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Re: Car Crash Loads

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Since I work on an accident investigation team with the Highway Patrol, let me add one slight correction to Steve's last sentence. It isn't the speed the kills... it's the stopping! ;-) I never thought that I would ever, ever, EVER use physics after I left school ** years ago, but now it has come back to haunt me!

Ken Reed, PE
Redding, CA

"Smith, Steven A" wrote:

> 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]
> > Reply To:     seaint(--nospam--at)seaint.org
> > Sent:         Thursday, February 25, 1999 12:30 AM
> > To:   seaint(--nospam--at)seaint.org
> > Subject:      Car Crash Loads
> >
> > 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.
> >
> >
> >
>