Re: [DML] crash tests / safety

["natesky1980" <ncskalsk@xxxxxxx>]

Sorry to hear about your friend.

You can disagree with it, but that is basically how it works.  I can 
certainly see where your intuition is misleading you; it is a bit 
counter intuitive.  I realize this is getting fairly off-topic.  
There are plenty of good introductory physics forums/resources out 
there (search for `collision physics' or `inelastic collisions').

You can't ignore direction where vector quantities are required.  
Yes, both cars are traveling at 40mph, but if you look at them as a 
system, one is traveling -40mph and the other +40mph, therefore 
assuming they have the same momentum (remember it is conserved here), 
the result will have them canceling each other out (the net momentum 
was the same before collision).. 

For example; say both car weigh 2700lbs and the are both traveling 
head-on @ 40mph.  Momentum p=m1v1+m2v2 or (2700lbs)(40mph)+(2700lbs)(-
40mph), which of course equals zero.  

So, really it's the `ride down' from one car's initial momentum to 
whatever the combined momentum is of both cars (in this case zero) 
where all the grindy grindy, bang, pop, screech, etc. occurs.  That's 
why we want cars to absorb the force, it spreads that `work' over a 
greater amount of time (the passenger's) thus reducing the 
peek `negative' acceleration (as does your airbag).  

Average Impact Force = (-(1/2)mv^2)/d where m=Mass, v=Velocity and 
d='distance traveled after impact, or `amount of car-crumple'. 

For our rigid barrier (Delorean hitting it at 40mph, and assuming 1 
foot of crumple) the average impact force will be Favg = 72Tons 
(converting from N) 

For a two car (Delorean) collision the Favg = 72tons for each car, 
assuming same crumple distance & velocity.

Keep in mind that if a Delorean is hit by something with larger mass 
traveling at the same speed (head-on), The Delorean will have to 
decelerate faster (over the same amount of time) than the larger 
vehicle , making up for its smaller `m'.  (change in momentum = m1
(DeltaV1)  = m2(DeltaV2)).  

Note: I have no idea how much of the Delorean would collapse, however 
it is likely more than one foot.  If d=3feet. The average force 
inflicted on the car is reduced to 24tons in a 40mph collision, and @ 
30mph it would `only' be 12tons (of course it may collapse less so 
this is not entirely valid).

Hope this helps, re-reading it I'm not sure I was all that clear..

-Nate
11501




--- In dmcnews@xxxxxxxxxxxxxxx, dmcvin6683 <dmcvin6683@...> wrote:
>
> I disagree on this. 2 cars traveling at 40mph hit each other would 
be  
> an impact of 80mph because they are both moving at 40mph.
>   I had a friend killed on a 55mph impact, he drove under a 
straight  
> truck traveling at 55mph. The police officer said the impact was  
> around 110mph with both vehicles traveling at 55mph.
> 
> Mark V
> On Nov 5, 2006, at 6:37 PM, natesky1980 wrote:
> 
> >  So a 40MPH
> > barrier test would be the same as two Delorean's colliding head on
> > while both are traveling at 40 (in which they would both crumple-
a-
> > bit and come to a complete stop).
> 
> 
> 
> [Non-text portions of this message have been removed]
>






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