>>According to Hill in an interview recently where he was explaining
>>the F1 physics he said a F1 car would go for 200->0 in 100 yards.
>>Seems damn hard to me, but those cars have brakes and then some.
>Not sure this has been mentioned (or how irrelevant!), but I've been told that
>by just getting off the gas, Indycars (and presumably F1 cars) get about a G of
>aerodynamic braking from 200+ mph! I guess I should be able to figure out this
>one. Let's see, 700hp=385000 lb-ft/s. At 200 mph, this converts to 1300
>lb-force. If at terminal-V, drop that power to zero and you have, in effect,
>that 1300 lbs in aero braking. If the car weighs 1300 lbs, that's a G.
>Sorry for the random meanderings...
Don't know if I followed your explanation there <g> but I'd say it's
not only the aerodynamics that contribute to negative acceleration
(this could also mean braking for some of you GP2 drivers <g>), but
it's also the torque of the engine. Don't really know how much torque
those V8's have, but having about 890 hp (talking about Indycars) I
can imagine an incredible torque around 11k-13.5k rpm.
A bit off the topic here. I remember, two years ago, I rode a
Suzuki 500 bike, normal street legal bike about 27 hp, While going
about 60 km/h I took out all throttle and felt the gas tank pushing
pretty hard but not too *** my "you-know-what" :). The point is,
I wasn't used to this type of idle torque. Before I rode a Honda 80
which had a 2-stroke engine. You let go off the gas, and the engine
keeps on roling... 4-strokes stop compared to 2-strokes.
So I can picture quite a (mechanical) drag created by the engine.
Take care,
Dave.
--
David Kadlcak http://www.racesimcentral.net/~kadlcak
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