[QUOTE]Originally posted by wc_one
[B]

Well I can just the same disprove your statement by keep delivering the same RPM at higher torque and you'll get more horsepower.

 

The very first post in this thread has all the equations you need.

Here is the reason I believe knowing the peak TORQUE of a vehicle and dividing it by the vehicle's weight is the best way to predict it's ability to accelerate.

T=Torque
F=Force
d=Distance
m=Mass
a=Acceleration

By definition, T=F*d

Therefore, F=T/d

By definition, F=m*a

Solving for a, a = F/m = (T/d)/m = T/(d*m)


Acceleration is equal to the net TORQUE divided by mass and gearing.

 

[QUOTE]Originally posted by pll
[B]
Make sure you file a patent when you make the car that produces variable torque at a constant RPM.

 

Well, your equations and your conclusion are not really the same.

Your equations show that the acceleration is equal to WHEEL TORQUE, divided by mass and wheel diameter. Nobody will dispute that.

The problem is that WHEEL TORQUE is not the same as ENGINE TORQUE. Wheel torque captures BOTH engine torque and GEARING, which in turn depends on the engine revving capabilities, so you can say that wheel torque captures both engine torque and engine revving capabilites. Just like horsepower does.


Go ahead, knock yourself out:

A car has about 150lb-ft of peak engine torque, and it weights about 2800lbs.

What does it run in the 1/4?



Hint: It could be a high 13's S2000, OR it could be a 17-second Golf TDI

 

Engine torque is measured in foot pounds. That's one foot. So d = 1. This is the same as the original post. Are we talking engine speed or wheel speed? It doesn't matter because you can get one from the other if you know the gearing.

 

 

[QUOTE]Originally posted by Destiny2002
[B]Everybody hold the presses!

 

 

 

Everybody hold the presses! We have a new expression for engine power! Power doesn't equal power! Power actually equals power/(mass*velocity)




ROFLMAO