Hi, First post and yes, I did do a search and found lots of discussions about torque steer, torque vectoring and the 85Ds impressive 864 ft-lbs of toruqe but nothing that explains this: The Model S electric motor makes 443 ft-lbs of torque at 0 RPM, right? That's driven through a 9.7:1 gear reduction, meaning that the Tesla has 4,297 ft-lbs of torque available at both wheels. Yet, even the 85D with 2 motors only shows up on a dyno as having 864 ft-lbs of torque. I thought maybe the wheel and tire acts like a gear but with a 1" shaft and a 30" diamter wheel, that would mean only 143 ft-lbs of torque at the point of contact between the tire and the road. Ok, there's 2 tires so that equals 286 ft-lbs but that's still no where near the 864 ft-lb dyno result. Would someone please school me as to how much torque is available from the motor right through to the wheel and why these results aren't showing up on any dyno? Thanks, Rob

Your calculation is good up to the point where you reduce the torque number because of wheel diameter. If you do that you get force that is pushing car forward at some exact speed and acceleration. dynos are mechanical devices filled with fudge factors that may have some correlation with how ice behaves, but have zero correlation with ev world. So, there are lies, statistics and dyno numbers. the best assesment of power is accurate measurment of car acceleration and speed curve and weight. Dynos are previous century tech but until previous century folks die out, it will still play its dying role. get an accurate accelerometer and good car scale and calculate acctual wheel power. Forget dynos.

A chassis dyno measures the torque at the wheels and then divides by the gear ratio to calculate the torque at the motor(s), which is the reported result. You calculated correctly that if the 2wd motor makes 443 ft-lbs, then multiplied by the GR it gives you the torque at the output of the differential, which is directly coupled thru the driveshafts to the wheels, so that is the total output torque at the rear wheels (the 4297) and each wheel carries half. The shaft and wheel/tire radius can be used to calculate the torsional force that the component must carry, e.g. used in calculation of material stress margins of the driveshaft, but the torque thru the shaft and into the tire is the same as what is output from the diff. The torsional force on your 1" shaft would be (4297/2)/(.5/12) ~52000 lbs, but at the contact patch of a 30" tire the force is only (4297/2)/(15/12) ~1720 lbs, and the total tire force would be twice that ~ 3440 lbs.