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TORQUE calculations for Brushless dc motor.

  1. Sep 21, 2008 #1
    OK hey guys,

    I am building an electric go kart for a final project, and i'm just trying to get my initial justifications correct, with respect to the classical physics involved.

    SPECS: i was the car to be able to reach 60km/h,
    The total weight of the car will be approx 200kgs.
    the size of the tires of the car are 13cm in diamator.
    (i know from research that conventional gokarts have approximately 70NM Torque and 550RMp at the wheels)

    so i need to calculate the torque... at the wheels , and then decide on the gearing ratio! this decision is obviously indicative on the paramators of the motor i have! and the goals i want to achieve! The Motor i have has a max RPM of 5200, and a peak torque of 7.56NM

    can anyone show me best how to perform these calculations ? i understand that i need acceleration ! can i assume constant acceleration ? if that is the case can i just say i would like the car to accelerate to 60km/h in 10 seconds

    ... any help will be much appreciated !
  2. jcsd
  3. Sep 21, 2008 #2
    It seems to me that you would want the motor to deliver the maximum amount of work per second. That means using it at a number of rotations per second where the number of rotations multiplied by the torque reaches the maximum value. So you would first need to make a graph of torque versus RPM. In this graph, the surface of a rectangle between the origin and any point of the graph is the power associated with that point. The point which gives the rectangle with the largest surface is the desired one. From this the gear ratio should follow.
  4. Sep 21, 2008 #3
    i was thinking more along the lines of :
    given i know what i want the car to achieve.. acceleration and speed
    i can work out the stall torque, of the system. Hence the torque to get the system moving initially.
    knowing this i can see if my motor will be able to achieve this via gearing ratios.. The motor has a set torque, i cant change that, we have been supplied it via the university. I'm just looking for confirmation if my requirements can be met.
  5. Sep 21, 2008 #4
    There really is no single equation that fits all states of a PMSM but this one works in general:

    Te = 1.5p[[tex]\lambda[/tex]iq + (Ld - Lq)idiq]

    p is the number of pole pairs

    [tex]\lambda[/tex] is the amplitude of the flux induced by the PMs in the stator phase

    Lq and Ld are the q and d axis inductances

    R is the resistance in the stator winding

    iq and id are the q and d axis currents
  6. Jan 31, 2012 #5
    hello my friend firstly I build same as your final project can you pls hepl to me about that.What I am going to do.What do you advise.I build body of the car and other part of the car but I can not do anything about electric motor.When I analyse your project I see the weight of the car and speed of the car are same.Which type of engine you select for this.By the way ıf you have any calculations.Can you pls send it to me my mail adress muhammerhalici@hotmail.com. Thank you very much
  7. Jan 31, 2012 #6


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    A crude approximation is that torque versus rpm curve is a straight line, with 7.56 NM torque at 0 rpm and 0 NM torque at 5200 rpm. Peak power occurs at 1/2 max rpm: 3.78 NM torque x (2600 rpm =) 272.27... rad/s ~= 1029.185 watts. You would probably want peak power somewhat below top speed.

    13 cm diameter, you sure this isn't the radius? That's a small tire.
    Last edited: Jan 31, 2012
  8. Jan 31, 2012 #7


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    Another couple of useful formulas...
    For a given acceleration [itex] a [/itex] you'll need a force of [itex]F = ma[/itex] and that implies a torque at the wheel of [itex] \tau_w = F*R = m a R[/itex].

    The peak torque is probably the starting torque for a brushless DC. So you can figure the gear ratio by the ratio of motor torque to wheel torque.

    For the 0 to 60km/h = 60,000/3600 m/s^2 = 10/6 m/s^2 and 200kg car that's 333.3 Newtons. 13cm diameter = .065 m radius gives you 21.67 N m wheel torque.
    You'd need around a 2.866 gear reduction from motor to wheel. Make it an even 3:1 and you'll have a bit of extra torque to start, acceleration will likely go down as your speed goes up.

    You can do the same for speed and RPM understanding that the actual max speed will be a power and friction question.

    At 60km/h = 1.666... m/s the wheel must be turning so this is your circumferential speed. Its circumference is [itex]C=\pi D = .13\pi m=0.4084m[/itex]
    V/C gives the rotations per second of about 4.0809 RPS = about 245rpm
    With a 3:1 gear ratio that comes to 735rpm at the motor.

    The max RPM is well above this but you'll never see max RPM under load, what you will be concerned with is whether the torque at the 735rpm is sufficient to overcome friction from gears, wheels and wind... also you may be more limited by your supply power than the motor rating here.

    Power is torque times RPM converted to radians per second: RPM * 2pi (rad/rot) / 60 (sec/min) = angular speed in radians / second. This times Newton meters = Nm/s = J/s = watts ( = volt amps for power supply calculation, You'll need extra supply power since the motor is not 100% efficient and neither is the transmission a fudge factor of 2 would be appropriate.)
  9. Jan 31, 2012 #8


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    60 kph = 16.666 m/s, so wheel speed ~= 40.81 rev / sec ~= 256.41 rad / sec ~= 2448.54 rpm. With a gear ratio of 1.062:1 (engine rpm : wheel rpm), engine rpm would be about 2600 rpm at 60 kph. A gear ratio of 1.1:1 would translate into 2693 rpm at 60 kph.
  10. Jan 31, 2012 #9


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    Ooops! Thanks for correcting my math error, that seemed a bit low. What I get for being in a rush to get to class.
    Yes so I'd think a 1:1 gear ratio i.e. direct drive would work fine.
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