How to calculate minimum voltage to start brushless motor with 37mΩ resistance, 305KV?

Hello-  I'd like to know how to predict the min. voltage required for a specific brushless motor to begin turning.  The motor specs that I know are:
    Internal resistance: 37 mohm
    Current capacity: 56A/60 sec
    Max efficiency: 18 - 40 A
    No load current: 1.1 A
    RPM/V: 305 RPM
    Recommended power: 6-8s Li-Po

Thanks in advance!

It depends on the load. If there is no external load then the only thing you need to overcome is the internal friction and friction-like components.  Since you don't know them we are left guessing. The one hint we have here is no load current of 1.1 amperes. Unfortunately that doesn't give the speed for the no load spec. You're interested in it at zero speed where it should be less unless there is a lot of sticktion (higher static friction than dynamic friction). In any event 1.1 amps across the resistance of the motor (assuming it is measured line to line) of 37 mOhm, would require 0.041 V across the terminals. So I would say that the voltage required to start the motor would be something less than 41 millivolts.

Thanks for your reply John-So... please pardon my ignorance, with no load at all,
this motor will start spinning with very little voltage -- except that it must overcome whatever friction exists.The reason for my question is that I plan to use a timing belt reduction to connect a motor to a system that I expect will require more start-up torque than the motor is capable of if connected at 1:1.  Since I do not have enough background in electromechanical design, I will likely end up using trial and error.

Hi, Dean,Yes, theoretically it should take very little voltage to start a motor turning with no load on it.  Just enough to overcome losses.  If the motor has a lot of cogging torque (torque required to overcome the attraction of the permanent magnets to the uneven shapes of the steel inside the motor), that will require more.  You can tell if the motor has a lot of cogging torque by turning it by hand.Is it smooth or do you feel torque ripple?I should point out though that it is really current that is required to get the motor turning.  Electrical current through a motor results in torque being generated at the shaft.  Yes, in order to get that current to flow, you will need voltage.Similarly, it is voltage that is required to achieve speed.  So when I say 41 mV to get it started, I mean, to just get it to turn.  Then to get it to really spin, you need more voltage.  This is because as a motor spins, it acts as a generator and generates a voltage, commonly known as back EMF (electromotive force).  To support a load, the applied voltage must be higher than the back emf voltage.  The difference between the two appears across the resistance of the motor which, by Ohm's law is what determines the amount of current that will flow to provide the torque needed to match the load.  So, you should not expect the motor to "spin" with a fraction of a volt, simply, it should move a little.  To get any appreciable speed, the voltage must be increased.You gave the spec "RPM/V: 305"  This likely means that for every volt applied to the motor it will go 305 RPM, no load.  Thus if you apply 10 V to the motor you should expect it to spin at around 3050 RPM.  If you use 6 to 8 Li-Po cells in series, the applied voltage will be  6 to 8 times 3.7 V or 22.2 to 29.6 V.  This should spin the motor at 6771 to 9028  RPM.You are worried about startup torque.  Though the numbers for brushless motors can be a little different than for brushed motors, lets see if we can figure out how much torque the motor will be able to provide. The back emf constant, Ke, is 1/(305 RPM/V) = 0.00328 V/rpm.  In SI units Ke is 0.0313 V/(rad/s).   It turns out that in SI units, the torque constant, Kt, is numerically equal to the back emf constant so Kt = 0.0313 Nm/A.  Since the motor spec says it can handle 56 A for 60 seconds, it should be able produce nearly 56 A * 31.9 Nm/A = 1.75 Nm for 60 seconds before overheating.  What do you expect your starting load torque to be?  Don't forget that the timing belt will add a little more to the torque loss equation.Best, -JD

Thanks, again, JD-I'm afraid the starting torque will be difficult to estimate.  I am trying to replicate a 1/6 scale flying version of the successful drive system used by the Wright Brothers in their airplanes of the 1907-1912 era.  It involves two contra-rotating propellers driven via roller chains and sprockets.  I've constructed most of the system, including 1/4-inch pitch nylon roller chains, sprockets, tubular chain guides and two scale propellers of 18-inch diameter.  I think I need to fully assemble the system and actually measure the torque resistance.

I am attaching four images: A 1908 Wright airplane, some of my components, and a sketch of a potential electric motor belt drive: