite expansion in the bore, we have waf w E A * ^ £or a compiete expansion of w 2g Ibs. of powder gas, hence a » /2gE (-*—)* A — 2 Now S has a value = 653 ft. tons roughly, and by experiment "_lJ. s 1/12 (approx.) Taking into account the various losses, it has been further found ex- perimentally that /2gE = 6823 for ordinary good powder. Therefore, the parameter "a" becomes, a * 6823 (— )t ATI ttf W 27.68 Now A a -n but with a powder chamber Vc, loaded with w Ibs. of powder, the specific volume of 1 Ibs. of powder evidently becomes, t^ ''ta * w assuming complete combustion of the charge, (2) that is the density of loading may be defined as the ratio of the weight of the charge to the weight of a 58 volume of water sufficient to fill the powder chamber, Hence the parameter a becomes, i __i a = 6823 (-£-y ,27.68 w.ta H f \ y / vc To evaluate the parameter b, we must consider the ac- celeration of the projectile, and the reaction of the ponder gases on its base, during its travel up the bore. The acceleration up the bore, becomes, jiv (b * u )a- av K a2 bu V"= (b + u )' (b * u)3 (3) since 3. V v = fr4.u from Leduc's formula, hence the pressure against the projectile, for a displacement v, becomes, w aabu P = — g (b+u)2 (Ibs) Further the maximum pressure occurs, when T— i.e. when, ] .4 .3 = -3u(b + u) + (b+u) (b+u)4 and u = - (ft) that is the maximum pressure in the bore occurs at a displacement equal to one half the parameter b or the parameter b - twice the displacement of the maximum powder reaction in the bore. We have, therefore, substituting v = ° in (3) 2 59 P ' - (Ibs) (4) The mean powder reaction on the base of the projectile during its travel up the bore, becomes, where VQ = the muzzle velocity ft/sec. UQ = the total travel up the bore (ft) The pressure against the projectile when the shot is about to leave the muzzle, becomes, a bu Hence to determine the parameter "b" we have the following equations:- 4 w a . m 27 g D f°-* \ where Pm VQ and UQ are * a*^_ ( known. 2 wv and PQ Pe, a and b are au v : b+tl ( unknown. ) Hence a solution is possible:- If A^ = Area of bore and Pm = a given property of the powder used Pm = 30,000 to 33,000 Ibs/sq.in. usually. Pm = p^ A for the max. powder reaction. Substituting atl e 2g (b+u0)2 o 2 (Ibs) hence a = 60 P . 4 » £ (its) bence a" • 27 g b 4 w Equating, we have, 2 Pe(b * uo)» _ 27 b Pm hence 27 b» + 2 b u0 * u*0 = o 27 t e + 12 - 3- — ) v b + u20 = 0 8 P. Solving, we have, (9 2? m 1 , S<9 "(2 " 8" Pi > "o ! /(2 ' g- F* • ' ' o - (ft.) (7) which determines the parameter b, in terms of the travel up the bore, the given maximum powder reaction and the mean powder reaction, being determined from the muzzle velocity and travel up the bore. To completely determine the velocity, powder force, and time against the travel up the "bore, we have — aa- (ft/sec) b + u 61 P = g (b+u)3 and the corresponding time of travel, becomes, , du (b+u) * - / y- ' (au) du b , 1 = a l°#e u * a u * Constant Now when u = 0, t = 0 and loge u = - a, and the constant cannot be evaluated without making some as- sumption. Since the initial powder reaction required to force the projectile into the rifling grooves is large and the displacement u » « , to Max. powder pressure is small, we can reasonably assume the powder reaction constant and equal to the maximum- powder re- action during the initial travel u = ^ . Hence as- 2 suming the maximum powder reaction to be reached at the beginning of the travel of the shot up the bore, and then and substituting Praax> from to remain constant up to u = hence t. = v-~ (-) (8) Sunstituting in the previous time equation, we have, '27" b b b b (-) = — log - •»• — - * constant a a 2 2a 62 and Constant = — ((/27 - 1 ) -2 loge ^ ] a\ (2.098 loge - ) 2 "a therefore 2u u = - (2.3 log — + - +2) (approx.) (9) a b b The powder reaction on the breech during the travel up the bore is somewhat greater than at the base of the projectrle due to the inertia resistance of the powder gases and charge. It has been shown previously that the breech pressure is augmented over that at the base of the projectile by either of the two following formulae:- v, + | Pfe = p (Ibs) w or Pb = 1.12 P (Ibs) The former is based on a theoretical assumption, and gives an idea as to the change in the pressure drop from the breech to the projectile with different ratios of powder charge to weight of projectile. The latter is entirely empirical and it appears that the ratio of the weight of the charge to that of the pro- jectile has no effect on changing the ratio of the breech pressure to that at the base of the projectile. Unfortunately the latter empirical value is somewhat limited especially for extreme ratio of the projectile weights but is, however, reasonably accurate for ordinary calculations. The former is more or less in error due to the assumptions made, but it gives the 63 characteristics for extreme ratios. Therefore, for extreme ratios of charge to projectile weights, the former formula should be used, while with ordinary ratios, the latter should be used. R e c ap i t ul ait ion of the Various Formulae Originating from LIDUC'S Formula - Let v = Velocity of projectile up "bore (ft/sec) u = Travel up bore (ft) v0 = Muzzle velocity (ft/sec) UQ = Total travel up bore (ft) t = Time of travel up bore (sec) to = Tine of total travel up bore (sec) W = Weight of powder charge (Ibs) w = Weight of projectile (Ibs) Vc = Volume of powder chamber (cu.in.) A = Density of loading P = Powder reaction on base of projectile (Ibs) P^ = Powder reaction on base of breech (Ibs) PJH = Max. Powder reaction on projectile (Ibs) Pe = Mean Powder reaction on projectile (Ibs) A<j = Area of the bore (sq.in.) pm = Max. given powder pressure (Ibs/sq.in.) (from 30,000 to 33,000 Ibs/sq.in.) Given :- Pm vo, Vc, w w To evaluate: — v, P and t 64 27.68 then, A = vc (i) i = 6823 (-)7 Al/l* (2) (Ibs) (3) (Ibs) (4) U° 16 pe 16 pe v = -^— (ft/sec) <6) b + u P = - g (b + u)» (Ibs) (7) w a 2bu (b+u)3 (Ibs) (8) w a2bun P0v = 1.12 - 2 )3 (Ibs) (9) t = - (2.3 log ^ t - * 2) (sec) (10) a bo - (2.3 log ^ + , a bo t0 = - (2.3 log + ,— +2) (sec) 3 ^ 2 vo v (approx.) Uec) (11) 65 DYNAMICS OF RECOIL The velocity and displace. DURING THE TRAVEL OF ment of the recoiling mass THE SHOT UP BORE with respect to the powder charge and projectile is ob- tained by the principle of linear momentum. Assuming, one half the charge to move forward with the projectile and the other half to move backward with the recoiling parts, we "have, ( «r * \ >Vf = ( » * \ )v and (wr = - )Xf = ( w + -• )x 2 £ Now the absolute displacement of the shot in the bore is related to the travel up the bore u, by the equation x = u - X Hence, we have, and w (w + - )u Xf = Since w and w are small as compared with wr, we have for a sufficient approximation w w + - Vf = 2- v (ft. sec) w_ X -iLlJl Af - ' *' U ft. The equation of velocity displacement and time of free recoil during the travel up the bore, becomes, 66 Vf = (wr ) b + u (ft/sec) Xf-C- (ft) t = - (2.3 log — + e * a b b With constrained recoil, assuming a recoil reaction X we have, dt Ubs) hence 1 Pbd<- Kt = v * pb <' f -r: — v but / -« vf therefore, Kt Vf - -- = V (ft/sec) mr Kt: 2m, X (ft) 2u » (2.3 log ^ + _ a b b + 2 (sec) In the several equations, it will be noted, that the common parameter is the time of travel up the bore in the gun. Hence if for various values of u, we ob- tain correspondingly values of time, the free velocity and displacement is obtained for the given time and the corresponding effect of the recoil brake during this time is deducted from the velocity and displace- ment respectively. Further it has been tacidly as- sumed that the powder reaction with constrained recoil is the same as with free recoil at the same time in- terval. This, however, is not strictly true since the 67 powder reaction is somewhat modified due to the slightly different motion of the gun with constrained and free recoil respectively. The effect, however, is entirely negligible as compared with the magnitude of the reaction and other factors involved, even with the most refined measurements and analysis. EXPANSION OF THE GASES AFTER The manner of THE SHOT HAS LEFT THE BORE the expansion of the AND ITS EFFECT ON THE RECOIL powder gases after the projectile has left the bore is very difficult to calculate, and various assumptions based on empirical data have beeri formulated, for calculations during this period. The following theory though imperfect gives an idea as to the manner of the expansion of the powder gases in the "After effect Period". (1) The momentum imparted to the gun during this period evidently equals the momentum iwparted to the powder gases: n>r<Vf * Vfo> ' S<vw - I ) where Vf = maximum free velocity of the recoiling parts. (ft/sec) Vf0 = free velocity of recoil when the shot leaves the bore (ft/sec) vw = mean velocity of the powder gases attained (ft/sec) i = mass of powder charge (Ibs) v = muzzle velocity of projectile (ft/sec) Since mr Vj0 =(n + - )v we have mrVf = mv + m vw In other words, the maximum free momentum obtained by 68 the gun, equals the sum of the total momentum of the projectile and the total momentum of the powder charge. It is important to note that the momentum relations are very nearly true provided we are able to calculate vw the mean velocity of the powder gases and can neglect the small effect of the air pressures exerted on the gases. (2) We have the following energy relations due to the expansion of the gases : (a) Initially the gases have a Kinetic Energy = * (-)v* 2 3 (b) The work of expansion of the gases in expanding from the pres- sure in the bore when the shot leaves the gun (i.e. the muzzle pressure) to the atmospheric pressure, becomes va We = / pdV a. fee y, ro where Vo = volume of powder chamber + volume of the bore of the gun. Va= volume of gases at atmospheric pressure . (c) The final Kinetic Energy of the gases may be approximately assumed equal to: 1 Sv^. 2 It is to be noted that the final Kinetic Energy of the gases is difficult to calculate due to the di- vergence or cone effect produced when the gases expand into the atmosphere The total Kinetic Energy equals 69 the sum of the Kinetic Energy of the center of gravity of the gases plus the relative Kinetic Energy of the gases relative to the center of gravity. From a series of experimental tests conducted by the Navy on the velocity of free recoil with guns of various caliber it has been ascertained that the momentum effect of the powder gases is equivalent to the weight of the charge times, a constant velocity of 4700 ft/sec. Assuming the divergence of the spreading of the gases to be similar at all muzzle velocities , it is possible to estimate the divergence factor and then in guns of very high muzzle velocities we may calculate the maximum free velocity by multiplying the work of expansion by the divergence constant and the solving for the mean velocity of the gases. The pressure of the gases rapidly falls to the atmospheric value or approximately this value, before the divergence of spread of the gases is appreciable, hence the maximum Kinetic Energy of the gases will be attained at approximately atmospheric pressure. The change in Kinetic Energy of the powder cases therefore, becomes, •jSvw---v = change in Kinetic Energy, and the work done on the gases, equals the work done by the external pressures po and pa and the work of ex- pansion pdV. Hence, vd p0V0 - Pava * J Pdv ~ totai work done. vo To allow for the relative Kinetic Energy due to the spreading of the gases, we may multiply the work done on the gases by a constant, and then equate this value to the changes of the translatory Kinetic Energy of the guns. 70 v -n v + f a PdV) =-mv2 -im.,2 o vp pa va J 2 w * q v V vo where K = the divergence constant to allow for the spreading of the gases at the muzzle. Now the work of expansion, becomes, ,Va Po vo ~ Pa Va We = / T> d V = : 'o where the expansion exponent k = 1.3 approx. Hence the total work done on the gases, becomes, Po Vo - Pa Va k . Po vo - Pa v« + ; = iT^I (povo ~Pava) K *~" X further, since prt V ^ _ n yk 0 o pa va, we have, P " rrr-fc" v° " Pa Va) = rri p° vo[1 - (-^ ) ^ i PO Hence the energy expression reduces to the convenient form, p ^L_I_1_ K[r— — r po »o\l - ' / k >J=__va PO 2 3 from which knowing po» VQ, pa m and v enables us to immediately calculate vw, the mean free velocity of the powder gases. To evaluate the dispersion constant, to take care of the relative Kinetic Energy of the gases after expansion, the ballistic data of the 155 m/m Filloux gun has been chosen, since assuming a mean velocity of the gases 4700 ft/sec., calculated and experimental results were found to check very close- ly. 71 Weight of powder charge w = 26 (Ibs) Volume of powder chamber S = 1334 (cu.in.) Total length of bore u = 186 (in.) Muzzle velocity v = 2410 (ft/sec) Area of bore Aj, = 29.2 (sq.in.) Weight of projectile = 96.1 (Ibs) Max. powder pressure pm = 35300 (Ibs/sq.in.) Mean Powder pressure = wyg _ _ 19200 (Ibs/sq.in.) " "e 644 uAv Twice Abscissa of maximum pressure .27 "m *i\ j. -i/^ i \ 2 IT = *^7 ^fl Muzzle pressure when shot leaves muzzle _27 2 _u 27_ 2 185.68 x 35300 P° = ~4~ S (e + u)3 P7n " T~ 57.38X (57.38 + 185. 68)3 10140 Ibs/sq.in. we have then, 0.3 K 32.16 [— « 10140 x U4 VIl - 0.3 V10140 = i 2 i 26 2 • - x 26 x x — x 2 4700 2 3 2410 I Solving, we have, 156 x 1Q« K V0 = (287 - 25) 10 6 = 262 x 1Q« 72 Hence K = = 0.430 - 3.915 cu. Hence the energy of translation is but 43* of the total Kinetic Energy of the gases after complete expansion. Therefore with guns of numeral ballistic relations, we may estimate the mean translatory velocity of the gases after complete expansion, by the formula: o where v = muzzle velocity (ft. see) w = weight of powder charge (Ibs) b = 1.3 approx. . pa = atmospheric pressure = 2116 (Ibs/sq.ft.) po = muzzle pressure of powder gases (Ibs/sq.ft.) VALLIERS The hypothesis of Vallier assumes, HYPOTHESIS that during the "after effect Period" in the powder period of the recoil, that the powder reaction on the gun falls off proportional to the time. That is, If Pob = the total breech reaction of the powder gases, when the projectile leaves the muzzle (Ibs) to ~ time of travel of the projectile to the muz- zle (sec) tt = total powder period (sec) P^ = powder reaction on "breech (Its) t = corresponding time (sec) then ?b = pob - c<t -to) VALLIERS HYPOTHESIS where p _ - p rob Pb rob c = How the momentum imparted to the recoiling parts by the gases during the after effect period, becomes, 73 tt Pfc dt » mr(Vf • _ Vfo) where Vfi - ^ax> free veiocity of recoil at end of * powder period. Vfo = Free velocity of recoil when the shot leaves the muzzle. / l [Pob f _t (t - t0 ))dt = «r(Vf , -VfQ) 1 Q t O Integrating, we have, P b( t-*t.) -i . «r(V£I VQ) 2mr(Vf, -Vfo) hence t, _ t ,, __L_J - H_ (sec) and pob C = 2(Vf. -Vfo)mr Therefore the powder reaction during the after effect period, becomes, o Pb = pob — ^ — ; (Ibs) 2(Vf. - Vfo)mr RECAPITULATION Of PRIKCIPLE FORMULAS OP INTERIOR BALLISTICS PERTAINING TO RECOIL DESIGN. The velocity and displacement of the recoiling parts during the travel of the projectile up the "bore have the following relations with the velocity of the 74 projectile up the bore and the relative displacement of the projectile in the bore. (Weight in Ibs. ) If m = massof projectilev 00 ., oc . la m = mass of powder charge mr = mass of recoiling parts" v = velocity of projectile (ft/sec) u = displacement of projectile in the bore from its breech position V = velocity of recoiling parts (ft/sec) X = free displacement of recoiling parts (ft) then =: ra (m + - )v m + ; V = 2 = £_ v Approx. (ft/sec) ra m m,. + •= mr • , ffl (m + - )u m + - X - = u Approx. (ft) The pressure on the breech, in terms of the pressure on the base of the projectile, becomes If Pb = breech pressure (total) (Ibs) P = pressure at base of projectile (total) (Ibs) m m * - P., = P = 1.12 P approx. (Ibs) m The mean pressure in the bore, becomes, I m + - 4 Pm = pb ^ 1 — ) (Ibs) 75 For building up the energy equation, we are concerned with the various equivalent masses of the moving ele- ments that the powder reacts on in terms of the major mass of the projectile. The equivalent mass of the projectile, becomes, if k = radius of gyration about its longitudinal axis (ft) p = pitch of the rifling " 9 = pitch angle of the rifling . 4k2tan2 9 , lh_ m = (1 + - )m = (1 + - ) a |i«t P2 d ' If we include the effect of the friction of the rifling we have, sin 9 + u cos 6 4 tan 9 k2 , Ibs . m " =[1 + ( - 5 - : - ^) - — - ] m ( - ) cos 9 - u sin € d2 The equivalent mass of the powder charge, for the energy equation = 35 3 <i*I) for the momentum equation = - - The equivalent mass of recoiling parts become, (m + =2 ffli t The differential equation for the motion of the pro- jectile up the bore becomes in terms of the mean pres sure in the bore Pm and the relative displacement u, (m + | )2 . _ 2 m . dv Pm = [m " *• - +, - ] v - mw du 76 If W » the potential energy of the gases at any instant, we have further, - dW • pm du * JdQr where fir»* neat lost "by radiation. The energy equation for the expansion of the gases, becomes, where w » ( pm du + J ) dOr (ft/seo)(Adiabaticexpansion) w » weight of gases (l*bs) cva specific heat for constant volume (B.T. U. per Ibs. per degree) Op* specific heat for constant pressure (8. T. U. per Ib. per degree) Ti » Initial temperature (degrees) Vi * Initial volume (i.e. volume of powder cham- ber) o where 0.7854df * area of bore. 0 . u * displacement of projectile up the bore. J * mechanical equivalent of heat = ( 778 ft. Ibs. per B. T. 17.) R » (Cp - Cv) J » the gas constant (ft. per degree) The torque reaction of the projectile in travel- ing up the bore becomes, 77 mv* 1.05 - r (Ibs.ft) where r « radius of tho bore (ft) v = muzzle velocity (ft/sec) g » number of calibers per revolution LEDUC'S FORMULA Leduc's formula gives results sufficiently accurate for recoil design. The formulas derived from it are compact and sufficiently short to "be readily used in ordinary practical design. These formulas have been used in the development of the various recoil formulas in the subsequent chapters. ?or recoil or gun design: let v = velocity of projectile up bore (ft/sec) u = travel up bore (ft) v0= muzzle velocity (ft/sec) UQ« total travel up bore (ft) t = time of travel up "bore (sec) tQ3 time of total travel up bore " w = weight of powder charge (Its) w = "of projectile Vc 'Volume of powder charge (cu. in.) A = Density of loading P » powder reaction on base of projectile (Its) Pb = " reaction on base of breech " Po}jsPressure on the projectile when the shot leaves the muzzle (Ibs) Pm =Maximum powder reaction on projectile (Its) Pe= Mean powder reaction on projectile " A(j= area of bore (sq.in.) Pm= Maximum given powder pressure from 25000 to 33000 Ibs/sq.in. (Ibs/sq.in.) Given, pm, VQ, Vc, w, w and UQ 78 To evaluate:- v, P and t, then, 27.68 w (1) A- __ (2) a = 6823 (->! A* " w (3) Pm = PmAd (Ibs) 2 W V (4) Pe = — £— » g = 32.16 ft/sec-2 . (6) v = ^L- (ft/sec) b+u w a2bu (7) P = (ibs) g (b + u)3 w aabu (9) g (b+uQ)3 (Ibs) vr'iC ;O «**a f- - (10) t = - (2.3 log — + £ +2) (sec) a b b b 2U0 U0 to = ~ (2.3 log — - + — + 2) (sec) a b b (ID t0 « - H® 2 VQ approx. The equations of velocity, displacement and time of free recoil during the travel up the bore, becomes, 79 w w + - p a*j Vf = ( ^)( ) (ft/sec) wf b + u » *5 Xf = ( -) u fft) wr :*jp»-~ee • • • .-•• 20 u , = 1 (2.3 lo? r- + 5- + 2) (sec) a With constrained recoil, assuming a recoil reaction K Kt V = V* - — (ft/sec) X = Xf - — (ft) 2mr t = (2.3 log + £ * 2) (sec) a bo Theexpansion of the gases after the projectile leaves the hore causes an additional recoil effect. The hypothesis of Tallier assumes the powder reaction to fall off proportionally with the time. On this assumption: If Vf ~ the velocity of free recoil at the end of the powder period. Vfo = the velocity of free recoil when the shot leaves the muzzle. p tt and t0 the corresponding terms --- (sec) ob .' " - P'01> 2(Vfl _ V)Br (IT,.) 80 » >y THE PARABOLIC TRAJECTORY The nucleus of exterior ballistics is the differential equations of the parabolic path of a shot projected in a vacuum. These equations then nay be modified for air resistance and gyroscopic deflections due to the angular momentum of the pro- jectile and air reaction: Let x and y be the horizontal and vertical coordinates of the trajectory. n * the mass of the projectile. Vo = the muzzle velocity. t » the time of flight. 0' = the angle of elevation from the horizontal of the axis of the bore. 0 = angle of elevation of the departure of the projectile from the muzzle. e * the increment angle or "jump" to the elastic deformation of the carriage and the move- ment of the gun in a direction not along the axis of the bore. 01- 0 r = angle of sight. Oaa line of sight. L * range to given target. LQ = horizontal range corresponding. • = striking angle from horizontal. m1 = angle of fall from line of sight. The differential equations of motion give: d»x d«y Integrating successively, we have, ~ » V0 cos 0 ~ = - gt + V0 sin 0 dt and . \\ 82 ^^^^^ x = V0 cos 0 t y = - - + VQ sin 0 t <Q Hence g x2 y = - - + x tan 0 (1) 2 V2Q cos2 0 The general parabolic equation of the trajectory in vacuum. For maximum range, x being a function of 0, we have — = 0, when y = 0 in (1) d0 that is, 2V2Q V2 x = sin"0 cos 0 = — °- sin 20 g g and dx 2V*Q — - = cos 20 = 0 d0 g Hence, cos 2 0 = 0 or 0 = 45° When air resistance is considered maximum range for ordinary guns is obtained at angles which may be from about 42 to 55 degrees. If x and y are the coordinates of some target point, we have for the equation of the line of sight, that y = x tan r Substituting in (1) X2 « COS20 + gx 7""* .., = tan 0 - tan r 2V_ cos*0 and ya » xa tan r and L = - xa2 - y which gives the coordinates and ranges in terms of the muzzle velocity, angle of departure and angle of sight, CHAPTER III EXTERNAL REACTIONS ON A CARRIAGE DURING RECOIL AND COUNTER RECOIL - STABILITY - JUMP. EXTERNAL REACTION The external reaction during re- coil may "be divided into two primary periods; that during which the force of powder pressure on the recoiling parts exceeds the restraining force or accelerating period and the retardation period. Again the period of powder pressure may be divided into the period of the shot traveling up the "bore to the muzzle and the after effect period of the powder gases ex- panding to atmospheric pressure. Considering the gun, recoiling masses and carriage as one sustem, the external forces are: (1) The pressure of the powder gases along the axis of the bore » ? (2) The torque reaction due to rifling * T (3) The weight of the recoiling parts » Wr (4) The weight of the stationary parts* W& (5) The balancing reactions exerted by the ground or platform on the carriage mount . If we sum these forces up into X and Y components and let Hr equal the mass of the recoiling parts, we have, noting the mass x acceleration of the stationary parts of the system is nil, ithat) «, dt* 83 34 If further, we assume our coordinates along and normal to the axis of recoil, we have X = M, daxr dt2 Y =0 Equation (1) may be written: (I1) (2') ZX - Mr d2x, dt; = 0 Hence, by the use of D'Alemberts ' principle regarding the inertia effect, that is, mass x acceleration re- versed, as an equilibrating force, we reduce the forces to a system of forces in equilibrium. Thus by including the inertia effect of the re- coiling parts as an additional external force, the problem is reduced to one of statics. This greatly simplifies the procedure of ac- curately and quickly obtaining certain overall effects in stability and the principal reactions throughout a carriage. i ;•: irii«9 tea if*t«fl $riiiaa«? ,Jejt. sd: - ,ioO EXTERNAL EFFECTS Considering now the external DURING RECOIL reactions upon the total system, (gun, recoiling parts, and car- riage proper) including the inertia of the recoiling -masses, we have the given forces as shown in figure (1), where Fig. 1 85 r P = Total Powder Pressure along axis of bore at any instant of Powder Pressure Period. Wr = Weight of recoiling masses "Mr" . Wa = Weight of carriage proper (includes stationary part of tipping parts ) H £*_=Inertia force of recoiling masses r d t Ha and Va = Horizontal and Vertical components or equivalent float reaction 7^ = Front Pintle reaction - Horizontal component assumed zero in order that the reaction may be determinate. B = Braking force, resultant hydraulic and re- cuperator reaction including stuffing box frictions. R = Guide frictions = Ri + R2 in diagram. Ka = Total resistance to recoil for recoiling masses equals B + R - Wr sin J0 at any instant during powder pressure period. Kr = Total resistance to recoil during any in- stant after P = 0. During the powder press.ure period, we have for moments about A, see figure (1) P(d + e ) - (Hr 1 )d - Wr Lr - Wa La + VbL - 0 d2x - (Mr 7~7>id * Pe ~ wrLr ~ ffaLa + vbL = ° hence (3) In like manner we have after the powder pres- sure ceases (4) Now considering the external reactions on the recoiling parts alone*, during the powder pressure period, we have figure (2) 86 Fig. 3. hence P - and when P » 0 figure (3) d'x k dt d'x — = * B + R - If- sin j dt (5) ,t B + R - Wr sin 0 = Mr ~ r dt (51) 87 Substituting (5) and (51) in (3) and (4) respectively, we have K»d + Pe - WrLr - WaLa + VfcL - 0 (6) Krd - VrLr- WaLa + VbL » 0 (7) Thus the external effect during the powder pressure period is always at every instant equal to the total resistance to recoil, that is, the sum of the total braking and guide friction, minus the weight component and a powder pressure couple Fe dependent upon the actual total powder force. In general, e is very small and usually for a first approximation the powder pressure couple can "be neglected. Further, for constant resistance to recoil Kr - Ka - K - B + R - VTr sin 0 (8) which is the average external effect during recoil on the total system. As shown in Chapter VI on the "Dynamics of Re- coil" i K « -* — E £ — b - E + VfT (9) where Vf * the maximum free velocity of the recoiling parts, that is _ W4700 + WQ wr (10) W * Weight of powder charge, W * Weight of shot and Wr • Weight of recoiling parts v0 * Muzzle velocity of shot u » Travel up the bore in inches b * Length of recoil in feet d a Diameter of bore in inches E » Unconstrained displacement of recoiling parts during powder pressure period. 88 T = Total time of powder pressure period. In equation (9) note that E = Kt Vf T and K o 2 v Substituting these values in (9) and solving for a wide range of artillery material and thus evaluating the variables as a function of the diameter of bore, muzzle velocity and travel up bore, Mr. C. Bethel has given the very valuable and serviceable formulae, and accurate to one percent. Mr Vf 1 = — = - - - b + (.096+. 0003 d) uvf vo >*|»H This formula holds only for constant resistance to recoil. It is important to note that the "total braking" sometimes called "the total pull" is not in general equal to the resistance to recoil, but is the total resistance to recoil plus the weight component, that is B+R=2Pa+ZPn+2Rs+2fig = K+Wr sin 0 (12) where 2Pa = Total recuperator reaction ZPj, = " hydraulic reaction 2Rs = " stuffing box friction 2R,< = Guide friction K = T M. V, ir vf (b - E + Vf T) To obtain the external reactions on the carriage mount, it is convenient to know d in the previous moment formulae about A, in terms of the height of the trunnions and the distance between the trunnions and a line through the center of gravity of the re- coiling parts arjd parallel to the axis of the bore. 89 Let H+ = height of trunnions above the ground distance from trunnion axis to line through center of gravity of recoiling parts and parallel to l>ore. moment arm of K about A nor. horizontal distance bet-ween reactions A and B. from A to center line of trunnions. As the gun elevates, we have two cases: (1) When the line of action K passes above A, see figure (4) (2) When the line of action K passes be- low A, see fig. (5) t s = d = 1 = c = -*- K Fig. 4 90 r Fig. 5 Fig. 5' 91 For case (1), we note that h' - (d sin 0 + c ) tan 0 = d cos 0 but s *' = Ht + ^I~0 and a E + _f d sln 0 _ c tan 0 = d cos 0 cos 0 cos 0 ht cos 0 + s - d sin2 0- c sin 0 = d cos 0 hence d = ht cos 0 + s - c sin 0 (13) For case (2), we note that h1 + d cos 0 = (c - d sin 0) tan 0 but n' = ht * — — ^ 1 cos 0 cos 0 cos 0 2 2 ht cos 0 + s + d cos 0 = c sin 0 - d sin 0 hence d = c sin 0 - ht cos 0 - s (14) If W = weight of the total system (gun, recoiling parts and carriage ), we have for moments about A *s ^s = vr Lr + wa La In battery or IB terms of the tipping parts = Wt and the top carriage alone (not including the stationary parts of the tipping parts = Wa Ws Ls = Wt Lt + Va La In battery where Ls = distance to center of gravity of' total sys- tem in battery from A If b = length of recoil, and 0 the angle of elevation., and Lg - distance to center of gravity of system out of battery, we have Ws 14 = Wr (Lr - "b cos 0)+ Wa La = ¥r Ly + Wa La - ¥r b cos 0 hence TTS L^ = Ws Ls - ^r b cos 0 Out of battery 92 Hence the external reactions at A and B on the carriage mount become in terns of the resistance to recoil, powder pressure, height of trunnions and distance between trunnions and line through center of gravity of recoiling parts parallel to axis of bore, For low angles of elevation, Taking moments about A, we have, Vb L + Kd + Pe - Ws Ls + tfr(x cos 0)= 0 hence Wg Ls - Wr(x cos 0)- Kd -Pe Pe disappearing for a finite value of x or in other words, when pe is used Wr x cos 0 may be neglected. And since Va r Ws + K sin 0 - V^ or directly from moments about B, noting that moment arm of K becomes d'= d + L sin 0 = ht cos 0 + (L-c)sin £l * S we have, Wa(L-Ls) + WP x cos 0 + K(d+L sin 0) - Pe V 3 Va L and as before Pe disappearing unless x is very small. Obviously Ha = K cos 0 and is in no way directly ef- fected by the powder force. For high angles of elevation, the moment arm Kd reverses, and d and d1 become respectively, d * c sin 0 - ht cos 0 - S and d1 » L sin 0 _ d See(fig.5) = (L-c) sin 0 + bt cos 0 + S Now taking moments about A and B respectively Wg Ls - Wr x cos IS + Kd - Pe b L and W8(L-Lg) + Wr x cos 0 + K(L sin 0 - d) * Pe 93 and Ha - K cos 0. For design use, the external reaction formulae be conveniently grouped. IN BATTERY: for low angles of elevation: j r Nvn* cos 0 + S - c sin 0) - Pe ) •» g Ul a ™ v ) V " "T~ ) WS(L-LS) + K(htcos 0 +(L-c) sin 0 +S)+Pe ( ) Ha » K cos 0 ( for high angles of elevation: WgLg+ K(c sin 0 - ht cos 0 - S ) - Pe ( L ) > Ws(L-Lg)+K[(L-c) sin 0 +ht cos 0+S] +Pe ( ( Va» ) (16) ) Ha a K cos 0 ( OUT OF BATTERY: for low angles of elevation: ( ya^s ~ wr k cos ^~ ^^ht cos 0 +S- c sin 0 ) ) ) WS(L-LS) + Wr b cos 0+K(ht cos j0+(L-c)sin 0+S) ) v ~ T""(17) c ) ) Ha = K cos 0 ( 94 for high angles of elevation: b cos 0 + K(c sin 0- ht cos 0- s WS(L-LS)+ Wrb COS0 + K(htcos 0* (L-c )sin#+S L ( Ha = K cos 0 ) These formulae are immediately applicable to platform mounts traversing about a pintle bearing as well as field carriages. In platform mounts, the horizontal reaction of the platform on the mount is usually taken at the pintle bearing which is usually located in the front or muzzle end of the mount. Hence in place of Ha we have HO = K cos 0. The reactions V^ and Va remain the same, Va now being the reaction of the platform on the traversing rollers of the mount. Very often V)j is divided into two equal vertical oomponents at the two ends of the traversing arc of the mount, and in such a case L is the horizontal distance in the projection of a vertical plane containing the axis of the bore from the pintle reaction to the traversing reaction, that is, if L is the actual distance from the pintle to the other end of the traversing arc, and 9 is the spread of the arc, then 6 L = L' cos - 2 In a field carriage, for a first approximation we may assume the horizontal and vertical reaction to be at the contact of spade and ground. These reactions are obviously Ha and Va of the previous formulae and Vfc is the vertical reaction of the ground on the wheels, and L the distance from the wheel contact to the spade contact with the ground. For split trails, 95 Va and Ha are obviously equally divided and if the gun is traversed, a horizontal reaction normal to the plane of Ha an^ Va is introduced; however, this re- action will not be considered until later, that is, the gun will be assumed at zero traverse. .A closer approximation to actual conditions in a field carriage is to regard H as acting at a vertical distance g from the ground line, usually when from 1/2 to 2/3 the vertical depth of the spade in the ground. The equations then will have an additional moment: Hag = K cos 0 g, which is substracted from the moments of the numerator in the expression for V^ and added to the mpments in the expression for Va- The general equations for field carriages are then, for low angles of elevation: WSLS- Wr x cos 0 - K.