## Doing the Math

Electric Motors and Controllers

### Re: Doing the Math

Ahh well, my Cabernets are crushed, but not my spirit! So Ã¢â‚¬â€œ moving right along, this means that we will have to be rigorous with our deductive testing

I might as well design from the outset with planned losses. Let me step back a moment then to gather new examples to study, for if the concept can be understood I will be better for it than hunting for boar guilelessly armed with a blunt stick

There are several questions now and I shall try to keep this organized; the proverbial can oÃ¢â‚¬â„¢ worms has been kicked:

Air Gaps and Filler:
Briefly I investigated what I think might be a more suitable approach than an initial idea of some additive to the bonding agent.
• Soft Core material: Ignored.
• Electric Steel Laminations: 4000X that of air. Good choice and commonly available even locally in my metro.
• Permalloy: Twice that of Electric Steel. No sources.
• Mu-metal: 5X that of Electric Steel. Expensive, though has my imagination twitching wildly.
This now brings up a question of construction, and methinks the most economical would be to have wedges that fit between the bar magnets rather than a challenging water-jet-cut cage. IÃ¢â‚¬â„¢m just thinking out loud; the wedges would be very cheap to fabricate, less waste of material, and better than a poke in the eye Ã¢â‚¬â€œ as in doing nothing. See Figure 1.

(Green) Electric Steel Lams between magnets

Magnetic Circuit:
Similarly, I got to thinking about completing the magnetic circuit as you well pointed out. On the left side of Figure 2 is a representation of the initial assembly: Two rotors with magnets separated by a cap at the top and bearings at the bottom (not shown), with a stator in the center and copper in-between.

The second idea to explore is using a horseshoe shape for routing the flux as displayed on the right of Figure 2. Fanciful, though I wonder how practical. Moving onÃ¢â‚¬Â¦

Double-Sided Rotor & Stator arrangements

Simplistically weÃ¢â‚¬â„¢re taught that the magnetic force of attraction is inversely proportional to the square of the distance. However in a double-sided Halbach arrangement, the calculations are expressed here:

I bought the book, but found the references on Google first. Actually the section begins on Page 107, with the Equation for Peak Flux at the top of Page 109 (3.42), and in the double-sided configuration with 3.45 & 3.46.

Bm0 = Br[1 - exp(-ÃŽÂ²hM)]((sin(Ãâ‚¬/nM))/(Ãâ‚¬/nM)) [3.42]
Solve for Bm0, where
hM = 6.35mm (1/4-inch)
nM = 4
ÃŽÂ² = 2Ãâ‚¬/la,
la = nM * (hM+ Ag) where Ag = median airgap between magnets = 10mm. This figure will vary depending on the ID of the rotor face and the number of poles in the array.
If I use an N52 cube magnet, Br = 1.48
Therefore, Bm0 = 0.61 T
We can use 0.61 T if you like, but I rounded down to 0.5 T for the sake of discussion.

The part where IÃ¢â‚¬â„¢m a little bit lost is ÃŽÂ²x and ÃŽÂ²z, however I have space between the discs (z) as 10 mm (again for the sake of discussion).

IÃ¢â‚¬â„¢m going to hold on the Winding Calculation until we sort out the Flux Density.

Have I lost anyone?
~KF
Last edited by Kingfish on Fri Mar 04, 2011 2:49 pm, edited 1 time in total.
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### Re: Doing the Math

Don't take anything I said to mean that there isn't value in analyzing a simple case like the one you used. Just have realistic expectations. Also, even a not-quite-right expression will (usually) give a result that's in the same ballpark as reality. That can be instructive, too!

Kingfish wrote:This now brings up a question of construction, and methinks the most economical would be to have wedges that fit between the bar magnets rather than a challenging water-jet-cut cage. IÃ¢â‚¬â„¢m just thinking out loud; the wedges would be very cheap to fabricate, less waste of material, and better than a poke in the eye Ã¢â‚¬â€œ as in doing nothing. See Figure 1.

