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Considering that KV is RPM/Volt, does it ever tell us anything about the torque? Wouldn't you need to be told the torque separate from the KV rating? If torque can be derived from a KV rating then why is a lower KV rating considered more powerful?

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  • $\begingroup$ The torque follows from the power. $\endgroup$ – CuriousOne Oct 30 '15 at 23:56
  • $\begingroup$ @CuriousOne Can you explain what you mean? I edited my question to clarify. $\endgroup$ – Johnston Oct 31 '15 at 0:04
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For a linear DC motor the "back emf constant" in units of volts/rad/sec is equivalent to the torque constant of the motor in units of N-m/Amp. But the manufacturers will usually specify the back emf constant in units of volts/RPM, so the numbers will differ.

Strictly speaking a brushless motor is no a linear DC motor, but rather a synchronous motor, and the equations of motion that model it are nonlinear, and much more complex than the linear DC motor. But when feedback is used to control the brushless motor, its behavior can approximate the linear DC motor and so sometimes they will specify an equivalent back-emf constant.

Judging by the units you write for "KV rating", I would suppose its the reciprocal of the back-emf constant. So a lower KV rating implies a higher back-emf constant or equivalently higher torque constant, thus a more powerful motor.

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Yes, you can relate Kv constant to BLDC motor torque by a very simple f-la: $$ \tau=8.3*\frac{I}{K_v} $$

or

$$ K_t=\frac{3}{2}*\frac{1}{\sqrt3}*\frac{60}{2\pi}*\frac{1}{K_v} $$

enter image description here

a thorough explanation can be found here

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All else the same, KV is inversely proportional to the stall torque of the motor. But, all else is seldom the same, and stall torque often isn't that useful a metric.

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There is another constant used to describe electrical motors that quantifies the torque, the torque constant Kt

$$ K_t =\frac{\tau}{I} $$

where $\tau$ is the torque, and I is the current.

Therefore, the torque a motor produces is $ \tau = K_t I $. The SI unit of Kt is the weber.

It turns out that:

$$ K_t = \frac{1}{K_v} $$

So a lower Kv implies a higher Kt; a higher Kt means the motor will produce more torque from a given amount of current.

Also, a given motor can be wound with more turns of thinner wire (higher Kv), or fewer turns of thicker wire (lower Kv). Thicker wire can handle more current; a motor that can handle more current can create more torque without burning out.

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  • $\begingroup$ So @nick, KV tells us nothing about torque? $\endgroup$ – Johnston Oct 31 '15 at 0:23
  • $\begingroup$ The motor constant is different than the torque constant. Check your units - the two equations you write are not consistent with the units. $\endgroup$ – docscience Oct 31 '15 at 0:38
  • $\begingroup$ The second equation is not right. You need to use the torque constant, not the motor constant. $\endgroup$ – docscience Oct 31 '15 at 0:40
  • $\begingroup$ This is only true when the motor is stalled. $\endgroup$ – Daniel Griscom Oct 31 '15 at 0:43

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