UNIVERSITY OF CALIFORNIA LOS ANGELES GIFT OF Col. Glen F. ORDNANCE: DERARTI^IEINT DOCUMENT NO. 2O3S-A THEORY AND DE5IGN OF RECOIL 5Y5TEM5 AND GUN CARRIAGES (CHAPTER 3ZHI) Prepared in the oT -the ChieF aF Ordnance JANUARY CONFIDENTIAL ENGINEER REF 3 RODUCTION F=l_ANT WASHINGTON BARR ACKS.O.C. UP LL ORDNANCE DEPARTMENT Document No.2035-A Office of the Chief of Ordnance WAP DEPARTMENT, Washington, February 1922. The following publication, entitled "Theory and Design of Recoil Systems and Gun Carriages", is published for the information and guidance of all students of the Ordnance training schools, military academies, and other similar educational organizations, The contents are strictly of a Confidential nature and should not be reputlished without authority of this Department. C. C. WILLIAMS, Major General, Chief of Ordnance, U.S.A. 523414 HYDRO PNEUMATIC RECOIL SYSTEMS. INTRODUCTION. The fundamental advantages of these recoil systems are smooth- ness of operation with conse- quent durability and ability to meet with requirements of service within space and weight limitations. Light and small recuperator forgings are made possible by the use of "high pressures, both in the re- cuperator cylinder as well as in the hydraulic. To obtain and hold these pressures, special sys- tems of packing are necessary and successful manufacture depends largely upon use of proper packing. RECOIL SYSTEM WITH This system consists usually CONSTANT ORIFICE of two cylinders; the hydraulic FIXED LENGTH OF brake cylinder and a recuperator RECOIL. cylinder containing a floating piston which separates the oil and air, this cylinder communicating with the hy- draulic cylinder by means of a large opening at one end of the cylinder. The floating piston has a long buffer rod which passes through a fixed diaphragm in the recuperator cylinder and enters a buffer at its end, serving a double purpose since it gives constant throttling through the diaphragm on recoil and serves as a throttling buffer on counter recoil. During recoil the constant orifice about the buffer rod at the diaphragm of the recuperator cylinder produces a throttling of the oil and a consequent drop of pressure due to the constant orifice. 3 During counter recoil the oil passes through a constant orifice produced by constant throttling grooves in the buffer chamber. Toward the end of recoil the depth of these grooves vary, causing a varying orifice. The throttling is so designed as to cause a sufficient drop of pressure from the air, to produce a pressure against the hydraulic piston just sufficient to balance the total friction and thus nearly maintain uniform motion through a considerable part of counter recoil. At the end of c 'recoil where we have the de- creasing variable orifice, the pressure is rapidly reduced to zero; thus the recoiling mass is brought to rest practically by only the total friction of the guides, stuffing box and piston frictions. A characteristic of % constant orifice in a recoil system is to give a peak in the pressure curve in the region of the maximum velocity of re- coil. The maximum velocity and the consequent maximum drop of pressure due to throttling occurs, fortunately at the beginning of recoil where the air pressure is at its minimum. As the gun recoils, the drop of pressure decreases due to the decreased velocity and consequent less throttling through the orifice, whereas the air pressure rises, and since at all times the hydraulic pressure is the sum of the air and drop of pressure due to throttling, we have a tendency to maintain a constant hydraulic pressure in the brake cylinder. This effect is especially true with a fairly long recoil. If, however, the recoil is shortened, the peak effect due to the throttling at the beginning of recoil is much greater than the increase of air pressure at the end of recoil. The type of system now under discussion has a single length of recoil for all elevations which is made sufficiently long for stability at zero degrees, while the ratio of final air pressure to initial air pressure and the constant orifice are so de- signed as to give a uniform, or nearly uniform pressure curve consistent with the stability slope. This type is ohviously restricted to a limited ele- vation. It is proposed now to discuss the theory de- sign both for recoil and counter recoil and then to approximate design formulae for the use in de- sign, Le t .A = effective area of the hydraulic or brake piston A a = area of the air side of the floating piston, A^ = effective area of the oil side of the floating piston, at, = area of the buffer rod. Ajj = area of the buffer chamber or buffer head. W o = area of constant orifice during the re- coil, W^ = area of counter recoil constant orifice. W x = variable orifice in buffer chamber during end of counter recoil. p = intensity of pressure against hydraulic pis- ton p t = intensity of ressure in recuperator cylinder back of valve or at entrance to constant orifice. P t -= drop of pressure in passing spring valve. P drop of pressure in throttling through constant orifice. R v = reaction of spiral spring a-fe lift "h" on spring valve. d -= spring constant of the spiral springs h o = initial compression of spring a s = area of hase of spring valve c = circumference of spring valve rrd If K equals the total resistance to recoil, we have during the powder pressure period, the total powder force being P t J 2 y F R P K F-K=M_ - hence r- At=Av or -r-At- ~ At=Av r dt 2 M r M r M r bat for the same time intervals when the recoil- ing mass if free, assuming the powder force to be unaffected by the slightly different motion of the powder charge and shell, we have F At=Av* where v* as the free velocity of recoil \i M r for the recoiling mass. Hence during the powder period, f{ A Vf - - At=Av and the corresponding displacement of recoil, becomes, A x = v A t. During the subsequent retardation, J 2 yr V - K=M r - or - - A t = A v, when K = pA+R-W r sin# r and R = the total guide and stuffing box and piston friction. Considering now figure(l) we have the pres- sure at the base of the spring valve equal to the pressure against the hydraulic piston, neglecting the slight velocity head generated at the base of the valve. From the law of continuity, the quantity of oil passing through the valves, becomes, Q v =AV-A b V-' where V = the velocity of the float- AV ing piston = A a .A b thus Q v = AV(1 - ) Now if we neglect the A dynamic reaction against the valve which is small, we have P j a s =R v =S(h+h o ) but P for the four valves, becomes, p = - = -- hence the valve lift 2800 (ch)* a s is obtained from the cubic equation: h*(h+b) * an d the corresponding drop of pressure by substituting the K obtained from the above equation in the formula P~Pi =P i = ~ < h * h o> a s In general this refinement may be omitted, since the drop through the valves is at a maximum, but 10* of the total drop, and the movement of the spring is at a maximum not greater than one third its total deflection to solid height, thus making a gross error of less than 3* if we consider the spring re- action that at assembled height. Consequently in the calculation of the pressure and velocity curves we will assume P t a=S h o =p-p i = a constant. Now the quantity of oil passing through the constant orifice, is equal to that passing through the valves plus the quantity of oil displaced by the annular area about