(d + g cos 0)-Pe vb = . W3(L-LS) + Hr x cos 0 + K(L sin 0 - d + g cos#)+pe Va = Ha = K cos 0 d - ht cos 0 + 3 - c sin 0 and for high angles of elevation: Y^L, _ Wr x cos 0 + K(d-g cos 0) - Pe _. *S . ^ _____ _______ ___ L WS(L-LS) + Wrx cos 0 + K(L sin0-d+g cos0)+ Pe Ha = K cos 0 d = c sin0 - ht cos0 - S where Pe disappears if Wr x cos 0 is used or vice versa. 96 Another class of mounts in which the previous formulae are not applicable, are known as pedestal or pivot mounts used on Barbette Coast mountings and for naval guns, as well. These mounts are attached to the foundation by bolts on a circular base usualljr equally spaced around the circumference. With such mounts the question of stability is of no consideration. The reaction between the foundation and mount and the distribution of the tension in the bolts, may be obtained approximately by considering the base of the mount as absolutely rigid. Then on firing, the front bolts become the most extended, the deflect- ions and corresponding stress being proportional to the distances measured from the back end of the base along the trace of the intersection of vertical plane, con- taining the axis of the bore with a horizontal plane, to the perpendicular chord connecting any two front bolts Thus if L0, L, etc. are the lengths from the base end to the perpendicular chord connecting a set of two bolts, and if jo, j, etc. are the deflections of the bolts, we have j0: jt: jt: » L0: Lt: L, Now if the bolts are of equal strength, the ten- sions are proportional to the deflections, that is TO : TI: T f * jo: jt: j8 = L0: Lt: L8— ~ that is TQ = C LQ, Tt* C Lt, Q C T C : Hence the moment about the back end holding the pedestal down, becomes, C ll + 2 C L2 + 2 C L* + C Lj = SM Considering now the gun and mount together we have, K d - WgLs - Wr x cos 0 = £M hence Kd -(WSLS- Wr x cos 0.) C = L* + 2L2 + 2L2 L2 Oil n and the maximum tensiqn to which the bolt at the farther end is subjected, becomes, 97 [K d - (WSLS- Wr x cos 0)]L0 L* + 2 ij + 21; — L; If the gun traverses 360° every bolt should be designed for the maximum tension, T0. The same method may be applied to various other combinations for holding a gun down on its foundation. BENDING Itf THE TRAIL In considering the strength AND CARRIAGE of a carriage body, the reactions at the trail, Va and Ha, subject the total carriage to a bending stress. This is of special con- sideration in field carriages of the trail type. The reaction Va causes bending while Ha decreases the bend- ing. Hence for maximum bending we should examine the conditions for maximum Va and minimum Ha. Now, WS(L-LS) + K[(L-c)sin 0 + h^ccs 0 + si +Pe •\f s -^— — ^_^__________^____^________________ a T : Ha = K cos 0 where L « horizontal distance between wheel contact and spade contact with ground (in) c = horizontal distance from spade to vertical plane through trunnions (in) ht = height of trunnions from ground, (in) s = distance from trunnion to line parallel to axis of bore and through center of gravity of recoiling parts (in) Lg = horizontal distance to center of gravity of total system, recoiling parts in battery (in) PS » powder pressure couple (in/lbs) K = total resistance to recoil (Ibs) With a field carriage, since the trunnion position is very close to the wheel contact with the ground, 98 (L-c)sin 0 is always very small compared with ht cos 0, hence, we have approx. Ws(l-ls) + K(ht cos 0 + s) + Pe Va = If Lx = distance from trial contact with ground to any section in the carriage body or trail hv = the height of the section from the ground y we have, for the bending moment at section xy, "xy = Va^x - H hy Substituting the value for Va and neglecting s being small, we have L Mxy = [WS(L-L^) + K'ht cos 0 + Pe^"r~ ~ K cos 0 hy = WSLX(1- ~) + K cos 0 (JL ht - hy) + Ps^i BENDINd \N TRWL 8r Fig. 6 Now from fig. (6) it is evident y- ht is always greater than hy, hence for maximum bending moment we must have cos 0=1, that is 0 = 0. Hence the maximum bending moment occurs at horizontal elevation, 99 At horizontal elevation, ht cos 0 + s = h henca W-(L-U) + K h + P. V, = a L but we also have critical stability at horizontal elevation, that is K h + Pe - WSLS = 0 (approx.) therefore, Va =- Ws (approx.) that is in virtue of the mount being just stable at horizontal elevation, or in practice approximately so, the vertical reaction at the spade equals the weight of the entire system, gun and carriage to- gether, yy L _ p Further Ha = K = S — (ibs) and the bending moment at section xy in the trail, becomes, h lui =WT ^ W T - P ^ * - ( -j n 1 K c ^ neglecting Pe as usually small compared with W$LS, ws have, Mxv = W_(L, -L. — ) (in Ibs) A jr o A. K For the maximum bending moment in the trail, we consider the section at the attachment of the trail to the carriage, then, Lx = Ls approx. and therefore, the maximum B. M. becomes, h - h Li — Iff f f .^^^^^^^^^^^^ l ill ro-v \T ~ ** c He* \ ) \-Li a most useful formula in a prelinary carriage layout It is important to note that if the recoil varies the above formula and analysis do not hold. When, however, the recoil varies on elevation the maximum bending moment in the trail is obtained at the minimum elevation where the short recoil COTD- mences, that is, when cos 0 is a maximum for the minimum recoil. If b_ = the short recoil at maximum elevation, then, we have, KS - maximum total resistance to recoil, then, 100 W.L^Cl - ~) * K cos 0 (^1 ht - hy) + Pe JL where Lx = distance from trail contact with ground to any distance in the carriage body or trail. h> * the height of the section from the ground. Pe 3 maximum powder pressure couple. EXTERNAL REACTIONS DOSING Counter Recoil may COUNTER RECOIL be divided into two periods, the accelerating and the retardation period so far as the external effects on the mount are concerned. During the accelerating period, the external re- actions on the recoiling parts alone, are the elastic reaction of the recuperator in the direction of motion, the guide and stuffing box frictions and a hydraulic resistance during the whole or part of the accelerat- ing period, together with the component of the weight of the recoiling parts parallel to the guides, oppos- ing the notion of counter recoil. Hence, if x = the displacement from beginning of counter recoil of the recoiling parts with respect bo guides. Ffa= the resultant accelerating force of counter recoil. Kra the resultant retarding force of counter re- coil. Fx * the recuperator reaction for displacement x trom beginning of counter recoil. R = the total friction. Hx3 the hydraulic resistance, if any, of throttling through recoil orifices or counter recoil buffer. Then, during the accelerating period 101 d x i i m j^j- = Fx - R - Hx - Wr sin 0 = Ka and for the subsequent retardation . t - m = R + Hx + W_ sin 0 - F = K' dt* Considering now the external forces on the total system (recoiling parts together with mount) the braking resistance for the recoiling parts then be- come internal reactions, and considering inertia as an equilibrating force, we have, as before the fol- lowing external forces, • d*x Ka = m- — - The inertia resistance during ac- celeration which is opposite to C'recoil. • » d'x *r md~T* ^ne inertia resistance during retardation which is in the direction of C'recoil. Wr * Wt. of recoiling parts Wa * Wt. of carriage proper Ha and Va Horizontal and vertical reactions of spade and float y^ * Front Pintle reaction - horizontal com- ponent assumed zero as before. During the accelerating period, obviously, Ka < Kr that is, FX-R-HX - Wrsin 0 < Fx+ R + Hx - Wrsin 0 hence, so far as stability and the balancing re- actions exerted i>y the ground or platforu on the carriage mount are concerned, the external effect during the acceleration period of counter recoil 102 need not be considered. If now, the inertia resistance is considered as an equilibrating force, we have Kr (d+L sin 0)-Wr [ (L-Lr)+b-x cos 0]- Va(L-La)+VaL= 0 Let d = d + L sin 0 = ht cos 0 + (L- c) sin 0 + S Hence the limitation for counter recoil stability, noting that Wr(L-Lr) + Wa(L-La) = Ws(L-La) becomes Krd' = Ws (L-LS; + Wr(b-x) cos 0 For a constant marginal counter recoil stability . moment O1 this equation becomes Krd' =[G'+WS(L-LS) + Wr b cos 0] - Wr cos 0 x and the stability slope for a constant marginal counter .ecoil stability is evidently , Wr cos 0 that is decreasing as the recoiling masses move into battery. Minimum stability is evidently in battery position and 0=0, that is Wf r r \ s \L> -L>s I h where h = d for 0=0 In ordinary field- carriages, the weight of the system in battery is very close to the wheel axle or contact of ground and wheel, consequently (L-LS) is very small . Therefore counter recoil stability is the primary limitation in the design of a counter recoil system. STABILITY The question of stability for field carriages is of fundament- al importance, it being a primary limitation imposed on the design of a recoil system. If a gun carriage is to be stable, then Kd - WaLa - *r(lr - x cos #) = 0 103 If we have a constant marginal moment G, that is an excess stability, we have Kd - WaLa - Wr (lr - x cos 0) = G K = -G + VTSLS - x v»r cos 0 = A - m x A = where — G + W I W cos 0 \* • » « L » !•«» w W Jv Thus the resistance to recoil to conform with a constant margin of stability decreases in the recoil proportionally to the distance recoiled from battery. In battery, the resistance to recoil, -G * WSLS Kb = A = — and put of battery, the resistance to recoil becomes, where b = total length of recoil -G + WSLS Wr b cos 0 d d consequently, fot a constant margin of stability, Wr b cos 0 Kb d From this we obtain the equations of resistance to recoil for constant stability against displacement, Wr cos IS -G + WsLg Kx = Kb x, where Gb = d In our Ordnance Department, KX = KQ = a constant during the powder pressure period. Thus if B represents the corresponding length of recoil, then for a constant stability moment G, -G + WSLS E Wr cos 0 K /-\ "~ v ~ , ' A ~~ 1 104 *r cos 0 and Kx - K0 (x - E) Obviously the stability "slope" or space rate of change of resistance to recoil for constant margin of stability, is ^ cos 0 ID = — — — — d where d = ht cos 0 + s - c sin 0 A3 the gun elevates, Wr cos 0 remains finite, while d decreases to zero at the elevation Si, where the line of action of the resistance to recoil passes through the spade point. Thus the stability slope "m" thereby increases to an infinite value at that same elevation. But it is important to note that the resistance to recoil out of battery is finite and increases con- siderably as "d" decreases so far as it is limited by stability. Obviously in design it is inconsistent to •ake the slope of the space rate of change of resist- ance to recoil consistent with the stability slope as the gun elevates, since the stability becomes sufficiently increased to allow a large resistance to recoil bo be used. We may, therefore, cause the slope to vary arbitrar- ily as a linear function from a maximum value at an arbitrary low angle of elevation, say some value from 0° to 6°, to zero at the angle of elevation where the resistance to recoil passes through the spade. Thus if, 0° = the initial angle or lower angle of elevation from which the slope is to decrease arbitrarily. £Jt * the angle of elevation corresponding to where the resistance to recoil passes through the spade. do = moment arm of resistance to recoil about spade point for angle to 105 8 SOO 0 .j* 8^8 * *J " Z g* L, tf m = stability slope for any angle of elevation 0 Wr cos j0 m = - = stability slope at lower angle o do of elevation. m = m0 - k (0 - 00) then, m = m0 • At angle, of elevation 0t, o = 0 hence •o * k^t - *o> °r k = TO°. 0-0Q hence m = mo - (-zr\ TT— ) (# - 00) or substituting for i " o Wr cos 00 0 - 00 Wr cos 00 0t- 0 Thus the variation of the space rate of change of resistance to recoil may be divided into two periods, (1) from 0° to 0Q w •• ~ C^™ ci Wr cos 0 OB = which is parallel to the stability slope (2) from 00 to 0° Wr cos 08 0-0 , where the slope ln W ~ & f\ is arbitrary. A graph of the variation of the space rate of change of the resistance to recoil against elevation conforming' to the assumption (1) and (2). If there is always to be an excess stability couple G we have from the previous discussion, fixed limitations for the resistance to recoil in and out of battery. Thus, from 0° to 0° o 106 - G + WSLS - G + WSL3 Wr b cos ft Kv = - : k = : - where throughout recoil G is a constant marginal stability couple, and from 0O to 0t - -G + WSLS »r b cos 0 the length of re- coil being as be- fore shortened as the gun elevates but if the stability marginal move- ment is never to be decreased for any part of the recoil below G, since the stability slope and space rate of resistance to recoil increase and decrease respectively as 0 increases from 0Q to 0 it is ob- vious that the minimum stability is in the position of out of battery. Therefore the resistance to recoil in battery is the resistance to recoil out of battery with a marginal moment G of actual stability, augmented by m b . That is, K = -G + WSLS Wrb cos 0 Wrb cos 00 ji + r J - 0 1 ] d d do < »J=*0 cos 0 cos 00 0- 0 f * 1 \ LENGTH OF RECOIL Obviously the overturning CONSISTENT WITH force, that is the resistance to STABILITY OF MOUNT recoil, is a function of the length of recoil varying roughly inversely as the length of re- v coil. Hence as the gun elevates the stability in- creases and the recoil may therefore be shortened. In a preliminary design it is desirable to know the length of recoil as limited by stability, from 0° or the lowest elevation wherein stability is de- sired to the elevation 0° where the stability slope 107 is made to change arbitrarily. Let Cs = the constant of stability = Overturning moment = where the overturn- Stabilizing moment ing moment = Krd and the stabilizing moment = WCL_ - W_ b cos 0 OO I We may consider the limiting recoil at various elevations (1) with a constant resistance to recoil as would occur in certain types of re- coil systems. (2) with a variable resistance to recoil using a stability slope as outlined in the previous paragraph- For a constant resistance to recoil : = K : The critical position of stability is obviously with the gun at the end of recoil out of battery. Then CS(WSLS - Wr b cos 0) K = - -r- >" -sj • :?f.Q -j^bwcq arfJ $.«iii;fe 1 va 2 mr Vf K = Ses "DYNAMICS OF RECOIL". Chap. VI. b-E +VfT Where E = displacement during powder period in free recoil. T = total time of free recoil. Vf = Max. free velocity of recoil, hence CS(WSLS - Wr b cos £5) b-B+VfT d The above equation reduces to the quadratic form Ab2 + Bb + C = 0 and its solution is, /—* - B ± / B - 4AC b = — - 2 A 108 Where A = Wr cos B = Wr cos 0 (VfT - S) - WSLS C = WsLs(VfT - E) For rough estimates, especially where the length of recoil is comparatively long, we may assume, ^ mrVf CS(WSLS - Wr b cos 0) - B + /B2 - 4 AC Cs Wr cos b . » and A . - C = - mr V£ For Variable Resistance to Recoil: - - -- -- ----- The resistance to recoil is assumed constant during the powder pressure period and thence to de- crease uniformly with a stability slope as given in previous article. Therefore, from the end of the powder pressure period to the end of recoil, the stability factor remains constant from 0 to 0O (i.e. to where 'the stability slope is made to change arbitrarily) . At the end of the powder period. (See Dynamics of Recoil): a Kd = G.(»3LS - Wp (E - -—-) cos 0 2m_ hence CS(WSLS - *rE cos 6) do I w -- COS 0 Now the resistance to recoil out of battery at the end of recoil, becomes, 109 KT* K - m (b - E + - ) (See Dynamics of Recoil) 2m r hence by the equation of energy VT* KT2 KT 9 [2K-m(b-E+ £-)] <b-E+ g-j - Mr(Vf ---- )2 /snip «•• nr Expanding and simplifying, we have the quadratic form: Ab* + Bb + C = 0 - B + B2 - 4AC where b = - — - H fr cos 0 0 and A = m = Cs — 7 - from 0 to 0 B = - 2K - 2mB ID. c = [2C 2VfT) K + C(WL - WE cos 0) From #o to 0t degrees, the stability slope is made to change arbitrarily, decreasing proportionally with the elevation from the stability slope at j0o to zero slope at 0^ where the line of action of the re- sistance to recoil passes through the spade point. The critical stability is obviously at the end of recoil, and the resistance to recoil in battery (K) is the resistance to recoil out of battery (k) aug- mented by the product of the length of recoil from the end of the powder period to the end of recoil multiplied by the arbitrary stability slope (m). From the energy equation, we have, 110 2 KT KT (K + k) (b - E + £L-> = m (Vf - — ) 2rar mr now K = k + m(b-E + — ) 2mr K = 5 — = the constant resistance to mT 1 _ recoil during the powder 2rar period, and CS(WSLS - Wr b cos 0) k = = the resistance to d recoil at the end of recoil. Substituting these values in the enery equation, we obtain a quadratic equation in "b!! A sufficient approximation and simplification can be made, by not- ing that 2 E - — — - 0.9 E approximately and 2mr KT Vf = 0.9 Vf approximately Therefore, (K +~k) (b- 0.9E) = 0.81 mr vj and K = k + m(b- 0.9E) Cs(WsLg - Wr b cos 0) = + m(b-0.9E) d substituting in the energy equation, we have, 2Ca (W_LS - Wr b cos J0) + m(b- 0.9E)(b- 0.9E) = 0.81 m_Vf d z Reducing and simplifying, we have the quadratic sol- ution, ,— -B ± / B - 4AC b = — 2A s where A = m - —7- Wr cos 0 2C, III 2Cf * °-9E wr cos 2 Q = 0.8l(mE - mrVf o from 0 to Wr cos 00 0t - 0 m , ( ) from 0 to 0 For a close approximation and when the resist- ance to recoil is not constant during the powder period, if K = the resistance to recoil in battery k = the resistance to recoil out of battery, we have, Kl- fe m V ? r R 111 M V f ( >b = (approximately) but K = k + mb cs and k = 7- 0»_L, - W_b cos 0) d as Substituting, we have 20 s [-•— (WSLS - Wr b cos 0) + mb] b = mrVf and the value b, becomes, -B ± /V - 4AC b = 2A where pp 3 nr A = m - ~r~*'r cos « .fee d 2CS B = -r WSLS c = - «rvf8 N Wr cos 0 ra = " or any arbitrary slope as desired. d The above formula is sufficiently exact for a preliminary layout with a variable recoil and resist- ance to recoil provided the margin of stability is chosen fairly large, that is when a low factor of stability is taken. JUMP OF A FIELD CARRIAGE When the overturning moment exceeds the stabiliz- ing moment, we have unstabil- ity and an induced angular rotation about the spade point. After the recoil period, the gun carriage is returned to the ground by the moment of the weights of the system. This phenomena is known as the jump of the carriage. For the condition of unstability, we have: K d ~ *s^s * *r cos > ° where as before, K = total resistance to recoil Wg= weight of entire gun carriage including gun 13= distance from spade contact with ground to center of gravity of total system in the battery position. wr= weight of the recoiling parts x = movement in the recoil of the gun. To analyse the motion of the system, consider (a) the recoil or accelerating period. (b) the retardation or return period. The recoil period may be subdivided into the 113 powder period and the pure recoil period. During the recoil period the gun and gun carriage are given an angular velocity which reaches its maximum at the end of recoil. During the retardation the angular velocity is gradually decreased to zero, but with increased angular displacement, the maximum angular displacement occuring when the angular velocity reaches its zero value. Further change in angular velocity results in a negative velocity and a corres- ponding angular return of the mount to its initial position. ttniliootn jo x-*Y"''^ The acceleration during the recoil period is not constant, even with constant resistance to recoil, due to the fact that the moment of inertia and the moment of the weights of the recoiling parts about the spade point varies in the relative recoil of the gun. Therefore, the angular acceleration is not constant during the accelerating period. Likewise during the return of the recoiling parts into battery. Further the effect of the relative counter recoil modifies the return angular motion. Consider the reaction and configuration of t"he recoiling parts and carriage mount respectively. See figure (7). 114 Let X and Y = the components of the reaction between the recoiling parts and carriage mount, parallel and normal to the guides res- pectively. M = the couple exerted between same. Ia= the moment of inertia of carriage mount about the spade point. Ir= moment of inertia about the center of gravity of the recoiling parts. dx = perpendicular distance from spade point to * i line of action of X. dx = perpendicular distance from X to center of gravity of recoiling parts. d = dx + dx = perpendicular distance to line parallel to guides and through center of gravity of recoiling parts from the spade constant with ground. Q = angle made by d with the vertical 0 = angle of elevation of the gun (in battery) x = distance recoiled by gun from battery position x0s distance from "d" to center of gravity of re- coiling parts in battery measured in direction of X axis of perpendicular to line d. r = distance from spade point to center of gravity 5 J of recoiling parts, e = angle r makes with vertical lr= horizontal distance to center of gravity of recoiling parts from spade contact with ground . Wa = weight of carriage proper (not including recoiling weights) ra = distance from spade point to center of gravity of carriage proper, o = angle ra makes with horizontal la = horizontal distance from spade point to center of gravity of carriage proper. Then lr = (xQ-x) cos 6 - d sin 6 115 la = ra cos (9 + a - 0) where in battery 6 = 0 and for any other angular position during the jump of the carriage, 6=0+6 B=a variable angle during the jump, For the angular motion about the spade point, For the carriage mount, without the recoiling parts, d 6 i " "u a~* "** d + 2 Xdx - Y(x0 - x)+ m - wala = Ia T— - (i) and for the recoiling parts, adding (1) and (2), we have, d*6 (3) Xd - Y(xQ-x)- wala = (Ia + Ip) — - Since the recoiling parts are constrained to rotate with the carriage mount, they partake an angular acceleration about the spade point combined with a relative acceleration along the guides. The acceleration of tne recoiling parts is AfT Jtfc> 9& divided into: (1) The tangential acceleration of the recoiling parts about the spade point; due to the constraint in the guides, dt2 a°d is divided into components in the x and y direction d*e /« j d2e ) - cos (e + e) = d — -— f dta dt2 116 (2) The centripetal acceleration of the recoiling parts about the spade point due to the constraint in the guides, s ,de . 2 rw - r (. ) dt and divided into components in the x and y direction. ,d9,* , N,d6v! r(— ) sin (9 + e) = (x0 - x)(— 4 r(- — )* cos (e + e) = d at (3) The relative acceleration of the recoiling parts d'x <*vr —— — - -T — along the x axis (4) The relative complimentary centri- petal acceleration due to the combined angular and relative motion of the re- coiling parts: de (5) The angular acceleration of the recoiling parts which obviously equals the angular acceleration about the spade point, that is d2e dt! For the motion of the recoiling parts along the x axis, we have dvr d% Pb - m_ - ard 4 wr sin e - mp(xQ- dt dt« 0 (4) 117 For the motion of the recoiling parts normal to the guides, x d*e de ,,de .a Y - ra_(xn-x) -- wr cos 9+2 nrvr - + m_d(— ) = dt» dt d t 0 (5) Substituting (4-) and (5) in (3) we have, 1 2 2 2 ur 1 — mi 1 + 9m v ( v — v ^ - - aL . _ at_ "r^r Wa1a 60Brv,Axo x; *9 d29 (Ia+Ir) — - = 0 (6) dt dt where Ir=(x0~x) cose - d sin 0 ) ' functions of the , . variable angle la =ra cos (e+a-6) ) e From equations (4) and (6), we have 8 as a function of t. An exact, solution of these dif- ferential equations is complicated and therefore an approximate solution must be resorted to. APPROXIMATE SOLUTION 09 THE JUMP Of A FIELD CARRIAGE. The static equation of recoil, that is the equation of motion of the recoiling parts upon the carriage is stationary, becomes, s >• W_ sin e -XR * 0 and the equation of motion of the recoiling parts along the guides when the carriage jumps, becomes, 9 - X - o,rd L£ -«r <x0-x> (A 0 118 Now the term mr(xrt-x)( — ) is small and may be neg- dt lected, but on the ,2 d 9 other hand the term m_d may be considerable. dt2 Furthermore the braking X and Xs may differ considerably as well. The term 2 Pb-mr - — -= Ks in static recoil. whereas with the jump of a carriage ,2 Pb- mr = cKs where c = 0.9 approx. dta During the pure recoil on retardation period of the recoiling parts, we have m_- — - = K, in static recoil. rdt2 whereas when the carriage jumps, ,2 m_ — - = cK, where c = 0.7 to 0.9 r dt8 Considering the moment equation for the movement of the total mount about the spade point, we have, (pb-rar — * )d-[mr(d2+(x0-x)2)+Ir+Ia] +2mrvr(xo-x) dt2 dt2 d6 _ **rlr _ „ I _ Q dt a a where lr=(xo-x) cos 6 - d sin 6 la=r_ cos (0 + a -v ) de The term 2mrvr(xo-x) — is always small and may be Q t neglected . If b = length of recoil, for the average during the recoil, let b v — y — Y — •«- *o x ° 2 119 Then, we have, for an approximate solution, 2 cKd-[mr(d%(x0- |)2 + Ir + Ia] i-i - wsls = 0 (7) where wsls = «ala + «rlr c = 0.8 to 0.9 and ws = wa + wr and approximately, if the jump is small, lr= (x0- -) cos 0 - d sin# _ . ~ _ . .. i ' t la= ra cos a 0 = angle of elevation of the gun b = length of recoil d = perpendicular distance of line parallel to guides and through center of gravity of recoiling parts from spade contact with ground. XQ= the perpendicular distance from d to the center of gravity of the recoiling parts ra = distance from spade point to center of gravity of carriage proper a = angle ra makes with horizontal Hence for the angular acceleration, d29 _ eKd- wsls _ rad ,2 x,2 D72 I I - dt mr(d +xo- -) +Ir+ Ia sec* If we assume a constant acceleration, we have, for the angular velocity attained at the end of recoil, d0 (cKd - w,,ls)t rad ( \ At 2 1 mr(d(x0- |)-Ir-Ia where t^ = the time of recoil we have approximately, cKs wv -H w 4700 where V = 0.9 120 w= weight of projectile (Ibs) w= weight of charge (Ibs) ws= weight of recoiling parts (Ibs) c s 0.9 approx. and tp= the total powder period ob- tained by the methods of interior ballistics. The angular displacement during the first period of the jump, becomes, t (cKd-wsl3)t* 6 = - B radius. - During the second period of the jump we have, the angular velocity decreasing but the angular displacement still increasing: then rad ~ Integrating, we have t «r(d»(x0- |)2)+Ir+Ia and for the angular displacement, SI *,!< (- -Ox* L)^m it - ws!3 t de - *)2) +1, +1, + ~ fc " 9l To determine the time of jump required to attain the maximum angular displacement, we have the angular velocity reduced to zero, whence, aet wsis ta » j i. ' ~ , from which we may determine ts> Therefore, the 121 maximum angular displacement, becomes, _I *rde mr^d2+(x0- ' 2 tf The effect of counter recoil is to increase t 2 and decrease the negative moment (- w0l ). S S RECAPITULATION OF FORMULAS: EXTERNAL EFFECTS AND STABILITY, Resistance to recoil: J Mf K = (Ibs) constant resistance through- b-E+Vfr out recoil. mp= mass of recoiling parts = wr (16s) I g = 32.16 ft/sec. a b = length of recoil (ft) E = free recoil displacement during powder period (ft) T = time of free recoil (sec) Vf= Max. velocity of free recoil ft/sec. BETHEL'S FORMULA . mrvf 1 (Ibs) Constant lD + (.096 +. 0003d )M Vf resistance. Vo M = travel up bore (inches) VQ= muzzle velocity (ft/sec) d = diam. of bore (inches) Assuming a gun carriage to be supported by a hinge joint at the rear (A) and a vertical support 122 in the front (B) we have the following equations for the reactions of the supports: Let Ha-= horizontal component at rear hinge support or spade of carriage. (Ibs) Va= vertical component at rear hinge support (Ibs) Vjj= reaction of front support assumed vertical (Ibs) L = horizontal distance between carriage supports (in) ht= height of trunnion above support (in) s = perpendicular distance from center of gravity to recoiling parts to line of action of the resistance to recoil (in) c = horizontal distance from rear support to trunnion (in) K = total resistance to recoil (Ibs) <& = angle of elevation of gun g = vertical distance from ground to horizontal .component of resultant spade reaction. IN BATTERY: For low angles of elevation: w<.L»-K(h+ cos 0 + s-c sin 0)-Pe ' :v - ' - K[ht+ cos 0+(L-c)sin ) Ha = K cos 0 123 For high angles of elevation wsL, +K(c sin 0 -ht cos 0 - s)- Pe ) ) Vb = ( ws(L-Ls) +K[(L-c) sin 0 + h + cos 0+s]+Pe < (Ibs) ( va - -r ~) ( Ha = K cos 0 ) OUT OF BATTERY: For low angles of elevation: w"3Ls - Wrb cos 0-K(ht cos 0 + s -c sin 0 ( WS(L-LS) +W_b cos 0 +K(b+ cos 0+(L-c)sin0+a> ) Va = - *— - ( ) ( ( Ha = K cos 0 ) For high angles of elevation: ( ) ) wgLs-»rb cos 0+K(c sin 0 -ht cos 0-s) ( ( Vb • - L - ) ) ( ( »?g(L-Ls) +»rb cos 0 +K(htcos 0(L-c)sin0+s) ) Ha » K cos 0 ( With a field carriage where the spade is in- serted in the ground, the center of pressure lies a distance "g" inches vertically down. The general equations for the support of a field carriage, therefore become, 124 For low angles of elevation: wsLs- wrx cos 0-K(d+g cos 0)- Fe ( b L ) ws(L-Ls)+wr x cos 0+K(L sin 0-d+g cos0)+Fe ( Va= )• ( Ha = K cos 0 ) ( d = ht cos 0 + s - c sin 0 ) For high angles of elevation: \ - *sLs wr x cos 0 *K(d-g cos 0)-Fe ; V^_ ( Va=w(L-Ls)+wr x coS0+K(L sin0 -d+^ cos 0)+Fe ) ) ( ( Ha= K cos 0; d=c sin 0-htcos 0 - s ) In certain types of Barbette mounts, we have the bottom carriage held down by tension bolts to a circular base plate. If we draw a series of parallel chords through the bolts on either side of the axis of the gun, and if we let the distance from those several chords measured from the rear bolt, be LQ, L - — - - I»n. we have, for the maximum tension induced in a tension bolt given by the ex- [Kd-(wsL3-Wr x cos 0)JL0 TO = — — : — : BEHDINa III THB TRAIL AND CARRIAflB. Considering the section at the attachment of the trail to the carriage, for a constant length of 125 recoil the maximum bending in the trail occurs at horizontal elevation and is given by the following expression: h_h B.K. at the attachment of trail to carriage. h = the "height of the center of gravity of the recoiling parts (axis of bore practically above the ground when the gun is in its horizontal position. hvs= the height of the neutral axis of the section above the ground. ws- weight of entire mount including the gun. Ls= horizontal distance from the spade to the center of gravity of the weight of the entire mount. When the recoil varies on elevation, the maximum "bending moment in the trail is obtained at the minimum elevation where the short recoil commences, we have, ^s kx Lx Mxv=wsLx(l )+Ks cos 0S( — ht-hv)+Pe — - L L !-• where Ks = maximum total resistance to recoil corres- ponding to short recoil "bs. 0S= minimum angle of short recoil. Ls= distance from trail contact with ground to any distance in the carriage body of trail. hy= the height of the neutral axis of the sect- ion from the ground. Pe = maximum powder pressure couple. STABILITY OP COUNTER RECOIL. In the design of a field carriage counter re- coil stability is a basic limitation. We have for counter recoil stability that. 126 The equation stability, gives, for variable resistance to recoil, for low angles of elevation consistent with the stability slope, -8+ /B* -4AC b = 2 A where ff_cos 0 o ^o A = • = Cs -*— (from 0 to 0O elevation) B» SSI - 2K-2mE mr and CS(WSLS- wr E cos 0 ) (Ibs) w T2 d - C3 — E- — cos J25 2rar After an arbitrary elevation 00 (approx.5 ) the stability of the mount greatly increases with elevations and therefore the stability slope is made to arbitrarily decrease with the elevation arriving at constant resistance to recoil at the elevation corresponding to where the line of action of the resistance to recoil passes through the spade point. To estimate the minimum recoil allowable for the various angles of elevation in this range, we have -B± /B2-4AC b = 2A . jsstaas jjaiJ BC 0 J0.