0
\$\begingroup\$

I have an E-flite Power 46, 670Kv outrunner motor (Kv is the shorthand for RPM/volts, it has 3.9A idle current, 40A maximum continuous current, 14.4-19.2 voltage range, 925 watt maximum) but I wish to use it for a low-speed robotics application where I don't need more than 100 RPM and I want as much torque as (cheaply) possible. For any other motor type I would simply attach a gearbox, but I can't figure out the reduction ratio that I need.

If I run it at 20V (round up for simplicity), that would be 13400 RPM, so I'd attach a 120:1 gearbox and that would be close enough to 100 RPM. However, that's 13400 RPM unloaded; the gearbox will be a load and the robotic parts will be a bigger load, but I can't find anything that specifies how much the speed will change based on load. So where/how can I calculate, estimate, or even dead reckon that?

Additionally, I'll usually want to run the motor at much lower speeds, but I've not found anything saying how much torque is available at those lower speeds - most of the stuff I've found just implies that constant voltage into the motor will mean it has constant power, and thus it will provide whatever torque is needed to deliver that power. But with 925 watts, if I want to get 5 RPM out of the motor, the formula "Power (kW) = Torque (N.m) x Speed (RPM) / 9.5488" says I'll get 1575.55 Nm - which seems like a ridiculously high number compared to similarly powered gearmotors (and in this case, the motor isn't necessarily geared). It seems unlikely that the motor (without a gearbox) could run at 5 RPM, so there must be some minimum operating speed for the motor. Is there any way to find or calculate a minimum speed for the motor, perhaps depending on load and/or voltage?

Also, I intend to run this off an AC->DC converter (variable voltage and current, up to 10A 30V, but I can get large constant-voltage converters cheaply), and I do have the ESC. I'm not certain if that power supply makes a difference - such as if a battery is better at handling rapid changes in load. I could add capacitors or batteries into the system, but it is a lot easier to handle and test the system if I don't need to worry about a battery running low.

\$\endgroup\$
  • \$\begingroup\$ 670kV? Are you sure? \$\endgroup\$ – Gregory Kornblum Mar 23 '18 at 19:46
  • \$\begingroup\$ Its actually "670Kv" which is the part number of the motor :P \$\endgroup\$ – Tyler Mar 23 '18 at 19:50
  • \$\begingroup\$ Kv means RPM/V not kV \$\endgroup\$ – Sunnyskyguy EE75 Mar 23 '18 at 19:54
  • 1
    \$\begingroup\$ Speed will reduce (reducing back EMF) until (Vsupply - back EMF) / Resistance = some current I which generates the torque you need. (note you can get the torque constant from the speed constant ). At lower speeds ... Forget constant power. The torque you get will come from the current you supply. That's it. \$\endgroup\$ – Brian Drummond Mar 23 '18 at 21:06
  • \$\begingroup\$ Motors that are designed for turning a model airplane propeller, often depend on the prop-wash for cooling. If you put 925 Watts of power into that thing without a big blast of air washing over it, it could get really hot, really fast. \$\endgroup\$ – Solomon Slow Mar 23 '18 at 21:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.