the buffer rod in the buffer chamber, that is Q -(AV-A b V ' )+ (Afc-a^V 1 = AV-a b V a b A' = AV(1 -- A check on this valve may be obtained by noting that of the total quantity of oil displaced by the floating piston A a V'=AV but (A a -a b )V' = A a V' passes AV through the constant orifice, and we note V = A a A a~ a b hence as above, Q Q = (- - )AV A a The drop of pressure due to throttling through the constant orifice, becomes, K* d p o = n n =P ~Pa Therefore, we have, p=P J *P o +p l To obtain p^ we must further consider the equilibrium of the floating piston, neglecting the effect of its inertia during the acceleration and retardation as being small. The forces in the direction of motion acting on the floating piston, are: (1) The air pressure on the air side of the piston resisting the motion. (2) The packing friction resisting motion. (3) The oil pressure on the oil side of the floating piston, (4) A resisting pressure on the annular area of the buffer. (5) The pressure on the base of the buffer. The pressure in the buffer chamber is assumed the same as the pressure in the hydraulic or recoil cylinder since by computation it is found that the drop in pressure due to throttling through the small set of holes at the buffer end of the recuperator cylinder is exceedingly small. Therefore, considering the equilibrium of the floating piston and neglecting the drop of pressure through the valves, we have ^ - p a A a -R f =0 A a R f, a b hence p a = =(p > )-p The pressure in the hydraulic or recoil cylinder becomes, K (A a -a b )AV A a R f a b p=Sh + ^ * p + - p " 175 A|w A a A a A a therefore a b K*A a 2 A 2 V a A a R f p(l + ) =Sh + - + p* where Pa=P a "~ + ~" A^ 175Aw A^ A a The resistance to recoil, becomes, since 10 a h A a )~ A a A a Sh o A a A " ^ 176A|w or in terms of p a , where R s = stuffing box friction of piston rod. R_ = hydraulic piston friction Rg = guide friction As will be shown in the theory of packing the stuffing box and hydraulic piston frictions de- pend upon the pressure in the hydraulic cylinder Further a recoil buffer action is obtained towards the end of the recoil by making W Q variable, i.e. substituting w x in the above equation for W Q . The total resistance to recoil may be simplified into the following equation: V* K=C j p a +C f -7 +(ZR+C g )-W 4 sin# Towards the end of recoil H O changes to w x i 1 * A 3 A K R(A * s The above equation is exactly equivalent to previous recoil equations with the fundamental difference that now we have a constant orifice during the greater part of recoil in place of the previous variable orifice. If the term C a is large as compared with the other terms in the 11 equation for the resistance to recoil, K obviously varies considerably with the velocity of recoil, having a large peak at maximum velocity. To reduce this peak in the resistance curve, we must have a larger orifice with less throttling and a conse- quent longer recoil together with a higher air pres- sure than would be used on other types of recoil systems. Hence, we may completely limit the ratio of the peak to the average resistance to recoil by suitable high air pressure, a large constant orifice and a long recoil. Thus we see the limitation of this system for high elevation and consequqnt short recoil. THEORY OF PACKING. To prevent leakage of air or oil under high pres- sures in the cylinders the packing. is designed to be subjected to the actual intensity of pressure of the air or oil as the case may be. The packing, however, cannot be regarded as subjected exactly to a hydraulic pressure since the pressure in one direction differs from the pressure in a direction at right angles by the amount of the previous pressure multiplied by Poisson Ratio. To compensate for the deficiency in oil or air pres- sure normal to the cylinder surface or piston rod, packing springs are introduced. These springs are designed for a slightly excess pressure normal to the cylinder surface in addition to compensating for the air or oil hydrostatic pressure. The total normal force multiplied by the coefficient of friction of the packing gives the total friction. Thus we see the friction may be considered as consisting of two components; the friction component due to the oil or air, and that due to the packing springs. The packing frictions, therefore, are a linear function of the pressure and may be represented by an equation as follows: F =c iP +c t 12 1 c t = gross length of outer flaps of flange K = Poissons 1 ratio for leather = 0.73 p a = intensity of air pressure Pg = intensity of oil side of floating piston p = intensity of pressure in leather on air side caused by packing springs. PQ = intensity of pressure in leather on oil side caused by packing springs. pq = intensity of pressure in grease caused by reaction of spiral spring on slide moving on inner rod of floating piston. R s = reaction of spiral spring on grease slide at assembled height. R b = reaction of belleville washers on annular area of packing on air side at assembled height. K b = reaction of belleville washers on oil side of packing at assembled height, f = coefficient of friction of leather = .05 f t = coefficient of silver = .09 Then in the metric system of units if all lengths or dimensions are measured in millimeters and pressures in kilograms per sq. cm. and spring reaction in kilograms, 400 R s 400 R b PO " . _ 400R b n p 1 = where -(d*-d 2 ) is the an- n(d*-d a -) 4 nular of the packing ring in sq. mm. and the total pressure in the packing, become, Grease pressure = P a + Pq ( considered a purely hydrostatic pres- sure) 13 I i 14 Air side packing pressure = p a + p o Kg per sq. cm. Oil side packing pressure = Pa * PQ If we multiply the packing pressures by K we obtain the pressure normal to the cylinder surface. Prom the diagram of packing contact with cylinder fig. (2) we nay assume = c-2e and a =c'-2 e f since the contact with cylinder is some fraction of the breadth of flap of the flange. The total friction of the floating piston pack- ing becomes, in kilograms, F f i f jPq ].01K + 7td[ (bf+af t * Trd[(bf+af i )(p +p )+2a i f i p q ] .01K+ nd [2(bf +af J ^a^pa .OIK Thus the packing friction reduces to Rf=C J +C 9 p a (kilograms) where C t - irdt (bf + af x ) (p o +p o =+2a t f t p q ] .OIK and C f = ndt2(bf+af i )+2 a t f 1 3 .OIK The drop of pressure due to the packipg friction of the floating piston, becomes, in Ibs. per sq. inch 2.205-R f (Pa"Pa^ =p f = ""7 - ver y roughly (See previous exact formula) Where A a = the area of the floating piston in sq. inches and the plus sign for recoil and minus for counter recoil. The above formulae are not strictly exact but give however, a very close approximation, indeed, of the friction drop due to the floating piston. In the design of the packing given later, tables are shown for proper width of packing and ex- cess pressures due to springs as well as a table for 15 Belleville given in the appendix. STUFFING BOX FRICTION. In the stuffing box packing the leather and flaps of the packing flanges, come in contact with the peripheral surface of the piston rod. A typical stuffing box packing is shown in figure (3). If, now, as before p o = intensity of pressure caused by bellevilles or packing springs in Kg. per sq. cm. p = pressure in hydraulic cylinder Kg. per sq. cm. d r = diam. of piston rod in millimeters. d o = outer diameter of packing ring in millimeters RJ.J = Belleville reaction at assembled load on annular area of packing ring measured in kilograms Then in metric units as before, 400 E b tr p = (A* _d2> where 7 (d*-d*)is the annular area of 7r\Q0~Qr ^ 4- the packing ring and the stuffing box friction, becomes, R s = nd[ (bf+af i )(p Q +p)+a 1 f t p] .OIK Kilograms = irdCbfp+afp + a ] .OIK Or as before R s =C i +C 2 p where C t = * dp o (bf +af ) .C1K and C 9 = 7td(bf+af+ a t t ).01K HYDRAULIC PISTON FRICTION. The packing in the hydraulic piston is similar to that in the floating piston. We have, if d = the diam. of the hydraulic cylinder in mm. d = the inner diameter of the packing ring, in mm. 16 R^ = the reaction of the belleville washers on the packing, in Kg. 400 R b P n(d a -d) or in the form R_ = C i + C Jt p C t = ndpoCbf + af^.OlK C f = nd(bf+af t + a^) .OIK CALCULATED RETARDED VELOCITY During the pow- CURVE OF A TYPICAL SYSTEM. der period, we have or in terms of any two given points, K(t -t ) v-v v fra v fn v m v n m r During the retardation, we have V = .'