-0 from<.00 to 3 * ^ A = D — - 5f_ COS 0 d 2Ca B = (waLs* P.9E wr cos 0)-1.8 mE 0.81 127 < • (L-t.) Rr " "~"h" where Kr= the total resistance of counter recoil at horizontal elevation. ws= weight of entire mount including gun. L3= horizontal distance from spade to center of gravity of ws. L = horizontal distance from spade to wheel contact with ground. Further 2 „ d x (4— f- may be obtained from the velocity curve of counter recoil towards the battery position). and Kr= Hx +R+wr sin 0 - Fx where Hx = hydraulic or buffer brak- ing at end of counter recoil. (Ibs) R = total friction resistance wr sin 0 =0 weight compound equals zero at horizontal elevation. Fx= recuperator reaction. (Ibs) RECOIL STABILITY The stability limitation of the resistance to recoil varies in the recoil due to the movement of the recoiling weights. The slope or rate of the variation in the recoil of the equivalent force applied through the center of gravity of the re- coiling parts and parallel to the guides that will just overturn the mount, is given by the following expression: wr cos 0 ra= from 0° to 0 128 where ra = the stability slope 0 3 angle of elevation d = perpendicular distance from spade to line through center of gravity recoiling parts parallel to the guides. wr=» weight of recoiling parts 00= the initial angle or lower angle of elevation from which the slope is to decrease arbitrarily. If from 0O the slope is made to decrease ar- bitrarily with the elevation, to the elevation 0, the angle of elevation corresponding to where the line through the center of gravity of the recoiling parts parallel to the guides passes through the spade point, we have for the stability slope wr cos 00 0t- 0 m = (— — ~r— ) where the slope is d 0 ~0 arbitrary. LBMQTH Of RECOIL COM3ISTBNT WITH STABILITY OP MOUNTS. The equation of stability, gives, for constant resistance to recoil, -mrVf _Cs(wgLs -wrb cos The solution of this quadratic equation for b, gives: -B± /B -4AC b = 2A where A= i»r cos 0 ) where all units 8= wr cos 0(VfT-£)-wsLs ( are in feet ..2s ) and pounds. _ i"r»i ci ) CHAPTER IV -soo INTERNAL REACTIONS. <: - ' ' In the design of the various parts of a gun carriage it is of fundamental importance that we have a coraplste knowledge of the stresses to which each member is subjected, and the variations of such throughout recoil and the position of elevation and traverse . We have already considered the external reactions on the whole system, and such reactions are useful in computing the stresses in the supporting structure for a gun mount as the strength of concrete emplace- ments for barbette mounts, or the strength of a rail- way car or caterpillar frame. The primary internal reactions within a gun and its mount may be classified as follows: (a) The mutual reactions between the recoil- ing parts and the carriage proper or gun mount. (b ) The mutual reaction, between the tipping parts or cradle and the top carriage. (c) The mutual reaction between the top carriage and bottom carriage. The mutual reaction (a) is between the moveable and statipnary part of th3 total system during" the recoil; that of (b) between the moveable and station- ary parts during elevation of the gun; and that of (c) between the moveable and stationary parts in traversing the gun. The mutual reaction (a) may be subdivided into individual or component reactions as follows: (.1) The reactions of the constraints due to the guides or clip reactions at 129 130 the two ends of the clips in contact with the guides, which may be subdiv- ided into friction and normal com- ponents . (2) The mutual reaction of the elastic medium connecting the recoiling parts to the carriage proper, that is, the hydraulic brake and recuperator re- action, together with the joint frictions. This will be known as the elastic reaction between the recoil- ing parts and carriage proper. The mutual reaction (b) may be subdivided into: (1) The trunnion reaction between the tipping parts and top carriage. (2) The elevating arc reaction between the elevating arc of the tipping parts, and the pinion of the top carriage. The mutual reaction (c) may be subdivided into: ,(1) The pintle or pivot reaction between the pintle bearing on the bottom car- riage or platform mount and the pintle of the top carriage fitting within this bearing. (2) The traversing arc reaction, that is, the reaction between the traversing arc of the top and bottom carriage. These are usually roller* reactions for platform or pedestal mounts, the rol- lers being either a part of the top or bottom carriage or else clip reactions field carriage and may be more or less distributed about the arc of contact. Let X and Y - the coordinates of the center of gravity of the recoiling parts along end perpendicular to the guides with origin at center of gravity of re- coiling parts. 131 xt and yt = coordinates of front clip reaction measured from the center of gravity of the recoiling parts. &t = Normal component to guides of front clip reaction. uQt = Frictional tangentional component of front clip reaction. x and y = coordinates of rear clip reactioa 2 2 ^^^ measured from the center of gravity of the recoiling parts. Q = Rear clip reaction normal component. u&2= Rear clip reaction frictional component. B = nQ + nQ = total guide friction. 1 2 B = elastic reaction (hydraulic breaking and re- cuperator reaction including friction of joints) assumed parallel to the guides. F = the total powder pressure on the breech of the gun which necessarily lies along the axis of the bore. e = the perpendicular distance from center of gravity of recoiling parts to line of action F, that is to axis of bore. Assuming as in Chapter III, the mount to be hinged at the rear or breech end to its support and resting on a smooth surface at the front end, and if d = perpendicular distance from hinge to line through center of gravity of recoiling parts, parallel to guides. djj= perpendicular distance from hinge to line of action of B. lr= horizontal distance from hinge to center of gravity of recoiling parts in battery. la= horizontal distance from hinge to center of gravity of stationary parts of system (includes stationary parts of tipping parts). From fig.(l) considering the reactions on the re- coiling parts alone, we have from the equations of motion: 132 Fig. 1 133 for notion along the x axis, 2 .F"— B — uQ - uQ + w*r sin 0 = mr— — (1) and since there is no motion along the .y axis, .-••; -ie* adi oj ; .- 'r^ i»fljft?j';^ i'-'ic'~ Qf- Qi = wr cos ^ (2) tdJ *c and taking movements about the center of gravity since there is no angular acceleration B(d-dh)-u(Q y -Q y )+Fe-Q x - Q x =0 (3) u 22 1122 Now in fig.(l) considering the gun carriage or mount including the stationary parts of the tipping parts, we have for the moments of the reaction of the recoiling parts on the gun mount about the hinge A, lr-x cos 0 1 - x cos 0 Qt(xt* + d tan 0)-Q2( +d tan 0-x ) oo: C°3 0 WxiEWUAiU^aaaal cos * ?bi?ib +uQ (d+y )+uQ (d-y )+Bdh= 2M_Q (4) 1 " 1 2 2 D ~ a but uQiyi-uQzya=B(d-db)+Pe-Qixt-a8x2 and uQ <-uQ =R 1 2 Substituting these values in (4), we have, lr-x cos 0 (Qt-Q2)( + d tan 0)Bd+Rd+Ae=SMra cos 0 that is, -Wr(lr-x cos j0)-Wr sin 0 d+Bd+Fe+Rd=»2Mr or simplifying and combining, we have, (B+R-«lr sin 0)d+Fe-Wr(lr-x cos 0)=2Mr (5) at maximum powder pressure, x is usually negligible and the equation reduces to: (B+R-Wr sin 0)d+F9-Wrlr= SMra (51) From this we observe that the reaction between the recoiling parts and the mount is equivalent in effect to a force (B+R-Wrsin 0), the line of action of which is parallel to the axis of the bore or guides and 134 passes through the center of gravity of the recoiling parts, and a couple of magnitude Fe, due to the powder pressure, together with a component equal to the weight of the recoiling parts and in its line of action assumed concentrated . Thus the reaction on the gun mount of the recoil- ing parts, therefore, is equivalent to a single con- centrated force, the resultant of (3+R-Wr sin 0), equal to the total resistance to recoil and a force equal to the weight of the system together with a couple Fe. Since a couple and a single force in the same plane are equivalent in effect to a single force, parallel to the former, and displaced from it equal to the couple divided by the force, the resultant reaction on the mount of the recoiling parts reduces to a single force; the resultant of B+R-W sin 0 and Wr, which be- comes, since B+E-Wr sin 0 = K, equal to J= /K2+yf* -2KWr sin 0 and the line of action of J makes an angle -l(Wrcos 0) -KW-cos 0) & =Tan *- =Tan * (K+Wrsin 0) (B+R) with the axis of the bore and is displaced a dis- tance Fe. frora the center of gravity of the recoiling J parts. It is however, more convenient in computation to resolve this resultant into its components, K and Wr together with Fe. If now we consider the equilibrium of the gun carriage mount, we have for moments about the hinge point, 2Mra- WaLa +Vb 1=0 that is, (B+R-Vfrsin 0)d+Fe-Wrlr -WaLa+Vb 1 = C (6) and since Wala + Wrir = WS18 135 136 Equation (6) reduces to (8+R-Wrsin 0)d+Fe-WsLs+Vb 1 =0 (6') which of course is the same as the equation obtained in Chapter III. It is, however, very often more convenient to regard the mutual reaction between the recoiling parts and carriage as divided into component .re- actions along and normal to the axis of the bore to- gether with a couple. See fig. (3). Let x and y be the coordinate axes along and normal to the axis of the bore or guides. Xr= the sum of the component reactions along the x axis. Yr= the sum of the component reactions along the y axis. Now by introducing a couple Mr between the re- coiling parts and carriage, it is entirely immaterial where we assume the line of action of Xr and Yr. Let r = the perpendicular distance from Xr to the center of gravity of the recoiling mass. z = the perpendicular distance from Yr to the center of gravity of the recoiling mass. Considering the equations of motion of the re- coiling mass, we have, ,2 P*wrsin /HCr=Mr~j- Yr»Wrcos 0 ) (7) » Pe +Xrr -Yr z - But Xr-Wrsin 0 = P-Mr 777 =Ka is the resistance to recoil during the accelerating period, and 2 Xr-Wr sin 0 = - Mr -jpr = Kr ia the resistance __ to recoil during the retardation period. In general Ka and Kr are 137 Fig. 3 138 different in value. Hence, let K = Xr~Wrsin 0 for any given displacement x of the recoil. If we now consider the reaction of the recoiling parts on the carriage proper, the moments about the binge A of this reaction, becomes, lp- x cos 0 Xr(d-r)-Yr( + d tan 0-z)+Mr=ZMra cos 0 Inserting values for Yr and Mr from the equations of notion of the recoiling parts, we have: Xrd-Wrlr+Wrx cos 0-Wrsin 0d+Pe = 2Mra hence, (XrWrsin 0)d-Wr(lr-x cos 0)+Pe= 2Mra ) or ( Kd-Wr(lr-X cos 0)+Pe= SMra ) which is the same expression as obtained before. We may regard the line of action of Xr and Yr to pass through the center of gravity of the re- coiling mass, together with a couple M=Pe, the powder pressure couple about the center of gravity of the recoiling parts. See fig. (4) Taking moments about A we have directly 1 - x cos 0 Xrd-Yr( ) + d tan 0)+Pe = 2Ura (8) cos 0 and since Yr= Nr cos 0. we have as before (Xr-Wrsin0)d+Pe-Wr(lr-x cos#)= 2Mra (9) where JCr=B+R The object of this analysis has been to show that so far as the external effect of the reaction of the recoiling parts on the carriage mount is concerned the exact location of rod pulls or the lins of action of the guide frictions, is entirely immaterial, though as we shall see immediately, the value of R, the sum of the guide frictions, does depend upon the line of actions of these pulls together with the friction line 139 140 of action of the guide friction itself and thus in- directly the external effect on the carriage mount is affected slightly. Further the location of the center of gravity of the recoiling parts may considerably change the guide frictions during the accelerating or powder pressure period. BRAKING PULLS The total resistance to recoil if assumed constant throughout the recoil is readily evaluated from the following relations: If K = total resistance to recoil (assumed constant) (Ibs) b = length of recoil (ft) Vf= velocity of free recoil at end of powder period (ft/sec) B = displacement of free recoil at end of powder period (ft) T= time of powder period (sec) Then from the energy equation for the movement of the recoiling parts after the powder period, we have, (ft.lbs) Simplifying, K = « mr f (Ibs) b-E+VfT With a variable recoil consistent with a stability slope "m", and assuming a constant resistance during the powder period, we have, If K= the resistance to recoil in battery k= the resistance to recoil out of battery instability slope = °"r h (h=height of axis of bore above ground) 141 2 then :^l{b-(E- —-)]= £ mr(Vf-~)* (ft. Ibs) m and k =K-m[b-(E- - — )] (Ibs) Combining and simplifying, we have, K = ~L^~ ~* (lbs) 2[b-E+VfT- - — (b-E)] in battery £ fflp at ettikjj si :£!!•.•<> r»i V"^^*^09 KT2 k =K-m(b-(E- - — )] out of battery II ) , r • B-the total braking pull (Ibs) R=the total guide friction (Ibs) Pk=tne total oil pressure on the hydraulic pis- ton (Ibs) p' rj)=the hydraulic reaction plus the joint frictions (stuffing box + piston) (Ibs) Pa=the total elastic reaction(due to compressed air or springs) (Ibs) p' ra=the total elastic reaction plus the joint frictions (stuffing box + piston) (Ibs) fti=the normal front guide reaction (Ibs) ^»=tbe normal rear guide reaction (Ibs) u=coef f icisnt of guide friction (0.15 to 0.25) Then K=B+R-Wr sin 0 (Ibs) Total resistance to recoil where B=P' + P' (Ibs) Total braking h a R= u(Qt+Q8) (Ibs) Total guide friction The stuffing box friction is usually assumed at from 100 to 150 Ibs. per inch of diameter of rod, and if du and da are the stuffing box diameters of the hydraulic and air cylinders respectively, we have 142 (Ibs) Pa=Pa+100da (Ibs) GUIDE OR CLIP REACTIONS The recoiling mass is constrained to translation V <• C 1 1 parallel to the axis of the jc. 3 (i r "bore by the recoiling masses engaging in suitable guides in the cradle of the top carriage. In general the re- coiling mass may recoil in a sleeve, a part of the cradle, or along guides considerable below the axis of the bore and the center of gravity of the recoiling parts . For the former case, considering the external re- actions on the recoiling mass, fig. (5) aa-at=Wr cos 0 Q xl + &8x2+u (Qtyt-Q-2y2)-Fe-B b=0 (moments about the center of gravity of re- coiling parts) where eb= d-db , then &txt+(at+llr cos 0)+u (Q^-Q^y.,- cos 0)-"Fe-Bb = 0 cos J0(x2~uy2)- Fe-Bb=0 Hence Fe+Beh-W_ cos 0(x -uy ) Qt= — * (10) Further Fe+Beh-W_ cos 0 x +W_cos 0 uy +Vlrcos 0 x +W_cos 0 x + U* 2* 2' 1* 2 Wr cos 0 uy -Vf.cos 0 uy i 'II "2 143 0(x +uv ) Hence Q2= - - - — (11) When the guide reactions are below the axis of the bore as in Fig. (4) y2 remains the same in the above formulae whereas yt reverses in sign. Hence for case (2) Fe+Beb-W_cos 0(x -uy ) Q= - — - - - - 2_ (12) and Fe+Beh+HLcos 0(x -ay ) ft= - loti - * - L_ (13) The total guide friction becomes, R=«(Q1+Q8) hence 2(Fe+Beb )-Wrcos 0 x2+Wrcos 0 ny2+ ¥rcos xt + x2 + u(yt - ya) 2(Fe+Beh)+Wrcos 0f (x -x J + ii(y +y )] (14) x +x +.u(y -y ) 1 2 w 1 ' S Now if M=x1-t-x,+ u(y1-y,!) for case I or M=x1+xs-u(yi+ya) for case II and if N=(xt~X2)+ u(yt+ye) for case I N=(xi-x2)+u(y2~y1) for case II we have therefore, in general that 2(Te+Beb)+Wrcos 0 N R= - - - - - u (15) M which gives the total guide friction. The value of the coefficient of friction u ranges from 0.15 to 0.20. The total braking evidently becomes, 2(Pe+Beb)+Wrcos 0 N B+ - u-K+Wr sin 0 M or B(M+2uh)=(K+Wr sin 0)M-(2Pe+Wrcos 0 N)u hance (K+Wrsin 0)M-(2Pe+Wr cos 0N)u BS . M+2Uh which gives the total recuperator reaction in terms of the total resistance to recoil. Denoting as before by Pjj= the hydraulic reaction plus the joint frictions (stuffing box and piston) P0 ~ the total elastic reaction plus the joint 3 frictions. 6^= distance from center of gravity of recoil- ing parts to line of action of hydraulic brake pull Pn ea= distance from center of gravity of recoil- ing parts to line of action of the re- cuperator reaction Pa The front and back clip reactions become, , a a r ,g ( and cos (17) y^ reversing in sign when the guide reactions are entirely below the axis of the bores. Combining as before and noting that F=u(Q +Q ) we have t 2 2Fe+22p' ea+2ZPu eh+Wr cos 0 N R- — h h r (18) M where M and K are the constants referred before. Now K= £Pa+2Ph+R-Wr sin 0 (19) 145 and combining (18) and (19) we get KM=(2Pa+2Ph -Wrsin 0)M+ n(2Fe+22Paea+22Pheh+Wrco5 0 N) Simplifying, KM=((2Pa-Wrsin0)M+ZP,!| x M+ u(2Fe+22Paea+Wrcos 0 or further simplifying, a+Wrcos 0 N) Hence, M(K_sp'+Win 0)-u(2Fe+2XP'ea+N Wnos 0) I at 0 a M+2U6J, which gives the gross hydraulic pull in terms of the total resistance to recoil, the gross air or spring reaction and the maximum powder force. APPROXIMATE FORMULAE Assuming the reaction be- GUIDE FRICTION tween the recoiling parts to be equivalent to a normal force passing through the center of gravity N, a couple M, and the braking and guide friction forces B and R having moment arms about the center of gravity of the recoiling mass equal to dv and r respectively where r is the mean distance to the guids frictions, we have, for moments about the center of gravity of the recoiling parts, Be^+Rr=M neglecting the powder effect which is usually very small and N=Wr cos 0 for the total reaction. Obviously the actual normal guide reactions, becomes, 0 x and . W_ cos 0 x. fl s -M+ i. * 1 1 where 1 = x +x also R=u(N +N ) 12 12 146 2M+lfrcos 0(x -x ) hence R=u ( • — ) Substituting the value of M, we obtain, u Wrcos 0(xt-x2) (21) l-2ur Very often xl~xi is small and in a preliminary design xt nay be assumed equal to xe Hence 2ufieb : l-2Ur (22) which gives an approximate value of the guide friction, useful in a preliminary design - u may be assumed from 0.13 to 0.25. Very often as in symmetrical barbette mounts, the value of Beb may be small due to a small value of eb and a certain limitation arises as to the use of the friction formula previously derived. When lf cos 0 x * 1 1 that is, Wrcos 0 xf = Be0+Rr = 8eb approx. we have continuous contact along the guides, the distributed guide reaction oalancing the weight comoonent normal bo the guides. For such a condition the guide friction, becomes, R=0.2 to 0.3Wrcos 0 (23) INCREASE OF GUIDE FRICTION If we assume the total DURING POWDER PRESSURE oraking B to be constant PERIOD. during the powder pres- sure period, the guide frict- ion R is augmented by the powder pressure couple together with the increased friction couple. Let B = the constant braking force fa the varying powder force 147 Nt= the normal reaction of the stationary part on the recoiling mass. R = the guide friction during the powder pressure period. M = the reacting couple of the stationary part upon the recoiling mass. RS and M2 are the corresponding values during the retardation period. Then, during the powder pressure period, we have N-Wr cos 0=0 (24) and during the retardation period, we have, -Wrsin 0=Mr d'x dt« N-Wr cos 0=0 (25) Further, let AM=M1-M3 and AR=Rt~R2 , then subtracting (25) from (24) we have AM=Fe+AR r (26) Now during the accelerating period the normal guide reactions, become, , Mt Wrcos 0 xf (27) U ' - *«- KrCOS 0 Xi and during the subsequent retardation 142 Wrcos 0 x? W*= T 1 (28 M2 Wr cos 0 x i Na= T~ "T~ 148 Adding the two equations in (27) and (28) respect- ively and subtracting (28) from (2?) and multiplying by the coefficient of friction n, we obtain obvious expression: 2AM *"• T U (28) Substituting (29) in (26), vie have AM=Fe+ ur (29) Fe Fel and AJ4- - = - 1_ iHE 1 -2ur (30) 1 and substituting in (29), ire have for the change of friction during the powder period, (31) Thus the guide friction is continuously augmented always proportional to the total powder pressure, providing the braking is assumed constant. We also note an additional cause of first class importance for the reduction of "e", that is, the importance of locating the center of gravity of the recoiling mass along the axis of the bore. Another cause for a change in guide friction during the powder period is due to the torque reaction of the rifling, Tr though the total guide friction remains the same. The normal reaction on the left guide, becomes, m M_ wrcoa 0*a Tr Ntl3 2, " 2dg (32) 149 and the same for the right guide, becomes, U W-COS 0X., Tr N s * E 1 + -£_ (33) *r 2X 2l 2dg Wrcos 0xt Tr 2l ' 2dg" where dg is the distance between guides. In the gun recoiling in a sleeve this torque must be balanced by the reaction of the key way. Noting that, R-w(Ntl*Ntl + N +Htr)»u(Nt + Nt) the total friction remains the same. It is important to note that the friction on the left guide over that on the right due to this rifling introduces a couple in the plane of t"he guides which tends to cause rotation about an axis normal to the plane of the guides. Therefore, it is always essential that small side grooves or flanges on the clips be introduced. The additional friction on the flanges is entirely naglsgible, but the normal reaction to the flange in extreme cases may be considerable. During the recoil the guide friction is seldom constant since the distance between clip reactions pro- gressively decreases in the recoil, that is the front clip approaches the rear part of the guide in recoil. When the recoil is long it is desirable on field carriages to have an additional clip near the muzzle which engages in the guide sometime later in the recoil. Due to this cause the guide friction continually increases until the engagement of the outer clip and then we have a sudden drop in the magnitude of the friction. When the braking pull remains constant and the powder pressure coupls is small and n o outer clips are introduced during the recoil, the clip reactions should always be designed for the condition of out of battery. To recapitulate in the limitations in design so far as guide is concerned, we note, 150 (1) The bearing pressures and consequent friction of the guide are reduced by increas- ing the distance between the clip reactions nearly directly, consequently for a given guide reaction and friction, we have a minimum distance between clip contacts on the guides. (2) The guide reactions are reduced by bringing the resultant of the rod pulls through the center of gravity of the recoiling parts. (3) The moment effect and consequent guide reactions are further reduced by bringing the resultant guide friction line through the center of gravity of the recoiling mass. (4) It is highly desirable to center the center of gravity of the recoiling mass midway between the guide reactions. This condition is usually impossible to attain especially out of battery, but may be compensated by increasing "1" the distance between the clips, by an additional front clip near the muzzle. (5) Proper functioning of the recoil may be srrtirely destroyed by having the center of gravity of the recoiling mass too far below the axis of the bore, thus introducing a powder pressure couple with excessive guide friction during the powder pressure period. This powder pressure couple may cause a "springing" of the guides and considerable heating as well. The center of gravity of the recoiling mass should never exceed 1.5" from the axis of the bore unless a friction disk for rotation during recoil about the trunnions is introduced. 151 COMPUTATION OP BRAKING We have seen from the POLLS previous discussions that the guide friction is not independent of the braking pulls due to the hydraulic and recuperator reactions. These pulls tend to cause rotation and thus augment the guide friction over that due to the weight component. The total resistance to recoil is given by:- (1) when constant during recoil, (lbs) b-E+VfT where Vf= maximum free recoil velocity (ft/sec) T = total powder period (sec) E=free recoil displacement during powder period (ft) b= length of recoil (ft) •r= recoiling mass (2) when variable consistent with the stability slope "m", mrVc+m(b-E)2 K= - ; - (Ibs) 2[b-E+VfT- ^— (b-E)] 2 mr where K=B + Rg-Wrsin0=Pn+Pa+u(Q1-»-Q2)-Hfrsin0 (Ibs) and 8=Pn+Pa= total braking Rg=u(Ql+Q9)= mean guide friction assumed constant 2uBeh We have seen -— Ibs.approx. where nf=0.15 to 0.2 e)j= distance from center of gravity of recoiling parts to line of action of B. (in) 162 B3 total hydraulic and recuperator pull (Ibs) I3 total distance between clip reactions (in) r= distance from center of gravity of recoiling parts to mean friction line (in) then, 2uBeb ' - " (Ibs) hence, (K+«_sin 0)(l-2ur) B« T - - l+2u(eb-r) Further since, (Ibs) 8in we have on simplifying ~~J"" (K+Wrsin j0)( (Ibs) l-2u(eb-r) Very approximately, Rg=0.3 Wr cos 0 and B=K+Wr sin 0-0.3 Wr cos 0 (Ibs) Pn»K+Wrsin 0-Fa-0.3Wr cos 0 (Ibs) INCREASE OF RESISTANCE TO During the powder RECOIL DURING POWDER PERIOD period, the powder pres- sure couple may be suf- ficient to cause a large increase in the guide friction, whereas the braking pulls due to the hy- draulic resistance and recuperator reaction are not 153 affected. Prom the previous discussion on guide friction, the increment in guide friction equals, 2Feu (lba) or more exactly 2Feu 2 M where M=xl+x2+u(yt-ys) Guides above and below axis of bore =xt+x2-u(yt+y2) Guides entirely below axis of bore. Hence the total resistance to recoil becomes during the powder period, K*=K + ARg (Ibs) and this value should be used in the computation of the trunnion and elevating gsar reactions. Strictly speaking, the value of K is slightly high, since the augmented friction due to a large powder pressure couple, will diminish the maximum velocity of restrained recoil and thus the resistance to recoil for a given displacement b. A more exact value of the resistance to recoil can be estimated as follows: Thus, (K+ARtf)T* i (K+ARff)T K[b-(E — )]= - mr[Vf * — ] 2mr 2 mr Simplifying, we have, * r f *' J 2m_ R. ?— (Ibs) 154 •bore 2uPae 2nFme £Rtf» ————— - - (approx)(lbs) "»+x+u(-y) l-2ur (lbs) Fro« interior ballistics, we have, ' w aVu P«*« 1.12 - « (b'+u0)8 w = weight of projectile (Ibs) uo- total travel up bore (ft) v = muzzle velocity (ft/sec) b'-u0[(|l-*-l)±/(l-ll-)a-l] <"> ~ (ft/sec) u where P,= total maximum powder force (Ibs) Unless the powder pressure couple is excessive, that is the center of gravity of the recoiling parts is considerably below the axis of the bore the above refinement in calculation is unnecessary. When e exceeds 1.5 to 2 inches the above effect becomes of consideration. INTERNAL STRSSS IN THE It is very important to RECOILING PARTS observe that the braking force 8 when treating of the external forces on the re- coiling masses as in the previous discussions refers always to the reaction of the oil in the hydraulic brake and the spring or com- 155 pressed air reaction of the recuperator. During the accelerating period the reaction on the gun lug nay differ considerably from the braking force B due to the acceleration of the piston and rods where these recoil with the gun or to the acceleration of the re- cuperator sleigh or slide when the sleigh recoils and the rods are fixed to the carriage. If now we consider a recoiling mass consisting of a gun together with a single cylinder recoiling with the gun, figure (4) and if we let, B3 the total braking force along the axis of the cylinder. B =the normal reaction of cylinder on the gun lug. j =the tangential or shear reaction on the gun lug M -the bending moment reaction on the gun lug. Neglecting the guide friction, let, QI and QS be the normal guide reactions xt and x? the coordinates along the axis of the bore of the clip reactions with origin at the center of gravity of the gun. x and x the coordinates parallel to the axis of the bore with origin at the center of gravity of the recoiling parts. xc and yc=the coordinates of the center of gravity of the recoiling cylind«r with respect to the center of gravity of the gun as origin. e^- distance from the center line of the recoil cylinder to a line through the center of gravity of recoiling parts parallel to the axis of the bore. .Mr and Wr= mass and weight of the recoiling parts. Mc and tfc* mass and weight of the recoiling cy- linder. j = the distance from the shear reaction on the gun lug to the center of gravity of the gun 156 itself. F» the maximum powder force along the axis of the bore. If the mass of the lug is negligible as compared with the mass of the gun, the coordinates of the center of gravity of the recoiling mass with respect to that of the gun becomes Vc Vc xr» -r — and yr=» e= — — "r "r Further *t - - - and y y - -^j— " " «.- s * — and *. 3 ^ Now considering the gun with its lug alone, the reaction of the recoil cylinder on the lug, consists of the pull B a bending moment Wc cos 0 (xc + j ) and a shear reaction Wr cos 0. Taking moments about the center of gravity of the gun, we have fi'(b+e) * \»c cos A j-Wc cos 0(xc+j )aQ8x^* B'eb*B'e-Wcco. 0 xc=Qt(xt+ -~^)*(ar wr wr Simplifying, we have, B'eb+B'e = Ot(xt+xt) + Wr cos 0 xa Considering the recoil cylinder alone we note that during the accelerating period, d'x B -B+Wcsin0 »MC - ° df 157 but from the recoiling mass, we have, d*x F-B+Wr sin 0 dt« Wr hence B' = B-Vfc sin 0 + r-(F-B+Wr sin 0) and substituting in the previous equation, we have, V (B-Wcsin 0 + --(F-B+Wrsin 0)(eb+e)=Qt "r "r5 but yc= 8t>+« and yr= e hence (ejj+e)= - — 'r Wre "c Therefore, substituting in the above equation, we have, 8(eb+e)-Wre sin 0+Fe-Be+Wrc sin 0 = Q1(xt+xt)+Wr cos £) ; hence Beh+Pe =Q (x +x )+W_ cos 0 x. Obviously if we consider the recoiling mass, fig. (4) we have, taking moments about the center of gravity X Qx tut Q-Qs W cos A Hence Be^ + Fe - Qt(xt+ xf) + Wr cos 0 xa the same equation as obtained above as of course we should ex- pect. The above discussion shows the importance of considering either the mass of the gun with its proper external reactions or the mass of the recoiling parts with its proper external reactions and not confusing the mass, of the gun and recoiling parts, and the co- ordinates of their center of gravities. The maximum stress in a section m - m, see fig. (4) of the gun lug obviously occurs when the bending moment due to the weight of the recoil cylinder is a minimum and the braking force B a maximum that is at maximum elevation, In the above discussion the normal reaction between the 158 piston surface and cylinder was assumed zero. This reaction obviously depends upon the weight and relative deflections of the rods and cylinders. If these weights were equal and at the same distance from the point of support, and with equal elasticity, this reaction becomes zero and we have the bending moment assumed; but since the rods are relatively very elastic as compared with the cylinder in general the moment Wc cos 0 (xc + j) if anything is augmented. If Imn is the moment of inertia of the section "« - m", AJJ-U its area of cross section, and y is the distance to the edge from the neutral axis of the section and "g" the distance from 6 to the neutral axis, we have for the maximum fibre stress cos 0(xc tJ)]y fccos 0 + - (34) c where B =B-Wcsin 0 + — (F-B+Wr sin 0) wr Since the weight components are small as com- pared with the powder pressure force and braking for a first approximation, we have, wc [B+ —-(P-B)lg Wr _ a-n>= (35) lm-m which is a useful formulae for practical design. TIPPING PARTS The tipping parts consist of all the parts that move in elevation with the gun. The two principje parts of the tipping parts are the recoiling parts and cradle, the one moving in recoil and the other remaining stationary. The cradle supports by its guides the 159 recoiling parts on recoil, it takes the reaction of the braking exerted on the recoiling mass and is sup- ported by trunnions resting in bearings in the top carriage and is further prevented from rotating about these trunnions during the recoil by the reaction between the elevating pinions of the top carriage and the elevating arc of the cradle. When a rocker is in- troduced between the elevating pinion and cradle for an independent line of sight it should, properly speak- ing, be included in the tipping parts. It is of fundamental importance to always balance the center of gravity of the tipping parts about the trunnion axis since with massive parts the elevating process must be done quickly and with the minimum re- action on the elevating pinion of the top carriage. Let x and y = the coordinates parallel and nor- mal to the axis of the bore. X and Y = the x and y components of the trun- nion reactions. F = the total powder pressure force. E = the reaction between the pinion and elevating arc. j 3 the radius of the elevating arc. 6ea the angle between the "y" axis and the radius to the elevating pinion contact with the elevating arc. The mutual reaction between the tipping parts and 'top carriage may be divided into the component reactions X and Y of the trunnions and the elevating arc reaction E. By D'Alerabert's principle, considering the inertia of the recoiling mass as an equilibriating force, we have during the powder pressure period assuming the gun practically in battery, for equilibrium of the tipping parts, that, fig. (5) ,t F-Mr — ~ -2X +Wt sin 0 +E cos Ge * 0 (1) d u for motion along the "x" axis, 160 9 ~.s?.