Smrekar's Dialog on Lightpoles and Momentum Options

[Smrekar]

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).

http://www.airliners.net/aircraft-data/stats.main?id=101

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.