Wedges sound reasonable. You're along way from construction, but a word of magnet caution - those wedges will be powerfully attracted as you try to put them in place! I wouldn't worry too much about using a high-end magnetic material, I doubt you would be able to tell the difference. The airgap will still dominate the magnetic circuit. At this point I feel it does beg the question - at what point does the effort of making the Halbach array make it worthwhile to do the simpler wedge magnets and flux ring instead?

Kingfish wrote:The second idea to explore is using a horseshoe shape for routing the flux as displayed on the right of Figure 2.

You're getting the idea, which is good, but this is not necessary and would actually be counterproductive! The complete magnetic circuit in this case does not run through the air around the stator, but instead through adjacent pairs of poles on the two rotors. That's the function of either the flux ring or Halbach array. A little flux bridge like you've drawn, if it had any effect, would likely serve to divert some of the useful flux around the stator. You're thinking in the right direction, though.

Your book looks like a good reference. In this case Bx and Bz refer to the two components of the flux vector. Bz is the useful flux perpendicular to the magnet faces. Bx is wasted flux parallel to the faces. The equations express them as functions of x, moving horizontally across the array, and z moving from one rotor to the other. Probably the most useful value for you would be the value of Bz centered between a pair of poles. In that case, z=t/2, so the two cosh terms will cancel and the sine will be 1 at x=0, so Bz=Bm0.
Eric

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### Re: Doing the Math

Probably the most useful value for you would be the value of Bz centered between a pair of poles. In that case, z=t/2, so the two cosh terms will cancel and the sine will be 1 at x=0, so Bz=Bm0.

I am very happy to learn that. One source in particular suggested the best transformer would be one where the wires were embedded with the laminations (not possible, but still the best one could hope for). Indeed, I was hoping that the flux density between the rotors would be similar in theory. I suspect that it would only grow stronger should hM, Br increase, or z decrease Ã¢â‚¬â€œ all of which is possible.

Flux Ring:
I presume you mean this sort of arrangement in Figure 3, yes?

That design requires a backing steel plate, or at least some way to capture the flux on the backside, maybe with a sheet of electric steel. Also there are only 8 pole-pairs, meaning high-rotation. LetÃ¢â‚¬â„¢s compare/contrast with for example a 9C 2806 hub, which is what I presently own Ã¢â‚¬â€œ and I should add that I find it to be quite a capable motor and a worthy challenge to betterÃ¢â‚¬Â¦

• 9C 2806: 23 pole-pairs, 51 Teeth. Magnets are 3 x 13.5 x 27 mm. K = 0.93 (according to ebikes.ca). I would be keen to know what the Tesla rating is for these magnets; for argument I presume that are 42N since that is commonly available, though perhaps the temp rating is higher.
Ref: http://www.ebikes.ca/simulator/
If you enter the custom values as 62V and 32A that is just about what I am achieving. On a flat at WOT, depending on wind I top out between 33-36 mph, and obviously faster on a downgrade. ItÃ¢â‚¬â„¢s plenty quick for urban, however I want and desire to go faster yet.

• Windmill: magnets Â¼-in x 4-in ID x 8-in OD, N42 is common, and Â½-in is available. 8 pole-pairs; about 1/3 the gearing of the 9C. Even with double-sided, I fear the rotation is much too fast. Definitely the easiest to assemble having the fewest poles and essentially no air gap between the adjacent poles. See Figure 3.

• Halbach Array: magnets 6 x 25 x 10 mm x 2 = one pole, with 20 pole-pairs & 21 teeth; that's 43% of the poles for the 9C 2806, and more than twice the Windmill. The airgap between adjacent magnets is 4mm at the OD. Planned Coil cross-section is 36 mm^2 within the 10 mm air gap (z). Leaves me room to add 1 mm carbon fiber facing/reinforcement on each side to retain the windings to the stator plate. See Figure 4.
EDIT: Corrected typo.

Horseshoe Revisited:
A little flux bridge like you've drawn, if it had any effect, would likely serve to divert some of the useful flux around the stator. You're thinking in the right direction, though.