. V -V m = The corresponding displacement during the interval, becomes, X mn = A X, and Pa = Pa~*JT s Friction Values: (1) Floating piston friction K = 0.73 R s = 950 Ibs. R B = 2150 Ibs. RjJ= 4080 Ib. p- = 190 Ibs. per sq. in. (calculated ) p = 880 Ibs. per sq.in. p,J = 1690 Ibs. per sq.in, 17 Hence from the previous theory on packing, the floating piston friction may "be expressed in the following formula: R f = .53 p a +420 Ibs. (2) Hydraulic piston packing friction R b = 2980 P = 2080 Ibs. per sq.in, Hence R = .121 p +280 (3) Stuffing box friction R b =2810 Ibs. p =1970 Ibs. per sq.in. Hence R S = .063 p + 140 (4) Guide friction neglected From (2) and- (3 ) we obtain for the total friction resistance during the recoil, SR=.184 p+420 Drop of pressure through check valve: a s = 0.0779 sq.in. Sh o = 13 Ibs. 13 Hence P = - = 167 Ibs. Use 150 Ibs. 0.0779 Air pressure curve: A = 1.314 U^ = initial air vol. = 188.7 cu. in. p a j = initial air pressure = 1675 Ibs. per sq. * U i 188.7 Paf = Pai( ^' 3 = 1675( > l Ui-Ab 188.7-l.314b = 2731 Ibs. per sq. in. where b = length of recoil in inches. TABLE OF AIR PRESSURES; A 188.7 18 O -in op fed \ J J 1 I J.33J Ml pit 133 T& b U R! - I I I 19 RECOIL CURVES X / / / S tn 20 a b c a e f h 182. 7 X 1.314X 188. 7 1 logd 1. 3*e lj p 1.314x 188. 7 1.675 tf -1.314X 188. "70 1. 000 .00000 3. 22401 1675 2 2.63 186.07 1.014 . 00604 .00785 3-23192 l7o6 6 7.88 180. 82 1.043 .01828 .02376 3.24777 1*769 14 18. 40 170.30 1. 105 .04336 .05637 3. 28038 1907 22 26.91 159.79 1. 180 .07188 .093*4 3.31745 2077 30 39. 42 149. 28 1. 264 . 1O175 . 13228 3- 35629 2271 38 49.93 138. 77 1.359 .13322 . 17319 3.39720 2496 42 55. 15 133.55 1. 412 .14983 . 19478 3.41879 2623 Throttling Contraction Areas: A x x 1 = floating piston displacement = A a Constant orifice from a to b = .0631 sq. in. x 1 = 9.685" x = 39.0" Variable orifice from b to c ; at b = .0631 sq.in. decreasing, a* c = .0302 sq.in. uniformly x'bc = .0787" x bc = 0.4" Variable orifice from c to d at c = .0302 sq.in. de- creasing xP*I? ' *T P ii A A SL ol Rf .01V 8 hence P f = .065 p a +-7 .0667(140+-- +1.0667p^ a o Table for pressure drop across floating piston R f .01V 2.065p a + ~r .0667(140+ ; +! Ka AJ w 2 X .065p a R Aa' V 2 01V 2 ut 2 O i.oep; .0667 2 109 250 762 1820 2190 4250 280 70 4 113 260 912 2300 2240 4680 310 80 8 117 260 918 2320 2320 4780 320 90 16 126 280 681 1720 2500 4300 290 120 24 137 290 445 1120 2700 3960 260 170 32 149 310 240 600 2930 3670 240 220 40 164 340 34 414 3250 3800 250 250 In terms of the intensity of pressure, we have W r sin0 R-W r sin0 r : nr Let a = b * a+ Sh ( c = b+p a * = c+p f e = 26 Prom these values the following table was coEputed for the various pressure drops. Re- coil RW r sin0 A a+sh b + Pa C + Pf d + P 2 660 800 2610 2580 4630 4 720 870 2610 2680 5130 6 730 880 2540 2720 5270 8 720 870 2670 2760 5230 12 700 850 2720 2830 4980 16 670 820 2760 2880 4710 20 640 790 2820 2950 4450 24 620 770 2880 3050 4250 28 590 740 2950 3160 4070 30 590 740 3000 3220 4000 32 580 730 3040 3040 3910 34 570 720 3080 3310 3830 36 560 710 3130 3360 3750 38 560 710 3180 3410 3660 39 550 700 3210 3430 3610 40 530 680 3210 3460 3900 41 520 670 3240 3480 3480 u" travel of shot in the bore ain_ 2046*7.5 _ 15340 1.377*7.5 " 8.877 n = b + n V o - 1730x16.75 1050 a 7.5 " 1730 max. pressure 89.87 1730 33.8 27.6 .0065 36000 Ibs 27 P b = 1.12P m 40300 Ibs. Pg = mean constant pressure of breech 12WV m 15.95x1730 = = 14200 64.4u"s 64.4x7.6x6.95 _ f/ 27 36000 ,. /~ 2736000., ,. 89.87[( x - i) + /(I - ) - 1] * IV 16 14220 - 1614220' 14.08 27 14.08 x89. 87x40300 PK * x . (103.86) 8 4315 2(V g -V W r ) ^ 2(33.8-27.6)1050 _ 2x6. 2x1.050 , ** P ob x s "xg " 4315x6.95x32.2 " 4.315x6.95x32.2 .01248 T = t t +t f = .0065+ 0135 .02 seconds CALCULATION AND COMPARISOS OP AVERAGE RE8I8TASCK. 89.87x16.75 12(W r +W+if) 12x1067.53 = .00005 + .372 = .372 E = .1175 + .372 = .49 T * .019 second E = .50' = 6 9 - mV| - 32.6x33.8 K J - -4 - = 5300 (Ibs) 3. 4-. 5+33. 8x. 019 -nVf - 32.6 x 33.8 K = 2 L = - = 5500 (Ibs) b 3.4 Average from the curve of forces 4230 28 A = 1.314 hence total force = 5530 Ibs. PEAK VALUES Curve of resisting forces gives an average 5550 Ibs. for K. Peak value of K from the curves 5280 x 1.314 * 6940 Ibs. 6940 Peak ratio d = = 1.25 with respect av.re- 5530 coil force from curve 6940 Peak ratio d' = = 1.30 with respect com- puted av. recoil reaction. RELATIVE SIZE OF ORIFICE UNDER DIAPHRAGM AND COUNTER RECOIL THROTTLING ORIFICE. Q f = A a x f = 5.303 x f Q h = Ax41=l. 314x41=53. 9 5.303 A = .3945 7087-. 6811 x .35 = ^^- x .35 = .00346 2.7953 2.7953 d x = .00346+. 6811=. 6846 in. A x = .366 Smallest area under the diagram . 3945 - .366 = .0285 sq.in. Largest throttling area = .00665 sq.in. 29 .-. the throttling take place not under the diaphragm REIATIVE DISPLACEMENT OF FLOATING PISTOK. A a :A::x r :Xf where x r = displacement of recoiling parts Xf = displacement of floating piston x f 1.314 5.303 Variable throttling begins =f = 5.3545 on recoil corresponding x r = 5.3028x5.166 = 20<80 1<314 take .305 in. Total floating piston displacement = Xf x41 = 10.1595 Variable orifice ends at 41-20.5=20.5 of counter recoil THROTTLING AREA. Dis- Depth Area Total Recoil Recoil tance of of Orifice Disp. in groove one inches gun 5. 9055 5.3150 .0469 .0453 .003325 .003212 . 00665 .00642 2.38 20. 50 22. 88 4.52*76 .0421 .002985 .00597 5.56 26.06 3. 7402 .0386 .002737 .00547 8. 74 29. 24 2. 9528 .0339 . 0024O4 .00482 11.92 32. 42 2.5591 .030*7 .002177 .00435 13.50 34.OO 2. 3622 .028*7 . 002035 .0040*7 14.30 34. 80 2. 1654 .02*72 . 00 1 9 2 8 .00386 15.09 35.59 1. 9685 .0248 . C01758 .00352 15.89 36.39 1.7717 .0220 . OO1560 .00312 16.68 37.18 1. 5*748 .0193 . OO1368 .00274 17.48 37.98 1.3*780 .0157 .001113 .00223 18.27 38.77 .9843 .0063 .0004467 .00089 19. 82 40. 32 . 5906 .0059 .000413 .00083 21.45 42. 05 . 30 = P a - Floating piston friction R f = .53 P a * 420 C 'Recoil p a R f Pi' 25*70 340 2,230 i 2530 340 2, 29O 2 2510 330 2, 160 3 2480 330 2,150 6 2390 320 2,0*70 12 2240 300 1,940 20 205O 290 1,*760 28 1890 2*70 1, 620 34 1*790 260 1,530 3*7 17<0 26o 1,480 38 1*730 260 1,4*70 39 1*720 250 1,4*70 40 1*710 250 1, 46O 41 1*700 2^0 1,45O VELOCITY OP COUNTER RECOIL. = AV r where V^ = velocity of floating piston V r - velocity of recoiling parts A = quantity of oil throttled = 31 K"a 2 P ' i = = P = hydraulic piston cull (where PI i-P a W 2 a a X R. PA -F=M r V f dx R = .1211 P+279+.0634P+138 = .1845p+417 w 175W* W* PA - .1845 P - 417 =M r V r dv r P(A-.1845)-417=M r V r K'V 2 dV r (P'i T - -)(A-.1845)-417=M r V_ - r x r dx K'V 2 ! (Pa 1 ~ "P )(A-.1845)dx-417 dx = 5- M(V 2 -V 2 ) X 12 AVgRAQB VELOCIT? OF AK IBTEGRAL PfPIOD. In computing the velocity of recoiling parts by the formula V.