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#### dtbaker

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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?

there's lots more involved.... real wheels are not solid discs which means that the 'r' for moment of inertia is not the rim. and then you are assuming that there is nothing attached to the wheel? you'd have to figure ALL the rotating weight you are accelerating AND the F=ma for accelerating the non-rotating weight thru space.

#### dtbaker

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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).

this all handles the non-rotating parts.... I was just trying to point out that to be accurate you would need to find moments of inertia for all the rotating parts too to predict the acceleration. Not insignificant with the wheels plus brake rotors, all the gears/clutch/flywheel in the drivetrain, and the guts of the motor itself.

#### dtbaker

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hence the need for 'a better lever' with a transmission.... besides a warp9, even at plasma-inducing amps isn't going to win any races in a 3000+# vehicle.

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