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Old Apr 04, 2007, 02:38 PM
IBCrazy is offline
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Creating the optimal motor

Now that you know how to rewind the motor, it's time to get a little more advanced if you want. Below is how to calculate the optimal motor as well as much of the math and physics necessary to do so.

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 values 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

Any questions or comments? Send me a PM or post if you need more details. I might have left something out. As a final note I'll give some examples of motors and how they are wound.

Tower Pro - bm2409-18 - 18 turn double wind Delta
Tower Pro BM2409-21 (BP21) - 21 turn double wind Delta
Just Go Fly 400DF - 8 turn WYE

Gobrushless triple stator:
- 4 turn Wye = 2100kv
- 6 turn WYE = 1550kv

Gobrushless double stator - 6TWye = 2150kv

I'll post more as I destroy more motors. It's only a matter of time

Also from experiment I have found that a speed control senses the position of a WYE wound motor more easily than a Delta. At higher rpms my delta winds get choppy than comparable KV WYEs. For this reason and the fact that WYEs are more efficient, I almost always terminate my motors as WYEs.
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Last edited by IBCrazy; Sep 19, 2009 at 11:30 PM.
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