GregoryUrich
Graduate Poster
- Joined
- May 16, 2007
- Messages
- 1,316
Sizzler:
My calculations take care of momentum AND energy conservation. In fact they are BASED on momentum and energy conservation! Each impact DOES slow the descent a little, but the downward velocity increases overall, (just not as much as the free fall case).
For free fall through one story height of 3.7 meters we have velocities of 8.52 m/s after 3.7 meters; 12.05 m/s after 7.4 meters; 14.76 m/s after 11.1 meters; 17.04 m/s after 14.8 meters... and so on.
Now with floor resistance leading to a drain on the KE of 1 GJ for each floor we have a reduction of the free fall velocity at each impact:
For WTC 1 for example:
For the first impact after a drop of 3.7 meters by a mass of say 58 x 10^6 kg the KE is 2.1 GJ. If the energy to collapse the first impacted floor is 1 GJ we still have about 1 GJ of kinetic energy remaining. Now we can work backwards and calculate the new velocity of the upper section as it continues on to the second impact. Using KE = 1/2 Mv^2 we have:
v = Sqrt { 2E/M} or v = 5.87 m/s.
The descending mass now falls the next 3.7 meters starting from an initial velocity of 5.87 m/s (Not zero!). Now we can use:
v = Sqrt{(5.870^2 + (2g x 3.7)} = 10.34 m/s
Hence the upper block strikes the 2nd floor with a velocity of 10.34 m/s
For free fall this velocity, as we have seen, would be = 12.05 m/s
So the collapse continues "at near free fall speed" even though the structure offered great resistance!
You see Sizzler, you HAVE TO plough through calculations like this to REALLY look at the collapse. Just saying I THINK the collapse should have been arrested, or I think it should have slowed is not going to help you understand the collapse...........
Apollo,
Why are you still using 58 x 10^6 kg, when I have carefully demonstrated that the value was closer to 32.8 x 10^6 kg? That makes me feel like I was wasting my time.
/Greg


