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#### Anaerin

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Okay, so I have a wheel of radius r (Meters), attached to a vehicle of weight M (Kilograms). If I apply a constant torque T (Newton-Meters), how much acceleration do I get?

My shaky math comes out to:

Inertia I=W*r²
Therefore Acceleration = (T/I)*r (In Meters/second²)

Right?

#### Anaerin

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I was trying to simplify things, and just assuming you had to move what was effectively a M-massed weight on the end of a r-length arm (Effectively, the force necessary to move the wheel against the ground). Essentially I need to calculate the force created by the motion of the wheel against the ground, and use that as the F in an F=Ma (Rearranged to solve for a, so F/M=a). Yay, more equations!

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

#### Anaerin

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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).
Okay, let's throw some numbers in here.

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

Vehicle weight: 1556KG