#### JimCovington

##### Well-Known Member

Here's why the ratio works. (PS - I tried other numbers, and they were within rounding error in every case. Let me know a specific condition you find that's not.)

Use the formula a = (v^2-u^2)/2s for both acceleration and deceleration. Simplify it to a = v^2/2s for both conditions, as you are either starting or stopping at rest (the G forces will be opposite signs, of course.)

v is the same in both conditions - the speed you reach after a freefall is the same as the speed you must decelerate from.

Use the following variables:

G= G

a = rate of deceleration

ff=distance in freefall

st= distance to stop

Solve for v:

v = sqrt(2 G ff)

v =sqrt (2 a st)

Substitute one v for another (they're the same value)

sqrt (2 a st) = sqrt (2 G ff)

square both sides (remember to check for the correct sign at the end)

2 a st = 2 G ff

solve for a

a = G (ff/st)

And there ya go....deceleration (in Gs) is your freefall distance divided by your stopping distance.

/In honor of this post, I've changed my avatar. No, that kayak's not flying and yes, it's about 6" above the water after falling 15'. I'm about .05 seconds away from a compressed spine.

//And no, I didn't say "Hold my beer and watch this" before I did it.