Ha! Now IÃ¢â‚¬â„¢ve got the Ã¢â‚¬Å“happy puppyÃ¢â‚¬Â syndrome Ã¢â‚¬â€œ with my tail all wagging. Early on I had considered adding a Radial Flux component to the design but discarded it because of the issue of trying to reduce the airgap between the winding and the magnetic face. In reference with Figure 5, is that what you meant?

Miles of smiles, KF
Last edited by Kingfish on Sun Mar 06, 2011 3:40 pm, edited 2 times in total.
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### Re: Doing the Math

Kingfish wrote:Flux Ring:
I presume you mean this sort of arrangement in Figure 3, yes?

That's one configuration, but not the only one possible. With the flux ring, it's no longer necessary (or actually desirable) to have the magnets tight against each other. The magnetic circuit is completed through the flux ring rather than directly from magnet to magnet. If the magnets are too close together, some of the flux will actually short-circuit to an adjacent magnet rather than across the stator. I believe the usual rule of thumb is to keep the separation equal or greater than the airgap. Since separation isn't an issue, you can use any convenient shape of magnet. You could make a large diameter, high pole-count rotor out of circular puck magnets, for example.

Kingfish wrote:Ha! Now IÃ¢â‚¬â„¢ve got the Ã¢â‚¬Å“happy puppyÃ¢â‚¬Â syndrome Ã¢â‚¬â€œ with my tail all wagging. Early on I had considered adding a Radial Flux component to the design but discarded it because of the issue of trying to reduce the airgap between the winding and the magnetic face. In reference with Figure 5, is that what you meant?

Er, you were on the right track but you must've taken a hard right turn while I wasn't looking.

For the axial flux design, the only useful flux is exactly that - in the axial direction directly from rotor to rotor. Are you familiar with vector cross-products? (For the moment I'll assume you're not, otherwise feel free to ignore me) Going back to your Lorentz equation, the direction of the force is determined by L x B, that is the direction of the current cross the direction of B.

First of all, we can easily show that the current flowing through the end windings (that is, tangential around the circumference) will never give us a useful result. A tangential L vector (theta-directed in engineering speak) can never produce a force that is also in the theta-direction. It could only produce a force in the radial (r) direction or axial (z) direction. Both of those do nothing but put stress on the stator.

Now let's look at the current in the radial direction. A flux in the radial direction will do nothing - r x r = 0. A flux in the theta-direction will produce force in the +z or -z, doing no useful work except stressing the stator. Finally, z-directed axial flux produces theta-directed force and thus useful torque. So, any flux that does not a)travel in the z-direction and b) pass through the stator windings does not do any useful work for us.

Radial flux such as you've drawn would only do useful work if current is flowing in the axial direction... would would be a standard radial-flux motor! I'm sure some sort of hybrid is possible, but I doubt that's a rabbit hole you want to go down.
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### Re: Doing the Math

Er, you were on the right track but you must've taken a hard right turn while I wasn't looking.

Gotcha. ThatÃ¢â‚¬â„¢s why I discarded the concept early on, but for a moment you had me thinking of something else. Was it perhaps the laminate shoes as indicated in Figure 6 as another type of Flux Ring? Hmmm, I prefer to keep it simple.

Air Gaps & Adjacent Magnet Spacing:
It seems to me that the Air gap between Magnets (Agw) and the surface of the Copper Windings should be closest. Are you suggesting that the distance between magnetic faces (z) should also be less than the air gap between adjacent magnets (Agm) as well so as to promote the inductance across z (Agz)?

With all these air gaps, I figure we better have some shorthand

BTW Ã¢â‚¬â€œ thanks for explaining this
~KF
Last edited by Kingfish on Fri Mar 04, 2011 2:51 pm, edited 1 time in total.
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### Re: Doing the Math

Kingfish wrote:Was it perhaps the laminate shoes as indicated in Figure 6 as another type of Flux Ring? Hmmm, I prefer to keep it simple.

That is a valid configuration. Not sure about the relative merits, but it would work.

