How wide is the barrel assembly on the M134? And how much does it weigh of the M134's 18.8 kg? If some one can give me a better estimates on weight and power use, I'll readjust my calculations accordingly. I'm using the 4KWt number and weigh from the guns.ru page.
Torque:
4000 Watts = Torque * 2pi *2000 RPM
1/pi = Torque
1/pi = r X F = r*F*sin(theta) Sin(theta) can be assumed for 90 degrees in this so
1/pi = r*F
I'm guessing from those pics that the M134 has maybe a 6 inch diameter cluster of barrels? so r = 0.0762 m (= 3 inches radius)
1/pi/0.0762 = F = 4.2 n = 0.94419756102 Lb force, which is pretty much nothing.
Centripetal force:
Assuming 75% of the mass of the M134 is barrels (it looks all barrels to me at least).
F=m*r*w^2=18.8*.75*0.0762*(2pi*2000)^2
F=1.07442 * (4000pi)^2 = 169665605 n = 170000000 n = 170 Mega newtons
And finally, Angular Momentum: (which should be the force which gyro stabilizes the gun)
L = I*w
I = 1/2*m*r^2 (assuming a solid cylinder, the closest approximation for me to use)
I = 1/2*18.8*0.75*0.0762^2 = .040935402
L = .40935402 * 2pi * 2000 rpm = 514.4... = 514 J*s = 514 N*m*s
err, now that I read this over, this is all really just me being bored. So I hope that the results interest some one here as much as the work kept me interested. And I hope my math isn't off.