> Nice formula! Doesn't the force increase proportional to the square of
> speed? I think if you double speed, the force goes up four times, AND you
> cover twice as much distance in the same period of time, so the energy per
> unit time is 8 times (force times distance).
>
> In any case, the formula and typical numbers are interesting! It really
> looks like the vast majority of the losses are aerodynamic. How much
> improvement can we get by using "perfect tires?"
>
> Harold

As Russell noted one formula is a derivative of the other. Speed costs
money, how fast do you want to go? has a whole new life when you look at
it this way. To get a good picture of what is doing what you really need
to look at the contribution to the whole. Its usually the little things
that add up to create large drag numbers. Some you can fix some you can't.
For instance the belly pan of a car is a pretty miserable place for drag.
All sorts of lumps, bumps, pipes drive shafts etc. You could cut down the
drag of any car by 35-45% just by fitting a smooth belly pan. Car makers
won't do it as no one looks under there. About the only half descent one
is the lowly VW Beatle the original one. Pusher engine so nothing under.
The only car I was ever in that could get stuck on speed bumps.

Low Mu or low resistance tires are about 20% less drag than a similar
sized normal compound tire. That would mean about 4% increase in mileage.
And the reason you don't see those on most cars is they will make the car
ride like a buckboard as the tires use higher pressure and harder rubber.

Dave

-- 
http://www.piclist.com PIC/SX FAQ & list archive
View/change your membership options at
http://mailman.mit.edu/mailman/listinfo/piclist