Frank,
To actually calculate power in Watts, the formula is torque in Newton-meters times the angular velocity. Angular velocity is the angular changeper second, expressed as radians (1 turn=360 degrees= 2*pi approximately =6).
Normally we talk about rpm==revolutions/minute, so to convert to revolutions/sec (rps) , divide by 60. To convert to angular velocity from rps, multiply by 2pi or 6. If you look carefully, the conversion is approximately rpm/10.
Now most of us can't measure torque, so all the conversion numbers are pretty meaningless. But what is nominally true is that the torque of an electric motor linearly decreases from it maximum when the motor is stalled out (0 rpm), to zero at rpm="k"*Vbattery. "k" is the value they often quote in the motor specs, typically kilorpm/volt. So for example, the AXI 2820/8 shows a "k" =1500RPM/Volt, so a 10 V battery would spin this motor-noload (with no prop at all) at 15krpm. At this rpm, the motor will generate an effective voltage (because it acts like a generator) which exactly cancels the Battery voltage---so not current flow. With no current, the torque is equal to 0. I am ignoring any friction or hysteresis loss here, but not the ohmic resistance of the motor coil, internal battery resistrance, or ESC resistance. I note that if I look at the specs of this AXI,
http://www.modelmotors.cz/index.php?lang=czech&id=en&nc=produkty_vypis&kategorie=m_neodym_ac&id_rady=axi_28&id_produktu=axi_2820_8&nazev_rady=AXI%202820/8%20it tells me that it actually draws 3.3 Amps at 10V under no load--this is just the torque on the motor due to friction--and other small losses--in other words the "real" world.
Since Power=Torque*angular velocity, and torque is dropping linearly with rpm, and angular velocity increases linearly with rpm the power ends up maxed at just 1/2 of the "k"*Vbattery value. For that AXI above, that would be about 7500 RPM with the 10V battery. This is of course the ideal case, but tells you something about these motors. It's power and torque you want when you pull the nose up.Since the prop rpm drops under load, the torque on the prop goes up. If you are running the rpm above the max power level, the power also goes up, so your plane gets that extra boost in the climb. Of course the efficience drops as the rpm drops (since the current rises, and ohmic heating in the windings, battery, and ESC goes up), but hey---you can't have everything!
That was all at fixed voltage levels. If you run your system at less than full blast, and also enable constant rpm load in the ESC, you even can get more! As the rpm drops, the ESC turns up the effective voltage to the motor, so you get even more power. And when you drop the nose, the ESC lowers the effective voltage, making the prop act as a big break in the downhills. Since we are apparently allowed to run ESC's in this mode, it sure sounds like a neat way to run motor control. The main point is that the plane needs to fly acceptably at the lower power settings for this to work.
What I kind of enjoy about electrics is that you can actually make nominal calculations on how the motor works--they are a lot easier to model and understand than the standard 2 stroke engine, where you need to understand airflow, mixtures, turbulance, motor loads. Of course there is still the complications of a real model flying on lines with a real propeller, so actual experimentation is the real way to find out what works.
Topic: how much of the battery gets used in a 6 minute run? (Read 3162 times)