Smrekar's Dialog on Lightpoles and Momentum
Oct 13 2008, 04:33 AM
Joined: 11-October 08
Member No.: 3,929
Looking at the momentum (p) in this alleged inelastic collision:
p = mass * velocity
530 mph ~= 777.3333333333 fps [or approx. 460.5574082073434 knots]
Since the amount of fuel or passenger load in AA77 can be disputed, let's use the operating empty weight for a 757-200 for a "light" estimate:
Operating empty with P&W engines 57,840kg (127,520lb), with RB211s 57,975kg (127,810lb). Basic max takeoff 99,790kg (220,000lb), medium range MTOW 108,860kg (240,000lb), extended range MTOW 115,665kg (255,000lb) or 115,895kg (255,550lb).
127,810lb * 777.3333333333 fps = 99350973.3333334 ft-lb/sec [of B757-200 momentum]
Let's "guess" a 1% momentum transfer in this inelastic collision (a quite conservative estimate):
99350973.3333334 ft-lb/sec * 0.01 = 993509.733333334 ft-lb/sec [of light pole momentum]
Now dividing by the assumed 200 lb. light pole mass,
993509.733333334 ft-lb/sec / 200 lb = 4967.5486666667 ft /sec [of light pole velocity]
Just for comparison, most rifles fire a bullet at 2000-4000 ft/sec.
Is a broken windshield all that we would expect to see happen to Lloyd's? cab?
I'm sorry, but this analysis is simply impossible from a physical standpoint. It assumes that an object traveling at 777 ft/s hitting a stationary object could propel the stationary object along it's trajectory at over 4000 ft/s. This is, I believe, quite impossible. (IMG:http://pilotsfor911truth.org/forum/style_emoticons/default/rolleyes.gif)
You're welcome to source me to an article that would say otherwise, of course ... though I do doubt there would be any.
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