--- "Michael R. Clements" <mrc01@flash.net> wrote:
> Q1: Is that a real gold star, or just an imaginary one? If it's a
> real gold star then I'll derive it and I promise not to peek.
> Otherwise I'll just look it up. Basically, the 33000 comes from
> the fact that power is work / time, work is force * distance, 1
> horsepower = 746 watts, 1 foot is 0.305 meter, 1 lb. is 4.45
> newtons, then you plug it all into the basic physics equations
> and you get 2*pi / 33000.
>
Actually, I was looking for 1hp = 550 ft-lbf/s and 60 sec/min to get the
33,000, with the real question being where the 550 came from (typical English
unit garbage). Two means to the same end. I just can't remember where the 1hp
= 550 ft-lbf/s come from. Thought you would.:-)
> Q2: YES. If you get the same torque at higher RPM, then you can
> use shorter gears at the same vehicle speed ('cuz the engine revs
> higher) and that means you get a bigger gear ratio multiplier and
> more torque at the wheel. Basically, when you can rev higher you
> can use shorter gears without sacrificing vehicle speed. The
> overall push of the car is the product of the gear ratio with the
> crankshaft torque, so both are equally important for
> acceleration.
>
NO, as you worded your blanket statement. Of course, once you take gear
ratios, wheel diameter, drivetrain losses, etc. into account it is possible to
*tune* them for optimal acceleration with a given powerplant. Stating that the
same torque at a higher RPM produces more acceleration is false (all else
remaining equal). As you well know, real world acceleration depends only on
how much torque is being transferred at the wheel to the ground. Be careful of
general conclusions.:-)
-Andy
Do You Yahoo!?
|