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To: antiRepublicrat

Can you run that through the graph program and show the parabolic track, as soon as the void opened, downward motion begins and as you said the horizontal impact distance would be roughly 2/3 of the edge of the clifftop; a sheer drop would put him about 45 feet past the cliffwall in the first second — arcing out to the terminal point.


81 posted on 07/14/2009 10:28:03 AM PDT by Old Professer (The critic writes with rapier pen, dips it twice, then writes again.)
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To: Old Professer

What was the car’s mass? What was the driver’s mass, and the mass of gas, fluids and anything else in the car? At what angle did the car fall and what is the drag coefficient and surface area at that angle? What was the angle of launch? Is it flat, downhill or uphill towards the edge?

Even with that I can’t do right. Cars tend to fall somewhat flat on their bellies, but shift around as they go, constantly changing the drag. My estimate of time, speed and distance is completely off if the terminal velocity is less than the straight gravity-based calculation.


110 posted on 07/14/2009 11:04:20 AM PDT by antiRepublicrat
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