mcaul

For a given rear wheel speed, the RX8 engine runs higher RPMs than other cars. I was wondering where the extra revolutions get taken out? Is it the transmission? Is it the differential?

I started to post the question, but decided to see for myself, comparing my 6 speed RX8 and 5 speed 1990 Mustang.

Here are the transmission gear ratios I looked up...
RX8: 3.76/2.27/1.64/1.19/1.00/0.84
Mustang: 2.95/1.94/1.34/1.00/0.63

Here are the differential gear ratios...
RX8: 4.44
Mustang: 2.73

This says, for example, in first gear, the RX8 engine has to turn 3.76 times to get 1 revolution out of the transmission. The Mustang engine has to turn 2.95 times. Similarly, the RX8 drive shaft has to turn 4.44 times to get 1 wheel rotation. The Mustang drive shaft has to turn 2.73 times to get one wheel rotation.

Its interesting to compare first gear all the way through,
RX8: 4.44 * 3.76 = 16.69 engine revs for 1 rear wheel revolution
Mustang: 2.73 * 2.95 = 8.05 engine revs for 1 rear wheel revolution

As a 'sanity' check the numbers can predict the speed of the RX8 in a particular gear and RPM. For example, given that the RX8 tires are 26" diameter, if we run at 3000 RPM in 1st gear, the wheels will be spinning about 180 RPM and we should be going about 13.9 MPH. Similarly, in second gear at 3000 RPM we should be going 23 MPH. It checks out, but that may not necessarily mean I'm sane.

Blue87Sport

But remember that the rotors are turning at 1/3 the speed of the eccentric shaft. So your 3.76 rotations is only a little over one complete rotation of the rotors = 3 combustion cycles (one per rotor face.

zoom44

now drop the Torque numbers in your equations there and see who has more T in 1st

mcaul

Well, one thing does lead to another... but, it looks like thats been talked out.

https://www.rx8club.com/general-automotive-49/horsepower-torque-49469/

Neat stuff.
M.