You just gotta love it. In response to the question "It’s been stated everywhere that there is no such thing as Stopping Power or Knock Down Power. So that being the case, what DO you call it?
We get.....
No offense intended to poster of this.
We get.....
AhhhUgggg. I said I'll stop, but I'll show you the numbers. I did these off the cuff.
Assumptions.
40 grain .22 lr traveling at 1,000 f/s.
A child rolling a 6 lb bowling ball at 10 mph (14.6 f/s) when it hits the pin - It will probably still knock down a pin even at 5 mph, but lets use 10 mph. The ball has a radius of 4.25" or .375 feet.
For the bullet the momentum is m*v where m is in slugs and v is in f/s. To convert 40 grain to slugs m=40/7000/32.2 = .000177. The momentum = .000177*1100 = 0.195 slug-ft/s.
The KE of the bullet is .5*m*v^^2 = .5*.195*(1100^^2) = 107 ft-lb.
The total momentum of the bowling ball = the transitional momentum + angular momentum. Transitional momentum = m*v. Angular momentum = Inertia (I) * Angular Velocity (w)
The mass of the bowling ball is 6/32.2 = .186 slugs. The Transitional Momentum is .186*14.6 = 2.72 slug-ft/s.
I = 2/5*m*r^^2 = 2/5*.186*(.354^^2) = .009 slug-ft^^2
w = v/r = 14.6/.354 = 41.2 rad/sec
Transitional momentum = 2.72 + .385 = 3.11 slug-ft/s
The bowling ball kinetic energy = Transitional KE + Rotational KE = .5*m*v^^2 + .5*I*w^^2
Trans KE = .5*.186*(14.6^^2) = 19.9 ft-lb
Rot KE = .5*.009*41.2 = 7.94 ft-lb
Total KE = 27.8 ft-lb
Bottom line...
A .22 with 107 ft-lb of KE and .195 slug-ft/s of momentum can't knock down a bowling pin, but can damage it.
A 6 pound bowling ball rolled by a child at 10 mph with 27.8 ft-lb of KE and 3.11 slug-ft/s of momentum can knock down a bowling pin and won't damage it.
I didn't double check the numbers, but I think they close. Like I said, I'm done.
No offense intended to poster of this.
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