Step 6: Actually Winding the Motor

If you've never wound a motor before, the diagrams of LRK windings are probably pretty meaningless. This is a time when you need to learn the nomenclature of motor hobbyists.

An example would be the dLRK winding:


or the classic LRK winding,


What? Did you just sing the alphabet song or something? Kind of. The three phases of the motor are referred to in this case as A, B, and C.

A capital letter indicates one winding chirality, a lower case means the other. For instance, if A is designated "make a loop of wire in the clockwise direction", then a means "wind the loop of wire in the counterclockwise direction". And a dash or space means an unwound tooth.

The general convention is capital letter equals clockwise loop, lower case equals counterclockwise loop. But, what is more important is consistency. If you do it one way, stick with it. 

So what does the above string of gibberish mean? Starting at any tooth (mark this as your index!), begin making loops of wire around it according to the designation. To wind two teeth Aa style, wind one of them clockwise, and the other counterclockwise (or vice versa - keep track of this.)

There is no "right method" to obtain clean windings, but the last thing you want to do is just bundle wires around the tooth with reckless abandon. For large motors, use latex gloves to ease hand abrasion and a wooden dowel to wrap wire around for extra leverage.

Unfortunately, I don't currently have any pictures of video of me winding a motor. This might change in the near future to save a thousand words of explanation.

Perhaps one of the most valuable resources available is the Combination Table. Input your number of stator teeth ("nuten") and your number of magnets ("pole") and it will automatically generate the correct winding pattern! The above table was generated by one of the Crazy German R/C Airplane Dudes, who seem to be the source of all technological advancement in the model motor scene.

Single Layered, Multi Layered

You may find that you can't get the N number you want by only winding one layer of wires on the stator. Simple solution: Keep winding and make a second layer.

Two to three layer windings are generally the limit of heating & cooling unevenness for small motors, and the Rm gets ridiculous as well. As more layers are added, the end turn effect will become more and more of a factor.

If you find yourself having to wind many layers, perhaps switching down a size of wire will alleviate that.

How many turns (N) do I need? 

The other killer question of small motor design. Given other motor parameters, you can backsolve easily for the minimum N needed to achieve a certain design goal, usually torque. Accounting for losses and assumptions, N should be above this number by a comfortable margin explained shortly.

Example (Updated 3/28/2012 to correct the math which has been wrong for over 2 years! I keep meaning to fix it, then never getting around to it. Ultimately, enough of you called me out on it, so congrats. Here's the fixed math using also the new torque constant factor m).

Let's say that I want to design a motor inside a 12cm (0.12m) wheel that will let me climb a 10% grade (or about 5.5 degrees inclination) at velocity v = 5 m/s (about 11mph), and I weigh m = 65kg. The force of gravity F pulling me back down the hill is

F = m * g * sin 5.5° = 61N, or thereabouts. I

I want to climb the hill at 5 m/s. Mechanical power is torque * rotational speed, but it is also linear force * linear velocity.

Thus Pm = 61 * 5 = 305 W

Seems reasonable, right? Assume the motor is a perfect transducer (it's definitely not). The electrical power required is also 305 watts.

Assume my battery is 28 volts, so i = 305 W / 28 V = 10.9A

To exert a linear force of 61N at a radius of 0.06m (wheel radius), the torque T is 3.66Nm.

Two variables, T and i, have now been established. The motor is a 12-tooth, 3 phase motor, so m is 4 (there are four teeth per phase). You can now reduce the equation to

T / (4 * m * i) =  N * L * R * B

R is ultimately limited by the size of my magnet rotor and inner diameter of my tire - a topic which is forthcoming. Let's say that my wheel choice has forced a  maximum stator diameter of 70mm, and the motor can't be more than 30mm wide to fit in my vehicle. 

B is my magnetic field strength. Let's assume it is 1 Tesla for now - we will see soon that this is not a bad guess if your motor magnets are reasonably thick.

T / (4 * m * i * L * R) = N

Let's see what this comes out to.

