K_\text{T} is the torque produced divided by armature current. It can be calculated from the motor velocity constant K_\text{v}. For a single coil the relationship is: : K_\text{T} = \frac{\tau}{I_\text{a}} = \frac{60}{2\pi K_\text{v(RPM)}} = \frac{1}{K_\text{v(SI)}} where I_\text{a} is the
armature current of the machine (SI unit:
ampere). K_\text{T} is primarily used to calculate the armature current for a given torque demand: : I_\text{a} = \frac{\tau}{K_\text{T}} The SI units for the torque constant are newton meters per ampere (N·m/A). Since 1 N·m = 1 J, and 1 A = 1 C/s, then 1 N·m/A = 1 J·s/C = 1 V·s (same units as back EMF constant). The relationship between K_\text{T} and K_\text{v} is not intuitive, to the point that many people simply assert that torque and K_\text{v} are not related at all. An analogy with a hypothetical
linear motor can help to convince that it is true. Suppose that a linear motor has a K_\text{v} of 2 (m/s)/V, that is, the
linear actuator generates one volt of back-EMF when moved (or driven) at a rate of 2 m/s. Conversely, s = VK_\text{v} (s is speed of the linear motor, V is voltage). The useful power of this linear motor is P = VI, P being the power, V the useful voltage (applied voltage minus back-EMF voltage), and I the current. But, since power is also equal to force multiplied by speed, the force F of the linear motor is F = P/(VK_\text{v}) or F = I/K_\text{v}. The inverse relationship between force per unit current and K_\text{v} of a linear motor has been demonstrated. To translate this model to a rotating motor, one can simply attribute an arbitrary diameter to the motor armature e.g. 2 m and assume for simplicity that all force is applied at the outer perimeter of the rotor, giving 1 m of leverage. Now, supposing that K_\text{v} (angular speed per unit voltage) of the motor is 3600 rpm/V, it can be translated to "linear" by multiplying by 2π m (the perimeter of the rotor) and dividing by 60, since angular speed is per minute. This is linear K_\text{v} \approx 377\ (\text{m} / \text{s}) / \text{V}. Now, if this motor is fed with current of 2 A and assuming that back-EMF is exactly 2 V, it is rotating at 7200 rpm and the mechanical power is 4 W, and the force on rotor is \frac{P}{V * K_\text{v(SI)}}=\frac{4}{2 * 377} N or 0.0053 N. The torque on shaft is 0.0053 N⋅m at 2 A because of the assumed radius of the rotor (exactly 1 m). Assuming a different radius would change the linear K_\text{v} but would not change the final torque result. To check the result, remember that P = \tau\, 2\pi\, \omega / 60. So, a motor with K_\text{v} = 3600\text{ rpm} / \text{V} = 377\text{ rad} / \text{V·s} will generate 0.00265 N⋅m of torque per ampere of current, regardless of its size or other characteristics. This is exactly the value estimated by the K_\text{T} formula stated earlier. ==References==