6ft/s = 1.829m/s
F = ma = 396000 x 9.81 = 3.884MN due gravitational loading only, i.e. "g"
Now, arresting ROD in half a second:
F = ma = 396000 x (1.829m/s)/(0.5s) = 1.448MN
This gives total force on aircraft of 3.884 ("g") + 1.448 = 5.332MN = 1.37g
Arresting ROD in a quarter of a second = 2.897MN = 1.75g
Why am I sitting here doing this instead of studying for my aircraft structures exam on Wednesday??
I don't even know if it's realistic, but seeing as we've come this far, assume a strut (well, oleo) length of about, I dunno, 1.5ft? = ~0.7m
Average speed of strut compression = half of original speed (vertically) = (1.829/2) = 0.9145m/s
so for a 0.7m oleo at that ROD it would reach full compression after 0.76 seconds.. so refining the original calculation with that time of deceleration would yield a g-force of about 1.25g..
AD
edit to add back of envelope calc for entertainment purpoes only!
Last edited by The African Dude; 2nd May 2006 at 09:16.