3.66 / (4 * 4 * 10.9  * 0.03 * 0.035) = 19.98 = N

This value is a reasonable first approximation for the number of turns per tooth you need.  Since hundredths-precision turn fractions aren't possible, take the closest integer: 20.

Fiddle Factors and Hand Waves

Every nonideality and inefficiency in the world will work to make your motor faster (read: less torquey) than what the number of turns alone would indicate. Therefore, it is good sense to consider this as the absolute minimum number of turns per tooth. The torque constant value derived from using NIBLR is generally 20 to 33% too high for average fractional-slot, permanent-magnet motors like the type we are considering.

Remember also that motors are not perfect transducers. The average efficiency of a decent BLDC motor is somewhere around 90%. So, if I want to perform this hillclimb at maximum efficiency, that's much different than attempting it at maximum power output. The efficiency of a motor at maximum power output is always less than 50%. This is something to be well aware of - if you are using this 'target output force' method to design your turn count, then you should take the speed to be somewhere close to your anticipated cruising speed. This makes sure that, if anything, you overdesign the motor for torque as nonidealities only take it away from you.

The above motor example is the motor for Project RazEr. In actuality, RazEr's motor has 25 turns per tooth - overspecified by roughly 25%.

To wrap up, R and L are mechanical constraints dictated by your vehicle's mechanical parts while m, B, N, and i are electromagnetic constraints dictated by your choice of magnets, wire, and coil layout.

