So first of all, hello! I have a question - how much of a torque a car needs to produce that a wheels would start slipping? Lets say i have a car which weights 3200 kg, and has a wheel radius of 0.4394m. So i've been told that to calculate that, i need to do the following procedures(thats very basic): 1.) Calculate normal force: F = 3200 * 9.81 = 31392 N 2) Multiply normal force by a friction coefficient(lets say its 1) and wheel radius: T = 31392 * 1 * 0.4394 = 13793,64 Nm 3) Divide by number of wheels: T/4 = 3448,41 Nm. And we get the result - to slip a wheel you need to send to the wheel 3448,41 Nm of the torque. Is that true? How accurate is this method if its true? What about if a car has a rear wheel drive? A wheel will still slip if it reaches over calculated torque? Or you need 2x amount of that(for both wheels)? I'm just confused here.. I need that because i'm calculating a set of gears, and i've calculated that a car at it's peak can send a total amount of 9000 Nm(after all gearing), and thats just crazy amount of torque - gears will be just too big if i would try to calculate them to withstand that. So i know that a wheel will start slipping much earlier and a wheel will never reach 9000 Nm, and thats what i need to calculate - how much of a theoretical torque a wheel can get in a worst case scenario with an LSD differential(i've been told that theoretically in a worst case scenario, one rear wheel of a rear wheeled car with an LSD differential could get a full torque from an engine).