



The thicker the wire, for a given/desired number of winds (Kv), the higher the efficiency.
Vriendelijke groeten Ron 




This thread would be a whole lot better if the more knowledgable people would contribute just one more redundant post about the fundamentals.
Example: How does the magnet size and dimension affect a motors performance. What about the airgap in relation to the thickness as well as the strngth of the magnets? The diameter of the wire is probably the most covered topic in this section, but it wouldn't hurt to cover it again. A little searching with reading will give you the answers eventually, but threads like this can really break it down and speed up the learning curve. So, lets post some basics and fundamental functions. Please, Thank you! 



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Good afternoon colleagues. I have a motor (stator2208,12 poles, 14 magnets)
How to calculate the maximal working current ({power}), and maximum theoretical current,with standard cooling. I know resistance, KV, voltage and сcurrent. Is there are any formulas for this?I'm know one formula IH=1000PH/√(ηHUHcosφH) bun not working wery well on 3Phase MODEL motor. For example I winding LRK 20 turns, resistance ~0.1om (100mOm), a current 1.4A, KV 1900. (3S Lipo) .How calculate maximum current(power).?I'm not want take calсulators, only theoretical, whenever possible would be desirable without the stand. Thank you very much. 



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"2208" stator class leads to a motor around 40 grams A "rule of thumb" gives an approximate power max (for some minutes) 3W per gram > 120W Of course for bursts 10 seconds you can go higher (140160W), it is a question of high quality magnet wire (HighGrade enamel) The rising temperature is a slope + asymptote Of course cooling air flow and/or fixingmount (aluminum calorific capacity) play an important role. Louis 




Which magnet dimensions fit best?
When you are in front of a magnet supplier's catalog like supermagnete in Switzerland,
http://www.supermagnete.ch/eng/magne...&switch_lang=1 http://www.supermagnete.ch/eng/magne...&switch_lang=1 you can be disapointed with your dimensions choice depending on the CDrom motor you have to stuff. This Excel sheet, in Excel 2003 format makes the calculation for you. It is degined for metric but feel free to convert it for imperial if you prefer! Enjoy, MrFixit 



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. 8085% 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: (33073.6) / 330 = 78% Efficient 10 Magnet: (33065.7) / 330 = 80% Efficient 8 Magnet: (33069.6) / 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 



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Do you have real world realizations for your 2200 rpm/V desired Kv , from Scorpion 3008 kits , to compare with 14, 10 and 8 magnets, windings to compensate for same Kv ? I am surprised that you could start from 78% to reach up 84% for the same rpm at the shaft on the same prop/load (approx same Kv, same voltage Vbat) I am interested by your results, anyway. Louis 




Hey! I just realized there is a formula missing.
Kt = T * L * D^2 Where T is a torqueconstant for the motor reliant mostly on the magnet strength and airgap. L is the length of the stator stack and D is the diameter of the stack. There is another Kt formula based on the actual values instead of just a constant. Kt = 1/Kv but depending on the Kv units it may become: Kt = 60/(2*pi*Kv) This turns a rpm/volts Kv into a radians/second/volts so that Kt becomes Nm/A. An interesting thing about the first formula is it shows that, as mentioned, The relationship between Kt and Kv is inverse proportional and that length of stack has a linear relationship with Kt and Kv. What it also shows is that the diameter has a squared relationship! This is important if you need more torque. For a inwheel motor, maximizing diameter will increase torque by D^2 while increasing width only adds L. 



Kreature
When you increase D, generally some parameters could be changed: * total volume of magnets (via width and thickness in addition to axial height L) * number of conductors per slot (or turns per tooth) * sometimes number of slots, number of magnets So Kv (or Kt = torque per A = 1/Kv) can vary for any diameter. It is difficult to say Kt = k L D^2 in a "generic manner" .. I don't say that your formula is wrong .. In your formula you have to indicate what remains constant or what is changed when diameter is increased or what is your coefficient T. But for a given lamination patternconfigdiameter , given turns per slot, given widththickness magnets, given airgap : Kt = k L sure ! (L = axial height for conductors AND magnets) One can say that torque/A = Kt depends of D 1) per the lever/arm (radius or diameter) = linearly 2) and secondly per the volume of magnets 3) per the number of turns and slots that could increase if D increases Of course the "possible max power" increases also with L D^2 that is also similar to total weight and volume. For a generic tendancy I prefer to say Power = K L D^2 instead of torque Of course for the same rpm : Power and Torque are proportional Louis 



Flyboy
You could add one formula for Rm dependence of copper temperature Rm = Rmref (1 + 0.004 *(T Tref)) temperatures T, Tref in Celsius degrees °C (or °K) Tref could be 20°C Pin = Pout  Iron_losses  Rm * I^2  ESC_losses Iron losses are function of rpm and can be easily estimated via no load testings I use an order 2 polynomial best fit (A,B,C coeff) for any motor I test : Iron_losses = A + B *(rpm/1000) + C*(rpm/1000)^2 Louis 



I did not say that for two given motors, k would remain constant.
It would change if pole parameters, airgap and such were to change. What I wanted to get clear was that if you can increase the diameter by a tiny bit, you would get it back squared compared linear when increasing the length of the stack. The formula itself comes from Dr Duane Hanselman. It can be used to estimate the benefit. For me, it has a big significance as I am working on a inwheel motor. By reducing the rubber around the motor and increasing the motor diameter I can gain significant torque. Also, the formula to get Kt from Kv is quite usefull as it is much more versatile. I know the torque I need maintaining a speed on the scooter so I can use the Kt to see if I have the amperage in the right magnitude for my batterys. Since I will be using 12 LiIon's in series I have a lot of voltage, but the cells do not like high discharge. You are ofcource right in that there are many more factors changing with D than with L. Also, it is important to realize that although the motors maximum power will be the same regardless of the number of winds for instance due to the rpm being related to torque and power it is also reliant on increasing the voltage to the motor or it will simply fail to draw enough power. 



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In a first approximation, if you increase the diameter D a little bit 1) Keeping L 2) Keeping the same magnets, turns, teeth .. You get only a linear increase of Kt = kD not squared (lever/arm influence) But of course if you change D the geometrical dimensions of slots/teeth vary a little so there could be some little influence regarding the magnetic flux reluctance. That secondary influence is not necessary linearly related to D, nor constant. Louis 

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