Quote:
Originally Posted by unclebilly
Why do high performance cars need a tranny cooler whereas commuter cars do not? Assuming both have the same parasitic losses in the trans...
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Quote:
Originally Posted by Tervuren
As force is increased, so is friction.
Think of sliding a 200lb object across ice, now try sliding a 400lb object across the ice. Even if you have the object on a low friction surface, the 400lb object will have more drag on the ground.
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Those two ^ together totally did it for me. I've often wondered. Assuming you have the same car, so a theoretical Veyron, and the "valet" key knocks the motor down to 100hp, so the frictional loss (15%) is 15hp, but then the owner gets in and has the full 1000hp, so 150hp loss. Yeah, sounds crazy and extreme, but it makes sense now. Granted, I'm assuming these are as stated, guidelines, and the graphs aren't a straight line, but still, pretty cool that I can visualize it now.
Quote:
Originally Posted by afterburn 549
Nah...I see what you are saying...but,...nah
Jump from 100 HP to 200 HP .
There will be no added 10 hp added loss.
SEE?
That's what I am saying....these arbitrary numbers floating around....
Just changing HP can not cause more drag.
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Yep.
Dragging a large steel cube over a large, lubricated steel sheet (MD Holloway is there). Together, the coefficient of friction is of steel on steel with lube is .16.
Ff = μ m g
First, a 100lb cube of steel.
Ff = .16 * 100lb = 16lb
Now, lets double the weight of the cube.
Ff = .16 * 200lb = 32lb
This is like doubling the horsepower, except I assume we would really be talking about the torque.
So the friction of 2 gears on each other with twice as much force applied would be double (you know, assuming very general, ideal type circumstances).