<p>This looks so cool, but after I read the why not to build list it got a little stale. </p>
<p>how large a motor would I have to build to power a bike light consisting of approx 50 led lights?</p>
<p>Could you just clarify for me the &quot;AC&quot; part of this whole thing.</p><p>Does the controller simply emulate the commutator of a DC motor, energizing the coils in such a way as to create a rotating magnetic field or are you saying it also converts the supplied DC to a true alternating current which it distributes to the coils in such a way as to produce a rotating AC field?</p><p>Thanks</p><p>Doug</p>
hi. i'm from tunisia. good job man !. can you contact me on jakefouly@gmail.com i need some help. thank you
<p>hello body. i have not came here for a while , if you need help i am in, but ask what you want in my mail: pejman_22000@yahoo.com or skype: kasra_Sa2</p>
<p>hello guys. i am new member. hope to could find good things for better future for succeses. tx all</p>
<p>Issues with your formula:</p><p>You pick Mevey's 2.30 equation:</p><p><strong>T = 2 * N * B * Y * i * D/2</strong></p><p>Where N is the number of turns per pahse, and '2' is the number of active phases. Then you redefine N as the number of turns per tooth, and define m number of teeth per phase. So (previous definition of N) = (new definition of N) * m. The result equation:</p><p><strong>T = 2 * </strong><strong>m * </strong><strong>N * B * Y * i * D/2</strong></p><p>Not four but two!!!</p><p>Next, in Mevey's 2.30 equation D is the diameter of the coil centers. Not stator's outer diameter, but diameter of the coil centers which is obviously smaller than stator's outer diameter.</p><p>P.S. Thanks for the article anyway. It is really motivating!</p>
<p>the most in depth view about brushless motors awesome saving this page for future guidence cheers</p>
<p>Hi, </p><p>good math derivation, T = 4 * m * N * B0 * (t / (t + g)) * L * R * i, <br>but this is exactly the double the torque you calculate using what reported on <br>LRK Motor Analysis Worksheet<br><a href="http://www.femm.info/examples/lrk40/lrk-bldc.pdf" rel="nofollow">http://www.femm.info/examples/lrk40/lrk-bldc.pdf<br></a>T = 4 * (rr+rs)/(rs-rr) * Br * N * L * t * ( I - I0)</p><p>where this formula is calculated taking into account that just two phase current are active on a trapezioidal drive, and that the current to be used should be just the active current (total current less free running current).</p><p>Which formula is the correct one???</p>
Torque should definitely depend on current. So, it should be T = 4 * m * N * B0 * (t / t + g) * L * R * i instead of T = 4 * m * N * B0 * (t / t + g) * L * R
<p>Or better, using correct parenthesis position:<br>T = 4 * m * N * B0 * (t / (t + g)) * L * R * i<br>but furthermore:<br>- for &quot;i&quot; you have use the useful portion of total current, i.e. &quot;i&quot; - free running i zero (taking into account the eddy current and friction losses);<br>- for &quot;R&quot; you have to intend the medium radius between stator and rotor (where the magnetic force act) which is in the middle of (t+g)=magnet thickness + air gap zone.</p>
Hi mate I'm making a hub motor for a skateboard, 100kv 15turns in wye dlark with 6 strands 0.34mm wire (this is all I can fit)<br>80mm diameter wheel, 50mm diameter motor, 40.7mm x30mm stator 12t 14p, with 40SH magnets 30x7x3, air gap between 0.5-0.7 depending on tolorance.<br>Will be using 29v batterys 8ah 30c.<br><br>I want to get up a hill of about 10-15% grade at about 20km/h<br>70-90kg. How much power do I need, is my wire cross section ok?<br>Iv been struggling with the math, and worried about the new winding not having the current capabilities I need, but it's hard to fit the copper inside of the stator, but maybe I'm just not good at it!<br>I'm also using hall sensors or optical sensors soon to get better start up as I seem to get a lot of cogging! From even a small push!<br><br>Each skateboard is using 2 motors on the rear.<br><br>I really need help!<br>My email is jacob.bloy(at)gmail.com<br><br>My build page.<br>http://endless-sphere.com/forums/viewtopic.php?f=31&amp;t=65636&amp;sid=32c74875705d1d55d0801eeae1381c11
<p>So I have a question about the equation to measure the theoretical torque. T = 4 * N * B * L * R * i in my case would be 10 turns per phase, 52 for the neodymium magnet and assuming you measure things using the imperial system the length of the stator would be 19.8 inches and the stator radius 3 inches. Putting through 42 amps would theoretically give me 5189184.0 torque. Now this can't be accurate because at 1500 RPM that would give me 1.4821e+6 horse-powers just to put things into perspective, which is a insane amount of horsepower.<br>t = 4 * 10 * 52 * 19.8 * 3 * 42 - Where did I go wrong in the equation?</p>
<p>If we rewire series connection of coils to parallel i.a.w. <a href="http://www.thebackshed.com/windmill/FPRewire.asp" rel="nofollow">http://www.thebackshed.com/windmill/FPRewire.asp</a> without changing position of hall sensors, does it influence on steering algorithm? How to include this wiring difference in one equation(e.g. torque equation)?</p>