} no tsdiexe fco -;*»oq for motion along the "y" axis, and ^ »Mt,w d*x bnrp(® + s)-Mr J^T 8-E j = 0 For moments about the trunnions, the weight of the tipping parts having no moment since the center of gravity is at the trunnions in battery. 1 i • • • d*x But F-Mr T~T =Ka fcns Votal resistance to recoi|.,T *pj ?. 1 -,..-, .dur ing th8 accelerating period. Hence equation (3) reduces to: Pa + K.s-E j = 0 and DO^ the reaction on the elevating arc in battery becomes, a ( 4 V ' J »fU lc eixs 5dJ p4 and the trunnion reactions in battery, becomes, (Fe+Kas) 2X=Ka*Wtsin 0 * cosee ) • ( fcn* nointq edJ ne>ev?ift<j isoiJoaci edi « 3 v (Fe+Kas) ty ^ (5) 2Y*Wt C03.0- "~~~ ' sin 9 ( tb« resultant trunnion reaction •y " S* / X*+Y* making ^tn angle tan"1 — with the "x" attWfc now if, fig. (5) "rf^ =-'!l-0tal weight of the tipping parts Wc = weight of the cradle :6f the tipping parts Wr = weight of the recoiling parts 3.T* 3d ; . It * fche horizontal distance to the center of • gravity of the tipping parts (in battsry) always assumed at the trunnions from the ' • i. f~ ! u r A n y § hinge point of the top carriage. • To account for contact of tooth rack with the pinion or worm of the elevating gear, the reaction B makes an gle approx. 20° with the tangent to pitch lines, erefore in above equations (eg. 5) *J • beoome«,"j o • and "oos Ce" beooaes » oos 1C +20;" and "«in Q " angle Therefore 20 beooaea •«in(«t+20) 161 fcoJrqqii : Js si L-___i_i!L:_iW *+. iio^i <?rfi lo v *— — "}' - V Oi ^ s s|*'i v^r . 5 162 1^ * the horizontal distance from hinge point of the tipping parts when the recoiling mass is at a distance "XH out of battery. lc = the constant horizontal distance from hinge point of the center of gravity of the cradle. Then for moments about the hinge point in battery, "t^t* "r^r* "c^c and for a displacement "x" of the recoiling mass from the initial position, we have, WtlJ = Wr(lr - x cos 0)+«clc Therefore, the moment of the tipping parts for a displacement "x" of the re- coil, becomes, Wt(lt - lj) »Wr x cos 0) (6) Hence, for any position out of battery of the recoiling •ass, i 2X-MP 2-7 -Wt sin 0 -E cos 9e» 0 ) ( (7) 2Y-Wtcos 0+E sin 6e = 0 ) for motion along the x and y axis, respectively, and t(r — S+»r x cos 0 -E j = 0 (8) dt» for moments about the trunnion axis. But the total re- sistance during the retardation becomes, .» Kr-Mr f -B+R-Hr sin 0 (9) Combining and reducing, we get for the reaction on the elevating arc, for a recoil "x" Kr s+Wrx cos and the trunnion reactions for a recoil "x" (10) 163 (Rrs+Wrx cos 0) 2X»K +Wt sin J6 + - : - cos ee ) -**' J ( (Krs+Wr x cos 0) ) (11) 2Y=Kt cos 0 - - ; - sin ee ( J ) which shows the trunnion reaction depends only on the total resistance to recoil and the moment effect of the recoiling weight out of battery. It is often more convenient to consider the reactions of the elevating gear and trunnion reactions between the tipping parts and top carriage as divided into horizontal and vertical components rather than along axis parallel and perpendicular to the guides. The elevating gear reaction will be considered positive when the line of action may be resolved into components horizontally to the rear and vertically upwards, that is when the radius joining the trunnion to the elevating pinion contact with elevating rack is measured from the vertical counter clockwise. Galling this angle ne, we have, 9e= 0 + ne whereas before 0= angle of elevation. We have then for the elevating components, measured horizontally and not vertically. He = E cos ne Ve =E sin ne and measured along the "x" and "y" axis, i.e. along and normal to the axis of the bore, X,» = E cos YP=E sin(0+n_) * More strictly to aooount for obliquity of tooth contact of elevating me o h a n i 3 n , j beoor.es j oos 20, 003 0. becomes oos (W +20) and sin O becomes s i n ( 0 * 2O ) . O o w 164 The horizontal and vortical components of the trunnion reaction become, in battery, (Fe+Ks) 2H=Ka cos 0+ — — - cos ne ) J P +K 2V=Ka sin 0+Wr - — • - sin ne J \ .v.ieJJed ito Ji ^ 9dj and out of battery, (Krs+Wrx cos 0) 2H»Kp cos 0+ - • - cos ne fcnoi* narii ierij£-t feSnenoqaioo IfiaiJiev bnadfiJocsi-jorf (Krs+Wrx cos 0) ) oi<" ; &•» erf ^s« noiJcs lo snil »riJ ae« and the resultant trunnion reaction becomes, no. "ot tuJ'fcsT sHJ aart* ai 3&di S = j/p* + 7* .*c INTERNAL REACTIONS OF It is important to observe TIPPING PARTS ~ that X, Y and E are the external ROCKER INTRODUCED. reactions exerted by the top carriage on the tipping parts which include a rocker if used. The total reaction on the trunnions include the reaction of the top carriage X and Y and the reaction of the rocker Xr and Yr . Hence the resultant reaction on th« trunnions, become, algebraically, l»oi X =X+Xr ) and ((12) = the shear components of the V =Y+Yr ) trunnion pins on the cradle. The reactions on the rocker, alone, therefore, become, the reaction of the top carriage pinion E, the reaction of the trunnion Xr and Yr reversed and the reaction of the cradle M reversed. In many cases an elevating screw is used between the cradle and rocker and when used, M reversed is the reaction of the elevating screw, 166 •~— — ^______^ -:i «• 10 fi>.7^nT~-- ;- fc{1 .* £fl * '5 « 5 j_/ ,13JJuOf Sri-* / / dd^ 1C 8S»4*M!ibT003 "^" bft* "x" &sntiL'to , *• . J.-- aQn««si& .! .- *di to ' ^ »*<! *jV t ** , -\ - -, - - - \ rl J 2 0R * 5l STdHw •J «Ml* (.-- its line of action since the screw is hinge jointed at its two ends necessarily lies along the axis of the screw. The screw reacts on the cradle with a force + M and is under a compression M during the recoil. Considering now the equilibrium of the rocker, we have, if • Wr= the weight of the rocker k= the perpendicular distance from the trunnions to the line of action of M which is the line through the axis of the elevating screw when used, or the normal to the contact surface of rocker and cradle when an elevating screw is not used. B= the angle between the line of action of M and the "y" axis. xa and ym= the "x" and "y" coordinates of the center of contact of rocker with cradle, or hinge joint of elevating screw of rocker on cradle. xr and yr = the "x" and "y" coordinate of the center of gravity of the rocker and is measured from the trunnions towards the breech and downward. For equilibrium along the "x" axis (see fig. (6) 2Xr+E cos 6 + *r sin 0~M sin 8 = 0 (13) for equilibrium along the Y axis 2Yr+M cos B-E sin 6C **r cos ^ = ° ^14^ and for moments about the trunnion, Ej-Mk-W (xr cos 0-yr sin) = 0 (15) where k = xra cos B+ym sin B. It is often convenient to replace Xr cos 0-yr sin 0 =hr the horizontal distance from the trunnions to the center of gravity of the rocker, then (15) reduces to 167 Ej-Mk-Wr hr= 0 (15 ') hence Ej_w' h M= (16) K and (Ej-w; hr) 2Xr= sin B-Wr sin 0-E cos ee ) k ( > (17) (Ej-Wr hr) < 2Yr-E sin 9 -Wr cos 0 - cos B ) K which shows the rocker reactions depend only on the elevating arc pressure. If now we consider the equilibrium of the tipping parts not including the rocker, we have, fig. (7) 2J('=Kr+(Wt-Wr) sin 0+M sin B ) ( 2Y'»(Wt-Wr) cos 0 -ti cos B ) (18) and Krs+Wrx cos 0-V»' hr M= — — — (19) k since the moment of (Wt~wr^ about the trunnions= -Wr hr but now from equation (12), we have 2X=lfX'-2Xr ) 2Y=2Y'-2Yr ( hence substituting the values obtained in (17) and (18), we have, * To account for tooth contact, more strictly, re- olaoe j to j cos 0.; oos 0. to oos.(0 +20) and sin. 0 to 168 w-t bnsqsb eueJ:49£Ai won . -. o ft. r i fit . "I .« i * tn & ( Ot » 169 ( 2X=Kr+(Wt-tfr) sin 0+14 sin: B-<H s&n B-W sin £J-E cos 6e) ( 2Y=(Wt,-wr) c^3 0"14 C50S ?~(? r*ae^-Wri cos 0^»CoQS B) Simplifying these values reduce' tV the- fofrnrer Values, 2X=Kr+Wt sin 0+E co« ee .(v.TSJtfid ni nu§) g^rseq. 2Y=Wt cos 0 -B sin'0e' and further sieriw n.'4o artj \o eujbfii J •. 1 * ,1 i i Krs +Wr x cos 0-IL. hr . EjHlL hr - — - - - - — • _ k k .<asc "io asjibAf n*eM B H and .therefore as. befor* ,0 floU36il9b e ^ .noi4*vele cBorclxfi* Je gaii ::Krs+Wr x cos 0 B= - . *> -* - iiu^nof; gnt-rqe thus checking the formulas derived for the rocker reactions. oa be;.. boil\£o s/iiiqe 8.Hj ni SfiiTqs tsriT TRUNNIONS LOCATED AT In guns shooting at -hijjjh J THE REAKr BALANCING^ r. ,elevAti<m s,uc.A ;^s; aqti- GEAR OR EQUILIBRATOR.. .T.,7cair;cra=ft ^ttn^,: mortars, and even howitzers, it is often necessary to Ipcate the trun- nions in the rear or near the breach of the gun in order to prevent the breech of the gun from striking the ground during the recoil when the gun is elevated. Ob- viously it is impracticable to balance the tipping parts about the trunnions without some sort of a balancing gear commonly known however, as an equilibrator An equilibrator should balance the tipping parts, since the center of gravity of the tipping parts are now displaced forward of the trunnions, at all angles pf elevation when the gun is in battery. It consists sometimes of a cam arc, a chain passing over the con- tour of the cam arc and connected with a spring cylinder oscillating about the trunnions fixed to the top carriage to take care of the small deflection due 170 to the change of radius of the cam. PROCEDURE TN DS3TGN OT B B U I L I B R AT OR : Let Wt = weight of the tipping parts. ht a horizontal distance from the trunnions to the center of gravity of the tipping parts (gun in "battery). ro = equivalent radius of cam at horizontal elevation. rn = final equivalent radius of the cam where the cam arc has turned through the maximum angle of elevation = 0. B = Mean radius of cam. dn = deflection of spring at zero elevation. d0 = deflection of spring at maximum elevation. C = spring constant. 0 = the angle of elevation expresses in rad ians . The spring in the spring cylinder is arranged so that it is in general under compression. As the gun elevates, the compression of the spring is decreased in virtue of the motion of the cam. The total motion of the spring becomes, R0 ' dh~do aPPro*' during the elevation 0 where ro*rn " for a first approximation and m c dhro= *tht for equilibrium about the trunnions for C dorn= *tht cos 0- all elevations. If now we assume dh=(- to -)d solid 11 3 4 and i t do =(— to -g-)d solid Then for a preliminary design, we have, 171 C = -i 5 (- to - )d Solid r0 Wtht cos 0 C = * * (- to -) d solid rn r + r and (- to -)d solid -(- to j)d solid = ( — - — )0 0 The unknowns in the above equatibns are C, d solid ro and rn; hence if we assume any one of these values the remaining values are determined. The equivalent radius of the cam may be obtained by a "point by point" method as follows'- The initial radius, becomes, g h "t L r°*5*r Now move the cam an increment angle A0 where nA0=0 the total angle of elevation, and we have, cos # and Ad = rnA0 C(dh-r0A0) then W^ht cos 0 ** C[d-(r0+rM07 >V<po+' Htht cos 0 rn~ C[dh-(r0+rt rn_,)A0] dh~do= ro+rt~"rn-i) A0 Strictly speaking the angle A0 in the above pro- cedure should be augmented by the angle r"n~ r° — radians- Where D = the perpendicular distance from the trunnions to the extremity of the equivalent cam radius. From the equivalent radius thus obtained the cam contour may be drawn by drawing in a curve always tangent to the 172 perpendiculars to these radius, drawn from their f. extremities. With a balancing gear or equilibrator, the trunnion reactions are modified and now become ___ - r it T = the tension in the chain. a = the angle T makes with axis "X", (takea *) j along the axis of the bore) Ks*WrXcos 0+Fe -a— iui)deos 0e*T cos • *>«3)»e-c onj- \n& ;,,^ bns > )sin6e *T sin a iT( J I v,d Jni:oq")B and soihsT IsiJ/ni: ftri'JY ) j -f"' ei^ns-Jnsaaeioni ne i!»8 Usually d = 0. and thus the Y component of the trunnion reaction is unaffected. DIRECT ACTING BALANCING Another form of GEAR balancing gear, for nedj balancing the tipping '3+ 'O~ i,Parts at. all angles of <, ' elevation about the trunnions which are located to the rear, consists of a spring or pneumatic oscillating cylinder and its rod directly connected between the tipping parts and the top carriage. In the position of the tipping parts at zero elevation, gives maximum moment and therefore re- quires the •axinua balancing reaction. $rtj moil ocf.flJaib IB i • jdW ~ To .ccount for tooth contact of .levatl.g ••ohaniSB, replace j to j COS 20, oos S to oos (Ot«20) »rd .In d, to .in(«e+20). • n - sri.t OJ 173 M ..-•? .noiievele ed.4 &i»jt«or> « ae eseeio Wt = weiSnt °f th8 tipping parts nt = horizontal distance from the trunnions tcbijr;nu} the center of gravity of —the tipping parts -11^0 %nii el j(g un- i^ b *Vt &ry.<\» r, o x n n 01 4 eri J §fil,»f.! .u^jfc *od yt - coordinates alon^ and normal to bore ; -oneisd orjfiiiiuffroift- trunjiion to center of gravity of tipping parts (gun in battery:)i.2.rq 5jni 0 - angle of elevation., Juov.*! VJ«nif«il9Tq s io1 0m = maximum elevation>.. erfj Ic sixs oornnot^ arii gni r = radius from the trunnion to the crank pin bft!j,llE:f*hich connects the tipping parts to the pi*-; vfciJfcon rod of the oscillating cylinder, (in.) iBi^cni R = reaction exerted by the balancing gear along the piston rod of the oscillating cylinder;. s) 3 iscfnos iiori TO Isi^iai siU : dt = moment arm of R about trunnion. (in) d = deflection of spring at horizontal elevation. do = deflection of spring at maximum elevation. (in) ,ooi**veJs uwsilxsa icl seasooed C = spring constant. RJ = initial, balancing reaction (0° elev.) Rt = final balancing reaction (0* elev.)(lbs) S = stroke of piston in oscillating cylinder. (in) pt - final ,air piressure j.^ pneumatic balancing cylinder (Ibs/sq.in.) ,|»£ = initial air pressure in pneumatic balancing cylinder (lbs/sq.in.1 ^,f, effective area of balaacing pistoa ..(sjj.>cinj._) j V0 = initial air volume (cu.in.) *\L ,'J3bniI^o Jniiq* 6 ri; :?•• At any angle of elevation 0, we must have, R dt » Et(xt cos J0-y-t sin 0) io? In general the center of gravity of the tipping parts lies approximately along the axis of the trunnions, and therefore, R d = W x cos 0. 174 If dt remains constant, the reaction R should de- crease as a cosinef unction in the elevation. Since it is usually impossible to decrease R according to a cosine function, we may so locate the trunnion of the oscillating cylinder so that the product R dt = Wtxt cos j0. By properly locating the trunnion axis of the oscillating cylinder a very close balancing is possible throughout the elevation either with a spring or pneumatic balanc- ing piston. For a preliminary layout, we may start by locat- ing the trunnion axis of the oscillating cylinder somewhere depending upon clearance considerations, along a line parallel to the chord joining the assumed initial and final positions of the crank and midway betwean the chord and middle of arc. The crank turns an angle equal to the total elevation & m* Then the initial or horizontal balancing reaction, becomes, r 0m R^ - (1 + cos — r)=Wt xt and the final balancing 2 <c reaction, becomes for maximum elevation, r *m Rf-(l+ coa T~-)3Wf (*t cos flfl+yt sin 0m) 3 Wt xt cos tm (approx.) Rf We have, therefore, cos 0m very roughly, and the Ri *• total stroke of the piston, becomes, S=2r sin -r— (in) Now with a spring cylinder, if, dh s 3 to '4 solid height of spring (in) do = - to - solid height of spring (in) then dh-d0= S (in) 175 and at 0° elev. R^ = c dh at max. elev. Rf= c do. Hence the required spring, may be approximated by the solution of the following equations, = = cos 0m ) dh-do=' 21 Cdh=- r(l+cos — ) ( • With a pneumatic cylinder, we have, since the expansion, may be assumed isothermal, pf vo —— = — — = cos J0jj, (aoorox.) from which we i| vo+ may determine the initial volume Vo hence pi cos 0^ V0 = AS = AS (approx.) (cu.in.) *f 1- cos 0m Now pf RiS TT Ri = p^A and V0= — -— [ ] (cu.in) Pi PP 1 --1 Pi We see, therefore, to decrease the tulk of the cylinder it is important to maintain as high an initial air pressure as possible. It is reasonable to assume the same initial pressure as used in the recuperator brake. Therefore, in the first approximation of the de- 176 sign layout of a pneumatic balancing gear, we may start with,- Pf (cu.in) V0 ' 2WtxtS ' PT 0 v r(l+cos - )p; 2 ] 1 - Pf 0 Pf where S * 2 r sin - , — = cos 0m (approx.) 2 Pi REACTIONS ON TIPPING PARTS With the trunnions to WITH BALANCING GEAR. the rear of the center of gravity of the tipping parts and a balancing gear introduced, we have a cantilever form of top carriage, and the reactions on the tipping parts are usually approxircataly in the position shown in fig. (8). Let x and y - coordinates along and normal to axis of bore. R » reaction of balancing gear (Ibs) 9r= angle between R and y axis. dts moment arm of R about the trunnions at any elevation 0. E = elevating arc reaction (Ibs) 9e= angle between the y axis and the radius to the elevating pinion contact with the elevating arc. J= radius of elevating arc from trunnions (in) X and Y = components along x and y of trunnion reaction. (Ibs) Wt » weight of tipping parts, x^ and yt * coordinate of center of gravity of tipping parts from trunnion (in) Wr = weight of recoiling parts (Ibs) He have, then, two positions to consider, the in and out of battery positions respectively. In 177 JPEAC7/CW5 ON T/PP/MG PA/?TS W/77J DIRECT ACr/N6 BALAMC/MG TPU/W//OM AX/S rig. a 178 the battery position we have the effect of the powder pressure couple, while in the out of battery position, we have the moment effect of the recoiling weight. Considering the Recoiling parts in battery: We have for the kinetic equilibrium of the tipping parts, for motion along the x axis, d% Pb- mr — — -2X+Ht sin 0 +E cos ee + R sin 9r = 0 Q X for motion along the y axis, 2Y-Wt cos 0 -E sin 6e + R cos 6r = 0 and for moments about the trunnion, , 2 Pb(e+s)-mr— 7-S+R dt-Wt(xtcos 0 -yt sin 0)-Ej= 0 d t» Since, however, 2 pb~rar = K and R d = -y+ sin 0) requirecl condition of the balancing gear, the above equations, reduce to,- Ks+Pbe E= - (Ibs) for the elevating arc re- ) action in battery ( 2X=K+Wtsin0+E cos ee+R sin 9r (Ibs) ) 2Y=lft cos 0+B sin 9e ~R cos er (Ibs) ) ( Wtxtcos 0 2Wtxtcos 0 ) where R = - - - = - - — approx.(lbs) dt 0m ( r(l+ cos - ) It is to be noted that 2X, 2Y, E and R are to be regarded as the reactions exerted by the top carriage on the tipping carts, a rocker if used being included * To account for tooth contact of elevating ••••••!••, substitute j cos 20 for j , B co»(C.-20) for I cos O. and E sin(0 -20) for I iin 0.. 09 o 179 as a part of the tipping parts. The resultant bearing reaction between the top carriage bearing trunnions, becomes, S = / XZ + YZ (Ibs) Considering the recoil parts out of battery if Wt= total weight of the tipping parts Wc= weight of the cradle of the tipping parts. Wr= weight of the recoiling parts. xro and yro = battery coordination of the re- coiling parts with respect to the trunnions. xc and yc = coordinates of the cradle with respect to the trunnions. Xfo and yfo = coordinates of the tipping parts in battery with respect to the trunnions . For a displacement "x" of the recoiling parts froB the initial battery position, we have, for moments about the trunnion, _To .SEC- I erii oJ v- rtiostu d x +WC sin 0 yc-E j = 0 U Utf v — W v +W v POW n-fXj.1" ffT%A_»-.~n.>A-, hence the moment equation about the trunnion reduces to ,» mr - .t-txtcos +tytsn +rx cos 0-Ej= 0 QTf but due to the balancing gear, we have, "txt c°3 ~yi s^n and further K=mr— - — — that is, the inertia resistance, d t equals the total resistance to 180 recoil. Hence K s + Wr x cos 0 For motion along the x axis, we have, ( 2X=K+R sin er+E cos ee+Wt sin 0 ( ) and for motion along the y axis, ( 2Y=«ft cos 0+E sin e+g sin9s -R cos er ) I where as before ; 2Wt xt cos 0 R= roughly. ) 0o> ( r( 1+cos -— ) f ) From the above analysis, we see, therefore, that the elevating arc reaction remains the same with or without a balancing gear, while the trunnion reactions may be increased or decreased according to the location of the line of action of the balancing gear. INTERNAL REACTIONS OF The rocker reactions depend TIPPING PARTS WITH solely on the elevating gear BALANCING GEAR - reaction, and with a balanc- ROCKER INTRODUCED. ing gear, the elevating gear reaction is independent of the eccentricity of the center of gravity of the tipping parts from the trunnions. Therefore, the rocker re- action on the trunnion is entirely independent of the reaction exerted by the balancing gear or counter- poise. In brief, the rocker reactions remain the To account for tooth contact of elevating •echanisB, substitute j cos 2O for ,i E oos(O.-20) for B oos 68 and B Bin (O0 - 20) for B sin O°. 181 same with or without a counterpoise or balancing gear. The reactions exerted by the top carriage on the trunnion do however depend on the mag- nitude and direction of the balancing gear. Therefore, the shear and bending at the section of the trunnion adjoining the cradle must also depend on the balancing gear or counterpoise re- action. . An analytical proof of the reactions is given as follows:- Let X and Y = trunnion components of the reaction of top carriage. Xr and Yr = trunnion components of the re- action of the rocker. X and Y = shear components of the trunnion pins on the cradle. Wr = weight of rocker. E = elevating gear reaction on rooker M = cradle reaction on rocker. j = radius to elevating gear arc. k = the perpendicular distance from the trunnions to the line of action of M which is the line through the axis of the elevating screw, when used, or the normal to the contact surface of rocker and cradle when an elevating screw is not used. B = the angle between the line of action of M and the "y" axis. xffl and ym = the "x" and "y" coordinates of the cradle hinge joint of rocker elevat- ing screw or the center of contact of rocker on cradle. xr and yr = the "x" and "y" coordinate of the center of gravity of the rocker and is measured from the trunnions towards the breech and downward. 182 Evidently for the shear at the cradle section of the trunnion, x'= x + xr ) ( Y' = Y + Yr ) For the angular equilibrium of the rocker, Ej-Hk-lfr(xr cos 0 - yr sin 0) = 0 if we let xr cos 0-yr sin 0 = hr, then Bj-Wrhr Ha 1 — , where k = xffl cos B + ym sin B, that is, the cradle rocker reaction depends solely on the elevating gear reaction. For the translatory equilibrium of the rocker, 2Xr= M sin B-E cos ee -Wr sin 0 2Yr=E sin (Je ~Wr cos 0 -M. cos B which shows the rocker reaction at the trunnion de- pends only on the elevating arc pressure and there- fore is independent of the counterpoise reaction. Considering the equilibrium of the tipping parts not including the rocker, we have, 2x' = Kr-i-(Wfc-1fr) sin 0 +M sin B*R sin 6r 2Y*=(Wt-Wr) cos 0-M cos B-R cos er Further if measured from the trunnion axis, lt=xtcos 0 -ytsin 0= the horizontal distance to center of gravity of tipping parts(re- coiling parts in battery) lt= the horizontal distance to center of gravity of tipping par.ts (recoiling parts out of battery) lr= the horizontal distance to center of gravity of recoiling Parts in battery. 183 lc= the horizontal distance to center of gravity of cradle . -hr= the horizontal distance to center of gravity of rocker measured in a negative direction from the i's. d^ = moment arm of the counterpoise reaction R about the trunnions. Then since the moment of the tiooing parts minus rocker about the trunnions is equal to the moment of the weight of the tipping parts minus the moment of the weight of the rocker, we have, W~h=?1~ x cos Now Ifrlr+Wclc= «Ttlt+Wrhr hence Wtl^+Wrhr=Wtlt +Wrhr~Wr x cos 0 =Wt(xt cos 0-yt sin £J) + Wrhr~Wr x cos 0 therefore, Krs+Rdt-Wt(xtcos 0-yt sin 0)-l?rhr+Wrx cos -Wk= 0 but for equilibrium of the tipping parts in battery Rdt=Wt(xt cos 0-yt sin 0) hence Krs+Wrx cos 0-1 M = • Since, however, 2X=2X'-2Xr ) 2Y=2Y-2Yr We have in substituting the previous values for 2X* and ZXp^j, 2X=Kr+Wtsin 0 +R sin 9r+E cos 6e 2Y=lft cos 0-E sin 9e~R cos 6r 184 Krs +Wrx cos 0 and E = — J In the preceeding analysis it is important to note, that the center of gravity of the rocker is assumed to the rear of the trunnions, and the elevating gear reaction is considered positive when the radius to the pinion contact of the elevating rack is measured counter-clockwise with respect to the "y" axis through the trunnions. Evidently when QQ is negative (i.e. clockwise from "y" axis, E cos 6e remains the same but E sin 6e becomes negative in the above equations. •ol' EFFECT OF RIFLING Cue to the rifling, the TORQUE ON TRUNNION torque exerted on the gun by REACTION the shell aust be balanced in considering the equilibrium of the tipping parts by an equal and opposite moment exerted by the top car- riage on the trunnions(assuming due to the ouch greater flexibility of the elevating arc and pinion that the elevating arc reaction is entirely unaf- fected). If the rifling is right handed, then in the di- rection of the muzzle, the Y component of the left trunnion is increased and the Y comoonent of the right trunnion is decreased by the amount equal to the torque of rifling divided by the distance between the trunnion bearings on the top carriage. Usually this affect is quite negligible as compared with the ther forces exerted. STRENGTH OF THfe TRUNNIONS. The critical section of the trunnions is usually where the trunnion joins the cradle. Let "«n" represent this section on the trunnion, see fig. (9). 185 xr.-s no n 186 a = the distance from "mn" to the center of the top carriage oearing. b = the distance from "mn" to the center of the rocker bearing . MX = the bending moment at "urn" in the plane of the X component reactions. Hy = the bending moment at "mn" in the plane of the Y component reactions. M = the resultant B. M. on the section, "mn" . f = maximum fibre stress. D = diameter of the trunnions at section "mn" I = the moment of inertia of the section about diameter. Then Mx= Xa+Xrb My= Ya+Yrb and M,/^J~" further nD4 MD I = and f = 64 21 32M 10.18M hence f = — — = r— nD D Usually the fibre strass is limited from — to — of the elastic limit of the material used, 2 3 and the minimum diarae ter of the trunnions becomes /10.18M D = 7 The shear stress is usually negligible as compared with the bending stress. LIMITATIONS ON THE EXTERNAL In considering the REACTIONS OP THE TIPPING external reactions on PARTS. the tipping parts, we have to consider the limitation imposed on the elevating arc reaction and the reaction on the trunnions by the top carriage. 187 ELEVATIHG ARC REACTIOH: (1) The elevating arc reaction is reduced by reducing the perpendicular distance between the line of action of the re- sistance to recoil, which passes through the center of gravity of the recoiling parts parallel to the axis of the bore, and the trunnion axis. When the line of action of the resistance to recoil passes through the trunnion axis, the reaction on the elevating arc in battery is zero if we neglect the effect of the powder couple and is equal to the moment effect of the recoiling weight when the gun is out of battery. The elevating arc reaction is re- duced proportionally to the increase of the radius of the elevating arc. The elevating arc reaction should al- ways be considered in the limiting con- ditions of in and out of battery, that is, with the maximum powder pressure couple acting and out of battery when the maximum moment effect of the re- coiling weight about the trunnions exists. (4) When the resistance to recoil does not pass through the trunnions the elevating arc reaction due to the short- ening of recoil is a maxim at max. ele- vation. TRUNNION REACTIONS: (1) The "X" component of the trunnion reaction (i.e. the component parallel to the bore), depends upon the total resistance to recoil and the component of the elevating arc reaction parallel 188 to the bore as well as the weight component of the tipping parts when the gun elevates. (2) The "Y" component depends upon the weight of the tipping parts and the component of the elevating arc pres- sure parallel to the "y" axis. (3) As the gun elevates the component of the elevating arc pressure parallel to the "x" axis decreases but the weight component increases and due to the shortening of recoil on elevating the resistance to recoil increases. (4) The component parallel to the "y" axis of the elevating arc reaction increases but in a negative direction, thus tending to decrease the Y component of the trunnion reaction. On high elevation because of the large re- sistance to recoil for a short recoil, the elevating arc pressure parallel to the "y" axis more than compensates the decreased weight component of the tiooin^ parts thus causing a reversal of direction of the Y component re- action of the trunnions. (5) Thus in general the X component in- creases while the Y component decreases, verv often reversing on elevating the gun and thus the trunnion bearing con- tact may shift 90° or over. STRESSES IN CRADLE OR The reactions on the RECUPERATOR FORGING. cradle or recuperator are: * (1) the trunnion reaction of the top carriage on the cradle: (2) the reaction of the elevating arc which is equivalent to a single force in the direction of the elevating pinion re- 189 action on the elevating arc together with an addition al moment: (3) the reaction of the recoiling mass on the guides: (4) a result and reaction parallel to the longitudinal axis of the cradle or the guides due to the various "pulls" exerted on the recoiling mass and (5) a distributed load which is uniform if the cross sections remain the same due to the weight of the cradle. In an accurate computation of the stresses in a cradle it is necessary from a preliminary layout of the cradle to locate roughly the neutral axis of each section and connect these points for a longitudinal neutral axis line. "We may then treat the cradle as a simple beam, talcing into account the bending moments caused by eccentric loads such as pull reactions off the neutral axis, guide frictions, etc. The trunnions usually are located considerably above the neutral axis and the X component of the trunnion reaction causes a large eccentric load with a consequent large abrupt change in the bending moment diagram. This is usually a characteristic in the bending moment diagram for all cradles or recuperators using guides. Let us now consider the various diagrams showing the characteristics for bending moment, direct stress and shear for the "Filloux" cradle as well as for the "240 m/m Schneider Howitzer" cradle representing typical cradles with guides (figures 10 and 11). Neglecting the weight of cradle as relatively small, and letting Mt= max. bending moment at the trunnions. Mc= max. bending moment at the point of contact of the elevating arc with cradle. Qt and Q? = the front and rear normal clip re- actions. x' and x' = the "x" coordinates of these re- 1 2 actions with respect to the trunnions, d and d = the distance of the friction components 1 2 of Q. and Q from the neutral axis. 190 -»«- SENDING STRESS 155 Mr^n GUN CRRRIRGE: MODEL OP 1916 FILLOUX CRRDLE SHE1RR Fig. 10 191 B = the resultant of the "braking "pulls" reacting on the cradle. djj = the distance from the neutral axis to "B" dx = the distance from the neutral axis to the trunnions . X and Y = the trunnion reactions on the cradle. Xg and Yg = the elevating arc reaction on the cradle . Me = the moment exerted by the elevating arc on the cradle. = the distanc , _^^ the distance from the neutral axis to Xe xe = the "x" coordinate of the elevating arc contact with respect to the trunnions. The bending moment changes at the trunnions by the amount 2X dx. Now for guns with "braking pull" reactions on the cradle in the rear as caused by the compression of the oil and air in the recuperator as in the 155 m/m Filloux, we have for the bending moments at the trunnions, Mt = Qt(xt + udt) just to left of trunnion. Mt = Qt(xt + udt) - 2X dx (just to right of trunnion) and at the elevating arc contact, MC = V1 - xe * Ud8) -Bdb As a check, we also have, Mc = Qt(x^ + xe + udt) + 2Yxe - 2Xdx - Me The bending moment MC is usually distributed caus- ing a parabolic curve as shown in the B. M. diagram of the 155 m/m Filloux.. For guns with "braking pull" reactions on the cradle in the front, due to the tension in the stationary hydraulic piston and recuperator rods, as in the 240 m/m Schneider Howitzer, we have for the bending moment at the trunnions 192 /"ft "1 ' » P J \ CRADLE 240 MM HOWITZER SCHNEIDER 1916 BENDlNCa MOMENT DIRECT STRESS SHERR rig. ii 193 1) -Bdj)( just to left of trunnions) and Mt=Qt (x| + udt)-Bdjj-2X dx( just to right of trunnion) Further Mc= Qt(x|*xc+udt )+2Yxe-2Xdx-Bdb or as a check Mc=Gz(x^-xc+ ud,) In order to compute the maximum fibre stress at a critical section, we must include the direct stress caused by the component of the reactions parallel to the X axis in addition to the fibre stress caused by bending. Therefore a direct stress diagram has been drawn for typical gun carriage cradles. For the case where the braking reaction it in the rear from the front clip to the trunnion, we have, A compression = uQt which is small and nay be neglected as compared with the bending. From the trunnion to the rear clip, we have at the trunnions (just to right of trunnionslsee diagram] a tension = X-uQ ) at the elevating arc contact section a tension. = 2X-uQ1-Xc Where the braking reaction is in front, we have from the front clip to the "braking yoke" on the cradle or ac a section through the point of application of the tensions of the rods on the cradle, we have a compression =uQ From the braking yoke section the compression increases to udt+B at the trunnions A compression =uQt+B(just to right of section) A tension =2X-uQt-B 194 At the elevating arc section.: A ^fltftfj.,to.