My shaky math comes out to:

Inertia I=W*r²

Acceleration = T/I (In Radians/sec)

Therefore Acceleration = (T/I)*r (In Meters/second²)

Right?

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My shaky math comes out to:

Inertia I=W*r²

Acceleration = T/I (In Radians/sec)

Therefore Acceleration = (T/I)*r (In Meters/second²)

Right?

Could I use the size of the wheel as a ersatz "Gear", and if so, how would I calculate it's ratio?

Okay, let's throw some numbers in here.The wheel torque is equal to the cross product of the force the wheels apply to the road (tangential to the wheel contact area) and the wheel radius, here simply equal the product r*F since the angle between r and F is 90 degree. The force F = m*a = T/r where m is the mass of the vehicle and a is its acceleration, so a = T/r*m. The product of this force and the distance the vehicle moves is the work done by the motor to move the vehicle against the drag and rolling resistance forces. It does not include the work done to increase the energy of rotating parts, or losses in the motor, controller, batteries, and friction in the drive train. The wheel torque equals the product of motor torque and overall gear ratio (trans plus rear end).

Example car: 2010 Chevy Malibu, stock weight, with a Warp9 in direct-drive through a 4:1 differential.

Vehicle weight: 1556KG

Wheel radius: 0.3345m

Motor torque: 321N-m

So, Final torque at-the-wheel is 1284N-m (4x mechanical advantage from the differential, 1/4 the speed. Right?)

acceleration = 1284/(0.3345 * 1556) = 2.4669. If that's in Radians/sec, then acceleration = 2.4669...*0.3345 = 0.82519 m/s^2

Either way, that's not much acceleration at all. 0.25G? Definitely not going to be winning any drag races with that. So where's my math wrong?

Do let me know if you spot any problems. I'm sorry about the Javascript being all on one line, but apparently Blogger loves to insert br tags everywhere. :/

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