<p>Excellent instructable! Thanks for taking the time to document and share your work. I need clarification on (at least) one topic. In the section discussing Magnet Length the author states:</p><p>&quot;Optimally, the magnet length is equal to the stator length (<em>L</em>).&quot;</p><p>In the same section, magnet width is mentioned. Would someone clarify for me the magnet dimensions that should be used for a given stator? Specifically, what is the stator length (L)? A diagram would be especially useful.</p><p>Thank you.</p>
<p>Jah mahn, danke mahn, huge, this is monster info bro.</p><p>Thanks you!</p>
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<p>You mention that you used 2 x 22AWG wires instead of 1 x 18AWG wire because it was hard to wrap and bend for 25turns. You said &quot;Use the wire gauge table to compare diameters!&quot;, now 22 AWG wire is 0.644mm in diameter and 18AWG wire is 1.024mm in diameter. So 2 x 0.644mm is 1.288mm and thats well over the diameter of the 18AWG wire. Now 24AWG wire is 0.511mm in diameter, and 2 x 0.511mm is 1.022mm which is a lot closer to 18AWG. I don't want to be annoying i'm just confused. If we compare the surface areas though 18AWG wire is 0.823mm^2 and the closest pair that would measure similar is a pair of 21AWG wires, at 0.411mm^2 x 2 = 0.822mm^2, BUT neither of those are the wire you said you were using. Should the diameter of the multiple strands not add up to close to the diameter of the single wire? Thanks for any help, just confused.</p>
<p>Hi, I thought I'd jump in here and clarify a few things for you: When doing anything with electrical wire, especially when said wire is going to be carrying a significant percentage of its maximum safe current, you should be aware that the determining factor in current capacity is the cross-sectional area of the conductor, not the outside diameter. Since the area increases faster than the diameter or circumference, a wire of half the diameter will have one fourth the cross-section and thus one fourth the current capacity. A wire 1.024mm in diameter has a section of 0.82 mm^2, where a wire of 0.511mm diameter has a section of only 0.20mm^2. A 0.644mm wire has a section of 0.32, which means a pair are up to 0.64, close to the original 0.82. If I were doing this, I'd use three strands of 22, for a section of 0.96, better than the 18 gauge.</p><p>TL;DR don't go by diameter, go by cross section. They don't equate directly.</p>
<p>Thanks for the formulas to make my motor, but your math was off by .01 on 3.66/(4*4*10.9*0.03 *0.035)=19.98.</p><p>it is really 19.99 (19.986)</p>
<br> I really love your <br> write-ups guys continue the good work. <br><p>http://www.zapelectricianbrisbane.com.au/</p>
<p>Being one of the last electrical and electronic engineering graduates from my school, before they dropped the &quot;electrical&quot; part, electric machines have always been a favourite subject of mine. This 'ible is one of the best I've ever read. Excellent work. </p><p>Incidentally, you can get tyres made by the guys who can retread forklift truck wheels. They vulcanise the tyre onto your own hub. </p>
<p>Being one of the last electrical and electronic engineering graduates from my school, before they dropped the &quot;electrical&quot; part, electric machines have always been a favourite subject of mine. This 'ible is one of the best I've ever read. Excellent work. </p>
<p>Oh my, that's a lot of work and thanks for putting it all up here.</p><p>I was looking for a motor I could pass my leg though instead of using a ring gear and a small motor to rotate it off to one side. The open motors would be used to rotate segments of a leg roughly depicted here: </p><p><a href="http://youtu.be/RV9fvg3C_fo" rel="nofollow">http://youtu.be/RV9fvg3C_fo</a></p><p>I'm still working out how many segments and at what angle and speed each segment should rotate at for the maximum comfort of the rider while still providing a good, natural leg motion. Seems making the motor would be beyond my capabilities and I'll have to settle on the ring gear driving by a motor or the like.</p>
very interesting very ( ty iv bine tring to find info on moters like this )( o and I Quote &quot; Their large outrunner motors are inexpensive enough to consider cannibalizing for stators. &quot; <br>well LOL!!! ) thank you for this it was very help full. :)
Can the stator core be plastic? Does it need to still be magnetic at all? I dont' understand why you wouldn't build it out of something completely non-magnetic
No not plastic. The material has to have a high permeability to concentrate the magnetic fields and at the same time reduce Eddy Currents.
I wonder if a motorcycle stator from the magneto would make a nice stator for a brushless motor? Used they are not too expensive.
Is it possible to melt down many cores in a foundry and then cast my own core? The core I need is huge and would cost a lot of money to have it machined and cast by a specialized group. Especially when I will need at least 3.