<ri'»3 -ax-ua -B-X- rfodffo s e-.- x c latUio'? Shear diagrams for the 155 ra/m Filloux and 240 m/m Schneider Howitzer show the variation of the shear in these typical cradles. To recapitulate if yt and yc = the distance to the extreme fibres from the neutral axis at the trunnion and elevating arc section, aod if At and Ac are tbej3£2yeo areas of these respective sections, It and Ic cor- responding moments of inertia, we have for the ex- nn9<j treme fibre stress, ;i,n the critical sections of the /sib ni si noJJsfesn fcniJUid erirf si^riw «»iso eriJ •;• qiio Jno"jl sdJ fnoil i«9i ( ft = - ,., - + — - , ,for,toe braking rep A action in the rear. _; gen oJ Jeuj,/ iaa(x+xc+uda)-Bdb]y or the braking re- ( *' action in the front. for the brak- ing reaction 9w .Jnoi'l ai KI no ln rear .AiftMO [Q (x -xc+udo)]yc uQa — —I — : - * + 8 ., , . 70 for the braking reaction C AC iu f i. in the front. .-0 fi iyoee »J(OH goi^sid «(< J With Barbette and Naval mounts the gun recoils in a cylindrical sleeve which forms part of the cradle, 195 aei . It is thus possible to conveniently locats the center line of trunnions along the axis of bore or rather along the line of action of the resistance to recoil which is usually slightly below the axis of the bjpre. Therefore the elevating reaction is simply due to the moment of the weight ofowponent of" the recoiling mass about the trunnions, though during the powder pressure couple we have a momentary reaction depending upon the distance of the center of gravity of the recqiling mass from the axis of the bore to the powder pressure couple. We may neglect this effect in battery as small and consider the reactions out of battery as the maximum stress con- dition. - In the design of -a sleeve cradle it is highly de- sirable to locate the .pulls symmetrical with the axis of the bore, this being completely done by grouping the two separate hydraulic and recuperator braking systems equally distant and symmetrically above and be- low the axis of the bore as in the 16.* model 1918 rail- way mount. In smaller mounts the spring or recuperator systems are symmetrically distributed about the axis of the bore and the hydraulic braking is affected usually in a single cylinder below the axis of the bore. The hydraulic and spring or recuperator cylinders are usually clamped in short distributed bearings to _______ the cradle proper. The reaction or thrust of the cy- linders is taken upon shoulders at the end of these bearing contacts. Hence a shoulder bearing surface may be regarded as the point of application of the eccentric load due to the pull reaction, on the cradle. Two typical stress diagrams for barbette sleeve cradles are shown for the 12" barbette mount, model 1917 and for the 16" railway mount, model 1918. Considering the reactions on the cradle of the .12" barbette carriage, model 1917 in figure (12) we have, for the bending moment diagram, from the front clip to the trunnion section, we have, a uniform increase in the B. M. from uQd to -• 196 DIRECT STRESS 5HERR BARBETTE MODEL. OF I9H CRRDUE. Hg. 197 In passing from left to right of the trunnion section there is no abrupt change in B. M. since the X re- action being approximately on the longitudinal neutral axis of the cradle is no longer an eccentric load. The maximum bending moment occurs at the section through the point of application of the eccentric load due to the reaction of the hydraulic cylinder against the cradle proper. If Xj!, * the distance to this load d^ * the dis- tance to the neutral axis, where Ph * the hydraulic brak- ing, the bending moment at this section becomes, to left of section, and Pndn-Qt)ud +x'+xn)-2YxJ) to right of section. In terms of forces to the right of this section the B. M. becomes, phdh~Qt(xi~xh~ud)+Ye(xe~xh)*xede to left of «ection» and Qt(x^-xn-ud)-Ye(Xg-xJ)-Xede to right of section. Since the elevating reaction is always very small in this type of cradle, its influence on the B. M. direct stress and shear is practically negligible . Prom the direct stress diagram, we note a maximum tension (sometines compression) between the trunnions and the section through the point of application of the eccentric load due to the hydraulic pall. In passing to the right of this section the direct stress drops in magnitude equal to the hydraulic pull. The shear diagram shows no reversal of shear along the cradle. Considering the reactions on the cradle of the 16" railway howitzer, model 1918 (fig 13) we find a. symmetrical distribution of the braking, the trunnions being located practically along the axis of the bore, thus reducing the elevating reaction to practically the weight effect of the recoiling mass out of battery and the neutral axis of the cradle through the trunnions and along the axis of the bore. There 19! } n 1 4 «i j. »- ,JW rr* v^" V i—^f- 3~ i * s , . >» uj,. $ ._ ^ ^ |_! ~t ~t -~-f-^- I" — t 1 * * 4 4 1 I i > r 1 „ c > -* « * fl p ^| • 4 0 ^ 6 5f! -•. U U E^i i • bn* O\RE1CT STRESS 4s ^c /•• — ^^ • aoiloee add 0^ I*ii?a r. ?> ol •rose aai^Bib it- ^^^^ SHERR _ r^ ^j ! &n i 9 d _ \NCH RRU-W/AY HOV/\TZE1R MODE.U OF CRADLE Fig. N3 199 being no eccentric pull on the recoiling mass we have a distributed load due to the weight of the recoiling mass as shown in the figure In the 8. M. diagram we find the B. M. at the trunnions to be relatively small it being merely equal to that due to a distributed load equal to the weight of the recoiling mass, together with ths friction caused by this loading. w Thus, the intensity of loading = x +x *'Y3 "50 HTCH Hence the B. N. at the trunnions, becomes, t . jtt Wr x Wr + u d !R*<jirt $*u oi tx xoaebqaJ Further between the trunnions and the application of the braking to the cradle by the brake cylinders thrusting against the shoulder bearings for the cylinders on the cradle, we have a tension equal to the total braking, thus causing a direct stress of tension in addition to the bending. To recapitulate, for barbette sleeve mounts, if y= the distance to the extreme fibres from the neutral axis at the section through the point of application of the maximum eccentric pull due to the hydraulic braking or at the trunnion section when the pulls are symmetrically balanced, A = the area of the cross section and I = the moment of inertia of the section, we have for the extreme fibre stress For barbette sleeves with eccentric pulls:— Section at eccentric load - • ', »x • leo (Q. (ud+x *xv ) + 2Yxy, )y X— uQ ft= — i ~ at left of section, I •> ( AT o - and (P^-Q^ud+x'+XhVzYx^jy X-Ph-uQt ^t= " - i at right of section. 200 ?or barbette sleeves with pulls symmetrically balanced about the axis of the bore: Section at trunnion: STRENGTH OP CYLINDERS AND The strength of a RECUPERATOR PORGI>K3S. recuperator forging is a matter of vital im- portance since in modern artillery the tendency is to use higher and higher pressures con- sistent with the various cylinder packings used, and to stress the forging higher with smaller factors of safety. Hand in hand with this goes the metallurgical side where improvements in the quality of the steel with higher ultimate strength are con- stantly being made. High stresses in the recuperator forgings as with high ultimate strength and low factors of safety reduce the weight of the carriage and its cost considerably . Weight of course is of fundamental consideration for mobility. Hence it is of importance to calculate the stresses in the cylinder walls to a considerable degree of accuracy. The maxim stress in a recuperator forging is a combination of the following: (1) A bending fibre stress normal to a plane section perpendicular to the longitudinal axis of the forging which is caused by the external reactions exerted on the forging during firing. (2) A radial compression stress along a radius of the cylinder or normal to a cylindrical surface which is equal to the pressure in the cylinder for the inner surface. 201 (3) A tangential hoop tension, which is normal to a plane passing through the longitudinal axis of the cylinder. Obviously these stresses are principle stresses and with the aid of Poisson's ratio we may arrive at the resultant intensity of stress. In a first ap- proximation however it is sufficient to consider the tangential hoop tension alone, the effect and magnitude of the other stresses small. Consider now a single cylinder 1" long, and sub- ject to an internal pressure p, and external pres- sure pt and of inside radius R0 and outside radius Rt respectively. Further let r = the inside radius to any differential lamina of the cylinder wall (in inches) dr = the radial thickness of the lamina pr = the radial compression at radius r (Ibs/sq.in) pt « the tangential or hoop tension (Ibs/sq.in) £ » the modulus of elasticity. Cj = the longitudinal strain. er 3 the radial strain et = the tangential or hoop strain. Then, for the equilibrium of a differential lamina dr, of length "1" along the axis of the cylinder and a peripheral length equal to the circumference, we have, 2prlr-2(pr+dpr)l(r+dr)=2ptldr hence dPr d(rp, -pp-*1 - — » E- d, Pt a ~Pr-r — - - -T-^ (D V Let us further assume plane transverse sections to remain plane under pressure. This assumption is reasonably close to actual conditions for plane trans- verse sections some distance from closed ends, and in the case of a recuperator forging where the intensity 202 of longitudinal stress, i. e. the bending stress on transverse sections, is relatively small, except for extreme fibres from the neutral axis of the transverse section. We are, therefore, not greatly in error in as- suming the longitudinal strain to remain constant over the entire cross section, hence, vswcd c.: ej = - (pi ) = constant^1 q°< . ii ; edJ lo where p^ = intensity of longitudinal stress, — ev nt oi Jost hence pt-pr=k (2) ^ MU( dPr dP Pt"Pr= ~2Pr"r cT^— or k + 2pr= -r ^7- therefore dpr ^ k+2Pr " ~ Integrating, log(k+2pr)= - leg r* + c or k+2pr= — - r c k hence pr = — - - - (3) ^ r* & ^c_ k «*•»«• qoc 2r* 2 , Substituting (3) in (2) where pr=po r =RQ inside conditions r = Rt outside conditions c k hence P Q- po 2 . c k P = ' - - oJ 1 2R2 2 i_ p _p _ _ fK\ -39 88T9V ro ri vo^ ,T a to 203 Now eliminating c and k, respectively, we find t „ c - -- - - and k = r i <i -3*1 )HoH I1 j3* ~"o3o S Substituting in (5) we have, — - - _ I _L^^^_J-_.-»_ T-. V_JT_«^«— r RZ-R2 R2-B8 r2 (6) ( Apparent ( stress. o . p = , — + _^ — (7) ( HJ-B* Rt2-R« r» )b, Uiq»i« The stress corresponding to the actual strain produced in the material, which is the basis of stress limitation imposed, (assuming m = 3 Poisson's ratio), becomes, .noiaoeJ qoo/i ecU n*rfJ **el «x*»i* •* ool»e»Tq»oc itJt-8T Eet=St=E(^+~ - £i) (8) £i Actual £ m£ raE ( stress ) corres- p,. p. P] ( ponding n o f f r * *\ /n\ ^891^8 «U-J , Eer=S_= -E(— + — + — ) (9) ) to actual ( strain Where (6) and (7) are substituted in (8) ami (9) and o2 i-i-ii. p ° F^-RQ Jotet«<4 bn* { .^f'J) the above expression reduces to Clavarine's formula. Birnie's formula is a modification of the above assuming p^ = 0, that is no external longitudinal tension. Usually pi is relatively small and hence •* A (8) and (9) simplifying to, Uw 1<,01»J*< eriJt lo lifiw co - ebiaJyo «dJ ^o *noiap9^ oocti sd* 204 Eet - S - - — + = -3-2 i- — (10) * 3 R«-Rt 3 R»-Rt r» rg 3 R.- R, The maximum hoop tension, therefore becomes, 2R«+4R« which gives slightly higher values than Lane's when simplified, that is, Rg * R; St* P o«-R2 Lame's foraula (13) Kt Ro From the above formulae it is evident that the radial compression is always less than the hoop tension. With large bending fibre stress due to external reactions on the recuperator forging, pj is no longer equal to zero The maximum stress which is the hoop tension becomes, (14) max THICKNESS OF WALLS BETWEEN Considering two parallel ADJACENT CYLINDERS. cylinders bored in one forg- ing (fig. ) and passing a longitudinal plane section through the center lines or axes of cylinders (1) and (2), we have either half of the forging above or below this plane section in equilibrium under the internal hydrostatic pressures (which now are external with respect either half) and the hooo tensions of the outside and common wall of the 205 two cylinders. Further if the two cylinders are under pressure pt and p respectively and neglect- ing the small variation of the hoop tension for dif- ferent radius, we have for a close approximation ptCT, + T, + «) - ptdt + ptdf where Tt and Tf » thickness of cylinder walls (1) and (2) respectively. w » total width of common wall between the two cylinders . dt and df= the diameters of the respective cylinders. pt = the assumed allowable hoop tension fibre stress. Simplifying, Pt For a correction due to the fact that the hoop tension is not constant but varies sligatly with the radius, we nay augment w by decreasing pt to 0.9 Pf Further due to symmetry Pd P2da and ^ " ~* and substituting in the previous equation, we have w = 1-8 Pt which gives the minimum thickness of wall between two cylinders under pressures pt and p8 respectively. Evidently the maximum simultaneous pressures in the two cylinders should be considered together. 206 ALLOWABLE STRESSES IN Though this matter will be CYLINDER WALLS. taken up in detail in practical design applications, certain limitations could profitably be mentioned here. Cylinders should be tested for strength at pressures considerably higher than would be used in service. It is imperative that even under test pressure the elestic limit is not exceeded. Test pressures should be at least 1 — and preferably twice the maximum service pressures and these test pressures should not exceed 3 the elastic limit of the material or 4_ proportional limit. TOP CARRIAGE The forces exerted on the top carriage are the reactions of the tipping parts and the supporting forces of the plat- form, or bottom carriage, or ground and axle for a trail carriage. The reaction exerted by the tipping parts on the top carriage may be divided inter- CD The trunnion reaction. (2) The reaction of the elevating arc of the tipping parts on the pinion of the top carriage. The tension of the equilibrator chain or rod where an equilibrator is used. These reactions are balanced by the supporting forces exerted at the base of the carriage. In figure (14) the reactions on the top carriage are shown. Considering now the reaction of the tipping parts on the top carriage assuming that an equilibrator is not used. Taking moments about the hinge point A(as in previous discussions), we have, when the gun has recoiled a distance X out of battery. 207 1 noiJBupe svcds sri-i ni ssui^v i«J ,§• -iR. 14- .III TsJqsriO aaS fa - 2+S nie^l-^ aoo^ri woM :ods 1H 'jo BT6 208 2X(ht-it sio 0)-2Y(lt cos 0+ht sin 0) - E cos(ee- 0)[ht-j cos(9e-0)]-E sin(ee-0[lt-j sin(ee-9)] 'ta Mo" (Kps+Wpx cos 0) 2X»Kr+Wt sin 0+ • cos 9( J (Krs+Wrx cos 0) 2Y=Wtcos 0- . sin 9e and Krs+WrX cos 0 in _ J Substituting these values io the above equation, Krs+WrX cos 0 r+Wtsin 0+ ( • • )cos 9e](htcos 0-ltsin J Krs+WrX cos 0 -!Wtcos 6-( : ) sin 6e](lt+ cos 0+btsin J [KP+Wtsin 0+ ( , )cos 9e](htcos 0-ltsin 0) 0) J Kps+WrX cos 0 cos(ee~ 0)(ht-j cos(9e-0)l -(Krs+WrX cos)sin(9e-0)[lt-j siD( Expanding and simplifying, the above reduces to: Kr(htcos 0-ltsin 0+S)-wtlt+»rX cos » Mta (1) How htcos 0-ltsin 0+S = d See Chapter III. where d is the moment arm of Kr about the hinge point A. 209 Further due to the displacement of the recoiling parts a distance X from the battery position, the center of gravity of the tipping parts is displaced a distance ffrcos ^x from the initial trunnion Wt position. The moment of the weight of the tipping parts about A, is therefore, r Wt(lt--r-X cos 0) = *tlt-Wrx cos 0 ™t Hence from equation (1) we observe that the reaction on the top carriage is equivalent to the total resistance to recoil through the center of gravity of the recoiling parts together with the weight of the tipping parts act- ing at a distance IE x CQS ^ from the trunnions. There fore the line of action, is equivalent in effect to the resultant of the trunnion and elevating arc reaction. This is almost obvious from first principles since the reaction of the top carriage on the tipping carts must balance the resultant of Kr and Wt: hence by the law of action and reaction, the resultant reaction of the tipping parts on the top carriage is therefore equal in magnitude and direction to the resultant of Kr and Wt. With a balancing gear we have in addition to the trunnion and elevating arc reaction on the top car- riage, (which now have different values from the pro- ceeding) the tension of the equilibrator chain or rod. By exactly a similar analysis as in the above, the reaction on the top carriage reduces to the resultant of Kr and Wt where the line of action of the component • Wt if now disolaced a horizontal distance __r_ ^ X cos /) - the distance which the center of gravity of the tipping parts in battery is placed backwards from the trunnion position, when the balancing gear is used. 210 In figure (14) is shown the various reactions on the top carriage together with a force polygon. Thus from the space diagrams of reactions obviously the lines of action of the resultant of the trunnion re- ?, action and elevating arc reaction intersect at a com- mon point which necessarily lies along the line of action of the resultant of Kr and Wt where the com- ponent of Wt is displaced a horizontal distance X cos <6 from the trunnion axis. * In the vector polygon of forces we note that by vector addition, K+»t-X+T+E Further for the equilibrium of the top carriage, X+Y+E+Ha+Va+Vb = 0 hence K+Wt+Ha+Va+Vb = 0 The above results are exceedingly valuable in graphical methods as will be outlined later for ob- taining the various reactions throughout a gun car- riage. on ic-. ;pe «i .noijse 1o enrl edi eiol •if- fioifmuiJ arfo SUPPORTING REACTIONS Top carriages have been OW VARIOUS TYPES OF q classified in Chapter I, into TOP CARRIAGES. l[l) the ordinary type with side frames and connected at front or rear by cross beams or trans- oms, which contain the pivot bearing, (2) pivot yoke type used on small mobile mounts and (3) trail carriage. The supporting reactions in the ordinary type of top carriage are the H and V components of the pivot -• v ct ri Kd/i H **> riiwi fi^£.f*f bearing which is usually in the front and the V com- ponent exerted by the traversing circular guides in the rear. Sufficient horizontal play is allowed so that the • reaction of the horizontal traversing guides is only • vertical, the H component being taken up entirely at the pivot bearing. As a typical class (1) top carriage we may illus- trate by the Vickers 8", Mark VII, British Howitzer. ; siU a-oil --SJ.TBCJ ni EJieq 211 Further let 1 = distance between supporting reactions measured horizontally in the direction of the axis of the bore at 0° traverse. A = the front pivot point. • r B = the resultant of the distributed vertical re- actions of the horizontal traversing arc guide. 1 k = the horizontal distance to trunnions from B in the direction of the axis of the bore o at 0 traverse. ht = height of trunnions above the traversing guides. S = the perpendicular distance from the trunnion center to the line of action of the resistance to recoil which necessarily passes through e center of gravity of the recoiling mass. = height of horizontal component of pivot re- action above the horizontal traversing guides. = weight of top carriage proper. = moment arm of W about B. Considering fig. (15) we have for the horizontal component of the pivot reaction, Ha=K cos /5 and taking moments about fl, the center of pressure of the traversing guides, we have, Kd-W^-t*Wrx'cos 0-Wtcltc+Val-Haha = 0 ( . Wtlt+Wtcltc-Wrx cos 0-K(d-ha cos 0) ) hence Va= • ' ' *-ha cos 0)+Wt(l-lt)+lftc(l-ltc)-»-H'rx cos 0 ( and Vb = where for low angles of elevation, d=htcos 0+S-ltsin d' = htcos ^-«-(l-lt)sin 0+S 212 Fig. 15 213 and for high angles of elevation, d*ltsin 0-btcosl 0-S d' = (l-lt) sin 0+htcos 0+S ymn = the horizontal distance along the axle from the center of the wheel bearing pressure. Considering, max. traverse, right handed, at max. elevation, the reactions on the axle to the left of the section, become, (1) The components of the trail con- necting arm reaction on the axle:- X,Y and Z together with a couple Mxy in the horizontal plane. (2) The vertical reaction exerted by the left wheel, Sa. Therefore at section "mn", we have, (1) Bending in the vertical plane: (2) Bending in the hor- izontal plane: <in lbs-> (3) Shear in the vertical plane : X+Sa (Ibs) (4) Shear in the hor- izontal plane: X (Ibs) (5) Torsion about the y axis, or in the x z plane: T=X ZBn (6) A direct thrust: Y (Ibs) Thus, we have bending in two planes combined with torsion, and a direct thrust as well. Then for a 214 round section, as would^ Dually be the case, we h*»»,jon6 f 3 - - + - _ Max% normal fibre °-78J thrust on outer layer •oil »fx* etu c,ns.t8ifa 1*5 (Ibs/sq.in) gniised Jaedw &dJ lo -iscfnso sr)J , .- Oj elxfi eri^ no iis-U od^ io -ralxs ori^ no noj ";s iniJoso The m-axi/num fibre stress, therefore, becomes _ ._an£tq Ifi-tnoxiiori erid ni f =-f + /¥**"£ Jo* en ]80lj18v ftbs/sq.in) 3 4 • a 2 * j • 2 which should not exceed — of. the elastic limit of the material to be used. As a typical class o'f pivot yoke type, consider the reactions on a Barbette or Pedestal mount, figure (16) and a pivot yo.ke top carriage used on a trail carriage, fig.ure (16A). In the first type, the lower bearing sustains both horizontal and vertical com- ponent reactions, whereas the upper is merely a floating bearing and therefore sustaining only a hor- izontal component and designed to prevent bending in the lower pivot. In the second type, the middle bearing has a tapered fit within the axle, and therefore sustains both horizontal and vertical components, but suffers no bending moment since the axle is free to rotate. To prevent the top carriage and mount from rotating about the axle a lower cylindrical vertical pivot fits within an equalizer bar below the. axle, the equalizer bar being attached to the trails. (See Theory of Split Trail - next section). In this type of mount it is more convenient to compute the supporting reactions in terms of the hor- 6 10^ nsriT .Ilsw 8£ Jsind.* joetifa £ b.~. 215 foul as o l«3JUi»tr bn* isJ-nosx aoioiq Sai<JBV8is 04 z^ibsi V.Q absa -lev isdJ diiw jip*i 2oiJ*voI» no REftCT\ONS ON P\VOT >TOKE. TOP cerfT 1 J II ^4^ IS 8OO ~n ale {- aie :®n — -ZH 216 izorual and vertical components of the trunnion. If ne = angle made by radius to elevating pinion contact on elevating rack with the ver- tical. j = radius of elevating rack Then for the horizontal and vertical components of the trunnion reaction, we have, Fe+Ks >K cos 0+ ( ) cos ne ) In battery j ( Fe+Ks ) <K sin 0+Wt-(— : ) sin ne ( J ) and Ks+Wrb cos 0 2H =K cos 0+ ( ) cos ns ) Out of J ( battery. Ks+W_b cos 0 2V=K sin 0+Wt-( ) sin ne ( For the elevating gear reaction, we have Fe + Ks in battery Ks+Wrb cos 0 j out of battery In the Barbette or Pedestal Mount, figure (16) let, xt = distance from center line of pivot to center of trunnions. yt = height of center of trunnion from bottom of rivot. T " radius of pivot bearing. 217 r*f = radius of floating bearing. y^ = height between bottom of pivot and top of floating bearing. Then, Va = 2V + E sin ne and 1 ( Hb= ~T ( 2Hyt+2V(xt-rp)+Etj+(xt-rp)sin ne-yt cos ne] "m and Ha=Hb+E cos ng-2H In the pivot yoke trail top carriage, fig.(16A), let x^ = distance from center line of pivot bo center line of trunnions yt = distance from center of axle to center of trunnions . ym = distance from center of axle to center of equalizer beam. Then, Va=2V+E sin ne and H(j= - [2Hyt + 2Vxb+E(o+xtsin ne-ytcos ne)] Ha=Hb+2H-E cos ne THEORY OF SPLIT TRAIL. The object of a split is primarily to give a large aperture between trails for the gun to recoil at maximum elevation and maximum traverse When split trails are used it is also desirable to dis tribute the bearing load on spades when the gun shoots at maximum elevation. T"his is accomplished by the use of an equalizer bar connecting the two trails, or more strictly the trail arms, beneath the axle, the equalizer laying usually in a horizontal plane and pivoted at its center by a vertical pin through the center of the axle. With split trail and equalizer, a pivot yoke type of top carriage s-hould be used. The elements of a slip trail mechanism are :- (1) The two trails with their spades which are connected by a vertical pin 218 n o I i s v 5 7 ' & 2| *» KT^ I ^ x i... / toannoojifid "^ "V i i n 219 to two trail arms or trail connect- ing pieces, at either end of the axle. (2) The trail arn or connecting pieces are free to turn about the axle in a ver- tical plane and are prevented from slid- ing along the axle by thrust shoulders. The moment about the axle of the trail reaction is balanced by the moment about the axle of the shear reaction ,,£t the equalizer bar. (3) The equalizer bar is usually de- signed to rotate in a horizontal plane about a vertical pin through the axle. Thus, with a split trail we have the two trails, their connect- ing pieces to the axle and equalizer and the equalizer bar, connecting the trail arms and pivoted about a ver- tical pin which passes through the center of the axle. We have the following possible motions:- (1) A free rotation in a horizontal plane of either trail, about the vertical pin in the trail arm. (2) A constrained rotation in a vertical plane about the axle of either spade, the constraint being due to the equalizer bar. (3) A constrained rotation in a horizontal plane of the equalizer bar about a vertical pin through the axle. MAXIMUM BLBVATION: Let x and y = horizontal coordinates in longitudinal and transverse directions respectively, 220 i. e. along and cross wise to the bore at zero traverse. Z - vertical coordinate. AXAVAZ = the component reactions at the left spade (positive direction towards muzzle (Ibs) 8xByBz = the component reactions at the right spade (Ibs) Sa and Sb = normal vertical reactions for left and right wheel respectively. (Ibs) Qa and QJJ = shear reactions of equalizer on trails A and B respectively (Ibs) dj, = horizontal distance or span of equalizer between trails which it connects (in) de = vertical distance from center line of equalizer to center line of axle through wheel hub. (in) £Mav = moments of the components of A about the axle. (in Ibs) SMQV = moments of the components of B about the axle . (in Ibs) jc_ = distance from spade to axle. (in) yo = distance from ground to center line of axle. (in) 2Ma[, = moments of reactions of A about vertical pin in left trail am. SMfj 3 noments of reaction of B about vertical pin in right trail arm. Taking moments about the center pin of the equalizer, we have, Qa 111 = Q. Ik. a 2 b 2 hence Qa = Qb = Q. Therefore, for moments about the axle, we have -Q de = 0 hence, 2Ma * 2Mb We have for unknowns, Sa Sb 221 ( Sight unknowns Equations for solution: ZX £M ZY= 0 ZMy =0 X = 0 = 0 2Mbh= 0 ) ) Nine solutions ( ) ( ) We therefore would expect either 2Mah or ZM^j, not zero This is physically justified since on extreme traverse one of the wheels and trails must be in contact. This is met constructively by usually introducing a show attached to the trail which comes in contact with the wheel upon traversing. If N = normal reaction between shoe and wheel, dD = perpendicular distance from vertical pin on trail arm to vertical plane through wheel, i. e. to line of action of N. We have for maximum traverse in a right banded rotation, ^bh=^ ^n *nus introducing an additional unknown N. The solution is, therefore, statically possible either introducing the above equation or omitting it entirely. METHOD OF SOLUTION Assume maximum traverse right handed turn, Let 0ffl = maximum angle of elevation. 8m = maximum angle of traverse. 222 Ks = resistance to recoil at maximum elevation ,,. . :v*ri eW (Ibs) Xy = horizontal distance to projection of center of gravity of recoiling parts measured from base line AB. (in) yg = 0 assumed approx. Zg = height of center of gravity in battery above ground, (in) g = vertical distance from ground line to center of pressure on spade . (in) Wg = weight of total system, gun + carriage. (Ibs) g = horizontal distance from AB to W ,, Li *7 y A'B1 = distance between vertical trail pins. Then, the resolved component of Ks through the center of gravity of the recoiling parts, become, /<MS Kscos 0m cos em, along the x axis ) Kscos 6^ sin Q , along the y axis ) ( r. i ri T K sin <L along the Z axis ) xp = distance from AB to either vertical trail pin. yab = distance between A and B, trails completely , neswJed tv i-ioa = n 11 , spread. Taking moments about A B, we have, : muaix hence, W 1 -Kscos 0 cos 9m(Z.