I'm not sure if I am looking at this in the correct way or not, but hopefully someone can confirm if this is right or wrong. I'm trying to establish the difference in torque multiplication with different rear gears.

I'm also wondering, if you take a car to the dyno with 2.73's and the car dynos 400 lb/ft at the wheels. You then take the same car and put in 3.23's and run it on the dyno again. Am I correct in thinking that the run with the 3.23's will show more torque than the 2.73 run? I presume therefore the horse power will increase also?

I've done the following calculations to try and show what I'm trying to say.

Transmission Ratio --------Rear Ratio ----Combined multiplication
3.06 ------------------------ 2.73 ------------- 8.35
1.63------------------------- 2.73 ------------- 4.45
1.0 ------------------------- 2.73 ------------- 2.73
0.70------------------------- 2.73 ------------- 1.91

So in first gear, the motor torque is mulitplied by 8.35 before it goes to the tires!?! Assuming the engine makes 400lb/ft, that would be close to 3340 ft-lbf!?!

Does this sound right to everyone one, or am I looking at this the wrong way?


Thanks for the time


Robert

 

The dyno takes the gearing into account, since it's looking at both engine RPM and dyno roller RPM. Otherwise you couldn't even drive a 4.11 car up on it after a 2.73 car and get meaningful readings. So, in theory, the readings won't change if you merely change gear ratios or tire sizes, oe even if you shift the trans into another gear.

Yes, the "torque multiplication" thing means exactly that. However it's accomplished on lower gears at the expense of RPMs: all that torque multiplication occurs at very low wheel RPMs. That's what horsepower is all about; high torque at high RPMs.

If you take the engine torque at any given RPM, apply the torque multilpication from gearing to it, and then divide that number (in ft-lbs) by 2 pi times the tire diameter, you'll come up with the force trying to push the car forward (assuming traction of course).