I need all the info about the crush zone!!!

[twinky64]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Jimmy01:
I think you are getting yourself confused with mass, weight and force.

The maximum average force(F)(to keep acceleration under 20G) which the attenuator can apply is very easy to calculate:

M = mass of 'car'= 300kg
a = 20 * 9.81 m/s2 = 196.2 m/s2

F = Ma = 58860N </div></BLOCKQUOTE>

I thought that 20gs of 300kg would be 6000kg? For instance, let's say you weigh 200lbs. At 2g's, you weight 400lbs, 3g's is 600lbs etc. Doesn't that same logic apply?

[JasperC]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by twinky64:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Jimmy01:
I think you are getting yourself confused with mass, weight and force.

The maximum average force(F)(to keep acceleration under 20G) which the attenuator can apply is very easy to calculate:

M = mass of 'car'= 300kg
a = 20 * 9.81 m/s2 = 196.2 m/s2

F = Ma = 58860N </div></BLOCKQUOTE>

I thought that 20gs of 300kg would be 6000kg? For instance, let's say you weigh 200lbs. At 2g's, you weight 400lbs, 3g's is 600lbs etc. Doesn't that same logic apply? </div></BLOCKQUOTE>
6000kg = 58860N, Einstein.

[Jimmy01]

Your logic is fine, but I don't think 'weight' should ever be used in an engineering context. Mass and forces are all you need.