+ g)+Kssin • x. - -, .pa «vo4s art* r- .:is and Az+ Bz= V Ks sin ^«-(Sa+ Sb ) (2) Next take moments in a horizontal plane about the left spade, and we have, . .t £ v a 1 s J^ab = . em. xg-Kscos Bf.cos 9m -j- = 0 223 K cos 0m hence Bx = - (0.5 yab cos 9n~Xg sin 6m) (3) yab Further Ax+Bx=K3cos J0m cos 0m hencs AX=KS cos Am cos 9m-Bx (Ibs) (4) For moments about the vertical pin in trail arm for left trail, we have, Ayx_-Ax 0.5 y/^'B1 ~ 0 0.5 AxyAnj» hencs Au = - (Ibs) (5') XP Now if we take moments about the axle, we have ZMgy^ Az *0-W*> = Bzxo-Bx^o^> therefors (A-B ) (Z+g) 'Z "Z X, but lience Az =~ 2x ~~^~ (6) Bz=W>»+Kssin e)m-(Sa+Sb) - Az (7) Let X, Y and Z = components of the reaction of the trail arm on the axle, MXV = moment reaction of trail arm on axle, in the X Y plane. Considering moments on the left trail and trail arm together about the axle, we have, AzxQ-Ax(Z0+g)-Q de = 0 hence, the horizontal shear reaction of the equalizer becomes, *. Next consider the various reactions on the trail arm, and we have, 224 = ° along the x axis, -Y+A = 0 along the y axis. -Z+AZ =» 0 along the z axis. and further, -Mxy+Ay(xo-xp) '0 In the x y plane Therefore, the reactions of trail arm on the axle, becomes, A_xo-Ax(Z0+*) X=A+A » z o * o + A., (Ibs) (9) do Y = Ay (Ibs) (10) Z « A, (Ibs) (11) Mxy» Ay(x0-xp) (in. Ibs) (12) Of AXLS MAXIMUM TRAVBR8B, MAXIMUM ELEVATION: This critical section of an axle is at a section near the center where the axle becomes enlarged for holding the vertical pivot of the top carriage. If the axle is made straight, we have no torsion on the section but mersly bending in a vertical and horizontal plane. If, however, the axle is underhung for clearance and lowering the top carriage, in addition to the bending, we have torsion as well, the nagnitude of the torsion depending upon the depth of the underhang. Let mn be the section under consideration near the center of the axle. xnn ^mn an^ zmn = ^9 component distances from the center of contact of the trail connecting arm and axle. REACTION BETWEEN RECOILING During the counter re- PARTS AND MOUNT IN COUNTER coil, we may distinguish RECOIL. between the accelerating and retardation period so far as the. reactions between 225 the recoiling parts and mount are concerned. The re- actions during the acceleration however are of exactly the same character as during the recoil only of less magnitude. Therefore, from either the point of view of the internal reactions or stability of the mount, we are not concerned with the acceleration period of counter recoil. Therefore let us consider the various recoiling parts and mount coming into play during the retardation period of counter recoil, - Let (see figure 18) xt and vt = coordinates, along and normal to bore, of front clip reaction with respect to center of gravity of recoiling parts. x and y = coordinates, along and normal to bore, of rear clip reaction with. respect to center of gravity of recoiling parts. Qt s front clip reaction. Q2 = rear clip reaction. Wr = weight of recoiling parts. 0 = unbalanced retarding force exclusive of f rict ion. ^0 - distance from center of gravity of re- coiling parts to line of action of 0. n = coefficient of friction = 0.15 usually. d1 = distance from front wheel ground contact to line parallel to tore through center of gravity of recoiling parts. lr= horizontal distance from line of action of Wr to front wheel ground contact. x = displacement along bore or guides from out of battery position. Ma = moment of reaction of the recoiling parts on mount about front wheel contact and ground. Then, for the motion of the recoiling parts, we hava. 226 REACTION BETWEEN RECOILING PRRT5 RND MOUNT IN COUNTER RECOIL RECOILING PRRT5 227 d x 0+n(Qt+Q2)+Wrsin 0=-mr — - (1) d t Qt-Q2=Wrcos 0 (2) and 0 d0-ftlx1-Qfx8+n Qtyt-n Q8y2 = 0 (3) Next, considering the reactions on the mount and taking moments, about the point of contact of the front wheels with ground A , we have, (4) - - d'tan 0+x J=MrA' 0 Substituting Eq. (3) and (2) in Eq. 4, we have immediately d"x rQ )d +Wrsin 0.d -W_l = * 2 or (0+n(Qt+Q2)+Wrsin 0)d '-Wrlr=MA Further from equation (1) (-mr ^ )d'- Wrlr= My Q t» Hence, the reaction on the mount during counter recoil is equivalent to the total resistance to recoil acting in a line parallel to the axis of the bore and through the center of gravity of the recoiling parts, together with a component in line and equal to the weight of the recoiling parts. If further, we let, Fy = the recuperator reaction. RQ = total guide friction RS+P 3 total packing friction. Bx = total counter recoil buffer reaction. Then 0 = B '+£„.. n - F,, aa ) = n Wr Cos 0 (approx.) and the overturning force, passing through the center of gravity of the recoiling parts and parallel to the 228 bore, becomes, i dv - [Fv-Wr(sin 0 +n cos 0)-Rs+p-Bx)= -rar v — - (Ibs) GRAPHICAL CONSTRUCTION AND Very often it is more EVALUATION OF THE REACTIONS convenient to evaluate IN A GUN CARRIAGE. the various reactions by graphical methods. Graphical constructions are of special value since they give a vivid picture of the relative magnitude of the various reactions. Further the method is comparatively simple and the closing of force and space polygons combined with overall methods gives an admirable check. The ac- curacy of the method even -with rough layouts is suf- ficient for the computation of the various reactions required. If we consider the kinetic equilibrium of any piece of the carriage, we have, by introducing the . kinetic reactions or inertia forces with the actual reactions exerted on the piece, a dynamic problem re- duced to a problem of statics. For equilibrium of the piece, we have, 2X = 0 ) ZY = 0 ( for a coplanor set of forces. ZM = 0 ) Now the considerations ZX = 0, ZY = 0 are met by the vector diagram of reactions having a zero re- sultant, that is the vector polygon of the piece closing. The condition ZM = 0, requires a consideration of the lines of action of the forces in a space diagram in addition. Since the moments may be taken about any point, there can be no resultant moment exist- ing. The condition 2X = 0, ZY = 0 implies the result- 229 ant force to be zero, but does not imply the existence of a couple. Condition ZM 3 0. indies that a resultant couple cannot exist. A graphical method, therefore, always consists of two sets of diagrams, (1) a space diagram of forces and (2) a vector diagram of forces. The space diagram requires a layout proportional to the actual piece under consideration and the placing on this diagram the lines of action of the forces. The force diagram requires a layout proportional to the direction and magnitude of the various reactions exerted on the piece. The two diagrams must be carried on simultaneously since the direction of a resultant required in a sp^ace diagram, is obtained by the vector addition of the forces which are the com- ponents of the resultant. Since Vector addition is commutative, the order of Vector addition is immaterial. REACTIONS ON THE RBCOILIMG PARTS The known reactions consist: (1) The powder force along the axis of the bore Pfc . (IJbs) (2) The inertia force along an axis parallel to the bore and through the center of d • gravity of the recoiling parts - - mr x's Zt r ^ (3) The weight of the recoiling parts acting vertically through the -center of gravity of the recoiling parts - - Wr. The unknown reactions consist: A a; ** ^ ** •*• u £ " * (1) The resultant braking force B Ibs. (2) The front and rear clip reactions Q, and Q2 Ibs. 230 The lines of -actions of these forces however are known or can be readily determined. Procedure: Layout a space drawing proportional to the dimensions of the recoiling parts, showing the assumed lines of actions of the various forces. See fig. (19;. 2 Since P^ - »r — — = K the total resistance to recoil which is assumed as known, we have the effect of PK and m ^ * equivalent to, b G "r dt2 (1) a couple Pfc % (2) a force K through the center of gravity of the recoiling parts parallel to the axis of the bore. Since a couple and a single force may always be reduced to an equivalent single force, we have (1) and (2) combined into a single force K acting at a distance above the axis through the center of gravity of the recoiling parts equal to P. eh p 9 (in) ( where CK is in inches) K The reactions 0 and Q due to the friction in 12 i the cradle sleeve make an angle u = tan -*f with the normal to the guides, where f = coefficient of friction Q = 0.15 usually. Hence u = 8.5 approximately. Referring now to the force polygon or diagram, lay off K in the direction and equal to the magnitude of the total resistance to recoil. Lay off K = a b From b lay off b c = WR, the weight of the re- coiling, in magnitude and direction. Draw K + Hf = a c 231 232 Referring now .to tha space diagram lay off K at a perpendicular distance °b9b K above the center of gravity of the recoiling parts and parallel to the axis of the bore. At the intersect- ion of K and Wp , draw J k parallel to a c until it intersects the line of action of the motion of the reaction Q2 at k. From c of the force polygon, lay off c d and fn the direction of the rear clip reaction QZ. Draw k c from k to the intersection of Qt and 8 in the space diagram. Draw a d parallel to k c in the force diagram until it intersects a d at d. This limits and de- termines the magnitude of 0 in the force diagram. From d, draw d e parallel to B and a e parallel to Q . The intersection of a e and d e determines B and Qt respectively. Thus from a combination of the space and force diagram we obtain Q2 B and Qt respective- iy- REACTIONS ON THE CRADLB. Referring to figure (20): The known reactions consist:- (1) The rear clip reaction Q2 (Ibs) (2) The front clip reaction Qt (Ibs) (3) The weight of the cradle Wc (Ibs) (4) The braking force B (Ibs) The unknown reactions consist:- (1) The trunnion reaction T (Ibs) (2) The elevating gear reaction E (Ibs) The direction of the latter being' known. Referring now to the force diagram lay off a b = in the direction and proportional to the magnitude of flf. From fa draw c parallel and equal to B the brak- ing force. Draw a c. Referring now to the space diagram J k from the 233 in the force polygon, to the intersection of Q± • In the force diagram, draw c d = Q and parallel to Q.t . draw ad. In the space diagram draw J L parallel to a d to the intersection of Wc. In the force polygon draw lfc equal and parallel to Wc the weight of the cradle. Draw a c. In the space diagram 1 m parallel to a e to the intersection with E at m. From m draw m n to the trunnion axis, which gives the line of action of the trunnion reaction T. In the force polygon draw e f in the direction of E and a F in the direction of T. The intersection at f determines the magnitude of E the elevating gear reaction and T tha trunnion reaction. t» yf "X REACTIONS OH THE TIPPIHG PASTS. Locate the trunnions along the resultant of the "battery position of Wr and Wc --- See upper diagram. Without balancing gear:- Considering the external forces on tipping parts, we have, the known reactions, (1) The total resistance to recoil K (l"bs) (2) The weight of the tipping parts Wt (Its) The unknown reactions, (1) The elevating gear reaction E (l"bs) (2) The trunnion reaction T (l"bs) the direction of E "being known. In the space diagram lay off X parallel to the bore and at a perpendicular distance from the center of gravity of the recoiling farts = pbe (. . In the force diagram, lay off ab = K and be = Wt. Draw ac. In the space diagram, lay off J k from the inter- section of K and Wt parallel to ac of the force diagram 234 235 I <L 6 U o o t- 0 -o & 5 \ u o i of ft: a: o 0. o K cu V) 236 a: Q O d I 0 U 237 £ <c s cr uJ u ot o u. I Q LJ o d 0. CO u a (3 m £ K ID N <t DO 0 iZ Q *^'pfe; k ' y 3 ^* ^ - nl/f r x b i / 7 '{ 8^ \ £ / A §5 \ \ i 0 c / \/\ i g /y ' t /7 ! s \ \ \ \ \ \ f / \ x/ / \ I / / / i ' / / */ ' '// ^-X'/ / / " / // /7 / Aw !' Its nr^ / A ^ > \ 1 Z o ^^-"' I r>- 238 and t o the intersection of E. Draw k 1 to the trunnion axis in the space diagram. The line of action of T is then along h 1 produced. In the force polygon, draw cd parallel to E and a d parallel to k 1 of the space diagram. Their inter- section at d determines the magnitude of £ and T respectively. With balancing gear:- Determi nation of the balancing gear reaction R. On the space diagram lay off Wj. the weight of the tipping parts in its battery position as well as the line of section of R. From the intersection of R and Wt draw o m. This must be the direction of the result- ant of Wt and F since the condition is that we have no moment about the trunnions when W^ is in its battery position. In the diagram below, lay off W. and R and draw o.m parallel to o m in the space diagram. This determines the magnitude of the balancing gear reaction R. Referring to the force diagram, lay off a b equal and parallel to K the total resistance to recoil, and be = Wt the weight of the tipping parts. Draw ac. In the space diagram K is at a perpendicular dis- D p tance _fc_- from the center of gravity of the recoil- ing parts and V^ at a distance Wrx cos 0 — * from its battery position, where x is the displacement in the recoil At the intersection of K and Wt draw j k parallel to ac to the intersection of R. In the force polygon draw cd parallel and equal to F. Draw a d. In the space diagram draw 1 k parallel to a d of the force polygon to the intersection of the line of action of E ac 1. Draw 1 m to the trunnion axis, thus determining the line of action of the trunnion re- action T. In the force polygon draw d e parallel to m E and 239 a e parallel to 1 m, thus determining the magnitude of E and T respectively, R3ACTION3 ON THB TOP CARBIAGB Without balancing gear:- The known reactions consist: (1) The weight of the top carriage Htc (Its) (2) The trunnion reaction T (Ibs) (3) The elevating gear reaction E (Ibs) The unknown reactions consist: (1) The horizontal component of the pintle reaction - H (Ibs) (2) The vertical component of the pintle reaction N (Ibs) (3) The front vertical clip reaction M (Ibs) The lines of actions of these forces are given from the construction of the piece. Referring to the force polygon fig. (25), draw ab =T equal to the magnitude and in the direction of T the trunnion reaction. Draw be parallel and equal to E the elevating gear reaction. Draw ac. In the space diagram draw j k parallel to T. At the intersection of j k and E produced draw k 1 parallel to a c in the space diagram to the intersection of Wtc. In the force polygon draw c d equal and parallel to Wtc- Draw a d Prom 1 in the space diagram 1 IE parallel to a d to the intersection of N produced. From m draw m n to the intersection of H M. Draw a e in the force polygon parallel to mn in the space diagram. We thns have d e in the force polygon = N and ef = M and ja= H. 240 Thus the pintle reactions H and N and the clip reaction are determined in magnitude and direction. With balancing gear:- The "known reactions consist: (1) The weight of the top carriage Wtc (2) The trunnion reaction T (3) The elevating gear reaction E (4) The balancing gear reaction B The unknown reactions consist: (1) The horizontal component of the pintle reaction H (2) The vertical component of the pintle reaction N (3) The front vertical clip reaction M The lines of actions of these forces are given from the construction. Referring now to the force polygon fig. (24) Lay off ab = T and be = E. Draw ac. In the space diagram from the intersection of T the trunnion reaction and E elevating reaction pro- duced at K. Draw k 1 parallel to ac of the force polygon. Con- tinue in the force polygon c d = R the balancing gear reaction. Draw ad. In the space diagram draw in parallel to ad and the intersect! on of W^ at m. In the force polygon draw de. Draw ae . In the space diagram draw mn parallel to ae to the intersection of N. Fron N draw n o to the intersection of M and H. In the force polygon draw a f parallel to o n. From E in the force polygon draw e f parallel to N to the intersection of e f . Draw f g and & a as shown. Thus we determine the reactions M, N, and H respectively. 241 REACTIONS ON THE ASSEMBLED CARRIAGE GUN ASP CARRIAGE TOGETHER. Location of the weight of the total mount:- Assuming a static reaction of 200 Ibs. under the spade, we lay off o'm = 200 Ibs. Then o N = Wg = 200 under the wheel contact. The resultant of o'm and o n » W3 obtained by the additional construction lines o'q and op. Hence we determine from the triangle of forces the line of action of *L. The external reactions on the as- 9 sembled carriage consists of :- The known reactions - (1) K = the total resistance to re- coil. (2) Ws = the weight of the total mount. The unknown reactions - (1) The horizontal spade reaction Ha. (2) The vertical spade reaction Va. (3) The normal reaction under the wheels The direction of these forces are known. Referring to the force polygon lay off ab = K the total resistance to racoii and be = weight of total system W3. Draw ac. In the space polygon from the intersection of < and Wg draw j k to the intersection of the reaction Va- Prom k draw k 1 to the intersection of Ha, Vb at 1. Referring to the force diagram draw ad parallel to 1 k of the space diagram to t"hs intersection of c e produced. We thus determine c d - Va, d e = Vb and e a - Ha< 242 Thus the reactions Ha, Ba and V^ are determined in magnitude. PROCEDURE IN THE CALCULATIONS FOR THE PRINCIPLE RE- A.CTIONS IN A GUN CARRIAGE MOUNT. (Illustrated by calculations on 240 n/m Hewitmer) REOUIRBD DATA. Type of Gun Howitzer Diameter of bore d (in) 9.45 Total Weight of recoiling parts Wr(lba) 15790 Weight of Powder Charge W (Ibs) 40 Muzzle Velocity v (ft/sec,) 1700 Travel of Shot in Bore u (in) 160 maximum 60° Angle of Elevation 0 ninimum 10° short 3.74 Length of Recoil b (ft) long 3>g0 Intensity of Powder Pressure p^dbs/sq.in) 32000 Initial Air Volume of Recuperator Vai 2970 (cu.in) Initial Air Pressure of Recuperator Pai 576 (Ibs/sq . in ) 243 INTKRTOR BALLISTICS. Maximum Powder Pressure on Breech 2,245,000 F » Pb = 0.7854 d2pm (Ibs) Maximum Powder Pressure on Base of 2,005,000 Projectile pm (Ibs) P°=A: (Us) Mean Constant Powder Pressure jQ * *7°Q 5.36 x 160 5*36U 1,350,000 1 = twice abscissa of Max. Pressure - ~ —)* - 1 3.996 POD = Muzzle Pressure on base of breech 622,000 Ibs. Vsl. of free recoil: 7f wVm + 4700 W = 50.25 ft/sec Wr Vel. of free recoil - Shot leaving Muzzle 0.5W Vm wr 40.50 ft/sec, Time of Shot to Muzzle t s — — i- 0.01175 sec. 1 2 12V,, 244 Time of Expansion of Free Gases - ob 32.2 0.01538 sec, Free Movement of Gun while shot travels to Muzzle •_ u"(w+0.5W) l~ 12(Wr+w + w) 0.31 ft. Free Movement of gun during Pow der Expansion P v t* 0.7179 ft. Total free Movement of gun; Pow- der Pressure Period: I • Z4 + X, 1.0279 ft. Time of Powder Pressure Period r - tt * tf 0.02713 sec, BRAKING PULLS AMD STRESSES IN CYLINDERS. x axis taken along bore: v axis taken normal to bore. Mass of Recoiling parts »r "r '' 32.16 15790 32.16 491 Constant of Stability C » 0.85 to 0.9 Calculations only for max. elov. Height of center of gravity of recoiling parts above ground h (ft) Calculations only for max. elev. Stability Slope elf. Calculations only for max. elev. 245 Total Resistance to Max.Elev. Recoil 491 * 50.75 Hor.Elev. ,2 K = 2(b-£+VfT) (Ibs) 2 (3. 74-1. 0279+50. 75*. 02713) 152,000 Variable Resistance to re- Calculations only for max. coil in battery (at elev. horizontal elev.) jnV K-- rf 2[b-E+VfT- - - 2 M, (Ibs) Variable Resistance to Re- Calculations only for max. coil out of battery (at elev. horizontal elev. ) k = K-m(b-E+ - ) 2m., Initial Recuperator Re- action, Pai = approx. 1.3Wr(sin 0m+0.15cos0m) Ibs. (unless given) 1.3 x 15790 (gin 60+0. 15cos60) = 19300 used 18800 Ibs. Total Initial Recuper- ator Pull, Pai = P^i 100 d. (Ibs) 18800+100x2.938 19094 da = diam. of recuperator rod. (in) 0. Effective Area of Recuper- 35.756 ator Piston - Aa (sq.in) Initial Air Pressure ai (Ibs/sq.in) 18800 32.6 576 246 Initial Air Volume Vai (cu.in) 2970 Final Air Volume Vaf(cu.in) vaf - Vai -12 Ab a Final Air Pressure P., "".'" Final Recuperator reaction 2970 - 12 x 32.6 x 3.74 = 1510 S76 1214 1214 x 32.6 = 39600 af = p af a air Paf= approx. 2Pai(lbs)metallic J_Distance from axis of bore to 3.038+ 3.850 mean guide contact r(in) 3 3.4444 Distance between clips 1 (in) 86.25 J. Distance from axis of bore to 16.365 center line of hydraulic pis- ton e^ (in) J_Distance from axis of bore to line of action, of recuperator 15.656 reaction ea(in) Assumed coefficient of guide friction u = 0.15 to 0.25 Guide friction constant 2u » Af l-2ur 0.15 0.15 86.25 -2x0.15x3.44 .00352 247 Total hydraulic Pull 152000+157908in60-18800( 1+.0635) (max. elcv.) 1+.0663 » 137500 UAfeh (Ibs) Total hydraulic Pres- 2 hydraulic cylinders: ^-100 dn ^ ' diam. of brake rod Ph»P^-100 dn ^ 68750-4.72x100= 68280 (in) Effective Area of Hy- 31.2 draulic Piston Max. Pressure in Hy- 68280 - = 2200 draulic Cylindsr 31.2 ' (Ibs/sq.in; Inside Diam. of brake 7.874 cylinder dih- 1, (in) (dn* diam. recuperator rod) Outside diam. of brake 9.450 cylinder Hoop tension in brake cylinder wall 2200(== — ) = 12150 9-45' -7.875' Ibs/sq.in. 248 Max. pressure in recuperator 1214 cylinder v Paf=?ai Ubs/sq.in) Inside Diam. of recuperator 7.087 cylinder dia = 1.13/A0+0.785d| (in) Outside Diara. of recuperator 8.267 cylinder doa (in) Hoop Tension in Recuperator Cylinder Wall ,d§a+d!a 1 d 2 -d * ' uoa uia . in) .8.267+7.087 12140==1) =8020 .267-7.087 Inside Diam. of compressed air storage tank d (in) 84-66 Outside diam. of compressed Air Storage tank doc ^in^ Hoop Tension in compressed air storage tank Paf ( 9.45 a — — __2 1214 C—^— * ' — ) 9.45-8.466' 11000 d2 -d2. oc ic (ibs/sq.in) Width of Wall between ad- jacent cylinders^ (in) Hoop tension between adjacent - - - - - cylinders p = phdih*Pafdia 1.8 w .i n) 249 GUIDE, ELEVATING GEAR AND TRUNNION REACTIONS: '. x axis taken along bore: v axis taken normal to bore. Coordinates from center of Xs 37.843 gravity of recoiling parts to front guide reaction yt*-3.038 xt and yt (in) Coordinates from center of xg » 48,4O7 gravity of recoiling parts to rear guide reaction ya= 3.86 x, and y2(in) J_ distance from center of 16,365 gravity of recoiling parts to brake piston rod axis e^ J_ distance from center of 15,656 gravity of recoiling parts to recuperator piston rod axis ea (in) •U^esa-r -:-™* U ( $* o Max. powder reaction P^T (Its) 2,245,000 (See Interior Ballistics) J_ distance from center of 6.13 gravity of recoiling parts to axis of bore e (in) Front guide reaction: gun re- coiling in sleeve: Fe+Pneh+Pa-Wrcos0(x2-uy2) Q 5S I - 0.15 to 0.2 (Ibs) 250 Rear guide reaction: gun recoil- ing in sleeve Q =• Fe+P£en+Paea+Wrcos (Ibs) Front guide reaction: gun recoil- 2,245,000x5.13+137500 ing in guide below axis of bore 37. 84+48. 41-0. 15><6. 91 Fe+P£en+Paea-Wrcos0(x8-uy8) t + x -u(yt+y Rear guide reaction: gun recoil- ing in guides below axis of bore Q xl6. 365+19094x15. 66 -7895x47.63 2,245,000x5.13+137500 37.84+48.41-0-15x6.91 x!6. 365+19094x15. 66 +7895x37.38 162600 Max. guide friction Kg » u(0t+a8)- (Ibs) u = 0.15 (approx. ) Weight of Tipping Parts Wt(lbs) Max. Resistance to recoil (dur- ing powder period) 2Peu Bg = 0.15(154800+ 162600)= 47,620 21,021 =137500 + 19094+47, 620- 13670=191000 =152000+ 5.13x0.15 2x2,245,OQQx 85.21 =192000 251 I distance from trunnion axis 3.73 to line parallel to axis of bore through center of gravity of recoiling parts s (in) Radius to pitch circle of 35.57 elevating arc. j (in) Angle between "y" axis and the 60° radius to elevating pinion con- tact with elevating arc 9e * 0 + ne Elevating gear reaction (in battery) E - Fe]K'a. (Ibs) Angle of E with horizontal Top carriage trunnion reaction (in battery with balancing gear) 2X=K+Wr3in0+Bcos99+Rsin9r (Ibs) 2Y=Htcos 0+Esinee-Rcos9r (Ibs) (E is sans with or without balancing gear) Top Carriage Trunnion reaction Not used. (out of battery with balancing gear) 2X=K+Rsin9r+B'cos9e+Wtsin 0 (Ibs) 2Y=Wtcos0+E'sin9e-Rcos9ii (Ibs) (E1 is same with or without balancing gear) 11,513,884*191,000x3.73 35.57 » 344,000 Not used. Estimated Weight of Bocker W- (Ibs) Neglected as small 252 Horizontal Distance from Trunnion to center of gravity of rocker hr (in) "measured to rear" Not used. Angle between line of action of rocker reaction on cradle and "y" axis. B I distance from trunnion to elevating sere* or normal to rocker cradle contact, k = x^os B+y^sin B (in) "xm and ym coordinates of rocker contact with cradle from trunnion to rear and down" . + 30 29. 43x. 866+15. 71 *0.5 - 33.35 Rocker Reaction on Cradle M= -^— — (Ibs) 344000^35.57 33 . 35 367000 Elevating Bear Reaction (out of battery) E1 = Ks+Wrbcos 0 (Ibs) Calculations max, elev. in battery, Top carriage trunnion reaction (in batterv)(X and Y components) 2X=K'+lft3in 0+E cos 6e (Ibs) 0-E sin 6 2X=197000+18200+ 17200=381200 2Y=10510-198,000= -287,500 253 Top carriage Trunnion Reaction (out of battery) (X and Y com- ponents ) 2X=K+Wr sin 0+ E cos 9e (Ibs) cos 0-Ein (Ibs) Calculation at max. elevation in battery, Vith balancing Gear: Distance from trunnion to center of gravity of tipping parts (in battery) along x axis: xt (in) Not -used. Radius of bell crank (balancing gear) r,, (in) Not used. Balancing Gear Reaction: 2Wtxt cos 0 ra(l+cos (Ibs) Not used. (very approx.) or calculated from layout 0m = max. elevation. Angle made by balancing gear: reaction with "y" axis 9, (See layout) Not used. Rocker Trunnion reaction (X and Y components) 2Xr =M sin B-E cos 8e~V* sin 0 (Ibs) 2Yr= E sin 9g -¥* cos B (Ibs) 183500 - 172000 11500 = 2Xr 297000 - 318000 - 21000 = 2Yr 254 Total shear reaction of trunnion 190600 + 5750 on cradle, - X'=X+Xr (Ibs) 196350 = X1 Y'=Y+Yr (Ibs) -143750-10500' 154250 = Y1 Total spring reaction of Top Carriage on trunnion sin (Ibs) cos 0 (Ibs) 10000 * .866 8660 = X3 10000 * o.S = 5000 - Ye Total rigid bearing reaction of top carriage trunnion Xb = X -Xs (Ibs) Yb = Y -Ys (Ibs) 190600 - 8660 181940 = Xfc -143750-5000= -149750= Y, Bending moment at cradle section 8660 x 5.5 + of trunnion 181940 * 2.9 + 5750 Mx= Xsa + Xbb + Xrc (Ibs) *.0.9 ~ 580780 My = Ysa + Ybb + Yrc (Ibs; 5000 x 5.5 - 149750 *2.9 - 10500 x 0.9 = -416,950 Resultant B. W. at cradle section of trunnion (in Ibs) /580,7802 + 416,960' 716,000 Max. fibre stress due to bend- ing a 10.18 M (Ibs/sq.in) 10.18 x 716000 —8* 355 n rJ evi 7.5 3.6 — TfcVJNN\QN PIN ~ n -. 256 SHEAR REACTION OF CRADLE. ON TRUMNtON PW : RERCT\ON OF ROCKER ON TRUNNION P\N : RERCT\ON OF TOP CRRR\RG£. ON TRUNNION \9O6OO RERCT\ON ON P\N \N X PURNE.. 257 258 CALCULATIONS FOR STRENGTH OP CARRIAQ1 AXLE Proposed 75 m/n St.Chamond 50° Elevation and 22- traverse: Maximum Peak Resistance to Recoil - - assumed at 20,000 Ibs. The resistance to recoil may then be divided into a horizontal and vertical component in the vertical plane of traverse. T"hen, the horizontal component in the vertical traversed plane, nay "be divided into a component along the horizontal axis of the mount and a transverse component at right angles to the longitudinal axis of the mount. The components in the vertical traversed plane are:- Horizontal comp. = 20,000 * cos 50° = 12820 Ibs. Vertical comp. = 20,000 * siii 50* = 15320 Ibs. The longitudinal and transverse "horizontal com- ponents are:- Horizontal comp. = 12820 * cos 22.5° = 11800 Its Transverse comp. = 12820 * sin 22.5° = 4900 Its. 859 Then, 15320 + 4000 = 19320 (Total Downward Force) S x 130 = 4000 x 120.25 + 15320 * 128.2 - 11820 * 47.2 4000 x 120.25 = 481000 15320 x 128.2 = 1970000 11820 x 47.2 = 19320 14550 4770 2451000 558000 1893000 S = 14,550 4,770 A, + B2 = 4770 »nef 260 12800 x cos *2 = 11800 12800 x Sin22- * 4900 2MA » 8X x 142.4 -11800 x 71.2 + 4900 x 128.2 » 0 11800 x 71.2 = 840000 4900 x 128.2- 629000 211000 H800 ... B • 2110°° • 1481 ) 1481 L42-4 ( A, = 10319 £M about vertical pin for loft trail Ay 126.38 * 10319 x 55.2 4900 .'. A, 4500 4500 400 By - 400 IM axle - A, 130 -10319 x 32 * Ba 130 - 1481 130C/1, -8Z) = 10319 x 32 - 1481 x 32 =283000 .'. AZ-BZ = 2180 Bz= 4770 2AZ - 6950 .*. Az = 3475 ) 8Z = 1296 j £M about left wheel base in Z Y plane: -4900 x 41.2 + 15300 x 30 + 4000 x 30 - 4900 x 6 + 3475 x 41.2 - 12.95 x 101.2 - Sp x 60 » 0 - 4900 « 41.2 * - 202000 49000 x 6 » - 29400 -1295 x 101.2 = - 131100 - 362500 261 15300 x 30 * 459000 722000 4000 x 30 = 120000 362500 3475 x 41.2=143000 359500 722000 Sg = 5980 ) SA = 8570 ) Reactions on Trail Axle. X and Y reaction on vertical pin of left trail: Ex = 10319 t Ey = 4500 B. ¥. in XY plane on axle: Ey x 10 = 4500 x 10 = 45000 " # XY plane: Thrust along X axis = 10319 ± Shear reaction of equal- izing bar. Thrust along Y axis = 4500 Thrust along Z axis = 3475 Shear reaction of Equalizer bar = 452000 - 331000 7.75 Thrust along K axis 15600 10319 15600 25919 EXTERNAL FORCE.S ON RXUE FOR SECTION -(m-n & -45OOO n 262 Section n-n 5" x 5" Torsion = 25919 x 2.2 = 57000 (" *) (B.M.zy) = 3475 « 12.2 + 8570 x 2'6 42300 22300Q B. M. jjTy = 265300 " * 265300 (B.M.zy) = 25919 x 12.2 - 45000 316000 45000 B. M. = 271000 (" *) ~27100(T f I \ * 30032- •= Ci > :'0---. * 01 * ^1 265300 x 2. . 12700 5 x 25 271000 x 6 13000 fy ' 125 " 25700 n 54 625 n 32 32 . 46500 •1.4 12850 + S\ x 257002 + 46502 12850 + 13620 = 26,470 BICAPITDLATION Qf gQ R HUT. AE OH THR TNT^BKAT. BBACTIOHS THROliaHnilT A GHH C*RBT«GB. F = Powder reaction (Ibs) B = Total braking force not including guide friction (Ibs) 263 "b = distance from center of gravity of recoiling parts to line of action of 8. (in) R = total guids friction (Its) r = mean distance from center of gravity of recoiling parts to guide friction (in) e - distance from center of gravity of recoiling parts to line of "bore. (in) Pn= total oil pressure on the "hydraulic piston. (Ibs) P'= the hydraulic reaction plus the joint frictions ^ (stuffing box at pistons) (Ibs) Pa= the total elastic reaction (due to compressed air or springs) (Ibs) Pa* the total elastic reaction plus the joint frictions (Ibs) Cj, 3 distance from center of gravity of recoiling parts to line of action of Pn. (in) ea = distance from center of gravity of recoiling parts to line of action of Pa. (in) d}, = stuffing box or rod diam. of hydraulic cylinder. "V5 (in) da - stuffing box or rod diam. of air cylinder. (in) Q = normal front guide reaction (Ibs) 0 = normal rear guide reaction. (l"bs) xt and yt = coordinates from center of gravity of re- coiling ]!>arts to front guide reaction, (in) 1 = distance between line of action of Qt and Qf (in) x and yf = coordinates from center of gravity of re- coiling parts to rear guide reaction, (in) !fr = weight of recoiling parts. (Ibs) 0 = angle elevation. u = coefficient of friction. X. and Y = component trunnion reactions (Ibs) Xr and Yr = component roc"ker reactions at the trunnion (Ibs) & = elevating gear reaction J = radius to pitch circle of elevating arc. (in) 9a= angle between "y" axis and the radius to elevating pinion contact with the elevating arc. 264 K * total resistance to recoil. (Ibs) s » distance from center of gravity of recoiling parts to trunnion axis measured along the "y" axis, (in) Total resistance to recoil on recoiling vass. becomes. K * B + R - Wr sin 0 (Ibs) but B = Pn + Pa where Ph = Ph + 100 dn ) assu>ing 100 1T)8. pCr and i i (in diaro. for frictions P. » P.+ 100 d. ) in stuffing box. hence K * P * ? + R - Vi sin 0 QUIDK OR CLI? »KACTIQ«8 TO QUIDS FRICTIOM. Gun recoiling in sleeve, front guide reaction, Fe+Bb-W_ cos 0(x. -uy. ) Qt - - - - ' - *— (Ibs) xt+xf+u(yt-yt ) and rear guide reaction. • Fe-»-BbCWr cos 0 (x±* uy,. ) /1V . \ (Ibs) Gun recoiling in guides below the axis of the bore. front guide reaction, Pe+Bb-W.cos 0 (x -uy ) Q = - - - - - 2_ (Ibs) and rear guide reaction, Fe+Bb+Wr cos 0(x -uy ) Q^ = - - - - - V * (Ibs) «,+ x4-u(yf+ya) If R * xt+xt+u(yt-ya ) for sleeve guides M = xt + xa-u(yi+y2 ) for guide below axis of bore and 265 H * x -x +H (yt+yt ) for sleeve guides * * xt-xt+u(yt-yt) for guides "below axis of bore. then the total guide friction equals, 2(Fe+Bb)+W cos 0 . N R = - - - u ,(lbt) and for the total braking force B4 (K+W_ sin 0)M-(2Fe+W_ cos fS N) u B = - 1 - 1 - (It.) X +2 u b In terns of tbe pulls, we bave for the clip re- actions, Fe + £Pa + 2Pe~lf cos ^(x~u) Q , - — — (Ibs) *t**t*u<*t-y«> Fe+IPaea+ 2Pv«h+WP cos 0(x +uy ) " - i - J- (Ibs) xt+ and tbe guide friction becomes, 2Fe+22P^eh+ 22Paea+ Wr cos 0 K R = - (Ibs) M and tbe hydraulic pull in terns of the total re- sistance to recoil and recuperator reaction, becomes, M(K-£Pa~Wr sin 0)-u(2Fe+22P'efl+ N *fr cos 0) , « A r - A. a . ., r_ , _< (Ibs) For approximate calculations, the guide friction equals, 2u8dr R " ~ From tbe foregoing analysis we observe, that tbe guide friction and bearing pressures are reduced: • (1) By increasing ths distance between the clips. 