RE: Fwd: Audi A8 Achieves Highest Possible Safety Test Rating

["Rob Winchell" <rbw@avs.com>]

I actually knew the answer (if not all the math).

My point is, that a test crash of a car into a stationary barrier isn't a
very good "real world" test for head-on collisions, since it's pretty rare 2
identical cars hit head-on.

What if one car was Torsen and the other was not...Oh god no!!!!!!!

Rob Winchell
91 200Q
87 4kCS (parting out)



> -----Original Message-----
> From: Robert Myers [mailto:rmyers@inetone.net]
> Sent: Wednesday, September 16, 1998 3:08 PM
> To: Rob Winchell
> Cc: Quattro list
> Subject: RE: Fwd: Audi A8 Achieves Highest Possible Safety Test Rating
>
>
> Gee, I had hoped we could let this thread die a mercifully quick natural
> death - kinda like spiders.  OK...  Here we go.
>
> At 02:12 PM 9/16/98 -0400, you wrote:
> >Doesn't that assume the cars are identical?
>
> Yes, it does.  See the initial setup below.
>
> >What if one car is 2000 pounds
> >and one is 5000 pounds?
>
> The lighter car and the larger car will exert the same forces on each
> other.  For every force there is an equal but opposite force...
>
> F1 = F2
>
> The accelerations experienced by the cars (and their occupants) will,
> however be different.
>
> Remember?  F = ma
>
> F1 = m1*a1; F2 = m2*a2 or --> m1*a1 = m2*a2
> Assuming the more massive car to be car 1 then 5000*a1 = 2000*a2
>
> (Yes, the masses are not 5k and 2k but the conversion factors cancel out.)
>
> This solves to a2/a1 = 5000/2000 = 2.5.
>
> Or... The lighter car and its occupants will experience accelerations 2.5
> stronger than the more massive car.
>
> Think of it this way.  Two identical cars colliding head-on at the same
> speed result in a zero terminal velocity for both cars.  Each car fully
> stops the other.  In the case where the cars are different, the more
> massive car doesn't come to an instantaneous dead stop.  It continues to
> roll forward on through the duration of the collision.  The less massive
> car comes to a full stop and then is forced backwards until the two
> velocities of the remaining hulks are equal.  Eventually friction
> forces or
> secondary collisions, etc, stop the two cars.
>
> Imagine a collision which is high enough energy to result in the occupants
> of the more massive car experiencing a 10 G deceleration.  The lighter car
> and its occupants will experience a 25 G deceleration.  Who do you suppose
> will be injured more severely?
>
> >> -----Original Message-----
> >> From: owner-quattro@coimbra.ans.net
> >> [mailto:owner-quattro@coimbra.ans.net]On Behalf Of Orin Eman
> >> Sent: Wednesday, September 16, 1998 1:19 PM
> >> To: four.rings@mcione.com
> >> Cc: quattro@coimbra.ans.net
> >> Subject: Re: Fwd: Audi A8 Achieves Highest Possible Safety Test Rating
> >>
> >>
> >>
> >> > An excerpt from the newswire post:
> >>
> >> > > To collect this data, vehicles are crashed into a fixed
> >> barrier at 35 miles
> >> > > per hour with instrumented dummies registering impact forces
> >> during the crash.
> >> > > The force is equivalent to a head-on collision between two
> >> identical vehicles,
> >> > > each moving at 35 mph.
> >>
> >> > No, it will be equivalent to a head-on collision between two
> >> identical vehicles,
> >> > each moving at 17.5 mph.
> >>
> >> In terms of kinetic energy that has to be dissipated, the 35 mph figure
> >> is correct...  A car travelling at 35mph, ending up at rest must
> >> dissipate the same amount of energy regardless of what it crashes into.
> >>
> >> Looking at forces, the concrete barrier provides whatever force
> >> is required
> >> to decelerate the car.  In the case of the head on crash, the
> >> same force is
> >> required... by each car.  If you then look at Newton's laws - equal and
> >> opposite reaction.  Each car supplies the reaction required by
> the other!
> >>
> >> Orin.
> >>
> >
> >
> >
> ___
>    Bob
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