Kingfish wrote:Air Gaps & Adjacent Magnet Spacing:
It seems to me that the Air gap between Magnets (Agw) and the surface of the Copper Windings should be closest. Are you suggesting that the distance between magnetic faces (z) should also be less than the air gap between adjacent magnets (Agm) as well so as to promote the inductance across z (Agz)?

With all these air gaps, I figure we better have some shorthand

Sorry, I'm probably being a little unclear in my terminology. When I say airgap in this context (a coreless axial flux machine) I'm referring to the entire distance from rotor to rotor. A better term is probably "rotor spacing", which would reserve the term "airgap" for the usual magnet-to-stator definition.

The important concept here is that flux will take the path of least resistance (like electric current, water, etc). Remember that all the stator materials here (copper, aluminum, carbon fiber, whatever) all equivalent to air in terms of reluctance, so for now let's pretend the stator is gone. All the flux cares about is going from N of one magnet, to S of another magnet, then from N of the 2nd magnet back to S of the first to form a complete circuit. It'll take whatever the shortest (least reluctance) path is to do so. If one of the adjacent magnets is closer than the opposite rotor magnet, some of the flux will tend to take that path instead of the useful one through the stator.

This isn't true of the adjacent magnets in the Halbach configuration, since they are oriented to form the desired flux path. You would want the opposite rotor pole to be closer than the next Halbach pole, though, for the same reasons as above.

For fun, a couple of FEM examples:
Planar Array1.png
Magnet-to-magnet spacing 2x airgap distance

Planar Array2.png
Magnet-to-magnet spacing 1/2 airgap distance
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### Re: Doing the Math

Understood. Neat FEA pics; did you make those?

OK - So image me as a dog with a bone; IÃ¢â‚¬â„¢m not quite finished chewing on this yet:

In a Halbach Array, the orientation will be

Ã¢â‚¬Â¦L-S-R-N-L-S-R-NÃ¢â‚¬Â¦
With L = Left, S = South, R = Right, and N = North, repeating etc.

In my design I can guarantee that the distance between a N and a S pole at the shortest leg on the ID of the Rotor face will be just about twice as far as the z-distance, the Agz, the opposite rotor pole. Therefore I think we are good here and in the pink. <nods>

In this case, does it matter then if in the Halbach array the adjacent magnets touch since we want the L and R flux to be directed towards N and away from S? See Figure 7.

I have another flux ring for you since weÃ¢â‚¬â„¢re talking out loud:
The top diagram in Figure 7 is a Halbach Array representation: Nice nÃ¢â‚¬â„¢ happy flux is flowing, we have a dandy lilÃ¢â‚¬â„¢ sine-wave, looks pretty good, all is swell.

The bottom image has pole shoes applied. Now, IÃ¢â‚¬â„¢ve seen a diagram like this on the K&J Magnetics website and itÃ¢â‚¬â„¢s also in the book IÃ¢â‚¬â„¢ve been using Ã¢â‚¬â€œ though the authors could be talking Swahili as I donÃ¢â‚¬â„¢t understand a word of the technical speak (recoil line, excitation current, permeance, and demagnetization action on pages 99-102). It seems to me that putting shoes on a magnet shorts out the circuit and reduces the force; in fact IÃ¢â‚¬â„¢ve seen horseshoe magnets stored this way with a little bar across to tie the field down. I wouldnÃ¢â‚¬â„¢t think weÃ¢â‚¬â„¢d want something like that in this motor unless maybe I am in the midst of assembly. Am I making any sense?

BTW - So far I havenÃ¢â‚¬â„¢t seen anything negative about the design other than it will be a bugger to secure the magnets.
Best, KF
Last edited by Kingfish on Fri Mar 04, 2011 2:53 pm, edited 1 time in total.
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### Re: Doing the Math

Just your friendly neighborhood tourist droping by to say keep the discusion going.....

