The Traffic Accident Reconstruction Origin -ARnews-


Re: Delta V

Gary Jackson ([email protected])
Fri, 15 May 1998 08:06:12 -0400 (EDT)

My response was not completed and got a little jumbled without formatting, here it is again:

Many people think of Delta V as merely the change in speed, which is misleading
if the final speed is not zero. I prefer to think of it as the speed for the
equivalent kinetic energy in MPH.

"Alternatively (different sceanrio)assuming an approach velocity of 60 kph
and a residual speed of 20 kph what would be the delta V for changes in angle
of collision departure relative to initial direction of 45, 90, 135 and
180 degrees respectively?"

For the same vehicle mass, the square root of the difference between the square of the
initial velocity and the sqaure of the departing velocity.

=============Departure
Speed === Speed / Angle === Delta Vx, DeltaVy
60 ====== 20 / 45 ======= (60^2-(20cos45)^2)^0.5 in the X , 20sin45 (14.1)in the Y

However, this assumes a perfectly elastic collision, the speed/energy
absorbed by the damage of the vehicle and the struck object must be considered.
Also, this is for the vehicle, occupants are not necessarily the same.
Gary Jackson, P.E.
Gary Jackson
[email protected]


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