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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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