I looked at Halbach arrays while playing with motor design.....my biggest issue was the increased pole count...& the issue of all the other magnets not contributing to the motor's power generation. by the time you add all the magnets for the effect, you are close to the same weight as an iron flux ring on my scale models ( I copped out & went low road)

The bigger isue also was flat/square magnets layed out to create the halbach effect on a radius that is in the vein of Axial topography.......I went looking for some custom magnet manufacturer to "zap" up a set with the N/S poles askew 45 or 60 deg. & in a modified "pie" shape(just like the launchpoint) but I am a lowly basment experimentor...with nothing but old ideas & bad habits to create with....so I side stepped(gave up untill I can manufacture my own magnets) & moved onto utilizing tangibles within my reach.

Given my status as a confirmed "dullard", I will be watching intently from the shawdows for some scraps of knowledge to fall within my reach. Where in I will pounch on it & incorperate into my database, & re-spew it in a build that will make me look far schmarter'r then I really am.
I assumed you have seen this thread KF, but never assume as a they say.
viewtopic.php?f=28&t=13957&hilit=axial+flux+discusion
get some......

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### Re: Doing the Math

KF, have you thought of a stator solution with PCB's ? The one on the picture is 780 Âµm thick and contains 140 gr copper. They can be stacked in unlimeted number.

/GÃƒÂ¶ran

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### Re: Doing the Math

After sleeping on the air gap between adjacent magnets (Agm), I would like to add another diagram in hopes of crystalizing the issue for better comprehension:

In Figure 8 we have two Magnetic Arrays; if these are of Halbach design, the flux density (ÃÂ¤) on the left will vary and weaken as the radius from the center expands, while the one on the right should remain consistent - at least in the ideal sense. We can likewise observe that the air gap widens on the left (left dimension), whilst the air gap on the right is consistent.

Conclusions:
• The Left array could benefit if the air gap was filled with more permeable material such as electric steel because it would condition and spread the field strength as the radius expands towards the OD. See Figure 1.
• The Right array does not require filler material since the geometry already propagates the desired affects.
• If the Left array requires filler material, does this material need to touch the adjacent magnets?
• What is a reasonable ratio of air gap to adjacent magnet?

I think once we get this resolved we can return to the windings aspect.

Thud: Oh gosh, IÃ¢â‚¬â„¢ve read and re-read your posts on the subject, and searched and plumbed the depths of ES on various aspects in the quest to comprehend this quirky twisty AF solution. The question I have for you is:
• When you asked for bids on the custom design of the shape, were you daunted by the quote or quantity? Can you briefly outline the issues that you faced? Perhaps together as a team we can overcome this, yes?

Goethe: Yes, a PCB stator has been conceived as one way to accomplish the windings. LFP has the same recommendation. We need to evaluate the characteristics of the windings relative to the flux density and the expected torque before evaluating which is the best approach to packaging. BTW Ã¢â‚¬â€œ that is an awesome pic of a PCB stator! Where did you get that? I can easily imagine a multilayer heavy-copper board. Can you imagine though the torque upon the trace face when going WOT? I wonder how thin we could make an internal layer?

Winding Material:
OK Ã¢â‚¬â€œ IÃ¢â‚¬â„¢m taking the plunge. Have you guys considered the multipath Litz wire to reduce eddy currents? I am imagining the Hz becoming something like 7 rps * Teeth/Phase-Count and then multiply that by how much faster than [30 mph/ 48,3 km/h] we wish to go. ThereforeÃ¢â‚¬Â¦

Solve for frequency (f) if the number of poles-pairs is 20, and the number of teeth is 21 for a 3-phase motor at 7 rps:
f = (21 teeth / 3 phase) * 7 rps = 7 * 7 = 49 Hz (previously solved on Sat Sep 11, 2010 12:38 pm).
If I want to go 1.5X faster then f would need to be 1.5 * 49 = 73.5 Hz. Nothing here suggests fear of eddy currents from high frequency. The part that is missing is how fast we need the microcontroller to pulse the FET stage, and I donÃ¢â‚¬â„¢t have an answer to that.

Regardless, I think itÃ¢â‚¬â„¢s worth evaluating all the winding options, and to do that we need to calculate the turns per winding next.

