Question

6 Problem 2 (20 pts) The SPAD XIII was a fighter plane used by various countries, including the United States, during World War I. It had a 97 kg wooden propeller that was 2.5 m long and 15 cm wide and 9 cm thick. A Vickers 7.7 mm machine gun mounted on the front, which used a timing belt to fire between the propeller blades at a rate of 500 rounds/minute. The gun is mounted halfway between the engine shaft at the center and the outside edge of the propeller, measured when the propeller blade is oriented vertically. The engine delivered 220 HP (164 kW) at 2300 RPM to the propeller. The rounds the Vickers machine gun fired are 7.7 mm diameter, have a bullet length of 22 mm, a case length of 56 mm for an overall length of 76 mm. Assume while spinning the propeller resembles a thi n rod.

a) Determine the torque applied to the propeller for the constant angular velocity specified. (4 pts)

b) Determine the time it takes for the propeller blade to completely clear the muzzle opening at front of the gun barrel. This time is needed in order to properly set up the timing chain, which allows the gun to fire and not hit the propeller. Assume the angular velocity is constant. (4 pts)

c) In a real dogfight the airplane may be accelerating or decelerating. How would the time it takes the propeller blade to clear the muzzle opening change, as compared to a constant angular velocity, if the propeller was increasing its angular velocity by 100 rad/s 2 ? (5 pts)

d) Assuming the rate of fire doesn't change with the timing, how many bullets can be fired during one full rotation of the propeller for a constant angular velocity? What angular displacement of the blade is required for each bullet? Will the bullet have enough time to pass between the blades? The bullets are fired at a constant velocity of 744 m/s and y ou can ignore the distance from the gun to the propeller.

Answer 1

6. given weight of propellor, m = 97 kg

l = 2.5 m

W = 0.15 m

t = 0.09 m

rate, r = 500 rounds per minute = 8.333333333 rounds per second

Power P = 164 kW

w = 2300 rpm = 240.85543 rad/s

diameter of bullet, d = 7.7 mm

length of bullet, s = 76 - 56 = 20 mm (case will not be fired along with the bullet)

a. moment of inertia I = ml^2/12 = 50.520833333 kg m^2

torque = T

power P = T*w

hence

T = P/w = 680.90638438 Nm

b. time taken to completely clear muzzle opening = t

t = (d + w)/wr'

r' = l/4

hence

t = (d + W)4/wl = 0.0010476012 s

c. if alpha = 100 rad/s/s

then

w = wo + alpha*t

and

t' = (d + W)4/wl

hence

dt'/dt = -(d + W)4/lw^2 *dw/dt

dt'/dt = -(d + W)4*alpha/l(wo + alpha*t)^2

hence

we see the time required to clear the muzzle decreases as time increases for constnat angular acceleration

d. time of rotation, T = 2*pi/w

number of bullets that can be fired = (T - 2t')*r = 0.1999308 bullets

e. u = 744 m/s

hence time taken for bullet to cross the propellor, dt = t/u = 1.209*10^-4 s

this time is lesser than T - 2t' = 0.0239918014

hence, the bullet will cross the propellor safely when synchronised properly