Close enough. Drag is proportional to the cube of speed, rolling resistance
is linear, and so is drive train resistance (except crank windage). You'd
wind up with some complex number for a given car, like ten to the 2.7657
power, but you're as right as you need to be.
Of course I just drove straight through from Portland to Monterey--about 780
miles, 15 hours. By myself--we'll, me and Patrick O'Brian (books on tape).
I'm not sure my mind is working at all. I'll probably look at this tomorrow
and wonder what I was thinking.
-----Original Message-----
From: owner-fot@autox.team.net [mailto:owner-fot@autox.team.net] On Behalf
Of Randall Young
Sent: Tuesday, August 10, 2004 8:56 AM
To: fot@autox.team.net
Subject: RE: speed vs hp question
> Assume for a moment that you have a car with a known hp at the rear
> wheels, and you also know the top speed attainable with that car.
>
> Does anybody know the math for determining how many additional hp it
> will take per additional mph on the top speed?
It's a cube law, so it depends on what your original top speed was and how
much you try to increase it. The formula, roughly, is
New Required Power = (New Speed/Old Speed)^3 * Old Power
Thus, if you were making 85hp at the rear wheels and top speed was 100 mph,
then to go 110 mph you would need
(110/100)^3 * 85
or
1.1 * 1.1 * 1.1 * 85
= 113 hp
Randall
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