+V P a A dx -- j; -- R dx = \ m r (V-v ). We V i* V 2 can take for value of ( ^) 2 as it simplifies 2 the equation to obtain V^ . It is nore accurate than taking either V i or V a as the velocit y f<> r the integral period. Its deviation from the actual average is very small. Proof Let V, = 2 and V 2 = 2.25. Actual average velocity of the intervals V V = 1 2 32 VJ = 5.0625 V ; +V J 9.0625 Approximate average = = - - = 4.5312 2 2 4.5312-4.5156 1.56 4.5156 COMPUTATIOK OF K. dv At max. velocity =0 dt pA = ZR K ^ v P =p a f "^T" x K V* ZR = .1845 p + 417 = 1845(P^. -- ~) + 417 W x K V 8 K V 2 .' PaA- dA =.1845(P, - ~j- a x x \ 4 *' * / \ f * y "" w x A w x A 417 K i v * "T (Pj' "" ...j ' ~ - - - - = 369 * i_ 'I 845 A Assuming max. velocity occurs on 16" of counter re- coil: 1850-370 = = 1480 *x -1*V fK(4.9714-.7006+.3314)| i ~|l 2v2 a 5.303 = 1480 175WJ 175 x (.0065) 8 x 33 (Kx4. 6022x1. 314) 2 3.28 " = 1480 175 x (.00665) x 5.303 1480xl75x. 00665 x 5.303 K 2 = - ; = ? K = .9048 (4.6022x1.314x3.28) Use K = 0.9 EQUATION OF FORCES OK COUNTER RECOIL. K iy> [K'Ua-Ab + a^V)* [^( "a W 175W* 175 x w (1.11x4. 6022x1. 314) 2 V* _ 9159 V^ (13.24x5.303) 8 W 10 6 W^ dx- 417 dx 10345 V (1.1295 Pit - -T-TT- rjj J.U "X (.06929Pi - 634>66?6 - 25.5828)dx=V?-V? . 317.33dx N , , _ 317.33, V l 10 e W a }= * U " 10*W + X X For variable orifice. For constant orifice V* (1-f .5979dx")=V(l-.5979dx")- (.005774?^ i-2.1316)dx" t i * 7 i 2 Ill 35 36 II PI ocMi-ir-rooox^ic\roHo x oovo^'CMocx> O iO CO O N O N COOOC~iC-X~X~C~- v OvovOvOVO lf\ lf> CMCMCMCMCMCMCMCMWCMCMCMCMCMCMCMCMCMCM ri tc\ o O oo tc\ ao ic\ir~oj-irivor~i>*-ir>roo -IVO CM IC < CM O - VO vO ^" V> CM i-t O OvoOX^t- *roroO^O^'rOv>ic\*r~iit~Oir^ | vo o rovoHiCNir>Oir\ocMrto\^vO ^rOHo x *~X~ irOrorr>rooox^j>CMCMri"i-lvOi-iOOr-vOic\'*rO-IOO> IO,OOOO C^OOOOOQOODOOOQOOODCQOOrO *--rox^i-ir~ oosri o co o cox~x-ir>CM 10 CM .c- vovO'jirort coiorotHOO N r^voiri^'ror i *OO x o o o o o oo i >ooooococooocoaoooao > - QVOO^OCMCM^CMHrOHOOVOlOOOICNO corvo f looi 1 "^ 1 rocMO o^oo.l-irv^' CMI-II-I CMCMWCMCMCMi-li-l-irirtOOOOOOOO oooooooooooooQoooo CMHCOC\fOOrooO^O ^"iHO^tvrOHCOVOlO CMCMrirlH OOO^OCI^OsCBflDCOOOOOC^X^X* OICMCMCMCMWfMf-liHi-liHrHrH" rH ri ri ri ri o ^o co o o^ ^* cjv o i*- ^* os ^ vOCM^vOvOlTti-l OOO^Os CM ^o^c^-^rlr^ lororoCMCM ooooooooooooooooooo OOOOOOOOOOOOOOOoOOO K^sOVOsOvO CMCM CMVOVOVOVOVOVOVOVOVOVO K\ vo CD o ^ CMro^mvor^ooovo O 37 QC B rt (4 9 Q 00 t-H II > i tCvCX O.C-ICt(OIC\lf>IOtM rH O .C- v> O> t- O - lftmmOIOMr-lr4rlrli-lrl H CM CM CM (N CM rHir>ir^O^O^O\OvO iCtlCXVO IC\ -} o. OOOOf-*O OIC(IC\CMIC\O rt O rS OJ JQ H W 1C\ VO CM W rt O O OS O rl rt i-l rS oooooooooooooooooo OOOOOOOOOOOOOOoOOo X CMCUCMCMCMCMCMCMCMCMWCM T3 I ic\ irxoirt CMinr^ooii^t^ocMinj^ K -in CJ O c\ (O 10 to rr K-I 39 LOG/.PITHMIC METHOD (Check on velocity of c'recoil) cV* dV F-,-W r (sin0-n cos 0)- ZR = roV W> dx cV 2 _ dV x cV 2 . dV W dx A=1.1295 P^i - 417 10345 1C 6 cV* cV* 2c(x 2 -x t ) log(A- )- log (A- )- 2>3mrW . cV* cV* 2c(x-x) T~ - A) - log(i - A) = W ' W 2.3 m g cV 2 2.3m W* ( - - A) w a "x 2c(x f -x l ) 2.3m W* 40 H-i v w$ * ^~ " As 2c .-*.> Antlg 2. m W V| = t - + A 1 2c(x-x) 2i Antlg 2.3 mW* 2CCX.-X. ) _ 2x10345 dx _ 276 dx 2.3mW* " 2.3xlOx32.6 W = 10^ W' .. . 276 dx . Antlg r| w* A = 1.1295 P^ - 417 10345 C x 276 dx a = " 10 6 W| b = Antlg a 41 ON ON CM o ITv in o o CO CM o 10 H O O <* o O to O CM co rl a\ rl VO CM ro ON rt CM CD O CD in rl m ON o : m CO CM o O O o co m in m in ro in ON .0 O in ON CM ON rl X- CM ro ON iTi rt ur\ ro ON O H m CO in CO CM CM rt ON CM vo 10 4 m X-- O rl m in o\ 09524 ON VO O ON o in m. x- vo X- ON in ON CM CSi rt JC- in ro ro vO O ON ro o o CM VO VO in VO 00 VO ON 00 CO CO ro CO ON O in vo in vO m rl rt rl rt rl CM rt CM CM rO vO X- r- * M O O o m O in O m O in m in in in m m 5 m X- 10 rt rt H rl rl rt in in m m m m m m (O vo rl VO VO VO VO VO m 00 o O oo o to CD O ro OB O ro CO o 0. co o to co o ro 00 o K CO o O O O CM o CM CM O CM O d o CM o CM CM o oli 00 NO ON in CO in CO Ox CM vO rt rl o oo CM X- * ON rl rl CM o ON CM O X- VO ro O CO O vo vo 11480 12880 13440 o CM VO ro rt o X- o rl ON VO CM X- * CM CM in CM 00 vo n vO m O ON rt co r- vO X- MX eo rl CM o CD x- in ? ro CM CM rt " X- VO X- vo CM CM rt H CM CM CM rt ro CM rt in CM CM rl rt CM CM rt ro CM rl VO CM CM rl ro CM CM rt rl CM CM rl CM rt in CM rt to rt CM rl ON - co ro vo vo VO CO in in in CM m in 1448 in ro O i rH O in vo CO CO __ o * rt 42 V dV - A dV 2 +A(b-l) Af bd C.B bd Af d dV* V V 35 1. 69 1.30 36 39, loo 63,300 76s 1296 1.65 1. 28 36; *j *7 4 10,070 859 1416 1.48 1. 22 z 37 12, 22O 14,360 985 1461 1.29 1. 14 t 22,720 23,070 162 1502 1. 08 1.04 38 53,7oo 45,700 420 1531 . 88 .94 38; 181, 40O 114, 000 978 1737 .64 . 8 39 1258,000 635, ooo 440 1560 .507 .71 * 126, 400 51,650 940 1490 .42 .65 4 1 400,000 133,000 67o 1542 .336 .58 S 9 4 22OO,OOO 550,000 80O 1610 . 252 .503 4O 32. 2x10 s 5.95 10* 640 1672 . 185 .43 1 4 28200xlO 2.98 10* 1480 . 106 .326 i 16. I6xl0 10 1.526 10* 12880 .095 .309 a 39.38X1Q 10 3.56 io le 13440 .0904 .301 41 50. lOxlQ 10 4.35 lo 13620 . 087 .297 C.R. V V 2 W*10 cV* W 2 "x ?* p/i P 1845] ZR ZR Ibs/ sq. in. o o 2570 2230 223O 410 830 630 1600 1.90 3. 60 44. 2O 750 2550 2210 1460 27O 690 530 84O i 2.54 6.42 44.20 1330 2530 2200 e7o 160 580 44O 230 2 2.87 8.21 44. 20 1700 251O 2180 48O 90 510 390 90 3 2.92 8.50 44. 20 l76o 2480 2150 380 7o 490 37o 10 5 2.93 8.51 44. 20 1730 2420 2100 370 7o 490 37o 10 7 2. 89 8.35 44. 2O 1690 2360 2040 350 60 480 37o 20 10 2.84 8.01 44. 2O 1630 2280 1920 290 50 47O 360 60 15 2. 73 7.09 44. 20 1530 216O 1850 320 50 470 360 60 20 2. 57 6.60 44.20 1430 204O 1750 320 50 470 360 60 23 2. 42 5.83 4O. 51 1380 1980 1670 290 50 47O 360 60 26 2. 23 4.96 33.99 134O 1930 1640 300 50 470 360 60 29 2.00 4.OO 28. 30 1310 1875 1580 270 50 470 360 60 31 1. 83 3.34 24. Ol 127O 1840 1560 290 50 470 360 60 33 1.62 2. 62 19.01 1260 160O 1530 270 50 470 360 60 35 1.33 1.79 13.32 1350 1780 15OO 300 50 470 360 60 1 cV 2 R :' Re- V V 2 W 2 p a Pa' P 3845p R coil e 10* X 1. 36 1. 15 1.32 10. 24 1180 I76o 148O 300 57 474 361 60 3*7 .96 .92. 7.29 1150 1750 148O 330 61 478 364 34 38 .72 . 52 4. 2O 1130 174O 14*70 340 63 480 365 25 39 .35 .12 1. OO 1100 1720 146O 360 66 483 368 8 40 . 29 . 095 .72 1080 1710 1455 375 69 486 370 5 41 . 25 . 08 .70 1050 I7oo 1450 400 74 491 374 26 DESIGN LAYOUT OF COUHTER RECOIL. Divide the total length of recoil into four parts: LQ = from velocity of counter recoil to max. velocity fa) L O -LQ L o = one-half of counter recoil distance (b) = last half of counter recoil distance - 6 inches (c) = last 6 inches of counter recoil (d) 44 cV 1 dv a w x dx (a) Assume a max. velocity of 2.95 ft/sec. When velocity is max. V -^ = dx cV* .' . (Pi ) A- 2R=0 W P . -ll - 5 a W 2 w o cV 2 2130 = 370 175 x W* Wx (i x4. 602x1. 314*2. 93) 2 _9 _ W2 (5. 303x13. 23x41. 