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978-1138055315 Chapter 10 Part 1
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 […]
978-1138055315 Chapter 10 Part 2
( ) ( ) ( ) ( ) ( ) 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 […]
978-1138055315 Chapter 10 Part 3
1 ( ) += p M x LC k CT 2 ( ) ( ) ( ) 565 10204.710154.1634.810208.6 −−− =+=T Now we can write ( ) ( ) ( […]
978-1138055315 Chapter 14
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 […]
978-1138055315 Chapter 16 Part 1
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 […]
978-1138055315 Chapter 16 Part 2
( ) = =s m 461 ft m 3048.0 s ft 512,1 s V The angle needs to be put into degrees […]
978-1138055315 Chapter 18 Part 1
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. […]
978-1138055315 Chapter 18 Part 2
( )( )( ) ( )( )( ) 2 10101 490.1896.7569.4896.7 uuuup −+−= 765.11530.23765.11077.36077.36 1 2 111 +−++−= uuup Then ( )( ) 842.47607.595.0530.23607.59530.23 1 1 −=−=−= = u du dp uu the rarefaction velocity is the negative of this […]
978-1138055315 Chapter 18 Part 3
s3.698s7.343s6.354 μμμtshock =+= If we approximated it using twice the length divided by the longitudinal wave speed we would have gotten ( )( ) ( ) s2.562 s mm 941.5 […]
978-1138055315 Chapter 19
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 […]
978-1138055315 Chapter 2 Part 1
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 […]
978-1138055315 Chapter 2 Part 2
( )( ) ( ) ( )( ) ( ) = […]
978-1138055315 Chapter 20
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 20º […]
978-1138055315 Chapter 3 Part 1
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 […]
978-1138055315 Chapter 3 Part 2
Problem 6 – Verify Equation (3.2.148) is valid for any θ. Solution: We start from ( ) ( ) ( ) − ==− + x x c x x c cc dxrpAdxpAxVxV w c […]
978-1138055315 Chapter 3 Part 3
( ) ( )( ) ( )( ) ( ) 2 22 11128 2fDfDfDDftc grain −++−++= (12) ( ) ( ) ( ) ( ) ( ) 2 22222 112112128 2fDfDffDfDDftc grain −+−++−++= (13) ( ) […]
978-1138055315 Chapter 3 Part 4
( ) ( ) ( )( ) ( ) ( )( ) = + + =22 […]
978-1138055315 Chapter 3 Part 5
+ = w c p pmuzzle S 3 1 (36) Inserting the numbers yields ( ) ( )( ) = […]
978-1138055315 Chapter 4 Part 1
Problem 1 – For the state of stress below, find the principal stresses and the maximum shear stress MPa 900 0415 01520 − =σ Answer […]
978-1138055315 Chapter 4 Part 2
( ) ( ) ( ) ( ) ( ) ( ) ( ) + − =2 22 2 2 22 in25.205.3 in […]
978-1138055315 Chapter 4 Part 3
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 […]
978-1138055315 Chapter 5
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 […]
978-1138055315 Chapter 6
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 […]
978-1138055315 Chapter 7
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 10º to the horizontal (assume zero azimuth) with a velocity of 3,013 ft/s. The […]
978-1138055315 Chapter 9 Part 1
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? x y V0 = […]
978-1138055315 Chapter 9 Part 2
( ) 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) […]
978-1138055315 Chapter 9 Part 3
5699.0 2 2 0=+= DDD CCC 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 ft/s. The […]
978-1138055315 Chapter 9 Part 4
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 let’s look at the third model where […]
978-1138055315 Chapter 9 Part 5
( ) ( ) 2 2 2 2 2inlbm768.78 ft in 12ftlbm547.0 = = P I t 803.1= t The drag coefficient is 2 20 DDD […]
978-1138055315 Chapter 9 Part 6
−+ = 0 1 21 x xVR t W t t going to hit the ground (y = 0). We’ll use. […]
978-1138055315 Chapter 9 Part 7
( ) ( ) ( ) ( ) ( ) ( )( ) ( ) − + […]
978-1138055315 Chapter 9 Part 8
( ) ( ) ( ) 513.05.31sin11cos1 3== i Which gives us We determine the direction of ω through the geometry ( ) ( ) ( ) ( ) ( ) +−= […]
978-1138055315 Chapter 9 Part 9
( ) 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 […]