OT Car crash Physics...

[jason@scirocco.org (Jason)]

At 03:57 PM 10/28/2002, Jason Adams wrote:
>With all the wealth of informed people on the list this shouldn't be too
>difficult...
>
>If I know the weight of the cars, the skid distance, assume complete
>inelastic collision. how do I work it out...

I'm confused here -- what are you trying to work out?

>Vf^2 = Vo^2 + 2a(dS)
>
>is there something wrong with my assumption for 'a' ?

Well, uK isn't a constant... the value you use is a "typical" value.  Every 
tire is different, as is every road surface.  And stopping distances on a 
car aren't that easy to compute -- you're dealing with 4 tires on an object 
that has suspension and that does not have even weight distribution.

"Typical" deceleration for a modern automobile is between 8.5 and 10 
ms^-2.   Of course, some are far below, and a few are above.

Using the 3 braking distance figures I have for the 16V from magazines (60, 
80mph from Road & Track, 70mph from Car & Driver), we can compute the 
average a for the 16v's braking:

60mph   150 feet        7.87g
70mph   196 feet        8.19g
80mph   257 feet        8.16g

The average of those 3 stops is 8.08g.  The 8V (158 feet from 60, 271 feet 
from 80), averaged 7.61g.

Of course, modern tires and brake pads improve those distances:  Two years 
ago I did 10 consecutive stops from 60 in my16V with Potenza RE71 tires and 
Ferodo pads and used my G-Tech Pro to measure the distance.  I discarded 3 
runs where the G-Tech did not provide an accurate reading, which left me 
with 7 good runs.  I discarded the best and the worst runs, leaving 5 
total.  The average of those 5 runs was 131.8 feet, which is an average of 
8.95g.

Jason