I read the link (thank you) and did more calculations anyway, for the sake of thoroughness. I applied the same data from post #2431, and assumptions as stated in post #2440, but varied Coefficient of Drag (Cd), from 0.6 up to 1.5. For each case I tabulated the time of transit (from detachment to impact), and relative impact velocity. As before, assumed 53.4 feet of axial travel from external tank attach point to wing impact. The data show that for higher Cd (more drag), the transit time decreases and relative impact velocity increases.
|
Coefficient of Drag (non-dimensional) |
Time of Transit (s) |
Relative Impact Velocity (mph) |
|
0.60 |
0.260 |
495.98 |
|
0.70 |
0.244 |
525.08 |
|
0.80 |
0.231 |
551.21 |
|
0.90 |
0.220 |
574.92 |
|
1.00 |
0.210 |
596.67 |
|
1.10 |
0.202 |
616.72 |
|
1.20 |
0.195 |
635.37 |
|
1.30 |
0.189 |
652.76 |
|
1.40 |
0.183 |
669.05 |
|
1.50 |
0.178 |
684.42 |
http://www.freerepublic.com/focus/news/838893/posts?page=15#15
Trying again, using 730 fps instead of 200 fps,
energy (in MKS units) is
J = 1/2 * m * v ** 2
m = 2.6 lbs = 1.2 kg (approx)
v = 730 f/s = 223 m/s
J = 1/2 * 1.2 * (223 ** 2) j
= 29837 j
Applying fudge factor (hardness ratio ~10:1):
J of tile = 1/11 * J
= 1/11 * 29837 j
= 2712 j
Using a formula to get ane earth gravity drop height for a 1 kg mass
E = mgh
g = 9.8 m/s**2 (MKS)
2712 = 1 * 9.8 * h
h = 2712 / 9.8 m
= ~277 m
so it seems possibly equivalent to dropping a 1 kg mass 277 meters.