# Is electric motor torque determined by Kv or by power (watts)

I'm trying to size a motor for an e-bike, but I'm confused as to what motor parameters determine the maximum torque that it can output (Kv or power in watts); I've found the following formulas:

Power (watts) = torque (Nm) * angular_velocity (rad/s)
also note that P = I(amps) * V(volts)

and

Torque (Nm) = Kt * I (amps)
Where Kt is the torque constant or (1/Kv), and Kv is rpm/volt

So let's say we have a 1000 watt motor with a Kv of 200, the motor operates at 24V. Let's also say that we're talking about a low rpm (60 rpm);if we multiply rpm by (2 * pi / 60) then we get angular velocity. Using P = I*V we can calculate the max current to be 41.6 amps.

So according to the first formula, with a low rpm of 60 we could get a torque of ~159 Nm at max power.

Using the second formula we get a torque of 0.208 Nm at max power

These are massively different, and would require changes to gear ratios, battery sizes, material choices, etc.

So my question is: which one is right? Is torque determined by the Kv of the motor? or is torque determined by the power rating of the motor?

• The first equation is correct. Can you provide a link to the second equation, as it seems to have bugs?
– r13
Sep 27 at 23:39
• Is Kv in the title & the body of the question a constant or kilo-volts (kV)?
– Fred
Sep 28 at 0:15
• @Fred $K_v$ is a constant for brushless motors which relates rpm to Voltage applied. Sep 28 at 0:19
• @r13 this is the link to the website. I saw it in a few locations, but this is the tab I still had open: learningrc.com/motor-kv Sep 28 at 2:30
• If you have read through the article you provided, you should have found it also states T = Kt*I, which is the same as shown in my answer, Kt = T/I - the slope of torque/current curve.
– r13
Sep 28 at 2:40

Your definition of $$K_t$$ seems a mistake. By definition, the torque constant is simply the slope of $$T (Nm)/i(amp)$$ curve of a motor, and It should be noted that the parameter $$K_t$$ is not related to the voltage under which the motor is operated. If you use the motor at 12VDC or 24VDC this constant will remain the same. This attribute of the motor is very useful as the motor is used in a motion control system. The overall torque output of the DC motor system is maintained by monitoring current while the motor’s speed output is varied by controlling the voltage into the motor.

• I think I have a better understanding. T = Kt * I is the correct formula, but most online hobby shops post Kv as rpm/Volt, but this needs to be converted to rad/Volt*second Sep 28 at 3:59
• This would give the above motor a torque closer to 2 Nm, which is much closer to what you would see on an electric scooter motor of that size. Sep 28 at 4:03
• @FidelG Kv = Speed/Voltage = (60rpm*0.1047rad/s/rpm)/24V = 0.26175; Kt = 1/Kv = 3.82; T = Kt*I = 3.82*41.7 = 159.3 Nm.
– r13
Sep 28 at 11:14

Kv is for back EMF. It is part of what you need to model the current, in addition to the resistive and inductive model of the motor windings. The current then gives your torque via Kt. This is all to a first approximation.

Here's a paper that describes the model, in Section II.C, equations (1-3) https://www.researchgate.net/publication/4260443_Modeling_and_Analysis_of_the_Technical_Performance_of_DC-Motor_Electric_Bicycle_Drives_Based_on_Bicycle_Road_Test_Data

As you note yourself, Kv = 1/Kt. 200 Kv (RPM per volt) = ~21 rad/sec per volt. Invert this to get a torque constant of 0.04775 Nm per amp. This is what you want to know. As you calculated, at peak power, the motor will draw 41.7 amps. Multiply this by the torque constant to get about 2 Nm of torque.

The issue I see is that you are assuming that all of the motor ratings are valid across the entire curve. They are only valid at one point. We don't yet know at what RPM peak power occurs.

Let's use your example of a 1000 W, 24V, 200 Kv motor: 1000W is the peak power. But you don't yet know at what RPM this power is achieved. For an ideal motor it is when the input and output impedances are matched. At 24 V, and 200 Kv, the motor would spin at 4800 RPM no load (zero torque). Peak power would occur at 2400 RPM. But be careful. Motors can't always operate across the full range of torque and RPM. 24 V is the max allowable voltage. But there is also a max allowable current (41.7 A). So at RPM below 2400 you MUST limit the current to 41.7 Amps. You will not achieve more than 2 nM of torque. And total power will be reduced. At RPM greater than 2400, you must limit the voltage to 24 V. Current (and thus torque) will be reduced, and again, total power will be reduced. You can see that there is only one place where you can actually achieve peak power. See the example image below.

Most motors are rated for operation at much higher speeds than peak power. Usually around peak efficiency. In this case the manufacturer should specify at what RPM peak power occurs.

Note that if you calculate power by your first equation using 2400 RPM (260 rad/s) and 2 nm, you get about 500 W. This is correct because the motor's power rating is the input power. When the input and output impedances are matched, half the power is dissipated in the motor, and half in the load (mechanical output). Sometimes motors are rated in terms of output power (usually for industrial or high performance uses), But then they MUST give you values for max allowable current as well as voltage. Reputable vendors will give you the exact torque curve and safe operating range.