View Single Post
      10-25-2007, 09:51 PM   #69
Lieutenant General
swamp2's Avatar

Drives: E92 M3
Join Date: Sep 2006
Location: San Diego, CA USA

iTrader: (1)

Hint - final drive

Originally Posted by ChitownM3 View Post
You would think that the high rpm would equalize that, but in fact, it doesn't! The fact that the Lexus has more hp means that it has more torque at higher rpms. While the Lexus can't rev up to 8400 rpms like the M3, the Lexus has tons more torque up to its redline of 6700 or whatever it is and will be pulling the m3 HARD until it has to shift at 6700 rpms.
Originally Posted by ChitownM3 View Post
Here's some simple math I did, and tell me how this is incorrect. I set up an equation where X/295 = 8400/6800 Pretty much we are solving for X. When you solve you get X to equal 364.41 torque. This is what you would need to equal the power of a car making 295 torque with an 8400 rpm. The Lexus makes more than 364.41, so if I am correct, if both cars have gear ratios in which their redline makes them shift at the same mph, the Lexus would be putting more power to the ground. I think ur program is flawed, no way the m3 should have more peak thrust than the lexus in first gear when according to your graphs it seems that they both shift at the same speed.
I'll give you a big hint.

Car, 1st gear ratio, final drive ratio, gear ratio product

That is a 50% advantage in torque multiplication for the M3! When you want to determine instantaneous accelerative wheel torque (and force and then F=ma for vehicle acceleration) the wheel torque is simply multiplied by the product of the gear ratios (there is some scaling as well but that is not important to do a simple A-B comarison). So M3 torque in 1st gear x gear factor = 295 ft lb x 15.6 = 4602 ft lb. IS-F torque x gear factor = 371 ft lb x 10.0 = 3710 ft lb. No surprises here! Add in the much higher redline and you will see more of the picture. Hope that helps.