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I have a brushless DC Motor that i am considering buying from hobbyking.com with 2100Kv (RPM/v) i am trying to follow the equation on Wikipedia's motor constants page i have included the equation here but i am having difficulty understanding whether i am using the calculation correctly. Motor Torque Constant

My calculation is 60/(2 x π x 2100) = 0.00454 Nm/A and this should be the value of Kt described above, also because Kt = Ke my back EMF constant will also be 0.00454?

Can anyone help me understand if this is correct?

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This is correct.

\$K_t\$ is approximately equal to \$K_e\$ (approximatly as \$K_e\$ is defined as open-terminal voltage and \$K_t\$ is defined at rated current & the machine may be saturating)

\$K_v\$ is the reciprocal of \$K_e\$

The units of \$K_t\$ are \$\frac{Nm}{A}\$

The units of \$K_e\$ are \$\frac{V}{\omega}\$ (NOTE: from a magnetic point of view this is peak line-line voltage...)

so... 2100\$\frac{rpm}{v}\$ == 219.911 \$\frac{rad/s}{V}\$ = \$K_e\$

Thus taking the reciprical of \$K_e\$

\$K_e = \frac{1}{K_v}\$ = 0.0045473

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  • \$\begingroup\$ your answer seems correct but i am unsure as to whether i understand it, where did you get the 219.911 what calculation did you do to get that and could i use the 0.0045473 in other equations such as to find torque at a given current 0.0045473*105=0.0154nm? \$\endgroup\$ – Chris James Dec 5 '17 at 14:55
  • \$\begingroup\$ 219.911 is the Kv expressing in rad/s/V as oppose to rpm/V. Basically 2100*(2pi/60). This is exactly the same as the wiki equation you have stated EXCEPT each step is very clear rather than a non-derived equation being used without any information as to what is being done. \$\endgroup\$ – JonRB Dec 5 '17 at 15:32
  • \$\begingroup\$ also 0.0045473*105 = 4.774665Nm.... ie 105A would produced 4.77Nm of torque. do you really mean 104Amps. \$\endgroup\$ – JonRB Dec 5 '17 at 15:34
  • \$\begingroup\$ Yes 105amps I know it sounds a lot but I'm working with RC hobby motors. Off topic I'm programming a calculator application for motors so it's easier for people to plug in values and get an answer. \$\endgroup\$ – Chris James Dec 5 '17 at 17:07
  • \$\begingroup\$ Well it isn't a lot, its a relative consideration. The reason I asked did you mean 105A is because your maths does not make sense. You wrote: " 0.0045473*105=0.0154nm? " while 0.0045473*105 = 4.774665Nm thus either your current was wrong or your maths was wildly wrong. \$\endgroup\$ – JonRB Dec 5 '17 at 17:10

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