95) 2 W Q =.00669 sq.in. from drawings W Q - .00665 Deviation .45* (b) Orifice is constant during this period W = .00665 cV 2 dv (P - -)A-2F=mV - equation of forces with one W dx unknown velocity Find V at "L o " (c) Calculate the unbalanced force at the beginning of the interval. This force is constant up to six inches from the bettery position. cV 2 dv (Pi - - )A-2R = mV - x i :"vV mV dv = f6 dx K " V - 32.6 32.6 f.. = /V 2 - . 307 A 45 C'Re- coil x" .3x V-.3x n v r V r diff . H 20. 5 6.44 2.54 2.54 00 23 2.5 .n 5. 64 2. 38 2.42 . O4 1.65 26 5.5 1.6-7 4. "77 2.19 2. 23 . 04 1.79 29 8.5 2. 60 3. 84 1.96 2. 00 . 04 2. 00 31 10.5 3. 22 3. 22 1. 8O 1.83 .03 1.66 33 12.5 3.83 2. 61 1.61 1.62 .01 . 62 35 14.5 4.5 1. 92 1.3*7 1.35 . 02 1.43 36 15.5 1. 14 1. 15 . 01 .8*7 (d) Velocity at the beginning of the period = V =1.14 -V if 2L V m Lra ' ( - 2x) 1 m IK V? - m = 82 72 = v, 2 - - (2x - 7,)= V - 2.51(2x - 24 x 2 ) 1 46 V 2 - 2.51(7 *"- - 0167 x *> = V* - 2. 51 (.167-. 0167) 1 g * = V* -.39 C.Rj X V 2 design drawing V -V ^Deviate V y 36 1 1.31 1. 15 1.15 o 37 1 .92 .96 .96 38 1 .53 .73 .72 .01 1,37 39 1 . 14 .37 .35 . 02 5. 40 40 1 o .09 .09 41 1 .08 . 08 Assuming orifice curve a parabola y 2 =Kx X *X W X Orifice by para- bola Orifices drawing a-b ^Deviation 1 8. 8< 2.98 . 0014*7 .00125 . OOO 22 16 2 17.65 4. 20 .00210 .00206 .00004 2 4 35. 3 5.95 . 00297 . 00320 . OOO23 7 6 53. O 1 ? 7. 30 . 00365 . OO410 . OOO45 11 8 70. 7 8.41 . OO420 .00465 .OO045 10 10 88. 45 9.45 . 0047O . 00513 . 00043 8 3 ? 106. 13 10. 30 . 00515 .00550 . 00045 8 14 123.82 11. 11 . 00555 .OO583 . OOO28 5 16 141. 51 11. 89 . 00594 . 00612 . OOO18 3 18 159. 20 12. 6l . 00630 . 00640 . OO010 1.5 20 176.89 13.30 . 00665 . 00665 . 00000 I I I W) L RECOIL SYSTEM WITH VARIABLE RBCOIL- IHPEPEHDBNT RECUPERATOR. In a mechanism of the type now to be discussed the brake cylinder contains a buffer rod fixed to the recoil cylinder with grooves of varying depth for throttling during the main recoil through the recoil piston. At the end of the buffer rod is a spear buffer which enters towards the end of counter recoil a smaller buffer chamber in the pis- ton rod, and brings the recoiling mass to rest at the end of counter recoil. The recuperator con- sists of two cylinders, the recuperator cylinder communicating through a passageway of suitable cross section to the air cylinder. The air cylinder contains a floating piston which separates the oil and air. A valve controlled by a weak spring in the oil end of the air cylinder opens wide during the recoil and thus the throttling through it during the recoil is negligible. In the counter recoil, the valve is closed by its spring and the return flow of the oil, and throttling takes place through small holes in the valve, thus lowering the pressure con- siderably below that of the air in the recuperator cylinder. During a greater part of the counter re- coil, the air pressure in the recuperator is just sufficient to balance the total friction. Towards the end of counter recoil, the recoiling mass is brought to rest by the reaction of the buffer in the hydraulic braking cylinder. EXAMPLE AND CALCULATION. As an example of a suc- cessful model, a com- plete calculation will be given. GEMERAL DATA. A|J = area of bore 29.2 sq.in. w = weight of projectile 96.1 Ibs. w r = weight of recoiling parts 9124 Ibs. w = weight of powder charge 26 Ibs. u = Pbm b = - v = S v Ph Pai Paf Vi Vf A = 49 travel of shot in bore 185.68 (in.) = max. powder pressure on breech 31,500 Ibs/ sq.in, height of center of gravity of W r from ground 5.7 ft. length of recoil 70.78 ft. angle of elevation O c velocity of projectile in bore = volume of powder chamber 1334 cu.in. = hydraulic pressure Ibs/sq.in. = total hydraulic pull Ibs. = initial air pressure Ibs/sq.in. = final air pressure Ibs/sq.in. = initial air volume cu. in. final air volume cu. in. effective area of recoil piston 190 0.7854 25.4 25.4 = 0.7854(56.955-18.755) 29.22 sq.in. A r = effective area of recuperator piston 82 a 40 .* '25.4' 25.4 = 0.7854(10.427-2.48) 6.24 sq.in. A a = area of air cylinder or floating piston 137 = 0.7854( ) 22.85 sq.in. 25.4 Free velocity curve By Le Due's formula Velocity of projectile in bore V = where u" b+u u = - = travel of projectile in bore - in ft. J C 50 a = 6823 x A A = w*27.68 a = log 26x27.68 1334 .5395 6823X.5395 vx.c a = log 6823 + ~ (log 5395- log 1000)+ J j 26- log 96.1) = 3.83398+ (3.73199-4.00000) + 7(1. 41497-1, 98272) = 3.83 398 + . 31100-. 33333 +. 707485- . 99136 = 3.52778 a ^ 3370 4W 1.12a 2 _ 4x91.1x1.12x3371 "27 X gA g P m "27x32.2x29.2247x31500 3371x154733 52161 = 6.1124- = 2416 ft/sec. 6.1124+15.4733 21.5857 Vf = free velocity of recoiling parts = V n (W+.5) _ P = free displacement = W, = .01196 V = .01288u u u au b + u V j_ . 01196? .01288V in. 1 12 3370 7. 1124 474 5.67 . 155 2 24 6740 8. 1224 832 9.95 . 294 3 36 10110 9. 1124 1310 13. 28 .464 4 48 13460 10. 1124 1330 15.91 . 618 5 60 16820 11. 1124 1512 17.99 .774 6 72 20200 12. 1124 1667 19. 84 .928 8 96 26900 14. 1124 1900 12. 61 1. 235 10 120 33700 16. 1124 2090 24. 87 1.545 12 144 40400 18. 1124 2230 26. 54 1. 854 13 156 43800 19. 1124 2290 27. 25 2. 010 14 168 4720O 20. 1124 2340 27. 85 2. l6l 15 180 50550 21. 1124 2390 28. 44 2.32 15.473 186 52150 21. 5854 241C 28. 69 2.392 51 CALCULATION OF S ANP T. '=t +t lQ B here t o = time, shot to muzzle t = time, gas expansion *o a u 16.473 V t f " * ' 2416 " ' 63 seconds ~ where V fjB = velocity of free P t S g recoil, mas. V Q * velocity of free recoil when shot is leaving muzzle W f = weight of recoil- ing parts P Q k = muzzle pressure at base of breech S ' = cross section of the bore in sq. in. W V-+4700 w = _ 96.1x2416*4700x26 . max - = - = 38.78 ft/sec. W 9124 2416(96.140.5x26) = 28.83 ft/sec. w r max. pressure on breech = 1.12 P m = 35,300 Ibs/ sq. in. mean coast pressure on breech W V 2 - = 19200 Ibs/sq.in. 64.4 . . .., (S !l - I). /! - *1 !)' - 1 ] = 67.38 16 P e 16 P e 27 u" 27 r--~ c 185.68x35300 p * e* PV x 57.38 x ob 4 (e+u") 4 (57375+185.68)* = 10141 Ibs/sq.in. p ob* A a x 8 2(38.78-28.83)9124 10.141*29.225*32.2 ^ .01899 T = i + t t = .009631 +.01 899= .0286 seconds. E = Xf + Xf i Q = Sf while shot travels to nuzzle + Sf gas cap. period. w+.05w 96.1+13 )=.1826 ft. +v o*. - 32. 2*29. 22x1 01 4Qx. pi 899 3.X9124 x + 28.83 x .01899 = .5476 ft. E = .1826 + .5476 = .7308 t = |(2.3 log Z* + Jl + 2.098) -c D D 6>1124 ' (2.3 log 0.327 u + " +2.098) 3370 6.1124 t = .00414 log 0.327 u + .000299 n + .00378 time up to (shot to muzzle) w ^ * f~ f\ ' ^ i = = will give time correspond- ^O P x^"xt$ a r ob ^a * ing to V f during gas ex- pansion period to = - .00412 0.0277 u" 0.0000249 u" 0.00378 BAG Xf q" 0.0277u" log A B c t 354 36 i. .0009864 . 00468 .566 48 1.3296 . 12385 . OO0513 .O011952 .00549 .920 78 2. 1606 .33465 .001385 .001942 . 00*710 1.345 114 3. 15*78 . 49941 .002068 .002839 .00869 1. 628 138 3. 8226 .58240 .002411 . 003436 . 00963 1. 840 156 4. 3212 .63558 .002631 .003884 .01049 2.053 m 4. 8198 . 68305 . OO2828 .004333 .01094 2. 191 185. 65 5.1425 . *71122 . OO2944 . OO4623 .01134 53 2W, (V f -V )= ,001912(V f -V ) V f V Vf-V *. t 3. 00 31. 30 2. 47 .00472 .01606 4.OO 33. 66 28. 33 4.83 . O0924 .02058 5- OO 35.40 6.57 . 01256 .02390 6.00 36.85 8.02 .01534 . 02668 "7.00 3*7.81 8.98 .01717 .02851 8.00 38.55 9.72 .01859 ,02993 6. 76 38. 68 9.85 .01883 .0301*7 Resistance to recoil. ^ Slope of the stability curve = - = R - r where b = length of recoil h = height of center of gravity of recoiling parts from the ground (5.70) W r = weight of the recoiling parts. R = max. resistance to recoil in battery _ r = max. resistance to recoil out of battery W r b 9124*70.78 R-r=- = 9450 h 68.4 Total variable resistance to recoil = K = * max 2b R-r b-E*V f T- - (b-E) T g I HrH K = t x 38.78 + 800(5.9058-. 