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Calculating Torque

  1. Oct 2, 2006 #1
    I am trying to calculate how much torque will be exerted on a half shaft on the rear axel of a car. I am a little confused as the power of the engine is put through a chain-drive and then turned by 90 degrees through the differential before it reaches the half shafts.

    Do you just assume the power is inline (parallel) with the axel and if so what information do you need to know? I will assume that the half shafts are only taking the torque of the engine and are not supporting any weight.

    I am a bit lost about where to start on this and need some help!


    Last edited: Oct 2, 2006
  2. jcsd
  3. Oct 2, 2006 #2


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    It's all about torque multiplication!

    The torque at the flywheel of the engine will go through the transmission (which depending on the gear, will be multiplied by the gear's ratio) and then the differential (which also has an associated gear ratio.) Also keep in mind the differential divides the torque by two, equal parts for each wheel.

    By the way, you mention a chain-drive in your driveline, is this "car" (truck) 4WD? Are you given operating efficiencies for each part of the driveline, or the driveline as a whole, or are you supposed to assume no losses?

    So, for an example, not taking into account efficiencies:

    Assume 350lb-ft at the flywheel of the engine (475 N-m).

    1st gear's ratio is 3.643:1, so after the transmission, the torque is multiplied to 1275 lb-ft (1730 N-m).

    The differential's gear ratio is 4.300:1, so the differential multiplies the total torque to 5482 lb-ft (7440 N-m), but this is then divided by 2 since the torque is split equally (assumption) between the two tires. So, each tire (assuming it's 2WD) can get a maximum of about 2741 lb-ft (3720 N-m) of torque.

    From this number you can calculate the maximum instantaneous force applied to the road as well, given the tire's diameter (assuming no-slip).
    Last edited: Oct 2, 2006
  4. Oct 2, 2006 #3
    it is actually a 2WD car-it is using a motorbike engine to drive the rear wheels using the chain.

    Thank you for your help-it all seems so simple now...I wasn't sure if there were other conversions to do because of the change in direction on the force, but what you've said seems to make sense.

    Thanks again for the help. I am one happy chappie! :smile:
  5. Aug 4, 2008 #4
    I have looked at several publications regarding calculating the final drive wheel torque and none of them mention dividing the diff torque by two to get the final drive wheel toruqe?

    The formula they give
    drive wheel torque = flywheel torque x first gear ratio x final drive ratio x 0.85

    So I'm a little confussed?
  6. Aug 5, 2008 #5


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    It's quoted that way because people aren't generally interested in torque at a particular wheel; they're interested in torque "at the wheels" (i.e. both of them).

    Good point though.

    (Also, be careful with that 0.85!)
  7. Aug 6, 2008 #6
    Vehicles only have equal torque to each wheel when they are moving in a straight line on level ground. When it turns the outside wheel must travel faster than inside one, therefore, the inside wheel carries more of the torque. Other things that can cause unequal torque are uneven tire size (one worn more than other or different pressures), banking and driving on a side hill. Also some no slip differentials sent total torque to one whee, if other spins out
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