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• To: "'seaint(--nospam--at)seaint.org'" <seaint(--nospam--at)seaint.org>
• Subject: RE: Car Crash Loads
• From: "Smith, Steven A" <Steve.Smith(--nospam--at)PSS.Boeing.com>
• Date: Thu, 25 Feb 1999 13:09:00 -0800

```Assuming constant deceleration (implies constant force) from a speed V to zero (totally plastic impact, no elastic rebound) over a crash deformation distance x (sum of car and wall deformation), the force would be F = mV^2/2x. For a car that weighs 3500 lbf, traveling at 88 ft/sec and a crash deformation distance of 5 ft, F = (1/2)*(3500 lbf / 32.2 ft/sec^2) * (88 ft/sec)^2*(1 / 5 ft) = 84,000 lbf. This solution is only a simple approximation to the order of magnitude of the forces involved. It also demonstrates that speed kills (V^2)!

Steve Smith, P.E.
Systems Stress
(425) 294-7681
M/S 02-56

> ----------
> From: 	Parkerres(--nospam--at)aol.com[SMTP:Parkerres(--nospam--at)aol.com]
> Sent: 	Thursday, February 25, 1999 12:30 AM
> To: 	seaint(--nospam--at)seaint.org
>
> To all:
>
> A client asked me today to give him a "simple" formula to translate a speeding
> car into an impact load.  For example, if a 3500 pound car is traveling at 60
> mph and hits a wall, what is the impact force?  Of course, my brain is so
> clogged with Code regulations, I only have vague recolations of the physics
> involved with solving this problem.
>
> Any input would be appreciated.
>
> Thanks,
>
> Bruce Resnick, SE
> Parker Resnick Str. Eng.
>
>
>

```