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Wheel force

  1. Jan 4, 2008 #1
    I tried to figure out the absolute maximum force that a car can create. But the numbers I got are obviously wrong.

    I started by trying it on the http://www.supercars.net/cars/1177.html" [Broken]. Torque is 649.4 Nm. First gear is 3.23. Then there's a "final drive ratio" of 2.37. I thought that this is first gear multiplied by the rear differential, and should be around 10. So I decided that this is the rear differential, and got a product of 7.6551. So the torque at the axle is 649.4*7.6551 = 4971.22194. Given a tire diameter of .315*.9 m + 17 in = 0.7153 meters, this means a force of 4971.22194 / 0.7153 = 6949.84194 N.


    "(60 mph) / gravity on earth = 2.73512362 seconds" - Google. F1 has a 0-60 of 3.2, giving 0.854726131 g. That by the mass of 1140 gives a force of 974.38779 kgf, or 9555.48002 N

    So, the theoretical force is 6949.84194, and the real is at least 9555.48002. Probably more, because the car can't be at peak torque the whole time. That is impossible.

    What did I do wrong? My guess is that it has to do with the final drive thing. Or maybe I left something out, like bore and stroke. I don't know what I need to do with them.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jan 4, 2008 #2


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    Gold Member

    Well, you're baically correct in your calculations, peak torque in the engine multiplied by the transmission and axle ratios should give you the maximum torque at the wheels. This would be the torque at a fleeting moment in time if the car was driving such that the engine is putting out maximum torque but the engine is not accelerating.

    EDIT: I just realized because the final drive ratio is the TALLEST gear multiplied by the axle ratio... maximum torque would have to be first gear multiplied by the axle ratio. It looks like the car's axle ratio is 2.55 if you take the final drive divided by 6th gear, so you maximum torque ratio would be 3.23*2.55=8.23; giving you a maximum torque of 5350 N-m... taking 5350 N-m divided by the rear tire radius (not diameter) gives you about 14,960 N.

    The basic problem is that looking at the peak torque and horsepower for the engine can't give you the whole picture of the accelerating car. For starters, the acceleration curve of a car going from zero to sixty is not a flat line, but your calculation of the car's acceleration assumes it is.

    The same problem is run into from a conservation of energy standpoint. The McLaren can do 0-60 mph in 3.2s so that means if we divide its kinetic energy at 60 mph by the total time taken to get there we should get the average power output from the engine for that run. Problem is, this gives you an average of 171.8 hp, far short of the McLaren's peak horsepower output...

    Basically, there are far too many dynamic issues in an accelerating car to solve for using simple equations.
    Last edited: Jan 4, 2008
  4. Jan 4, 2008 #3
    "radius (not diameter)"

  5. Jan 4, 2008 #4


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    I remember an incredible e-mail that a racing friend sent me a while back. It was the numbers for power, torque, acceleration, fuel consumption, etc. for top-fuel dragsters. Unbelievable. So if you want to know the max that you can get, look into the numbers for top-fuel dragsters.

    I tried a little googling to see if I could find the info, but didn't exactly find it in my brief search. Did find this interesting article about torque and horsepower for cars and dragsters at Yahoo answers, though:


    And this dragstrip ET extimation calculator:

  6. Jan 7, 2008 #5
    60 mph is attained in X direction, 'g' acts in negative Y direction.
    Taking m~1200kg and force~7000, acceleration is (7000/1200)
    Now apply (Final velocity/Acceleration in the direction of motion) = Time.

    And the downward acc is far more greater than 1g for a formula car
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