Thread: Discussion Motor formulas View Single Post
 Jan 20, 2009, 02:32 PM Engineer for Christ Amherst, VA Joined Jun 2006 11,093 Posts Calculating an optimal motor from existing parts Most of us need to use existing parts to build up a good motor. I will use an example and step through the process and formulas to making an optimal motor. My example motor will be a Scorpion 3008 Kit Scorpion 3008 12N14P kit. I want an efficient 2200kv powerhouse from this motor. The formulas are not exact, but they will get you very close. Much closer than a blind guess. A motor runs most efficiently when: Rotational loss = Copper loss Mind you, sometimes this occurs outside the motor's power handling capability, so it's up to you to decide whether it's possible to make it work in the real world. It can be estimated a motor dissipate about 1 Watt of heat for every 2 grams of motor. Obtain original motor specs Starting with the Scorpion 3008 - 32 specs we have: 1087 kv Io = 1.14 amps 16 turns delta. (Scorpion uses wires per slot notation which is # turns * 2) Determine field rotation # magnets/2 = field rotation factor Field rotation factor * kV = magnetic cycle/V So with 14 magnets, field rotation factor = 7, thus field rotation = 7609 cycles/v For 2200 kv: 14 magnet - 2200 * 7 = 154000 cycles/V 10 magnet - 2200 * 5 = 11000 cycles/V 8 magnet - 2200 * 4 = 8800 cycles/V Determine rotational loss Rotational loss can be estimated: New rotational loss = (New rotation speed/old rotation speed) ^ 2 Again for 2200 kv: 14 magnet - (15400/7609)^2 * 1.14 = 4.67 Amps That's 56 Watts on a 3S! 10 magnet - (11000/7609)^2 *1.14 = 2.38 Amps or about 28 Watts - half that of 14. 8 magnet - (8800/7609)^2 * 1.14 = 1.35 Amps or 16 Watts. Mind you less magnets might mean less rotational loss, but you also now need more turns which increases copper loss. # of turn calculation Original field rotation speed * # turns = new field rotation speed * new # turns Thus: New Wind = (Original field rotation speed / Desired field rotation speed) * Original # of turns Again targeting 2200 kV we calculate # turns for 14,10, an 8 magnets 14 magnets - (7609/15400) * 16 = 7.9 - We'll round it to 8 turns 10 magnets - (7609/11000) * 16 = 11.05 - We'll round to 11 turns 8 magnets - (7609/8800) * 16 = 13.8 - We can estimate about 14 turns. Calculate Rm Copper loss may be estimated by two methods: Method #1: New Rm = ((New # turns/ Original # turns) * Original Rm) / (1+ % fill change) Where % fill fill change is an estimate of how much more or less filling of copper you plan on fitting in the stator. This can be calculated or just estimated. Method #2: Rm ~ (Length of wire per phase (in feet) / resistance per foot at 60 degrees C) * Termination Factor Where Termination factor is 1.5 for WYE terminations and .66667 for Delta terminations. Using method #1 assuming I can wind with 25% more copper fill: Rm of original = .092 ohms 14 magnets - (.092 * 8/16) / 1.25 = .0368 ohms 10 magnets - (.092 *11/16) / 1.25 = .0506 ohms 8 magnets - (.092 * 14/16) / 1.25 = .0644 ohms Calculate Copper loss Copper loss = I^2 * Rm At this point you need to make an estimate of what amount of power you want your new motor to produce. I'm going to target 30 Amps for simplicity's sake. Copper loss: 14 magnet - .0368 * 30^2 = 33.1 Watts 10 magnet - .0506 * 30^2 = 45.5 Watts 8 Magnet - .0644 * 30^2 = 58 Watts Calculate Total Loss Total loss = Rotational loss + Copper loss As a motor is loaded up, it's rotational speed drops. Therefore to calculate this value, you need to estimate what percentage of no load speed your new motor will be running. 80-85% is a fair estimate for most motors. I'll use 85% speed. Remember that rotational loss is proportional the magnetic field rotation squared. Loaded magnetic loss ~ % no load speed ^2 * no load loss Loaded magnetic loss: 14 Magnet: 56 * .85^2 = 40.5 Watts 10 Magnet: 28 * .85^2 = 20.2 Watts 8 Magnet: 16 * .85^2 = 11.56 Watts Total Loss: 14 Magnet: 40.5 + 33.1 = 73.6 Watts 10 Magnet: 20.2 + 45.5 = 65.7 Watts 8 Magnet: 11.56 + 58 = 69.6 Watts Calculate %efficiency % Efficiency = (Power in - Total Loss) / Power in My 3S Lipo should be around 11 Volts when discharging at 30 Amps. So my Power in is 11*30 or 330 Watts. % Efficiency: 14 Magnet: (330-73.6) / 330 = 78% Efficient 10 Magnet: (330-65.7) / 330 = 80% Efficient 8 Magnet: (330-69.6) / 330 = 79% Efficient Thus we can conclude the 10 magnet configuration is the most efficient. Calculating peak Efficiency As stated in the beginng of this post, peak efficiency happens where copper loss = rotational loss. We can estimate that rotational loss will remain about the same. Calculating where peak efficiency occurs: Peak eff Amperage = (Rotational loss/Rm) ^ 1/2 14 Magnet: (40.5/.0368) ^ 1/2 = 33 Amps 10 Magnet: (20.2/.0506) ^ 1/2 = 20 Amps 8 Magnet: (11.56/.0644) ^ 1/2 = 13.5 Amps From here you put you obtained vaues into the % efficiency calculation and you get: 14 Magnets: 78% Peak Eff @ 33 Amps 10 Magnet: 81.5% Peak Eff @ 20 Amps 8 Magnet: 84% Efficient @ 13.5 Amps This should get you fairly close to the actual values you will see in your motor if your estimations are close. There will be more dramatic of a spread in motors with a lesser quality stator. -Alex Latest blog entry: Project Covert Ops: Long range ground...