I'm trying to calculate the maximum acceleration a car can achieve with the current gear ratio. I am ignoring drag forces and friction to keep it simple.

I'm doing this by:

  • calculating the torque on the wheels: $T_e/G$ (engine torque / Gear ratio)
  • Calculate the Force exerted on the ground: T_w/r (wheel torque / wheel radius)
  • then finding the acceleration using $F=ma$

However acceleration values are too small. Is there something wrong with my reasoning?

My car's engine Torque is 500Nm, gear ratios: [2.97,2.08,1,0.86,0.56], car's mass = 2500Kg


1 Answer 1


The gear ratios are gearing down, i.e. they reduce the angular velocity at the wheels and increase the torque. The correct formula is:

$$ T_{wheels} = T_{engine} G $$

Plus don't forget that there is a differential gear as well as the gearbox, and this reduces the angular velocity and increases the torque as well. The value of $G$ in your equation is the gearbox ratio multiplied by the differential ratio.

Response to comment:

I don't know what the differential gearing is in your car, but differentials are generally around 4:1 so let's take this value. In that case the torque at the wheels in first gear is:

$$T_{wheels} = 500 \times 2.97 \times 4 = 5940Nm $$

I would guess the tyre radius is about 0.3m, so the force at the contact between the tyre and ground is:

$$ F = \frac{T_{wheels}}{r} = \frac{5940}{0.3} = 19800N $$

So the acceleration is:

$$ a = \frac{F}{m} = \frac{19800}{2500} = 7.92ms^{-2} $$

That seems pretty good to me. 0.8g!

  • $\begingroup$ hmm ... thanks, it makes more sense. Yet this makes my acceleration values even smaller. I'm expecting values of around 1.5m/s^2 yet this gives me acceleration values like 0.0028m/s^2 which is way to low. :/ $\endgroup$
    – Jonny
    Mar 23, 2013 at 18:23

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