266 (2) By balancing the pulls about the center of gravity of recoiling parts or bringing the resultant pull closer to the center of gravity of the recoiling parts. (3) By "bringing the resultant friction line of the guides closer to the center of gravity of the recoiling parts . (4) By reducing the powder pressure couple Fe, that is by reducing the distance from the center of gravity of the recoiling parts to the center line of bore. The distance from center of gravity of the recoiling mass to the center line of bore should never exceed 1.5 inches unless a friction disk is introduced with angular notion about the trunnion. Stress QB Let Wc = weight of piston and rod or the weight of recoiling cylinder. (Ibs) d- = distance from center of recoil pull to section "mn" adjacent gun of the gun lug. (in) Imn = moment .of inertia of section. (in)' y - distance to extreme fibre from -neutral axis. Cin) fnn3 nax. fibre stress (Ibs/sq.in) then, W' [B+ -2. (F-B)]dgy -n (Ibs/sq.in) Trunnion and Elevating gear reaction: When the gun is in battery the tipping parts are balanced about the trunnion axis. This condition 267 implies that with the gun in battery, the center of gravity of the tipping parts passes through the trunnion axis. When the recoil is limited to a short movement under the breech when the gun is fired at high elevations the center of gravity of the tipping parts is placed forward if the trunnion axis and the balancing gear or counterpoise is introduced, balancing the weight of the tipping parts about the trunnion. The trunnion reactions are modified by the introduction of a balancing gear. Trunnion and elevating gear reactions when no balancing gear is used: (a) During the' acceleration period, ,Fe+Ks, 2X=K+Wt sin 0 * (— J ) cos 9 (ibs) (ibs) (b) During the retardation period, Ks+W_x cos 0 2X=K+Wtsin 0+( • )cos 9e (Ibs) Ks+W x cos 0 2Y=Wtcos 0-( ) sin 9P (ibs) J Ks+W_ X cos 0 E * (Ibs) J where x = the recoil displacement out of battery. Rocker Reactions: T^s reactions on the rocker are primarily three: (1) The reaction of the trunnion upon the rocker, Xr and Yr . (2) The reaction of the elevating gear, E. 268 (3) The reaction of the cradle , M, and the weight of the rocker, 1fr. If k = the perpendicular distance fro« the trunnions to line of action of M. B = the angle between the line of action of M and the "y" axis. h'r= the horizontal distance to the center of gravity of the rocker from the trunnion. J = the perpendicular distance from the trunnion axis to the line of action if the elevating gear reaction (i. e. equals the radius of the circular elevating rack on the rocker). Then, the cradle reaction on rocker, becomes, Ej-Wrhr Fe+Ks-Wrhr M = = (in battery) (Ibs) k k Ks-Wrx cos 0-Wrhr = (out of battery)(lbs) k K» approximately M = — k The rocker trunnion reactions become, 2Xr = M sin B -W^ sin tf-E cos 6e (Ibs) 2Yr =E sin 9e -Wr cos 0- M cos B (Ibs) Layout of Balancing Gear: Two types of balancing gear have been used ex- tensively in gun carriage construction: (1) A cam with chain type for small field mounts . (2) A direct acting balancing gear. For type (1), let Wt = weight of tipping parts. (Ibs) hj. = horizontal distance from the trunnions to the center of gravity of the tipping parts 269 (gun in battery) (in) ro = equivalent radius of can at horizontal elevation (in) rn = final equivalent radius of the cam where the cam arc has turned through the maximum angle of elevation = 0 (in) R = niean radius of can. (in) dn= deflectioa of spring at zero elevation (in) dQ= deflection of spring at maxim-urn elevation (in) c = spring constant. 0= angle of elevation expressed in radius. If ds - deflection of spring at solid height, take dn » (J to j)d solid ) ( d =(^ t° i) d solid ) then _ *tht " rodn rndo and dn-d0=( To layout the radii of cam, we have 0 divided into n parts, then, tht Wh cos rtnt r = c(dn-r0A0) Wh cos 0 cos 0 With a balancing gear of this type, the trunnion reactions are modified and now become, 270 if T = the tension in the chain d = the angle T oalces with the axis X(taken along the axis of the bore) Ks+Wrx cos 0+Fs 2X=K+"Vfc sin 0 + ( - • - ) cos 8Q - T cos d (Ibs) J Ks+¥_x cos 0+Fs = 2Y = ¥tcos0-( jsin 6e + T sin d (Ibs) J The elevating gear reaction obviously remains as before that is, Ks+\frx cos E = - £ - - - (Ibs) for type (2), 1st Tft= weight of tipping parts (Ibs) ht = horizontal distance from the trunnions to the center of gravity of the tipping parts (gun in battery) (in) x^ and yt = coordinates along and normal to bore from trunnion to canter of gravity of tipping parts (gun in battery) 0 - angle of elevation. £5m = max. elevation r = radius from tbe trunnion to the crank pin which connects the tipping parts to the piston rod of the oscillating cylinder, (in) R = reaction exerted by the balancing gear along the piston rod of the oscillating cylinder. (Ibs) dt= moment am of H about trunnion (in) d^= deflection of spring at horizontal elevation d]j= deflection of spring at maximum elevation (in) c = spring constant Hj = initial balancing gear reaction (0° elev.) Rt = final balancing gear reaction (0° elev) 371 S = stroke of piston in oscillating cylinder (in) pt = final air pressure in pneumatic balancing cylinder (Ibs/sq.in) p^ = initial air pressure in pneumatic balancing cylinder, (Ibs/sq. in) A = effective area of balancing piston (sq.in) Vo = initial air volume (cu.in) With a metallic balancing gear, the dimension of the spring" may be approximated by the solution of the following equations: = cos 0_ ) from nrhich we may obiain d0, dv, Q, ( s and c of the spring. a S r(l+ cos — ) ^ 2 ) S = 2 r sin - With a pneumatic balancing gear, we have, for a pre- liminary approximation, Pf 2Wtxt S ( Pl ^ ( ) ( _ — • r ( 1 +c< £lDJi \ ) (.CM . in / ) 3S2 1 ( Pi ) S = 2r sin *• f. . Pf (in; — = cos 0. ( 2 Pi (approx) ) With a direct acting balancing gear, the trunnion re actions are modified and become, Z72 if R = balancing gear reaction (Ibs) qr- angle between R and y axis dt = moment arm of R about the trunnion at any elevation 0 (in) when the recoiling parts are in battery: 2X«K+Wtsin 0+E cos 8e + R sin 9r (Ibs) 2Y=Hft cos J0+E sin 9e -R cos 6r (Ibs) Wtxt cos 0 21*txt cos ^ R » : = a (Ibs) Ks + P^e when the recoiling parts are out of battery :- 2X»K+R sin 6r+E cos 6e+Wt sin 0 (Ibs) ) ( 2Y»Wt ces U + E sin 6ft-R cos er (Ibs) ) ( 2Wtxfc cos t R = . (roughly) (Ibs) ( r(l+cos— -) ) 2 ( Ks+W_ x cos 0 Is (Ibs) ( J ) It is evident that th« elevating gear reaction remains the same with or without a "balancing gear while the trunnion r«actions are modified both by the position and Magnitude ef the balancing reaction. 273 Strength of the trunnions The critical section ef the truT>',i«»s is usually where the trunnion joins the cradle. L«t, "«n* represent this section. [See fig. (9)]. a = distance fro» "mn11 to center of top carriage bearing . b * distance from "mn" to center of rocker "bearing MX = the bending moment at "mn" in the plane of the X component reactions. My= the bending moment at "mn" in the plane ef the Y, component reactions. M « the resultant tending moment on section "an". f = aax. fibre stress (Ibe/sq . i'ff) D = distance ef fhe trumnien at section "mn" then Mx= X.a+Xrb (in Ibs) and M = My= Y*+Yrb (in. Ibs) hence /10.18 M D = / (in) Stresses in cradle or recuperator forging: Let Ql and QZ = the front and rear normal clip re- actions . xt and xg ~ the x" coordinates of these re- actions with respect to the trunnions. dx and da = the distance of the friction co»- ponents ef Q± and QZ from the neutral axis . B = the resultant of the braking pulls re- acting on the cradle. d-= the distance from the neutral axis to "B". 274 It = moment of inertia of a cross section at the trunnions. yt = distance of extreme fibre from neutral axis at trunnion section. ft= fibre stress due to bending and direct pull or thrust at the trunnion section. Ic = nonent of inertia of a cross section at the point of contact of the elevating arc with cradle. AC = area of cross section, at the point of con- tact of elevating arc with cradle, y = distance to extreme fibre from neutral axis of elevating arc section. fc = fibre stress due to bending and direct pull or thrust at the elevating arc section. A^ = area of a cross section at the trunnion, then )yt UQ- + — • — for the braking reaction in the rear, ft = — - — i— + — i for the braking ^ reaction in the front . *x, * x, • "" ~^«*~ •" w j^ • UVA for the brak- ing reaction in the rear. U°2 f = " ' " = ^^ + — i- for the bralc- T A Ac c ing reaction in the front. - -Jfc'^SJ APPENDIX----- APPENDIX CHAPTER IV- INTERNAL REACTIONS. BKACTIOHS AMD STRESSES IKDPCSD IK ELEVATING AMD TR«VERS- IH8 MECHANISMS: STRESSES DUE TO The reaction exerted on the FIRING. elevating mechanism due to firing equals, In Battery, Out of Battery Fe + Xs ?%v Ks*Wrx cos J0 J cos 20 J cos 20 where F = max. powder force K = Total resistance to recoil YT = weight of recoiling parts, r x = displacement out of battery. J = radius to pitch line of elevating arc fron center of trunnions, e = J_ distance fron axleof core to center of gravity of recoiling parts. S a J_ distance from line parallel to axis of gun through center of gravity of recoil- ing parts to center of trunnions. It is highly desirable to reduce the reaction E, since it stresses the teeth of the elevating mechanism. To reduce this, we may, (1) decrease "e" by so distributing the mass of the recoiling parts as to bring its masses ss near coincident with the axis of the bore as possible. (2) decrease "S" by bringing the trunnion axis along a line through the center of gravity of the recoiling parts and parallel to the axis of the bore. (3) increase "J" whenever feasible in a construction layout. 275 276 In certain types of oounts as those contain- ing a recoiling cylinder, the piston and rods "being fixed to cradle, the center of gravity of the recoiling parts is necessarily considerahly lowered froa the axis of the bore and therefore "e " is in- herently large. With large mounts, counterweights or bob weights are sonetines introduced to decrease "e". In this type of mount without a counterweight or "bob •eight a friction clutch or hand brake are often in- troduced on the elevating gear shaft or adjacent gear shaft. Then E becomes limited to that required to overcome the friction of the clutch or brake and a large reaction on the elevating mechanism is thus re- duced. With a cone clutch, we have, uPr E = : — - , where P= total spring load. r = mean radius of clutch re = pitch radius of gear or pinion, n - coefficient of friction = 0.15 approx. 2« = cone angle With a dislc clutch, we have, ), where P = total spring load. r2= outer radius: rt= inner radius of dislc. k = total no. of friction surfaces . n = coefficient of friction - 0.15 approx. FRICTION OF TRUNNIONS Tn elevating, or traversing AND TRAVERSING PIVOTS. a gun, a large amount of the energy needed is that required to overcone the friction of the pivot about which the gun is traversed. 277 Trunnion friction: During the elevating process the load on trunnions equals the weight of the tipping parts, when the trunnion is sufficiently free from binding, the con- tact is along a narrow strip. Then u t nR sin 0+R cos 9 = -r— ) u= coefficient of where tan 0 * n lO friction. Wt ) R = »or»*l pr«s- . * .R (sin 0 tan £J+ cos 0)=~r — ( SUP« ) r = rsdius of trwa- ( nion. and the friction moment Mt= R tan 0 .r = — r sin 0 m Since 0 is small, tan 0 = sin t approx. hence W Wt M+ = n — —r = 0.15 — •— r approx. $o l 2 2 In starting n nay be as great as 0.25 an& proper allowance should be nade. Since the load brought on the trunnions during firing is greatly in excess of that on elevating the gun, the bearing contact may be divided, one part to carry the major of the firing load and the other to carry merely the weight of the tipping parts. This is ac- complished constructively by allowing play in the bearing which sustains the firing load, and holding the tipping parts for elevating or transportation merely on a spring cushion, the reaction of the spring, for a deflection just sufficient to lift the tipping parts just clear from the firing bearing, being equal to the weight of the tipping parts. Thus it is possible to reduce the friction by using a 278 smaller trunnion diameter, in tliat part of the bearing that is spring borne since the bearing surface for a nominal bearing pressure can be greatly reduced. Pivot friction in traversing: This friction will vary considerably according to the type of bearing used. We will consider three types of pivots, 1* flat circular pivot, 2° flat hollow circular pivot, and 3° conical pivot. To esti- mate the load brought on the pivot, let, Va = pivot reaction or load (vertical ) V^ = normal load of traversing guides (vertical) Wt = weight of tipping parts. 1 = horiiontal distance between Va and Vv d U l+= horizontal distance from W+ to Vv v u U Wc = weight of top carriage. lc = horizontal distance from ¥t to V^ then ff 1 +W 1 Va = - load on pivot during traversing. If Kt= the friction couple exerted at the pivot during the process of traversing we have for the various types of bearings, 1° for flat circular pivot: The friction on an elementary zone = 2 re r dr The moment of this friction about the center = — j2nr*dr L 27a ro 2 2Vanro The total friction = — <g — n/ r dr = — ro o 3 Therefore for a flat circular pivot, letting n - 0.15, Mt » 0.1 Va r0 279 2° for flat hollow circular pivot The total friction evidently becomes, — I * r'dr = 2Van r 2Van r'-r' 3 hence, letting n = 0.15 3°for conical pivot: The intensity of vertical pressure on the projected area of the bearing = y n(rf-r*> V If the cone makes an angle 2«, and pn equals the in- tensity of normal pressure, then, rd6dr rd9dr the normal pressure on area — — = p_ — . since B fJin «t the vertical component of this pressure = rdedr rd 6dr sina but the pressure on the projected area rd6dr = p rd9dr hence p¥ = Pn = ?. 2nrdr the friction on a differential zone = n 2 the total friction moment, therefore becomes, If then we let n = 0.15 280 VELOCITY RATIOS OF ELEVATING Elevating and travers- AKD TRAVERSING MECHANISMS ing mechanism consists usually of a train of spur, bevel, helical screw and worn gears. l°)-Velocity Ratio of spur gear: Since *trt = v»tra ) * = angular velocity ^ r = radius to pitch line. we nave -± = — L - _L ( n = no. of teeth. 2°)-Velocity Ratio of Bevel gears: Again w r = w r where r and r are the outside 1 t Z 2 1 2 radii of the gears: The angle of coning for the first gear, equals, r. " * tan 6t «— ( 6t = 1 angle of cone, ) or the second gear tan 62 = — (62 = - angle of cone) hence w 1 2 1 = tan 9. and - = tan 0 w Therefore we may take any two common radii in ob- taining tlie velocity ratios, again 3°)-Helical screw gears: Tfe have for the velocity of the common normal, w r cos r2cos 9, r cos 9 but also, Pn = then = pcos cos 9. Pn cos 9. 281 ) 0t = angle be- l*(F* tween axis ) of gear fl ( and perpend- icular to ( common normal. . ) 98 = angle between ( axis of gear ) #2 and per- ( pendicular to ) common normal. ( pn = common normal ) pitch. ( pt = circuraferent- ) ial pitch gear ( #1. ) n = circumferent- tial pitch gear #2. n = no. of teeth gear fl. n2 = no. of teeth gear f2, Hence — = — = r cos9 r cos9 i 1 If 0 = the total angle between the axis of the gears in mesh, then since p = p cos Q = p cos 9 cos 9t = cos e = "• si" ~fP2~2 PtP2cos therefore 282 Further the axial pitches, become, TBX = pt cot 6t and m2 = pa cot 9f 4°)"Velocitv Ratio Worm gears: Though a worm gear is a specified type of nelical screw gear when Si = 90°. it is convenient to consider this type as a separate classification. When 0 = 90°, cos 6. * = sin 9 therefore the axial pitch of one equals the cir- cumferential pitch of the other. The worm of a worm gear has one to two or three threads while the gear has many threads. Now, for a single thread worm, rwsin 9 r., cos9 - — tan 9 Directly, we have. % P^T but 2n = tan9.w»r wlw ; *g= ang. velocity ( of gear wheel . ) ww- ang. vel. of ( worm wheel. ) rg - pitch radius ( gear. ) rw= pitch radius ( of worm ) p = axial pitch of ( worm ) 9 = angle of helix. — = — tan 9 Thus the ratio of angular velocities depends upon the angle of the helix of the worm. With a "n " threaded worm, : nwP — £n g r and — = n -^ ww rg tan 6 283 p In terms of the number of teeth, since = tan 9 w p nw n... ian 9 - — 2nr ng and for a single threaded worm, since nw = 1 * - Velocity ratio in gear trains: Combining the previous equations from one pair of elements to the adjacent pair, we finally arrive at the velocity ratios of the first and last wheels of the trains in terms of the number of teeth or radii of pitch circles: In this combination, it is always preferable to set the general equation up in terms of the number of teeth rather than the radii of pitch circles, for then the relations are independent of the type of gearing and velocity ratios between a meshing pair are inversely as the number of teeth or threads. Thus assume worm #1 to drive worm gear #2, while bevel gear #3 on same shaft as gear #2, drives bevel gear #4, then helical screw gear #5 on same shaft as gear #4, drives helical screw gear #6 and finally gear #7 on gear shaft #6, drives pinion #8. Since 2 and 3, 4 and 5, and 6 and 7 are on same shafts, we have, then, wwwwnn nn 13 5794 88 hence — * — * — * — = — * — * — * — 284 wt nt n4 n n therefore — » — x — x — x — w n n n n 8 1 3 5 7 If Tt =» torque on worm shaft #1 and T the torque on pinion shaft and e the efficiency of the total gearing, then T§wa = e T^ hence -p n xn xn Mn Tt * — ( ) where Tt = required power torque and T$s load torque at end of train. REACTION BETWEEN GEAR PAIRS:- The efficiency of EFFICIENCY. spur and bevel gears is hifh compared with helical screw gearing, especially of the worn gear type. The very large force and velocity ratio attainable by the latter makes this type preferable. 1° Spur Gears: For approximate calculations, the normal reaction between 'the teeth will be taken at an angle of 20* with the tangent to the pitch circles. The effect of friction between the teeth is to cause the resultant reaction to make an angle of 25° with the tangent to the pitch circles. Therefore if T is the torque to be transmitted, the reaction between the teeth R, becomes, T x 12 where T is measured in (Ib.ft) r cos 25° , . . r is neasured in (in.) Ifhen smoother running is required with high velocity ratios helical spur gears have been extensive- ly introduced. If B = the angle between the normal to a tooth surface and the tangent to the circumference (i. e. normal to axis of rotation), then 285 T * 12 r cos 25 cos B If b = tooth rim breadth, the mean pressure is dis- tributed along a linear element - b sec 6 and therefore the pressure on an element becomes per linear inch, proportional to T * 12 r cos 25. b the same as in ordinary spar gearing. 2* Bevel Gears: The reaction between bevel gears takes place at the intersection of the common pitch circles of the cone elements of the gears, and this intersection is in the plane of the axis of the gearing. The neutral reaction between the teeth makes an angle approximately equal to 20° with the normal to this plane due to the contour of the tooth. The tangential component pro- duces no axial thrust. The component parallel to the plane = P tan 20, where P is the tangential component. This component is also perpendicular to the common intersecting line of the two cones. If the cone angle of gear #1 equals 28 then the cone angle of gear #2 = The axial thrust for gear fl becomes, P tan 20° sin6 The axial thrust for gear *2 becomes, P tan 20° cos6 Further the radial reaction "between the teeth and there- fore the radial bearing loads for gear #1 and gear #2, becomes, R' = /p2 4. (p tan 20° cos8)2 = P /l + (tan 20° cose)' Where T x 12 P = ; and 28 - the cone angle of gear #1 r 71 2(- - 8)= the cone angle of gear #2. 286 3° Helical Screw Gears: Assuming the axis of the gears to make an oblique angle £J t the angle 6 between the contact line of the teeth and axis of gear #1 is given by the expression P2sin 0 cos 9t = while the angle 9 between the contact line of the teeth and axis of gear #2, is given by the expression p sin 0 cos 9. = ,. . * .. where pt and pz are the respective circumferential pitches of the two gears. The reaction between the teeth makes a resultant angle i with the normal to the contact line, where tan i = n the coefficient of friction Then, if Tt is the external torque exerted on gear #1, we have T± = R cos (9t - i).r± while if Tg is the torque on gear #2, Ta=R cos(92+i).r. Work expended = T w w ' r cose Work delivered = T8wt Then the efficiency E becomes, T w cos(9 +i)cos9 E = -2-2- ? - tt cos(0t-i)cos92 The reaction on the teeth is given by Tt r.cos(9 -i) T sin(6 -i) R sin(et-i)= ^ 287 and the thrust along gear shaft f2, is t R sin(8 + i ) = — r cos(9t-i) The total radial bearing load for shaft of gear #1 balances, T cos(6 -i) T R cos(9 -i)= -^ _— i-— = -1 rt cos@f-i; rx and the total bearing load of gear shaft #2 balances, ^ R. cos(9 +i)= — - 4° Worm Gear: Though, worm gearing is a special case of 3°, a separate analysis will be made due to the greater use of this type of gearing as compared with helical gearing when the shafts are not at right angles. Let xx and yy* be the coordinate axis along and perpendicular to the axis of the worn in the plane perpendicular to the radius of the pitch line of the worm through the common pitch point as origin. Let S = the angle that the contour of the tooth makes with the normal to the xy plane at the pitch point, and 6 = the angle of helix. Let R = normal component between worm and gear tooth, nR » friction component between worm and gear tooth, then the axial thrust along worjn wheel is X - R cos S cos 9 - nR sin 6 and the turning component on the worm is Y = R cos S* sin 6 + nR cos 0 and the thrust tending to separate the teeth is Z = R sin S . It is to be noted that tan S = tan S cos 6 288 If T_3 torque applied to worm gear Tg=» torque on gear wheel then, Tw=Yrw and Tg=Xr^ rw= radius of worm gear rg= radius of gear wheel To determine the efficiency, .vs hava but — = — tan 0 then „! cos S cos0-n sin8 . e = — — — — — — — — — — tan 6 cos S'sin0+n cos6 n tan0 cosS1 -) tan 0 n cosS1 tan e n e= ; — ~~ where k = tan"1— — — - tan(e+k) cos S' and tanS'-tanS cos 6 COMBINING THE REACTIONS In gear transmissionhaweA FROM ONE PAIR TO ANOTHER, between two elements, #1 and «2, f m w = angular velocity hence ;r- = 8t — — Likewise between gear elements Ti wt #3 and #4, -p w Then if gear #2 is on same shaft as gear |3, we have T = T and w = w hence 23 ? 3 T T w w 42 's-1 - x — = 6 — E — — T3 Tl *4^4 = e e — ia 34 289 w Now the velocity ratio — may be obtained as outlined w in previous discussipn on velocity ratios. In the proceeding discussion the inertia effect of the gear elements has been neglected in comparison with the friction developed between the gears. TORQUE AND POWER REQUIREMENTS In elevating ertravers- FOR ELEVATING AND TRAVERSING ing a gun, we nave three MECHANISMS. important periods :-(a) accelerating period, (b) the period of uniform motion and (c) the retardation period. The maximum torque obviously occurs during the acceleration and power is continued through period (b), while the friction of the mechanism brings the system to rest during period (c). Let 1^ = moment of inertia about the trunnions of the tipping parts. 1^ = moment of inertia about the vertical traversing pivot of the tipping parts and top carriage. E = elevating gear (tangential reaction.) J - radius of elevating arc. r = radius of traversing arc. Mt~ friction moment of trunnions M^= friction moment of traversing pivot. Then during the acceleration, J t t dt« for eievating the gun E.r-M'=l' ill fc * dt for traversing the gun Now MI and M^ are constant depending approximately on the weight on the bearing, while on the other hand E and E' depends on the elevating or traversing motor characteristics. 290 Neglecting the inertia of the gear elements, we have, the torque transmitted varying directly as the number of teeth, that is between any two gear elements, T i for gear pair 1-2 3 — = -- - for gear pair 3-4 Ten 4 34 « for gear pair 7-8 If gears 2 and 3, 4 and 5, 6 and 7 are assuned on sane respective shafts, T, = T3, T4= Ts, T6= T7 then Ii . L . L. , !i . L = A „ i. , L. „ ^i,J_ . ^_ , T. T, T. T. T. °2 "4 ". ", Si '.4 e t 5 • 7t Now TV = E re and = * * * e e e e e t t 34 S 8 78 then gr n. n n n T = — - ( — - x -i x -5. x — 1) « na n4 ne "« hence T „ n n n d20 e — — ( x — i x — i x — S)J-M 4. = !+ f°r elevating re ni na n, nr dt the gun. 291 l 2 4 s « II e — ( — * — * — * — )r-M* = It — - for traversing ' * z the gun. dtz and for the sngular velocity ratios, we have, J and w = — wt; - for spur or bevel gears: (elevating; r i * = — *t:~ ^or 3Pur or tevel gears: (traversing) re = -2— wt : for worm gear in contact with e np elevatin arc (elevating) 2nr . = -TP— wt : for worn gear in contact with • n p •• traversing arc (traversing; CHAPTER V. RECOIL HYDRODYNAMICS. OBJECT. The modern recoil system is essentially a hydropneunatic device for dissipating the energy of recoil by so called hydraulic throttling losses, and returning by means of the potential energy stored up in. the compression of air, the recoiling mass into battery. The potential energy at the end of recoil required to return the piece into battery is relatively small compared vritb the energy dissipated by the hydraulic braking. Further the potential energy of counter recoil is in greater part dissipated by the hydraulic counter re- coil buffer in the return of the recoiling mass into battery. In the design of the braking system misunder- standing may result due to incomplete comprehension of the fundamental principles underlying the hydraulic throttling and the various hydraulic reactions. Hence, in this chapter a resume of the essential principles underlying the hydraulic phase of recoil design will be attempted. ELEMENTARY HYDRAULIC Consider an ordinary tension BRAKE brake (fig.l) the oil being throttled through apertures in the brake cylinder from the front or rod side of the piston to its rear. Let ax = area of the variable apertures or orifice . An = effective area of piston on rod side. A = total area of cylinder. ar = area of rod. Pn * total hydraulic pull. 293 294 u Dp L J CvJ 00 L 295 Vx=velocity of recoil at displacement x. vx= velocity of oil through apertures. D = weight of fluid per unit volume^ p = p^= intensity of hydraulic pressure. C = contraction coefficient of orifice. K = reciprocal of contraction coefficient. For a displacement dx, the mass of liquid moved by the displacement of the piston, becomes, D Ah dx and due to the contraction of the liquid g in the throttling aperture or orifice, its effective area is reduced to C ax, therefore, the mass is accelerated to a velocity A v \r A w Hh vx Hh Yx 1 vx = = , since K = -5— now the energy SLy £L« of the jet, * D Ah dx . vx becomes, dissipated by a loss due to sudden expansion in fhe rear part of tlie cylinder, where we find a void equal to: (A-Aj1)x= arx . By the principle of virtual work, evidently x A, v - D Ah dx *h vx p* d«- -J^- (T^-> hence 1 D Au V* g c a x that is in terms of the liquid pressure : D K« A> V Consider again a brake where the throttling occurs between the hydraulic cylinder A and a re- cuperator cylinder B containing a floating piston which is contact with the oil on one side and the air on the other. See fig. (2). Let p = pressure intensity against hydraulic or recoil piston. 296 Aj, = effective area of hydraulic piston. ax = throttling area between the two cylinders which we may assume is controlled by a spring. vx = velocity through orifice. Vx = velocity of recoil. Va = velocity of floating piston. Aa = area of floating piston. pa = pressure intensity against floating pia- ton. x = displacement of floating piston. Then by the law of continuity, A^ dx = Aa dx Due to the contraction and sudden expansion of the liquid from the throttling apertures, the loss due to eddy currents becomes, D A dx «h "x By the principle of virtual work, we have, r D Ah dx Ah Vx „ , ,1 2 i) / U A \l Ph Ah dx -paAadx = - ( - ) g C ax Neglecting the slight change in the total kinetic energy of the liquid in its virtual displacement. Simplifying, we obtain, p K' v g a x which gives the drop in pressure through the orifice, or the so called throttling drop, Obviously, Pn=Ph*h» as before, i - D K*AV (4) PRINCIPLES OF (1) Though in the analysis HYDRODYNAMICS. of recoil brakes, liquid viscosity is an item of importance, the viscosity effect in modifying pressures is, with a few exceptions, small, and therefore, for a first 297 approximation we will consider an ideal fluid, that is a liquid with no viscosity. (2) It may be shown by simple analysis in the consideration of a small tetrahedron or triangular prism that the pressure intensity on all planes at a given point within a fluid is the same, the bodily forces such as gravity, inertia resistance etc. in limit being eliminated since they are functions of high- er order (three dimensions) than the surface pressures (two dimensions). (3) By higher analysis it may be shown that fluids flow in so called stream lines and therefore the variation of pressure with velocity at various points along the stream line as well as the change in such due to the acceleration of the fluid as a whole, may be determined by a consideration of the pressures on continuous differential elements. Due to the mutual action between differential elements, we nay, by simple integration along a stream line determine the pressures at the extremities of a stream line tube, that is the end pressures as well as the terminal velocities. Consider a differential element A 8 C D along a stream line, of cross section w of length ds and a circumferential perimeter c. Let, the intensity of pressure on A D be p, the weight per unit volume be G, then for the pressures on the surface A B - C D and the wall of the tube, we have dp pw-(p+ — ,ds)(w-dw)-pcds sin <*-D " ds sin J0 = ds — but cds sin <x = dw. Simplifying and g dt dividing through by w, we have, - dp-D ds sin 0 - 2*1*1 (5) g dt dv s dv dv dv dv but dv = — dt + — ds hence — = — + v -— -which dt ds dt dt ds shows the acceleration is both a time and space 298 function, inserting in (5) we obtain, Dds dv dv -dp-D ds sin 0 (77 * * 7~> <6) g dt ds Integrating from (1) to (2) along a stream line, since the mutual reactions between contiguous particles can- eel out, we have, ; i 'd°<zi>* 4?1 1 ,dv v"v Obviously, / * ds sin 0 =Z -Z hence dv The term / ds — is of special interest and when it dt occurs the motion is not steady. This tern is theoretical, always existing in a recoil brake, since the fluid in addition to a space variation of velocity due to changes of sections, is on the whole ac- celerated as well. . , dv dv To evaluate / ds — it is necessary to express -— dt dt as a function of s. If now we assume the same stream lines to exist whether accelerated or under uniform steady motion, we have, by the equation, of continuity, wiVi = *2V2= *3 V3 and dvt dv^ dva 1 dt dt ' dt hence knowing the acceleration at one section, dvn w! dv —-— = — r— for any point "n", hence if w is a con- dt wn u t tinuous function of s, we have dv 1 dvt, 1 wt - hence the line integral dt w dt dt of the acceleration along a stream lines, becomes, 299 dv d_v ds 3 dt " "ldt f(s) The line integral of the acceleration may be obtained to a sufficient degree of exactness by dividing stream lines into a linear group of columns of various sections, obtaining the proper acceleration. To form (8) and multiplying by the length of the res- pective columns and then adding these columns together. 1 dv The term - / ds (— — ) is found usually to be relatively g dt small compared with the pressure drops due to throttling and the changes of pressure due to changes of section. Hence (7) reduces to the energy equation for uniform or steady flow, known as Beraoulliis theorem, that is, t t P T D 2 D z 2g p v* The term — *• +Zt+ — x is known as the total head at section (1), composed res- pectively of a pressure head, gravitational head and a velocity head, (4) When friction, viscosity or turbulent motion occurs Bernoulliis theorem is modified by a friction head hf. Considering a tube of a stream, we have for steady motion Dw ds Pt*td«^t!ftd«,+D"t<Ut(*t-Zt)-ar = * * (v*-v*) (10) 2 2g Where d Wf corresponds to the differential work due to friction for a differential quantity of flow d ft. By the equation of continuity, dQ = wtdsi= ",dsf hence (9) reduces to Pt-pt+D(zt-zt)-^»-(vX> (ID dWf i but - •>- = hf i , known as the head loss due to dft 2 friction between 1 and 2, hence 300 t i <r * 5 * z.