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Nascar Gets a Catcher's Mitt

Posted January 02, 2007 2:23 PM

From PopSci.com:

When it comes to stopping speeding cars without injuring the driver, a cement wall just won't do. During a race on Virgina's Richmond International Raceway in 2003, Nascar driver Robby Gordon lost control of his car as he roared into the pit at 55mph. He smashed sideways into the concrete slab that separates the crew from the pit road and wrecked his car. Luckily, he didn't do the same to his body. Had he hit the divider head-on, the collision could have transferred a 100G-force jolt to his body, more than enough to kill him.

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Associate

Join Date: Oct 2006
Location: Princeton, NJ
Posts: 48
#1

Re: Nascar Gets a Catcher's Mitt

01/03/2007 12:28 PM

Just curious - how does one calculate G-force knowing only the velocity of a moving object coming to rest in a case like this? An energy equation? I suppose a mass would have to be assumed.

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Guru
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#2
In reply to #1

Re: Nascar Gets a Catcher's Mitt

01/03/2007 12:59 PM

Just as a guess, mass would certainly have something to do with it, but I believe the major contributor would be how the vehicle self destructs as it hits the wall. If the vehicle were a solid block of aluminum when it hit the wall, the g forces would be horrendous, because the deceleration would take place almost instantaneously.

Modern passenger vehicles are designed to fold up like an accordian in a crash, but this takes time. Extending the deceleration over time means that the G forces are less for the occupant(s) of the vehicle, resulting in less broken people.

I do long for my old '56 Chevy though, with real bumpers, with which I could push and pull things without having to worry about breaking it.

Bill

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Anonymous Poster
#3
In reply to #1

Re: Nascar Gets a Catcher's Mitt

01/03/2007 1:37 PM

Force = Mass X Acceleration

Force X Time = Mass X Velocity

Substitute for Force

Mass X Acceleration X Time = Mass X Velocity

Mass drops out

Acceleration X Time = Velocity

Solve for Acceleration

Acceleration = Velocity X Time

Gravity is either 9.81 m/s squared or 32.2 ft/s squared

So your really just need the speed of the car at impact, and how long impact occured.

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Associate

Join Date: Oct 2006
Location: Princeton, NJ
Posts: 48
#4
In reply to #3

Re: Nascar Gets a Catcher's Mitt

01/03/2007 6:45 PM

Okay - with this approach you assume a delta t, delta v is known, and accelleration is easily calculated (assuming it is constant). Still wondering how the author came up with 100G - an arbitrary assumption of delta t for a crashing race car? But I get his point.

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Anonymous Poster
#5
In reply to #4

Re: Nascar Gets a Catcher's Mitt

01/03/2007 10:57 PM

And there you have the issue. How do you absolutely time the impact? Until everyone agrees with t0 and tf you'll get 100G answers.

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Guru
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#6
In reply to #3

Re: Nascar Gets a Catcher's Mitt

01/04/2007 1:19 PM

Just for the heck of it, lets look at numbers. 55 mph corresponds to 80.666 feet per second. Now lets assume stopping times starting with 1 msec (.001 sec). We can make a table showing time, deceleration in feet / second2, and using the 32.2 feet/sec2 mentioned above, translate into g forces.

TIME (sec) DECELERATION G FORCE

.001 80666 FT/SEC2 2505 G

.01 8066 FT/SEC2 250.5 G

.1 806.6 FT/SEC2 25.05 G

1 80.66 FT/SEC2 2.505 G

Bill

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