Ballistics: Theory and Design of Guns and Ammunition 3rd Edition

10.1 Linearized Pitching and Yawing Motions 10.1 Linearized Pitching and Yawing Motions 10.1 Linearized Pitching and Yawing Motions Problem 1 - An 155mm M549A1 Projectile has the following properties and initial conditions 876.0 26 024.0 28.4 13.0 3.0 = =+ = = = = p q p M MM l M L D C CC C C C C = = = s ft 000,3 mm155 ft lbm 0751.0 3 muzzle V d lbm96 in-lbm610,6 in-lbm5.505 2 2 = = = m I I T P At an instant in time after launch

( ) ( ) ( ) ( ) ( ) Hz.130 s rad 813 in ft 12 1 mm in 4.25 1 mm105 s ft 1500 ftlbm547.0 ftlbm377.5 019.0 2 2 min = = =p Problem 10 - A right circular cylinder is to be fired horizontally for an impact test. If the cylinder is made of steel ( = 0.283 lbm/in3) and it is 0.5 inches in diameter and 0.75 inches long.

( )+=pMxLCkCT 2 ( )( )( )565 10204.710154.1634.810208.6 =+=T Now we can write ( )+=pMxLCkCT 2 ( )( )( )565 10204.710154.1634.810208.6 =+=T Now we can write ( ) += p M x LC k CT 2 ( ) ( ) ( ) 565 10204.710154.1634.810208.6 =+=T Now we can write 1 ( ) ( ) ( ) 962.0 10498.1 10204.72 4 5 = = d S Then

14.1 Non-Linear Forces and Moments Problem 1 - If the projectile in problem 9 happens to be flying at a limit-cycle yaw of 4 degrees with a spin rate of 130 Hz and velocity 1,764 ft/s. What would the non-linear pitching moment have to be for the projectile to

16.1 Penetration and Perforation of Metals Problem 1 - A German 280 mm armor piercing projectile weighs 666 lbm and is about 34 inches in length. It strikes a British warship in the 1/2 thick vertical side plating at an angle of 12 from horizontal along the path depicted

( ) = =s m 461 ft m 3048.0 s ft 512,1 s V The angle needs to be put into degrees so we write ( ) deg214.0 MOA deg 60 1 MOA85.12 = = We shall now determine the thickness assuming the impact velocity is the limit velocity from equation (TB-34) ( ) + = s m 1sec000,4 75.0 sec 75.0 3 15.0 d t le d t m d d l V The simplest way to do this is to iterate

18.1 Shock Hugoniots 18.1 Shock Hugoniots 18.1 Shock Hugoniots Problem 1 - An experiment is set up in which a steel slab is shocked from both ends. The pressure generated in the left going shock is 20 GPa. The pressure generated in the right going shock is 10 GPa.

( )( )( ) ( )( )( ) 2 10101 490.1896.7569.4896.7 uuuup += ( )( )( ) ( )( )( )2111 0.1490.1896.70.1569.4896.7 uup += ( )( )( ) ( )( )( )2111 0.1490.1896.70.1569.4896.7 uup += ( )( )( ) ( )( )( ) 2 111 0.1490.1896.70.1569.4896.7 uup += 765.11530.23765.11077.36077.36

s3.698s7.343s6.354 tshock =+= If we approximated it using twice the length divided by the longitudinal wave speed we would have gotten ( )( ) ( ) s2.562smm941.5mm674,12 tshock = Which is different but is was not what was requested in the problem. Even though we are pretty

19.2 Taylor Angles Problem 1 - A Bangalore Torpedo was a device built by the United States during the Second World War to clear beach (or any other) obstacles. It consisted of a long tube filled with explosive that was detonated on the end. Assume that we have a

2.1 The Ideal Gas Law Problem 1 - Assume we have a quantity of 10 grams of 11.1% nitrated nitrocellulose (C6H8N2O9) and it is heated to a temperature of 1000K and changes to gas somehow without changing chemical composition. If the process takes place in an expulsion cup with

( )( ) ( ) ( )( ) ( ) = = == s m 661 lbm kg 2.2 1 lbm25.0 s mkg 800,242 lbm kg 2.2 1 lbm25.0 kJ8.242 22 2 m W Vp This is over twice the speed of sound! To get a more realistic answer lets look at the ideal gas equation of state. If the projectile doesnt move, the volume behind it is ( )

20.2 Shaped Charge Jet Penetration Problem 3 - A conical liner is to be fabricated from copper and filled with Composition A-3 as the explosive. The thickness of the liner is to be 0.10 inches. The length of the 20 conical liner is 4 inches and the charge

3.2 Lagrange Gradient Problem 1 - You are asked to analyze the pressure of a charge zero (igniter) firing in an M31 boom for a 120mm mortar projectile. You decide to examine it as a closed bomb first. Assume we have 59 g of M48 propellant (properties given below).