730) 5 . 9058- . 730+3 8 . 78* . 0286-1 600 (5 . 9058- . 730 ) (.0286)' 2 32,2 K = 37200 Ibs. in battery. 54 R r KT 8 K 1 out battery K - :( ) (b-E+ ) =37200-1600(5.9058 2M r -.730+. 0538) K 1 = 29600 Ibs. out battery Construct a curve showing K at each point of recoil. OALCPHTIOJ OF AIR PRESSURE. A = effective area of recuperator piston V^ = initial air volume 1620 sq.in. Vf = final air volume = V^-A X p a i = initial air pressure = 1420 Ibs/sq.in. af final air pressure = Pi ( ) 1420 1620 1 .9 1620-6.25 log P af = log 1420+1. 3[log 1620-log (1620-6. 25x)] log P af = log 1420+1. 3[log 1620 og (1620-6. 26 x)] P af = 1420 log 1.3 e Re- 6. 25 1620-6^ c leg P .f o ' r e- o e 1 1 X I 1.13* ooil X log o e 1.3* 1 6 1614 3. 20790 .OO162 .00211 1.0049 143O 7o.7e f> 31 1589 3.20112 . OO84O .01092 1.0255 1460 65.78 10 63 1557 3. 19229 .01723 .02240 1.0529 150O 6o.7s 15 93 1527 3. 18384 . O2568 .03338 1.0799 1530 55.78 20 124 1496 3.17493 .03459 .04497 1. 1090 1580 6o.78 25 155 1465 3. 16584 .04368 .05678 1. 1405 1620 45.78 30 186 1434 3. 15655 .05297 . 06886 1. 1750 1670 40. 78 35 217 1403 3. 147 06 . 06246 .08120 1. 2055 1710 35.78 40 248 1 372 3.13735 .07217 .09382 1. 244O I77o 30.78 45 279 1344 3.12743 . 08209 . 10672 1. 2788 1820 25.78 50 313 1 307 3. 11628 . O9324 . 12121 1. 322O 188O 20.78 55 341 1279 3. 10687 . 10265 . 13345 1.3595 1930 15.78 60 375 1245 3.09517 .11435 . 14866 1. 4080 2OOO 10.78 65 406 1214 3.08422 . 12530 .16289 1.4550 2 >7o 5.78 68 425 1195 3. 07 I 73'7 . 1J215 .17180 1.4850 211-> 2.78 "70 438 1182 3. 0"7262 . 13690 .1*7 1 79*7 1.5 )6g 2140 .78 70. 78 439 1181 3. 07225 . 13*729 .1*7845 1.5^80 214O o Velocity of retarded recoil. During T V r =Vf -t where K = retarding or breaking force constant, during T. = 37800 Ibs. V r - V f = 37200x32.2 - t = V f = 131.28 t v X r =Xf = - t 2 where X r = retarded displacement &M* 37200x32.2 66 _ x9124 t t 2 *f 65.64t* *r V f 131. 28f V r . 01085 . 000122 2. 19 .008 2. 1 28.83 1.32 27.51 . 0128 .000164 2.954 . oio75 2. 83 31.5 l.6e 29.82 . 0148 . 00022 4. "734 . 01445 4. 56 35. 1.94 33.06 . 02984 . 00089 8.76 . 0585 8.06 38.78 3.91 34.87 After T, R-r K = K + (b-x) = 29600*1600(5. 9058-x) = 39 060 A b -1600 x -J V - 37200+ K, (x-. 737) 283 56 x" X 1 K x K**X n r X-.73 J *5 v r 9 .75 36860 265 .02 5. 30 1215 34. 9< 19 1. 585 36520 261 . 855 223 99*7 31.6 29 2.418 35200 256 1. 688 432 "788 28. 39 3.25 33860 251 2.52 633 587 24. 2 49 4.O8 32530 246 3.35 825 395 19.9 59 4.91 33 21O 242 4. 18 1010 210 14.5 69 5.75 29860 23*7 5.02 1190 30 5.4 70. "78 5.9 29610 236 5.17 122O C A V, 13.22 C = 1.39 A = 29.22 *r V r Ph v^ 13.22/P^" CAV r w 2. 1 27.51 955 30.9 408 1110 2.72 2. 825 29. 82 950 30. 8 407 12OO 2.95 4. 561 33.06 9*5 30.78 406 1330 3. 28 8. O6 34.87 940 30. 62 405 1405 3.47 9. 34.90 936 30. 60 404 1410 3.49 19. 31. 60 885 29. 78 393 1275 3. 24 29. 28. 815 28. 50 377 1130 3. 39. 24. 2 738 27. 2 360 975 2.71 49. 19.9 662 25. 8 340 8O3 2.36 59. 14.5 577 24 318 585 1.74 69. 5.4 492 22. 2 394 221 .752 70. 87 o 477 21.82 289 As the total oil displaced by the recoil piston does not pass through the recoil throttling grooves, a part passing into the buffer chamber the above throttling formula, is not exact and becomes C(AV r -Qb) w = _____ where Qb= the quantity of 13.22/P n oil passing into the buffer chamber. With this value of w. C should be given a value of about 1.5. 57 FRICTION CAI/CULATIOH. Floating piston friction = -a :137&* a .069 o ^r in ii TS ^37&-*.l37S*~ 66CX)7 Rr = K A P f P = air pressure (ibs/sq.in.) belv. spring pressure 42F= load 7700 Ibs. 7700 .7854(5.38 - 4.37 ) = 1000 Ibs/sq.in. P = belv. spring load 42E= 6600 lbs.= 860 Ibs/sq.in. K = poisson ratio f = coefficient of friction A = area of frictional surface P t = intensity of pressure 1 to A R f = d[(bf + af )P t + 7 af'P fl ]K+ ivdl(bf + af ' )P'+7af' P tf ]K A w z 5 ^25 d = 5.4 in. b = .197 in. a = .1376 in. K = poisson ratio = .73 P t = air pressure + blv. spring pressure (P a + P o ) = p a + - 1000 Ibs/sq.in. a 3 P' = oil pressure + blv. spring pressure (P^*P0) = P, I 86 = spiral spring pressure + P & - a* .7854x5.4 58 = 130 + P a Ibs/sq.in. Assume P a =P a R f = 3.1416x5.4[(.197x.05+.1376x.09)(P a +668)+.069 *. 9(130+P a )]. 73+3. 1416x5. 4[(.197x.05x. 1376 x.09)CP a +573)+.069x.09(130+P a )].73 = 16. 95[ ( .00986 + .01238)(P a +668)+.0062(130+P a )] x .73 + 36.95 L ( .00986+ .01238) (P a +573 )+ . 0062 (130+P a )] x . 73 12.38(.02224P a +l48.4+. 8 .0062P a )+ 12 .38C.02224P. +127+.8+.0062P_) 12.38(.02844P a +l48.4+.02844P a +127) R f = .352P a +3430 Ibs. RBCUPIRATOP PISTOK HEAD FRICTIOK R .1375 J375J / a lit a b f o Sf ^NO FRICTiON 2 e)60 * " - ^ Z37 < i R t - K A P f = d(af f +bf )(P+P )K = tr 3.225C.1375 x.09+.237x.05)(980+P).73 = 10.14* .0242(980+p).73 0.18 P +175 Ibs. 59 RECUPERATOR PISTON ROD PACKING FRICTION .1375 J l 6160 .156 1 -a R 3 = rr x 1.57(.1375x.09+.158*.05)(860+p 6 ). = 4. 94x. 0203(860 + p).73 = 63*. 073 p R 3 = 60+. 073 p a Ibs, RECOIL BRAKE STUFFING BOX TRICTIOH H. .1375 ,1375 J375 m a a * 1 a ' \ } ^ ' b K ^ 60* i AS 4-/D' ii o A D = .7854(32.85-19.4) = 10.6 4150 30.6 R = n x 4.4K- .1375*.09+.158x.05)(394+P e ).73 4-2 o 60 = 10. IK. 0186 +.0079)( 394+pg) R 4 = 105 + .268 p h Ibs. GUIDE FRICTION - R Rg = .15W r = 9124*. 15= 1370 Ibs. R = 1370 FORCES ACTING CORING RECOIL B = total braking pull = Ph+P a P^ = hydraulic pull P a = load on air recuperator pull P a p o =*=*= ARE1A Of PISTON A u II Fig. A. R g GUIDE FRICTION I p' = (F a + ) = intensity of recuperator pres- sure, Ibs/sq.in. R + = friction = R +F + R, + Rg T. 2346 +R R f = 0.18 R* = .073P a +60 R 4 = .268P h +105 Rg = 1370 2R=R t - .253P a +.268P n = 1710 I Ibs. 61 p Ph - A+.268 1710 26.268 26.268 22.85x26.268 26.268 R f .0381 p - 0.2472 p a - .2376 -- 65 22.9 but R t = .352 p a +3430 Rf - = .0154 p a + 150 hence p h = .0381 p- 0.2626 p a - 215 (Ibs/sq.in) P = K+W r sin ft P h = .0381 P - .2626 P a - 215 Re- P .0381P Pa .2626P a P h soil i 37200 1417 1430 376 826 5 37200 1417 1460 383 819 10 37ooo 1410 1500 394 801 15 36400 1387 1530 401 771 20 35800 1364 1580 415 734 25 35200 1341 1620 425 7oi 30 34600 1318 I67o 439 664 35 34000 1295 1710 450 630 40 33500 1276 1770 465 596 45 32900 1253 1820 478 560 50 32300 1231 1880 494 522 55 3l7oo 1208 1930 506 487 60 31200 1189 2OOO 525 449 62 65 30600 1166 2O70 544 407 6s 30200 1152 2110 555 382 70 30000 1143 2140 563 366 7b. 7fi 29600 1128 2140 563 352 R g = ZF = fa * 975) .180 p a + 175 .073 p a + 60 105+.268 p h 1370 .263 p a * .268 p h + 2685 Re- coil .263p a .268ph R phA A Pa 5 380 540 3605 21250 9130 15 400 540 3625 20000 9550 25 425 530 3640 18200 10110 35 45 450 478 530 520 3665 3683 16350 1455C 10700 138OO 55 510 520 3715 12650 12050 65 545 520 3750 1060O 12920 10 560 510 3755 9530 13380 70. 8 1 ? 560 510 3750 9160 13380 COUNTER RECOIL CALCULATIONS. Oil Ports Recuperator valve oil ports: see fig. (A) Page 59, Ports in 58B1 = 2*. 