The finer points of brake cooling

[Russ Noble]

These figures are proportional to speed so take half for the average dissipation for braking to rest.



Dave

Dave,

Ignoring the aerodynamic drag factor, I thought the kinetic energy involved which had to be dissipated by the brakes went up as the square of the speed! I could be wrong though, it's a long time since I studied physics at school. If that is true it takes four times as much energy to stop from 150mph as it does from 75. Maybe someone a bit more knowledgable than me would care to comment.

Cheers

 

[Gary Gibbs]

Correct on energy.

Sorry Ron for my quick post earlier. I think the Mk II's at LeMans had a problem of tremendous speed on the Mulsanne and thus great cooling followed by tremendous input of energy thus rotor seeing huge change in temp resulting in thermal shock and failure.

 

[Lynn Larsen]

Russ,

F=MA (force = mass x acceleration)

Acceleration is the total area under the curve if you graph speed vs time, which is the integral of velocity and therefore that variable is increased one order of magnitude, that is squared.

 

[Dave Bilyk]

Russ,
you are quite correct, the kinetic energy increases as the square of the speed. But power is the rate of change of kinetic energy, and increases proportional to speed. This is implicit in llarsens description too.
Look at it this way, if you accelerate (or decelerate) a body at a constant rate (0.9g in this case), the velocity at any moment is v, the instantaneous power is mass x 0.9g x v (i.e proportional to speed), and if we integrate the instantaneous power, we get the instantaneous kinetic energy 1/2 x m x v^2 (proportional to speed squared). So you got it dead right, but need to distinguish between Energy, and rate of change of Energy which is Power.

Dave

 

[Russ Noble]

OK. Thanks Lynn and Dave. I now understand the difference between energy dissipation and power dissipation.

Cheers