>- (r * if * z.) • h<; It is to be especially noted that in the flow of a liquid through an orifice as in a recoil brake, the •ajor loss is in the nature of a frictional loss of head due to the contraction and sudden expansion of the liquid through the orifice, thus u " * = in equation (3), that is p,-p. r "' — - - = - 7 since Z = Z approximately, also *a t x Vt V2 — and - are relatively small. 2g 2g It may be shown by a somewhat similar analysis that in the consideration of friction of or turbulent loss of head due to throttling, that, from (7) we have 2 8 where if wo = the area of the jet VQ = the velocity of the jet. where hfi has the form, Hence anaxact expression for a stream line passing through a jet, and the whole stream line itself under acceleration, becomes, D 2g D 2g g dt w0 (5) The pressure variation across a stream line may be obtained by a consideration of a cylinder 301 the end faces of which are in the outer and inner boundary surface of a stream line tube, and the axle .is perpendicular to the stream line axis. We have, if w= cross section of differential cylinder r = radius of curvature of stream line d = height of differential cylinder 0 = angle between r and the vertical that a Dw v (p+dp)w-pw+Dwdr cos 0 = — y dr — (15) 2 V wdp = Dwdr( -- cos 0) gr j g hence -7*- = D (- -- cos0) (16) dr gr which given the rate of change of the pressure across a stream line with respect to the radius of curvature. Neglecting the weight component, we have, £ - — dt gr Hence for circular or vortex motion, the change in pressure along the radius, becomes, dr In particular if the total system acquires the sane angular velocity1"'- , we have, — dt 1_ = r (*i)« r dt since the total head at any point in a fluid equals, 2 p v H = ~ + Z + r— the variation of head across D 2g dp vdv a stream line becomes, dH. - —— +dZ+ - L * vdv \ (2o) = r*- +dr COS0+ - ) D g Substituting Eq.(16) in (29) we "have, 302 _• vdv dH= — dr + (21) 8* 6 which is the general equation for the change in head across a stream line. (6) When the flow is radial, evidently the flow outward fro» circumferences of various radii, becones, Q » 2n rv = 2* rv , hence voro»vr or _ v ro — = — and for steady motion, we have, v_ r o x 2 !°. + Is + - L + L. , z D 2g 0= D + 2g hence * * °/, *i /««x P-P0= 2ja~ 7») (22) In terms of the total head H, we have, * 2 P vo ro r • H - -3^- (23) (7) A free circular vortex occurs when the total head of any annular stream line of the vortex is the same. That is, for any annular stream line, t p v fl = -jj— + — — = const. To find the distribution of pressure, we have, dp vdv ~~ "* ° and for the flow slowly out- u g ward radially, we have, P-Po vo"v* — - — « •»- (Neglecting friction) D 2g Thus the pressure variation is exactly similar to that of ordinary radial flow. How from (21) since dH » 0 303 v vdv dv dr — dr + = 0 hence — = and from the la* gr g v r of continuity for the flow outward, v r » v r hence v - JL r Likewise the flow outward is exactly similar to ordinary radial flow. THE EFFECT OF THE The viscosity of a fluid is VISCOSITY OF FLUIDS. the shear stress to the distortion of the fluid and this stress is measured by the coefficient of viscosity times the rate of distortion. In other words the viscosity or coefficient of viscosity, becomes, s dv u = — or s = u jr where v * velocity of a — lamina flow (ft/sec) h = normal to flow lamina (ft) s 3 shear 1° Flow between flat surfaces: r dv S = u — .bl dh d2v dS*u rrr dh.bl dh' Now considering the- forces on a lamina of thick- ness dh, breadth b and length 1, we have, for a constant pressure head (pt~p ) (p ~P ) bdh-dS = 0 for uniform flow 304 (p -p )bdh-bln ^-7 dh * 0 dh d'v (pt-Pt>-lu —7 dv — » dv (Pt-P. Integrating, us have — » ' + C dv when — = 0, Ct = 0 (p.-pX Integrating again, v = — - — - - * C. when v * 0, 2ul (Pl-P,)Hf and Cg = - - Hence the distribution of velocity across a section is given by the equation, <Pt-Pt> * ^ v = ^ul - (^ "" 7"^ (ft/sec) as measured from the center. For a dif- ferential flow, we have P -P, 2 H* dQ = vbdh = ~ — - (h -- ) bdh and for the 2ul total flow, summing up oo both sides of center line, we have, Q = — - bH Therefore the drop of pres- 12 ul sure between flat surfaces in a rectangular channel becomes. 12ul A ,., N pt-p2 = - ft (Ibs) bH» For the particular case of a square section, 12ulQ Pi-Pa = "TT" (Ibs) n 305 2° Plow through a circular section: p, 1 p 1 1 1 1 1 2- I rh 1 <L i dv The viscosity shear becomes, S = - 2nrlu — dr (r ID dr dS = - 2nuld — dr dr Considering the forces on a cylindrical lamina of thickness dr and length. 1, we have, (pt-p2 = 2iirdr - dS1 = 0 for uniform flow AI d^ T— ) (pt-p,)r*ul dr r = 0 Integrating, we have ^p _p )r* r — = - — - — + C dr 2ul which may be written dr 2ul r dv C when — =0, r = 0, hence — * 0 i.e. C =0 since dr r r may have any finite value. Integrating again v = - — — — — * C-2 when v = 0 for the boundary 4ul surface, r » R 306 t, and v = -i-i (R*-rf) which gives the variation of the velocity over a cross section as a function of the radius from the center. For the total flow per second, we have r Q » / 2nrdr.v o / (R'-r')rdr 2 ul o (pt-Pt)nR4 8 ul Hence for the drop of pressure through a small orifice where there is no abrupt change in section, 8 ul Pi~pa = — 4~" Q where u = coefficient of viscosity. 71 R that is the drop of pressure varies as the length and inversely as the 4th power of the diameter of the orifice. PRINCIPLE OF MOMENTUM The various formulae AND DYNAMIC REACTIONS. previously developed de- pended upon the ap- plication of Bernoulli's theorem, or the energy equation of hydro dynamics. A theorem of equal importance is the principle of momentum and from it with a com- bination of Bernoulli's theorem, we may compute the various dynamic reactions, that occur in hydro dynamic problems. If P = the unbalanced reaction on a mass of water and the velocities of the mass is changed in 307 time t, from vt to V2 ft/sec, then Pt=ra(v -v ) . In the application of this principle (1) the mutual reactions between the particles of water are obviously en- tirely neglected. (2) the velocity components in the direction of motion are only to be considered. DERIVATION OF THROTTLING FORMULAS. (1) Loss of Head due to sudden expansion. In the flow of a fluid through an orifice the drop in pres- sure or loss of head, is primarily due to the sudden expansion or abrupt change from a small to a large section, of the fluid flow. The loss of head is due to the formation of eddies due to the sudden expansion of the flow and the consequent dissipation of energy. If the cross section of the flow is gradually enlarged from that of a small orifice to a large section, no eddies are produced and we have no loss of head. Thus in the Venturi meter the fluid passes from a large section to a very small section and then back again to a large section but since the change in section is gradual, we have no drop in pressure. Hence we have a very important principle that is fundamental in the design of recoil throttling orifices:- A throttling drop in pressure cannot be pro- duced without a sudden change in section of the flow of the fluid. Y*\//t///jfa//l 308 Consider a flow of fuild passing through sections, ab, mn and cd respectively. Let the cross sections of the stream be w at ab w at cd 1 12 and the corresponding pressures be pt and p res- pectively. Let pQ be the pressure at mnt (Ibs/sq.ft). From the energy equation, we have, t * P! Vl P2 Vt — + — = — + — + hj assuming a continuous D 2g D 2g uniform flow. From the principle of momentum, we have • Q Piwi*Po(w2~wi)~P2W2= g~ D (v2~vi) where Q = the rate of flow (cu. ft/sec) D = the density of fluid (Ibs/cu.ft). Now from the experiment it is found that po = pt hence the momentum equation reduces to, ft _, Pi P2 T2(V2-Vl) (P-P" D(v-v) and— — = at hence g (v-v)2 which simplifies to tt hf= - or in terms of the area of the orifice 2g and the enlarged section, since wtvt = *2v2 «' 2' T W V W hf = -i- (1- -i)2 = -i (-1 - 1) 2* w2 2g wt (2) Loss of Head due to sudden contraction, When the cross section is suddenly diminished beyond the reduced section we have eddying of the flow with a resultant loss of head. This too is really a special case of (1) since just beyond the contracted section, the stream becomes even more contracted, followed by a sudden expansion until the stream reaches the cross section of the con- tracted area. 309 Therefore if w = the area of the contracted section, then the cross section of the contracted stream be- comes, c w where c = depends upon the preceeding area w and the area of the orifice. In terms of the velocity of the orifice, the loss head equals 1 2 V W V hf= — — ( - 1) = E-r — where if w = the cross 2g ex 2g section before the sudden contraction of section, the experinents of Weisbach give for o.i o. 624 o. 360 0.2 0.632 0.340 0.30 0.643 0.310 0.40 0.659 0.266 0.5O 0.681 O.220 O.6O O.V12 0.162 o.7o o.'/BS 0.106 O.8O 0.813 0.053 0.90 0.892 0.014 l.OO l.OO O. IB the special case when — — = 0, that is the area of the orifice w± is entirely negligible with the flow from the large cylinder as in a flow from a resevoir, c = 0.6 and E = 0.445. It is to be noted that the above analysis holds only when the length of orifice is sufficiently long to allow the contracted stream in the orifice to ex- pand and completely fill the orifice before expanding in the region beyond the orifice. Hence the loss of head due to sudden contraction only holds for long orifices or entrances into long channel parts. 310 LOSS OP HEAD AND PRESSURE Assuming uniform flow DROP THROUGH RECOIL ORIFICE. from the recoil cylinder of effective area w, through an orifice of cross section wt dis- charging into a cylinder or channel of cross section w . Then from section w to the mid section of the orifice, v* P v2 5 + jj- - -5 + jj + hfc (1) hfc = loss of head due to sudden contraction and from the mid section of the orifice w to the rear of cylinder or channel cross section vr , 2 a ^ * 5? = ^D + 55 + hfe (2) hf e = loss of head due to sudden expansion. Adding (1) and (2), we have P / P2 v* — + — = — + — -«• hfc + "f e D 2g D 2? Very often v = va approximately and usually the heads corresponding to v and v are small compared with the pressure and throttling heads and therefore the velocity heads may be entirely neglected. We have then, P - P. 1— = hfc + hfe that is the drop in pressure through an orifice is equal to the total head lost due to sudden contraction and ex- pansion. Now ?' 2' V W .V VI . hf - -i(-± - D'= ~U- -1)2 2g w Xg w2 311 *f* * *T* further AV = w^* wfvf, (where V- the velocity of the recoil piston.) (w=A = effective area of recoil piston) hence A*Vf< w ^r(1-^ When the orifice is in grooves in the cylinder or through orifices in the piston, we have )since *t=A approx- iraately, v =V approximately and "^ * c7" aPProximate1^ ( )the effective area (of the recoil pis- )ton and wt=effective j^ Therefore if n - — ,we have V . ,.t' — (m-1) (area of orifice, )that is the contracted (flow through the )orif ice. (c = coefficient of )contraction of "* (orifice. Usually the loss of head due to sudden contraction nay be entirely neglected as compared with the loss of head due to sudden expansion, hence we have, for a very close approximation of the pressure drop, (1) With throttling through grooves in cylinder or piston, p-p V8 2 (m-i)» D 2g or .*'«*' A V cw both forms having useful applications. of contraction, 312 (2) With throttling through an orifice froa the recoil cylinder to the re- cuperator, »i « area of orifice w - area of channel leading from orifice. ANALYSIS OF THE MUTUAL In an ordinary brake REACTION IN A RECOIL BRAKE. cylinder we have a groove in the cylinder or at the circumference of the re- coil piston. As the recoil rod pulls out, a pressure is created on the front side of the piston, due to the forcing of the fluid through the orifice groove. The pres- sure is by Berboull's theoren, obviously lowered in the vicinity of the orifice due to the increased velocity of the flow. Hence, with an orifice in the piston, the sure is not uniformly distributed over the effective area of the piston. Therefore, the brake reaction is not equal to the product of the recoil cylinder and the effective area of the recoil piston. Let X = the total reaction of the fluid on the re- coil piston. (Ibs) p * the pressure in the recoil cylinder. (Ibs/sq. ft) A « effective area of recoil piston. = 0.7854 (D» - d£ ) (sq.ft) where Dr = dian. of recoil cylinder (ft) dr » diam. of recoil rod (ft) Ar * area of recoil brake cylinder = 0.786 Isq.ft) V » velocity of the recoil piston (ft/sec) v * velocity of flow through the orifice, (ft/sec) D » density of fluid (Ibs/cu.ft) w » area of orifice (sq.ft) 313 c = contraction factor of orifice. Assume the recoil rod to only extend from one end of the piston. In this case, we have a void in the rear of the piston due to the volume displacement in the front of the piston being less than in the rear of the piston. Assuming the pressure in the orifice to be small, we have, for the reactions on the fluid from front head of cylinder to a cross section at the center of the orifice:- (1) and the reactions on the fluid from the orifice to the rear head of cylinder, becomes, The reaction on the piston = X to the rear The reaction on the cylinder = pA - Y to the front Adding (1) and (2) we have, pA> - X-Y = 0, which is immediately obtained since there is no change in the total momentum of the fluid, as we should expect from first principles, since the fluid acts as a medium for the transmission of the reaction between the re- coil cylinder and the recoil piston. Hence pA - Y = X. which gives the actual reaction exerted on the "brake piston. Since C vr v = A V, by the law of con- tinuity, then £ 2 AV Cwv , v D A v , . v - — and V = - and X = pA - - (ibs) cw Dv* now p = , from Bernoulli's theorem, 314 Hence X = — — (A - 2 cv») 2g = ° \ I (A-2 cw) (Ibs) 2gc w D v* DA* Y2 but p = - = - hence X = p(A-2 cw) 2g 2gcawa That, is the reaction on the piston equals the product of the pressure in the recoil cylinder and the effective area of the recoil piston, where the effective area of the recoil piston equals the an- nular area betneen the recoil cylinder and piston rod decreased by twice the contracted area of the orifice. A physical explanation i$ that due to the pressure of the orifice, we have the pressure lowered around the orifice. Hence we must not only subtract the area of the orifice, but also an additional equivalent area which is to account for the lowered pressure about the orifice. Since c = 0.6 approx., then 2 cw = w approx., and therefore for practical calculations, the an- nular area of the recoil piston is merely decreased by the total throttling area through the piston. Ifhen the rod is assumed to extend through both ends of the recoil cylinder, we have a continuous rod in the cylinder and therefore no void is pro- duced during the recoil. Assuming the same symbols as before, we have, since the total change of momentum of the fluid is nil, pA - X - Y = 0. Hence X = pA - Y and the fluid merely transmits the mutual reactions be- tween the recoil cylinder and recoil piston. Let pw = the pressure in the orifice. (Ibs/sq.ft) ) Xf = total reaction ( on front of re- ) coil piston. for the momentum of the fluid contained in the front 315 part of the cylinder to the orifice, and DAY Y-pww-Xr= - v ) Xf = total reaction on ( rear of recoil ) piston for the momentum of the fluid contained from the orifice to the rear end of the cylinder. Now X = Xf-Xr= total reaction on recoil piston. Due to the sudden expansion of the fluid after leaving the orifice, the pressure on the rear face of the piston, becomes, pw(A-w)=Xr (assumption from ex- periment - sudden expansion), hence Y -pwA = D A V v and X. = Xf-pw(A-w) DAY =(p-pw)A -- - — v Dv* Applying Bernoullis' theorem, we have p-pw = - but by the law of continuity c w v = A V therefore Dv* = — — (A-2-cw) (Ibs) Since pw is negligible *f compared with p, we have Dv2 P - Pw - P = 2g hence, as before X = p(A-2cw) That is the total reaction on the recoil pis- ton equals the product of the pressure in the re- coil cylinder and the effective area of the re- coil piston, when the effective area of the recoil piston equals the annular area between the recoil cylinder and piston rod decreased by twice the contracted 316 area of the orifice. Since c = 0.6, 2cw = w approx., and therefore again for practical calculations, the annular area of the recoil piston is merely decreased by the total throttling area through the recoil piston. DERIVATION OF RECOIL We may consider the throttling THROTTLING FORMULAS, effected in either of the follow- ing manners: (1) throttling through grooves in the cylinder wall or through a variable orifice in the piston itself, -(2) throttling through a stationary orifice. (1) Throttling through a variable orifice in the piston or grooves in the cylinder walls. Let A = effective area of the piston, i. e. the cross section of the cylinder minus the cross section of the rod. (sq.ft) p = the intensity of pressure at the pressure end of the cylinder (Ibs/sq.ft) D = the density of the liquid (Ibs/cu.ft) V = the velocity of the recoil (ft/sec) w = the area of the orifice (sq.ft) v - the velocity of flow through the orifice. (ft/sec) X = the total fluid reaction against the piston (Ibs) Then, we have, (neglecting the small pressure in the orifice) D A v pA -X » v - - - for the momentum generated in the jet, Dvz and p = _____ for the energy of the flow in the jet. AV = cwv ----- from the law of continuity of the flow, then 317 DA 2cw 2 X = ~ (1- — ), 3 2 DA V , 2cw, (1 r-) (Ibs) 2gc*w2 A Since the reaction on the cylinder is the difference between the force pA at the pressure end and the reaction of the jet D A V - v flowing from the orifice we have the reaction on the cylinder also equal to D A V DA3V* 2cw as would be expected from the equality of action and reaction. Ulith a continuous piston rod through both ends of the cylinder we may neglect the pressure through the orifi*ce and since by experiment the pressure on the rear face of the piston is practically that through the orifice, the reaction on the piston remains the same. Here again the reaction on the cylinder is DAY .. D A V pA-p A = pA -- v, since p A - - v as would 5 3 be expected from tlie equality of action and reaction. The reaction X on the cylinder may be written Y PA^2 2cw X = 2gc*wa A ' Dv2 Further since p = = — • — , , we have also. Zg 2gcaw* X = p(A - 2 cw) = p(A - w) approximately. Thus, knowing the pressure in the pressure end of the recoil cylinder to obtain the reaction on the piston, we must multiply this pressure by the ef- fective area of the piston minus the area of the re- coil orifice. (2) Throttling through a stationary orifice. With a stationary orifice, the throttling 318 usually takes place between the recoil or brake and recuperator cylinders. The loss of head or pres- sure drop is mainly due to the sudden expansion of the flow from the orifice, though with a relatively long orifice the loss due to sudden contraction may become appreciable. If w = the area of the orifice (sq.ft) A = the effective area of the recoil piston (sg.ft) V = the velocity of recoil (ft/sec) v = the velocity through the orifice (ft/sec) c = contraction factor of the orifice. H = the area of the channel leading away from the orifice, (sq.ft) Then from Bernoulli's theorem, we have p-pa ) where p = the pressure in the ~~~ = ^t ( recoil cylinder. ) pa = the pressure in the ( recuperator. Mow ) hf = total head lost due to hf=hfc+h.fe ( throttling. ) ^fc= l°ss °f head due to ( contraction. ) nfe~ l°ss °f head due to ex- ( pans ion. T* Now hf., = £ — where 5 may be taken 0.35 to 0.5 and gf and * z v , cw.a v ... cw.* f. hfe= "^ (1 ~ "1L) hence hf= ~~[(1 > * *] 2« 2g W In recoil mechanisms W is usually made from 2.3 to 3.0 tines w. Then, we have, if c is taken approximately = 0.65 (1 --J.)* = 0.515 to 0.614 For flow from an orifice into a large reservoir 319 4 > 0 and (1 - £-)* < 1 n * Hence usually cw « [(1 - -T- ) + &] = 1 approximately, D A*V» -55 hence p-p. = - for the drop of 2gcawz pressure through the orifice. The reaction on the recoil piston is, D X = pA = - — . . + p.A ** In recoil design, it is customary to measure areas in sq. inches and pressures in Ibs/sq.in. Further the average specific gravity of the re- coil oils used in our service may be taken at 0.849 and therefore the density D becomes, D = 62.5 x 0.849 (Ibs/cu.ft). The recoil throttling formulas become, therefore (1) For throttling through a variable orifice in the piston or grooves in the cylinder vralls:- X = P = 6 K2A»V2 (Ibs) w = (Ibs/sq.in); KA*V /6~ (sq.in) (sq.in) 175 w2 13.2 /x KAV 175 w* where K = — = 1.6 to 1.3 approx. 6=1 •—: c =- 0.6 to 0.8 approx. (2) For throttling through stationary X = orif ices:- KAV 175 320 ~P* * 175«* (Ibs/sq.in) CW ft where K = 1.6 to 1.3 approx. 6 =(1 -- ) + E VARIATION OF THE THROTTLING ffe have seen the CONSTANT IN THE RECOIL total braking on the recoil piston may be expressed, when throttling through a variable orifice in the piston or through grooves in the cylinder, as K« A» V« and when throttling through a stationary orifice, as XIL I IV A V f •* * \ = a - + p.A (Ibs) 175w« . 2cw .i cw.a _ where 6=1 -- - and o = (1 -- ) + £ Since w varies throughout the recoil, 6 and 6' must also necessarily vary in the recoil. Calculations with the omission of the term 6 or 6* have been found slightly in error and this error has been ascribed to variations in the contraction factor of the orifice. The contraction factor may also vary but it seems more probable that the error is due to the omission of the term 6 or 61 . With stationary orifices -^ and 5 can very- H often be neglected and therefore the variation in the throttling constant can be neglected. With throttling through the piston or by grooves in the cylinders -2filL is small but not negligible^ hence with this type of throttling variations in the orifice are more marked. 321 For a preliminary design 6 and 6' may be assumed equal to unity; but on recoil analysis and careful tests 6 and its variation in the recoil should be taken into consideration. CHAPTER VI DYNAMICS OP RECOIL. ELEMENTARY PRINCIPLES. The object of the recoil is to reduce greatly the stresses induced in the car- riage. Without recoil, the reactions brought on the various parts of the carriage are direct functions of the maximum powder force, which would require a very massive carriage for guns of large caliber. The mutual reactions created by the powder gases between the gun and the projectile is of very short duration compared with the time of recoil and for a rough approximation nay be treated as an impulsive reaction. Neglecting the mass of the pow- der gases, we have /Pdt = mv and /Pdt = MV. Therefore mv = MV, where m = mass of the projectile M = mass of the recoiling parts v = velocity of projectile V = velocity of recoil /Pdt = impulsive reaction of the powder gases. The momentum generated by the action of the pow- der gases in the projectile and gun is the same, as is immediately obvious from the principle of con- servation of momentum. It is to "be further noted that finite forces, as the resistance to recoil, can be neglected in the consideration of impulsive actions, and since the generated velocity of recoil acts for a differential time, the recoil displacement during the impulsive action can also "be neglected. The kinetic energy of the recoiling parts, after the impulsive action, is A. ~± MV' Since V = , the recoil energy in terms of the 323 324 IB velocity of the projectile becomes, A = — (- mv ). Hence the energy of recoil is but n - of the energy of the projectile. The total energy generated by the impulsive action of the powder gases, is, therefore i m - (i . 5>« Obviously the greater M, the smaller the energy of recoil. The reaction R between the gun and raount for a recoil displacement b, is - MVa R = ^ or in teras of the velocity of the projectile _ "* , * * \ ~H (; BV } The reaction is thereby reduced proportionally to the increase of ths recoiling mass M. Hence to reduce the recoil reaction we increase the recoiling mass 14 and the length of recoil h . The dynamical relations for an elementary recoil analysis in terns of the relative velocity of the projectile with respect to the gun vp can "be readilj obtained as follows:- Vp = v + V assuming V measured in the direction of recoil from the conservation of momentum m Vg MV * mv = m(vR- V): hence V = - M + m The energy of recoil is and the recoil reaction If the recoiling parts are hrought to rest hy friction alone, R = u Mg 325 1 V2 hence b = - — • 3 2 ug DOUBLE RECOIL SYSTEM: When a gun is mounted on a movable mount as a car body or itself rolls along a plane, we have virtually a doubl.e recoil systen, the upper recoil being between the gun and mount, and the lower between the mount and plane. As a first approximation we will neglect the resistance between the mount and plane as small com- pared with the upper recoil resistance. Let MR = mass of upper recoiling parts MC = mass of lower recoiling parts ra = mass of the projectile vo = the muzzle velocity of the projectile V = the initial velocity of the recoiling parts v = the velocity of combined recoil Then, during the impulsive action, neglecting the mass of the projectile, we have, T for the projectile / Pdt = mvo (1) T T for upper recoiling parts / Pdt - / Rdt = MV (2) o o Where F is the vertical reaction between the upper and lower recoiling parts. T How R is a finite force, .*. / Rdt - 0, if t is o very small. Further the displacement of the upper and lower recoiling parts inappreciable, since T T / Vdt = 0 and / Fdt = 0 respectively o o Hence, nvo = MpV with no appreciable displace- ment of either the tipper or lower recoiling parts and no moraetitura imparted to the lower recoiling parts. During the recoil, after the impulsive action, we have 326 T for the upper recoiling parts / Rdt=MR(V-v) o T for the lower recoiling parts / Rdt=Mcv o hence, the combined velocity of the system when the relative recoil between the upper and lower recoiling parts ceases, is MRV v = T. : — If the mutual recoil reaction R between upper and lower recoiling parts is made constant, then v" c n V R = Mc -— or T = — - where T is the time of the relative recoil. The relative displacement Z is, ,V+v v" V ? 99 £t & 6 Substituting for T, we have McMR V* Z 3 — — — • for the relative displacement MR + Mc 2R The relative displacement can also be obtained from a consideration of the energy relations in the recoil. We have V 1 X _* T) = — **R(V -v ) for the upper recoil- parts V i a -T~ T) = - Mcv for the lower recoil— parts Subtracting: RZ = J MR(Va-v2)- f Mcv* that is the energy of recoil, j MRV = is dissipated in friction and throttling (RZ) and 327 remainder is the kinetic energy of the combined masses. Now since, M v - R = MR+*c we have » * i MRMc t • « MR+MC Therefore as before, the relative displacement becomes MR+MC 2R ELEMENTARY RELATIONS. During the travel of the projectile in the bore of the gun, neglecting for a rough approximation the mass of the powder gases, a mutual reaction is created "by the powder gases between the gun and projectile, which generates equal momentum in both projectile and gun provided no extraneous forces are exerted on the gun. The resistance of the recoil brake is very small compared with the powder force, therefore its momentum effect is negligible. After the projectile leaves the bore, further expansion of the gases take place and the reaction due to the momentum generated in these gases causes an addition* al increment in momentum of the gun. This additional momenta is commonly known as the after effect of the powder gases. Assuming free recoil of the gun, if m = mass of projectile M = mass of the gun or recoiling parts P = total powder reaction v = absolute velocity of projectile V = absolute velocity of gun in the recoil u = relative velocity of projectile in bore then during the travel up the bore / Pdt - mv = MV but u = v + V for the relative velocity of the pro- 328 jectilc, hence m(u-7)=MV and the velocity of recoil becomes . m . mv V = ( )u= — m + M N Since m is snail compared with M, we are not great- ly in error in assuming u = v in approximate cal- culations. At the end of the travel of the projectile up the bore, we have mv and 7( After the projectile leaves the bore if P = the reaction exerted by the gases, then ** / it / Pdt = M(7f - V0) = nv where v = the mean tg velocity of the gases "m" after expansion. For a first approximation v will "be assumed a function of the muzzle velocity v0 and we will place BV =cvQm Hence MVf=(m+cin)vo. For computations c will be taken equal to 2.3. The energy of free recoil becomes hence T. = i — M How the recoil brake exerts a resistance R through a recoil displacement b, "hence Rb= -MV* roughly, and R = 2M.b The recoil reaction R is a measure of the stressed condition of the carriage and very often for a given carriage m, u, vo and b may one or all be changed. To compare the recoil reactions, we have for the sane gun, t t Rt (•t*cit) v0i b, ,,_ and for R =R =R, then — = — where c = \ (-.*«,)• v*, 2.3 approx., and for bt = bf, = "b, then Rt (in^+cl^)3 v0± r~ a ~, ., — : — where c = 2.3 approx. R, (»,+co>t)* v«, These equations are important in order to estimate with a given change in the ballistics of a gun, the necessary change in either the recoil or recoil brake reaction. The energy of recoil nay "be expressed as m +cu, t _. a . E = 1 r (m+cm) v0 } M r jM-f mvo) very roughly m = - (muzzle energy of the projectile) (approx.) M Therefore, to decrease the recoil energy M should "be made as large as possible. Since further The recoil reaction varies inversely as the recoil- ing mass, and therefore to decrease R, M s"hould "be made large. EFFECT OP POWDER GASES The effect of the pow- ON THE RECOIL. der gases on the recoil may be considered during two periods:- (1) while the pro- jectile travels up the bore, (2) after the projectile leaves the "bore and the ex- pansion of the gases ta"kes place. In either case an approximate assumption is- necessary in order to represent the phenomena with sufficient simplicity. During the travel of the shot up t"he bore it will be assumed that the gases expand in parallel lamina, and the motion of any differential lamina to be a linear function of the distance. from the "base of 330 the "bore to the lamina in question, that is i v + V v = c s + c where c =» - V and c = u v = velocity of projectile V = velocity of recoil u 3 travel of projectile up the "bore hence with free recoil TO I mv + 2 - v = MnV during the travel up the bore u but „ » i m ,u i . m(v-V) 2-v =-/ vds= — u u 0 2 The equation of momentum of the system during the travel up the bore becomes, therefore, - (Y~V) (m*0.5S)v • v + m — - — = MV or V = MR+0.5m Further since the relative velocity of the projectile is ~ = v + V then, l»+0.5D(?r- V) =(M+0.55)V at at du therefore (m+0.5m )T~ and for the displacement of recoil in terms of the relative displacement of the projectile, (m+0.5l) M+V+I If P = the reaction of the powder gases on the "base of the projectile Pfc3 the reaction of the powder gases on the base of the "bore of the g"un then, for the powder gases, we have I d(v-V) 5 d_v _ ra d_V ( ^>" " 2 " dt Z dt ~ 2 dt for the motion of the recoiling parts in free recoil, 331 Pb - *R ST ia4**' °ai^*a'1 ^" and for the motion of the projectile If tbe gun moves backwards a displacement X, while the projectile moves forward an absolute displacement x, then X * / Vdt, x = / vdt (4) Prom (2) and (3) in (1), dV dv . i. dv _ £ dV MR dt " dt 2 dt ~ 2 dt hence (Mp+0.51)— = (B+0.5 1)^ (5) Integrating, we have as before, (MR+0.5ii)V=(m+0.5i)v (6) and. (MR+O.SijX'dn+O.SSJx (7) For the relative displacement u = /l (V+v)dt or du = (v+V)dt o du (NR+0.65)d-u V+v . v . .... r / . . X - / Vdt = / ( _-) du o o Mp+tn+n hence m+0 . 5n X * .. as was obtained by direct sub- Mo +ra+m stitution of displacements. With a constant powder pressure during the travel •up the bore, the time of travel becomes, 2u0 2u0(MR+0.5in) (2WR+m) UQ * = — — = i = -••^^-•^ v+7 (t Actually since the powder reaction varies during the