Stators are not cast. If you look at one, you'd notice they are many thin and fine layers. Each of those actually are insulated from each other. <br><br>A cast stator would basically be a big magnetic brake and would be extremely inefficient and heat up quickly due to eddy currents. I think you should look into motorcycle alternators and washing machines for large-ish (5&quot; - 6&quot; - 12&quot;) stators.
Where can we buy one of these motors (not the wheel) as a kit to put together and learn? It's easy to get the wire, but not the pieces the wires get wrapped around :-(
it looks to me that the torque should be proportional to the square of the radius. At constant magnet induction and current density the force per unit circumference length would be constant so the torque would be proportional to the radius and the length of the circumference, in turn proportional to the radius , hence the radius square.
Im an electrician, and house wiring is done in 14, 12 and 10 gauges mostly. Winding a motor in 18 gauge must be a chore! But im sure chris farley would say, &quot; It builds dexterity!&quot;
That's a lot to read but I read it anyway I can't make one of these. Because I don't have the tools nor the supplys to build it but awesome job
very nice and educative. learned a lot from this.
So there is probably something stupidly wrong with what I am about to write, but I am tired and can not get this idea out of my head, so on with it. <br> <br>What is to stop you from taking a standard dc motor, like the ones used in toy scooters, and reinforcing the !#$@% out of it, namely in the (casing? or is it a shell?) itself and the axel, welding a rim to the (reinforced) casing of the motor and using that as a hub motor with the motors axel acting like the axel of a bike wheel, with everything revolving around it? <br> <br>Would the motor just plain not have enough torque?, or is there some other blatantly obvious issue that I can't think of?
here is a mild example of your concept and a solotiuon using a standard dc brushed motor as an axle or pivot point. and having to add a gear reduction to it to get it to move. <br> <br>http://www.instructables.com/id/6-AXIS-ROBOTIC-ARM/ <br> <br>check it out. <br> <br>and vote for me <br>
You would not have enough torque, even on scooters with small diameter wheels the motor is usually geared down at least one to five, on a bike you will need a gearing of at least four times that! I hope this helps.
Sort of like this, or this.
On this page you have a picture of some small car hub motors. Can you tell me where the came from?
how much cost for four wheeler hub motor?.
i have a question (yes, i read the entire instructables). If the thickness of my magnetic field isn't determinant can i build the stator as thin as i want? or there is some equation that relates the power (Pe) with the area of stator across each coil? <br>because when i study electronics there was a equation for transformers that related that. <br>i'm an electronic technician, but i never design a brushless motor (sorry for my english man, i do my best) <br>Thanks.
Has anyone investigated the use of motorbike / scooter magneto stators for converting the bldc motors? The smaller / basic bikes without alternators appear to have magnetos with stators that look very similar to what is needed for the centre of a bloc motor. There seems to be heaps of cheap 18 pole and 8 pole magneto stators and a few 12 pole stators. From the bike sites, it seems these fail fairly often on bikes, so there would be heaps of old stators being thrown away.
Absolutely - this has been done by a few people, most notably:<br><br>http://wattsdottime.blogspot.com/ (both as motor and as a generator)<br>http://amymakesstuff.com/2011/06/07/pf-hub-motor-complete-mostly/<br><br>The only downsides to the bike stators are their large bore and narrow teeth - can't stuff as many windings on it as you otherwise would be able to, and the teeth potentially will saturate earlier. But, those are all parameters you can design around.
Here's a (very simplistic) thought - just <em>how</em> important is it that these &quot;electrical steels&quot;/&quot;transformer steels&quot; <em>contain</em> silicon <em>as an</em> <em>alloy</em>?<br> <br> What I'm thinking is laminating&nbsp;very thin steel sheets with <em>very thin</em> sheets of silicon to mimic the effects you outline... Or maybe just laminate the steel with a silicon adhesive...<br> <br> There'd probably be a minimum size beyond which you'd lose too much efficiency, but would larger, slower-turning, motors be feasible?
Where would you get these &quot;very thin sheets of silicon&quot;?
Sorry, my bad. I put the idea down as it came, without&nbsp;checking up, so I don't even know if they exist... - that's partly why I mentioned the adhesive.<br> <br> Thinking about it now, anything thin enough would probably not be available in small-enough quantities to&nbsp;be affordable for the DIY'er (yet, anyway).<br> <br> So, we're back to the adhesive - and you'd probably need 100% pure silicon (like aquarium sealant is, I think).<br> <br> &nbsp;Maybe an idea for someone else to tinker with?

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