7854 x *x = .03896 sq.in 63 m m ^~ m 92'2 HI * - - ; KV* - (.253p. + 100 + 1710)= m v V ~ 68 .352 p a +3430 - 1810 KV dv 6.25 p a -.0963 p a -938 - .253 p a - 1810 = m y V ~ KV 2 d? 5.9 p a 5 275C = m v V - o KV 2 i 5.9 p fl dx -~ dx - 2750 dx = -m(V 2 V a ) W SI 2x5.9P a dx ^2_K7^ 2x2750 dx 284 " 284 w 2 dX " 284 =V 2 -V 2 .0416 P a dx -- V 2 +V 2 dx - 19.35 dx = V - V ^"o .6064 dx V 2 .6064 dx V 2 .0416F_ dx -- " -- i- - 19.35 dx=(V-V ) 10w 2 + )= . ^ _ 18>35 10W 10W 2 W = .0389 = .001513 V|(l + .334 dx") = V 2 (l - .334 dx") + .00347 p a dx - 1.611 dx" 69 C'Fe- dx" Pa .334dx .00347PJ .00347p a dx l.Sllax V .25 .25 2. 160 .0835 7.5 1.87 402 1. 165 .50 .25 2. 150 .0835 7.45 1.865 . 402 1.58 1. . 50 2. 145 .167 7.43 3.72 .605 2.07 2. 1. 2. 125 .33* 7.37 7.37 1. 611 2.54 4. 2. 2. 100 .67 7.27 14.27 3. 222 2. 84 6. 2. 2. 060 .67 7.15 14.30 3. 222 2. 87 10. 4. 2.OOO 1.3 3 6.94 27.76 6.45 2.82 15. 5. 1.950 1.67 6.76 33.80 8.05 2.77 20. 5. 1. 90O 1.67 6.6 33. 8. 05 2.72 25. 5. 1. 845 1.67 6.4 32. 8. 05 2.66 30. 5. 1. 800 1.67 6. 25 31. 25 8.05 2.63 35. 5. 1. 730 1.67 6. 30. 6.05 2.55 40. 5. 1. 700 1.67 5.9 29.5 8.O5 2.53 44. 4. 1. 6*70 1.35 5.72 23.19 6.45 2.51 48. 4. 1. 600 1.33 5.55 22.8 6.45 2.48 51. 3. 1. 600 1. 5.47 16. 64 4.83 2.43 V* 2 (l+.334dx n ) . Cl- .334 dx") .00347p a dx"-1.611dx" C' Re- coil dx" a b Pa c d I V s V 51 3 200 1600 16. 64 11. 80 11. 80 5.90 2.43 54 3 200 1580 16.45 11. 61 11. 60 5. 8C 2.41 56 2 1. 668 .332 1560 10. 81 7.59 9.52 5.71 2.39 58 2 1.668 .332 1540 10.67 7.45 9.35 5. 6c 2.36! 59 1 1.334 .666 1530 5.32 3.71 7.44 5.57 2.36 BUFFER IK ACTION - RETARDED COUNTER RECOIL. The dynamic equation during the buffer action period, becomes R f K*A$ v* mm KAgv dv Kv : K v ; I i\v i Simplifying, we have (p ~ T~) A v~ "Is It fl w w dv 70 Since ZR=253 p 8 +100+17lO we have .352p a +3430 KV K t V a Pa A v A v ~ 253 p a ~1810= 22.85 - -* dv dx . m r V KV* t dv 5-9 p a - - --- = 2750 - r V w o x 2 *6- 9 P a 2 KV 2 2 K , v * 2x2750 - dx --- dx --- l dx - = dx =V 2 -V J 284 m w* m w 2 284 o K .041p a dx - ID x V 2 + V 2 dx V a +V 2 dx - W 2 * m 2 " 2x2750 284 dx - V 2 - V s 6064 6064 VV + V 2 ) .041 Pa dx -- dxV 2 -- dx V - -2-Z - 1 dx 1C 6 W| 10'W 2 raw 2 = IS. 35 dx - V 2 - V 2 3 106 6064 6064 3804V 2 0416P a dx -- V'dx -- dx -- ^- dx - 10w 2 1C'* 2 10w 2 3804V 2 ... - 1 dx - 19.35 dx = V 2 -V* 10 e W 2 2 1 + .0416 p. dx -19.35 dx 71 00 ro X -o Cu 8 O CO CO X a M n > O CO o o o CO MX * X pt cv> I CfS .- CM m CM i-i o m oo con r~ NO ON n ON ^* O"* I" 4 NQ ^" ^* * K1O CM ON in CM o>- * ^ON CM o QO x*- o ^"* irtM o ON to H i-t o NO * o o o m o *^mO in o o in o CMOio, o CM * * ,,>*.. . . . . IO IO i-l ON ON to n o ON m m H H NO > \o ON^~ oo or NO \o oo OD <* CM CM in^ 3 ooo NO o CM * * * . ri ri H i-**"* I-KM i-i CM CM m in rt oo r^ NO ON tO ^ ^ Id in \o GOO IO T-l O ON ON . * . rH H CM in X 1 - X^ H i-l x- i 1 to os-ito** m m in o o 1-4 ri m oo * * 000 OOO Oo O O O O O CO CO l-M CM 1-l| CM i-1J * OC -H - o O 72 _ K AV 1.11 x 6.24 178* 38 1.23*36.94 48.07 ' 2646 175*' 175x6. 24w ZF .352p +3430 .253p a +1810 * 22.66 6.24 .0154p a +150.11+.0405p a + 280 = .056p a +440 .352p a +3430 .253p a +.268 +1710 22.86 6.24 .0154p a +150 + 043 p h 0483p a + 275 .0637p a +.043p h +430 C 1 Re- coil ?a Ph 063 7p 043 Ph i 2150 830 138 36 604 2 2130 825 136 36 604 4 2100 820 135 35 600 6 2060 815 132 35 ^00 8 2030 810 130 35 500 12 1980 800 12"7 34 500 20 1900 V30 122 31 583 30 1800 660 116 28 5*74 40 1*700 600 109 26 565 50 1580 520 101 22 553 60 1510 450 97 20 5<7 65 1480 410 95 IV 542 68 1450 380 93 16 539 70 1440 3*70 92 16 538 73 COMPONENT FORCES C'Fe- coil p a V 2 KV* Ph-KV* W|10 K t V .5 2150 2.50 4. 20 i7oo i. 2150 4. 28 790 1360 2. 2125 6.44 1180 950 4. 2100 8. 15 1500 600 6. 2060 8. 24 1510 550 10. 2000 8.01 1470 520 15 1950 7.70 1420 530 to 1900 7.45 1370 530 25 30 35 40 1845 1800 ilio 1700 7. 20 6,00 **ii 6.43 1320 1270 \m 530 530 IK 45 6. 20 50 1580 6.00 1040 480 55 5. 74 60 1500 4.53 830 67o 2350 320 350 62 1490 3.72 690 800 1560 420 380 64 1460 2. 86 530 930 930 540 390 66 1450 1.82 340 1110 484 660 45C 68 1440 .8"? 160 1280 175 870 410 69 143O . 42 80 1350 72 1010 340 CONSIDERATION OF VARIOUS ELEMENTS OF DESIGN - HYDRAULIC BRAKE LATOUT. In the type discussed, the throttling is ef- fected through grooves in the counter recoil buffer rod, there being two sets of grooves, one for horizontal elevation and long recoil and the other for max. elevation and short recoil. The recoil throttling is not effected materially by the filling in buffer since the filling in process is not continuous throughout the recoil. Hence the throttling grooves are computed in the ordinary manner. The maximum throttling groove occurs at horizontal recoil near the be- ginning of recoil. max the maximum allowable hydraulic pres- sure (assume 400C Ibs/sq.in) 74 C = the reciprocal of the contraction factor of the throttling orifice (assume C = 1.38) V r * the max. restrained velocity of recoil (assume V r = 0.9 V f )(ft/sec) wv+4700w Vf = = max. free velocity of recoil " r (ft/sec) where w = weight of projectile (Ibs) w = weight of powder charge (Ibs) w r = weight of recoiling parts (Ibs) v = muzzle velocity of projectile (ft/sec) K = total resistance to recoil at horizontal or min, elevation (resistance to recoil should follow stability slope ) (Ibs) K s = total resistance to recoil at max. elevation (Ibs) (made constant throughout the recoil) P = total hydraulic pull (Ibs) h F v ^ = initial recuperator reaction (Ibs) F y f = final recuperator reaction (Ibs) Fvf ~ = ratio of compression (1.5) F vi n coefficient of friction (0.2 to C.3) Then F y j = 1.3W r (sin /#* n cos /$) the max. hydraulic pull at short recoil, max. elevation. P hm = K s +w r (sin m ~O.S cos rn H p v i (Ibs) the irax. hydraulic pull at long recoil, horizontal elevation., P = K -0.3W -F.i (Ibs) ho h The required effective area of the recoil piston then becomes, p ho A = sq.an. Hi max Hence the roax. throttling orifice, becoves, C A V r w = (sq.in) h maX 13.2 75 and the minimum throttling orifice, becomes, C A V r W = (sq.in) n mm - ,,, ax The throttling areas decrease approximately in the recoil as a parabolic curve. For a satisfactory layout, for the cross section of the brake cylinder the inside cross section of the hollow recoil rod should be made approx. three times the maximum throttling orifice, that is three times the max. area of the four grooves in the counter recoil rod. Hence tne inside diam. of the hollow piston rod, becomes, / h max in. 0.7854 The diam. of the spear buffer chamber in the hollow recoil piston rod, is obtained by a consideration of the counter recoil stability and the maximum allowable buffer pressure. The maximum buffer reaction on counter recoil becomes, W s^s B ' = F vi +C s - ' 3 W r where F v j = the initial recuperator reaction (Ibs) Cg - constant of counter recoil stability = 0.85 W s = weight of entire system (Ibs) W r = weight of recoiling parts (Ibs) lg = horizontal distance from V & to wheel and ground contact (in) The max. allowable buffer pressure, p b ' m = 1500 to 2500 Ibs/sq.in. Ha nee the area of the spear buf- fer chamber in the recoil rod, becomes, B 1 / * (sq.in) Pb'm When counter recoil stability is not of con- sideration it is customary to make the buffer stroke 76 about 17.6* of the length of horizontal recoil. Assuming an entrance velocity into the buffer at 2.5 ft/sec., we have, f nn r v* B' * F v i * where v = 2.5 ft/sec. 0.175 b h =F vi 0.357 W r where W r = weight of the recoil- ing parts, and the area of the spear buffer becomes. A further consideration requires the diam. of the buffer to be approximately a given ratio of the inside diam. of the recoil rod, in order that a proper layout