Problem 6 Verify Equation (3.2.148) is valid for any . Solution: We start from ( ) ( ) ( ) == + x x c x x c cc dxrpAdxpAxVxV w c w 22 1 1 3 1 2 We now substitute for r(x) using our definition ( )++=lxlxxrc This gives us ( ) ( )( )( ) ( ) ++=++=+xxccxxcccccdxlxlxpAdxlxlxpAxVxVwcw2211312 The integral on the RHS can

( ) ( )( ) ( )( ) ( ) 2 22 11128 2fDfDfDDftc grain ++++= (12) ( ) ( ) ( ) ( ) ( ) 2 22222 112112128 2fDfDffDfDDftc grain +++++= (13) ( ) ( ) 2222222222 2221212128 2DffDDfDDDffDfDDftc grain ++++++= (14) ( ) ( ) 22222 231010 2DDDffDfDtc grain ++++= (15) ( )( )( )(

( ) ( ) ( )( ) ( ) ( )( ) = + + =22 in lbf 290,25 94.12 583.0 1 94.13 583.0 1 in lbf 442,26p (31) Now we use equation (LG-177) with this number to obtain ( ) + + =lx lL pLp c c (32) Then ( ) ( ) = + + = 2 21.1 2in lbf 317,13 735.6428.38 735.670 in lbf 290,25Lp (33) The muzzle exit pressure is actually that acting on the projectile base which is obtained by

+ = w c p pmuzzle S 3 1 (36) Inserting the numbers yields ( ) ( )( ) = + =22 in lbf 904,10 in lbf 485,13 8.313 1 672,11 muzzle S p (37) To plot the pressure behind the projectile at both peak pressure and muzzle exit we must use equation (LG-59) w c x x ppp g ss 2 12 2 += (LG-59) Where xg is the position behind the projectile varying between

4.1 Stress and Strain 4.1 Stress and Strain 4.1 Stress and Strain Problem 1 - For the state of stress below, find the principal stresses and the maximum shear stress MPa 900 0415 01520 = Answer MPa57max = Solution: We can use minmaxmax 21= but we need

( ) ( ) ( ) ( ) ( ) ( ) ( ) + =2 22 2 2 22 in25.205.3 in lbf 193 in25.205.3 1 Cu =2inlbf869,1Cu The von Mises stress is then found through ( ) ( ) ( )2213232221++=E If this value is greater than our allowable the band will discard properly. For our problem we have =2inlbf869,1Cu

So there are two pressure calculations; the chamber assuming 5,000 psi and the muzzle assuming 2,500 psi with a linear taper beginning at the peak pressure location. We shall assume that there actually is an axial force (assuming the projectile gets stuck) so for each location we

5.1 Fatigue and Endurance Problem 1 - It is desired to construct a 75 mm gun for a pressure of 43,000 psi. The chamber diameter has been chosen to be 3.1 inches. If we use AISI 4340 steel with a yield strength (SY) of 100,000 psi, determine the outer

Problem 1 - You are asked to analyze the mass of the charge require to fire a 100mm recoilless system. A 15 pound projectile is to be fired at a muzzle velocity of 1,500 f/s. The propellant properties are listed below. Using Equation 3.171 you may assume a

7 Introductory Concepts Problem 1 - In a test range a 0.50 caliber M33 ball projectile has a velocity vector which is at an angle of 10 to the horizontal (assume zero azimuth) with a velocity of 3,013 ft/s. The initial pitch and yaw angles are 1.030 and 1.263,

9.1 Vacuum Trajectory Problem 1 - A target is located at 20 km. A projectile muzzle velocity is 800 m/s, assuming a vacuum trajectory, at what QE should one set the weapon to hit the target?

( ) s02344.0 s ft 1820 0 0 0x x xx x x xV x V x VV x V V V x t= === Thus our table is simply Range (yards) V (ft/s) t (seconds) 0 1820 0.000 200 1597 0.352 400 1388 0.755 600 1194 1.221 800 1015 1.766 1000 850 2.411 We actually only needed the values for 600 yards but I just copied the last table. From

5699.0 2 2 0=+= DDD CCC ( )( )09.4 4sin55.05699.0 222=+= ( )( )09.4 4sin55.05699.0 222=+= ( )( ) 09.4 4sin55.05699.0 22 2=+= DD CC Problem 14 - It is desired to develop a close protection system using a 0.50 caliber machine gun. The muzzle velocity of the weapon is 2,950

100 2 = D DD C CC Error est Putting this in a table yields V [ft/s] CDest Error2[%] 1500 0.430 -21 2000 0.322 -5.3 2500 0.258 5.9 3000 0.215 14 Finally lets look at the third model where we see that a V CK D = 3 Also aK m S k33 2 = If we make

( ) ( ) 2 2 2 2 2inlbm768.78 ft in 12ftlbm547.0 = = P I ( ) ( ) 22222inlbm288.774ftin12ftlbm377.5 ==TI The total angle of attack is 222 sincossinsin += ( ) ( ) 22222inlbm288.774ftin12ftlbm377.5 ==TI The total angle of attack is 222 sincossinsin += ( ) ( ) 2 2 2 2 2inlbm288.774 ft in 12ftlbm377.5 = = T I The total angle

+ = 0 1 21 x xVR t W t t Our table is then Finally the height of impact can be determined noting that in the initial calculation we were Range (yards) V (ft/s) t (seconds) [Vx] (ft/s) [t] (seconds) [y]R (ft) 0 2600 0 2600 0 200 2447 0.238 2451 0.238 400 2295 0.491 2302 0.490

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( )( ) ( ) ( )( ) ( )( ) =+++22sft375,1sft00s0128710s0001.0s1002440s130.0 ( ) ( ) ( ) ( ) ( ) ( )( )( ) ( ) ( )( )

( ) ( ) ( ) 513.05.31sin11cos1 3== i Which gives us 321 513.0191.0837.0 eeei ++= We need to determine the angular momentum vector so first we need the instantaneous angular velocity vector. We know that the nose is rotating up at 2 rad/s. We must define this as a

( ) xVgV V ++= p ND C V d SVmVSC dt d m 2 1 2 1 In this case x is the ball axis which will have to point up to cause the ball to move left. With that knowledge we can write three equations as follows: 2 21xDxVSCdtdVm= mgVVSCdtdVmyxDy= 21 knowledge we can write three equations as follows: 2