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Re: crash tests
Huw could be half right. Some of us (me) are forgetting about the
elasticity of the collision. I don't think relativity matters.
Half the energy of the car travelling at 60 (or 20m/s in my example) could
be transferred to the stationary car resulting in energy by the moving car
of only one half (which in my example would be 200KJ - the same as a head
on collision). However, the sum of the energy in this state would be 400KJ
right? Which would mean both cars now travelling together at a little over
42mph (V=(2*400/4000)^.5). This is elastic right?
If inelastic, the moving car would transfer all it's energy to the
stationary car, and Huw loses.
For some reason Newton's Cradle keeps popping into my mind.
Huw only loses outright if there is a barrier involved :-)
I'm sure Huw will post the results of his tests (because we know his Audi
will keep him safe :-)
At 05:21 PM 12/13/1999 -0500, Huw Powell wrote:
>> Ok, for arguments sake, one car has a mass of 2000Kg (we'll call it
>> cadillac), and a velocity of 10m/s (36Kph) = 100KiloJoules (I think that's
>> the right unit). And that cadillac at 20m/s has 400KJ.
>> That means a head on collision with another cadillac of similar mass and
>> velocity has 200KJ Joules to release instead of the 400KJ of one cadillac
>> at 20m/s running into a parked cadillac (negating any solid barricade
>> argument). However, it might be less severe to run into a solid object at
>> 10m/s than a deformable one (eg: car) at 20m/s.
>> To recap, 1+1=2, but (1*2)²+0=4.
>> Make sense? It's been 10 years...what am I forgetting?
>You are measuring V relative to an external frame of reference. The
>subjects in question, the cars, only have energy relative to each
>other. in the 0/60 collision, if it were not for friction bringing the
>mangled clump to a halt eventually, the two damaged cars would keep
>travelling at about 30 mph (less shrapnel/fragmentation/deformation
>energy losses) in the direction the moving car was travelling
>indefinitely - this is where the "missing" energy ends up. Filmed from
>one car the accident would be indistinguishable from the 30/30 accident.
>This is why it is different than the barrier accident, where the
>terminal velocity relative to the observer is always zero.
>One problem here I think is that people imagine that the 60/0 collision
>takes place at a fixed point in their frame of reference. It doesn't.
>(unlike the barrier versions where everything stops dead and the 60 mph
>collision is twice as bad...)