for the filling in holes to the buffer may be made. Then, d ib 2 , = - or d ib = - d where 0.7864 d* b = A b The proper thickness of the buffer rod is based on a consideration of both the maximum longitudinal tension during the acceleration or powder period, and the compression stress due to an external pressure on a hollow cylinder, the internal pressure being zero. The max. allowable fibre stress in the brake rod will be taken, f ffi = - to - elastic limit (Ibs/sq.in) The outside diam. of the rod based on max. tension becomes 0.7854 where d_ = outside diam. of rod inside diam. of rod weight of piston + that portion of the rod (estimated )to the farthest hollow section from the piston. weight of recoiling parts 77 P = max. powder reaction on breech b P = max. hydraulic braking. h W P To allow for the acceleration load (P - B) we may w r b increase P to 1.3 P , then ~~ ~~~~ which is sufficiently close in a preliminary lay- out. If we now consider the compression due to the hollow piston rod with an external com- pressive pressure, we have, f m (d -d.j)= 1* d i approx. hence d Q = - where p bm = max. pres- 1 -3phm sure in the brake f m cylinder (Ibs/sq.in) The diam. of the recoil brake cylinder may now be calculated. We have 0.785 d* = A+0.785 d* here -0.3 cos A = Ph max. K S = total resistance to recoil at max. ele- vation. Ph max ~ max - allowable pressure in recoil cylinder consistent with packing = 4000 Ibs/sq.in. RECUPERATOR LAYOUT. (1 ) Ratio of compression m : The ratio of compression is based primarily on a consideration of proper counter recoil functioning. A high ratio of compression decreases the size of the recuperator and accelerates the counter recoil at max. elevation, while on the other hand a large value cf m causes undue heating which is injuraous to the packing in 78 the floating piston, and further limits the horizontal stability. m should be taken at from, 1.5 to 1.7 (2) Ratio of recuperator cylinders: If A y = effective area of recuperator piston (sq.in) .A a * cross section of air cylinder (sq.in) Paf n = * ratio of compression = 1.5 to 1.7 Pai A a r = = ratio of recuperator cylinder A v 1 = length of air column in terns of cross section area of air cylinder. j = - = length of air column in terms of re- b coil stroke Then, the initial volume, i k V. = A v b - where k = 1.3 a t and since = r = * = J .' * Let j 1 approx. hence r = = - To compute A V we have F vi * 1.3W r 1.3 W r hence A^ = (sin^-m cos# n ) sq.in. p ai where P ai * 1500 Ibs/sq.in. (approx.) Then A a = A y r sq.in. A further limitation for r, is that y r = - where V, nax = max. vel. of recoil 1 i 79 12 ft/sec. = max. allowable vol. of floating 1 piston LAYOUT OF CALCDLATIOI8 FOR COUHT1H RECOIL. (l ) During the constant orifice period, B f c'Av dv A v < Pa ) ZR* r v A a 176 dx where A V = effective area of the recuperator piston (sq.in) p * air pressure (ibs/sq.in) Rf floating piston friction (Ibs) A a = area of cross section of air cylinder (sq.in) C = reciprocal of contraction factor of constant orifice = - w = area of constant orifice in recuperator system (sq.in) 2R = total friction - guide + packing (Ibs) The max. velocity counter recoil is attained shortly after the beginning of counter recoil. If p a f = final air pressure (Ibs/sq.in) p a j, = initial air pressure (Ibs/sq.in) Then Pa - p ai +0.9( Paf -p ai ) p ai = (9m+l)=0.95 m p a j approx. 10 Paf where m = = 1.5 to 1.7 dv At max. vel. m r v =0 Assuming a max. velocity ^ x at horizontal recoil = 3 ft/sec. CAv 8 R f CAv - J = A v^Pa ^ ~ 2R and ^o = - 176w A a R f 175tA v (p a - -p)-2 A a 80 hence w o = . sq.in, 13 f~ R f .2/A v (p a - )-] The constant orifice should be designed for counter recoil at horizontal or minimum elevation. As a numerical illustration, the value of the constant orifice in one model has been worked out as follows: Rf yua (P a - )A V - = 2R A W 1 *a w o .352P a +3430 K t AS v* ( p )6.24 - =.253 p a + 1810 22.88 175W 6.24P a -.0961P a - 936.68 = .253P & +1810 o 5.891P,-2746.68 = 15.4267 (P =2080 Ibs) w o 15.4267 12253.28 - 2746.68 = W o .0016 W Q = .04 sq.in. from drawings = .03896 Percentage accuracy 2.57 Compute velocity of the recoiling parts when buffer is entering V = 2.35 ft/sec. (2) Retardation period of counter re- coil. In the retardation period, we will assume a constant unbalanced force acting in the retardation. Let = total unbalanced force in counter recoil retardation (Ibs) v o = entrance velocity into buffer chamber (ft/sec) k = factor of safety 0.95 C'Agv 2 B 1 = 175 w b l b = length of buffer (ft) x b * displacement in buffer (ft) p a = pressure on oil side of floating piston in 81 air cylinder (Ibs/sq.in) p * pressure in recuperator (Ibs/sq.in) A v s effective area of recuperator piston (sq. in) C'*A 2 v 2 then J0=B'+ZR- Pv A v B'+2R -(p - -^ )A V 175w S'+ZR-p a A v 175 w a - m ' "xb w o I^r 7 ! m r Further /J = -rr Clbs) and 6 xb = r(v -v| b ) 20 x b v 2 hence v* b = v* -- = rT"^ k 1 b" x b^ Hence the m y K JLv equation of counter recoil during the retardation period, be- comes, m v Pa AV-R The only variable being w xb which therefore, becomes a function of xb as we should expect. To solve for w xb we have the following expres- sion in terms of x b , the displacement in the counter recoil in the buffer: C' 2 Af(k I b -x b )v A (kl-x) B v C'A^Vp ( k l b - x b hence W xb z ^-^ / ^ ^ 13 - 2 175 82 (sq.in) F O * o If 1 1_ R f Obviously p 1 = p a , as a function of x b and its proper value must be used for a given Computation of v_: Knowing w o from (1), we may readily calculate v o by the logarithmic method, as follows: . C' g A*(x-x) " 200 % wS where A = p a A v - 2R Only few intervals are needed, the mean air pressure being taken for these intervals. The following is an illustration of actual calculations made for the entrance into the buffer on counter recoil. LAYOUT CALCUL ATIOHS OF BUFFER. V o velocity of counter recoil when buffer enters R b c A - ~m a V* = energy to be dissipated during buffer action where b = length of buffer in inches = 12 in. R * a factor of safety .95 -a V* -284*2.38 dv jtf = - MV m dx - , . 2x73^5 x b =;(V-V) 2.35 284 65 5.67 - .517 x b C'recoil *b V V 60 1.22 2. 22 2.18 1 % 8 62 2. 1.9V 1.93 1.8 64 2. 1.69 1.69 66 2. 1.36 1.31 3. V 68 2. .89 .93 5. 70.78 2.68 FORCES (Buffer forcee) K'AgV KAV 2 (.253P a +1810)+ K'A|V ,362P a +3430 22.85 0.253P a +1810+ 15x4 " 175 - (6.24P a -0.0963P a K -938 ) 175 V*' V* 0.263P.+1810+1.0804 (6.3363P.-938 ** -. .. l.l*x 6.24** v 0.253P a + 1810+1.0804 6.3383P a +938+1132.74 V 1.0804 68.78 * - 6.0833P.+2740 + V + 1132.74 V* W* "x 1.0804 V 1 6.08P.-2570-1132.74V 8 84 C'R V V 2 1.08V 2 Pa 6.08P a 1132 747V| *i W K Actual areas . 60 2. 22 4.53 4. 90 152O 9250 5550 1030 .004*75 .0438 .0395 62 1. 98 3. 92 4. 24 1500 9130 444O 2020 .00210 . 0438 .0395 64 1. "70 2.89 3. 13 1480 9000 32*70 3000 .001O2 .0319 .0305 66 1.35 1.83 1.9*7 14*70 8940 2060 4210 . 00047 . 0217 .0213 68 .95 *. 90 .9*7 1450 8820 1020 5150 00019 .0138 .0130 FILLIKG IN OIL HOLES FROM RECOIL CYLINDER INTO BUFFER CHAMBER. Area of the filling in oil holes must l>e large enough to fill the hydraulic cylinder during recoil period . P = average pressure = average hydraulic pull = 720 Ibs/sq.in. d = weight of cubic ft. of hydraulic oil 53 Ibs/cu.f t. max maximum velocity of recoil 35 ft/sec. b = length of recoil 5.9 ft. C - contraction factor = .85 A 4 * effective area of recoil cylinder 26 B