A bit confused about relation of torque-force

Discussion in 'Physics' started by Manjaks, Mar 17, 2012.

  1. Manjaks

    Thread Starter New Member

    May 2, 2010

    I am doing a little research about electric motors in EVs and I am a bit confused about this:

    As i find out:
    max torque of Tesla roadster motor is 270 N*m,
    the mass of vehicle is 1235 kg,
    the acceleration 0-100 km/h is in 4 sec.

    Using the formula M=F*R
    I find out that the "pushing" forward force of motor is 675 N (the letter R in formula is radius of rotor, and I put 0.4 in it, although it is already TOO much but never mind)

    After that I use formula F= m * a (mass times acceleration, which is 6.95 m/s^2)
    I found out that to achieve this kind of acceleration the output force must be 8577 N.

    IMHO these two formulas have equal solutions, doesn't it?

    So the question is why they differ?

    Thank you!
  2. mlog


    Feb 11, 2012
    Assuming a loss-less system, the power transferred is equal but the forces are not necessarily equal. If the motor generates "P" watts, then the wheels touching the pavement receive "P" watts. Power (watts) = Torque (N-m) * Angular velocity (rad/sec). If there is any kind of gearing to affect the angular velocity, then the torque will change. The force applied by the wheels to the pavement will depend on the size of the wheels too. If the motor radius and the wheel radius are different, it wouldn't be a surprise to find the forces tangential to the circumferences are different.
  3. strantor

    AAC Fanatic!

    Oct 3, 2010
    There is gear reduction in the axle and also in the transmission. I don't know what's in the Tesla, but I'll use the numbers from my old trans am.
    Rear end gear reduction: 4.11:1
    transmission gear reduction:
    1st 2.66:1
    2nd 1.78:1
    3rd 1.30:1
    4th 1.00:1
    5th 0.74:1
    6th 0.50:1
    My car had 475 ft*lbs of torque @ the engine. So, in 1st gear (475 * 2.66) I had 1263.5ft*lbs (assuming no losses) after the transmission. Then it's further reduced by the rear axle (1,263.5 * 4.11) to 5,193ft*lbs (7040N*M). Then the final force that's applied to the road is the axle torque divided by the tire radius.
  4. praondevou

    AAC Fanatic!

    Jul 9, 2011