Best, KF
Last edited by Kingfish on Fri Mar 04, 2011 2:53 pm, edited 1 time in total.
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### Re: Doing the Math

Kingfish wrote:Understood. Neat FEA pics; did you make those?

Yes, using a software package called FEMM. It's freeware and very useful for this sort of thing. It does require a little bit of knowledge to use, but they provide a few helpful tutorials and I believe more info can be found on the web.

I'm not really sure what pole shoes are, but Google seems to tell me that's referring to the wider "face" part of a stator tooth. I don't think that's applicable here; as you say I think it would tend to divert the flux in a non-useful way.

Talking of air gaps, spacing, and so forth:
The important bit to understanding here is to draw and visualize the complete flux circuit. In the Halbach case, you already drew the picture:
Kingfish wrote:Fig-7.png

You got it right here, although the arrows going from rotor-to-rotor are in the wrong direction (should be N-to-S). If you draw them in the correct direction you can easily see the circulation patterns that are set up. The total reluctance around that path determines the flux that flows. Just like V=IR, here we have F=R*phi, where F is called magneto-motive force (MMF, script F), R (script R) is reluctance, and phi is the flux. R for magnets and magnetic materials is very low, R for air and non-magnetic materials is very high. The flux path flows from S-L-N and N-R-S through the adjacent magnets, so any spaces here which are not filled with magnetic material will increase R and thereby decrease flux.

In the non-Halbach case, using regular N/S magnets and a flux ring, now the flux path flows down one magnet, through the flux ring, through the next magnet, across the stator and so forth. Now the flux ring is carrying flux from one magnet to the adjacent one, so a gap between magnets does not increase R. In fact, as discussed earlier, we want to have some spacing here. Tracing out the path lets you see which gaps are bad and which gaps are good (or neutral).
Eric

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### Re: Doing the Math

The PCB stator is a slick solution, but power handling would be limited I think. FR4 is a lousy conductor of heat. If you stacked multiple boards you'd have to leave an air gap between or the inner layers would be very effectively insulated. You'd face the same problem using the inner layers on a multilayer board. Also, even if you used 2 oz copper, the traces are pretty thin so they'd need to be wide to have a reasonable resistance.

I think it'd be great for a small, low-power motor, but probably not well-suited to something for high-power. I remember this was discussed here previously but I don't remember if there were any other pluses/minuses brought up.
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### Re: Doing the Math

Flux, Air Gaps, ReluctanceÃ¢â‚¬Â¦

Fig. 7:

Corrected; I thought it was off. Thank you!

Fig. 8:
• OK, so if these are NOT Halbachs then weÃ¢â‚¬â„¢d have to use a high-permeability/low-reluctance back plate, and air gaps are not a large factor Ã¢â‚¬â€œ in fact we want some gaps; gotcha.
• But if they ARE Halbachs, then we want them to touch, or fill the void with high-permeability/low-reluctance material.

If I have this correct, letÃ¢â‚¬â„¢s call this section done, finished, baked, closed.
Thanks, much appreciated KF
Last edited by Kingfish on Fri Mar 04, 2011 2:54 pm, edited 1 time in total.
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### Re: Doing the Math

I believe you have the idea.
Eric

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rhitee05
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### Re: Doing the Math

Cool. WeÃ¢â‚¬â„¢re baby-steppinÃ¢â‚¬â„¢ makinÃ¢â‚¬â„¢ progress

Halbach and Windmill magnets - Compare/Contrast on Mass:
Thud brought up a good point on the mass and I wish to run a quick review. Given Figure 7 Ã¢â‚¬â€œ Right side, letÃ¢â‚¬â„¢s presume that we have two arrays, one Halbach and one Windmill, and that the total pole counts are the same as is the Tesla rating.

Windmill requires an attached plate or permeable rotor to complete the magnetic circuit, and Halbach does not.
I would wager that by virtue of the elimination of the backing plate the Halbach array will be lighter. Without physical evidence we canÃ¢â‚¬â„¢t be certain, however as much can be inferred by review of Shane ColtonÃ¢â‚¬â„¢s thesis and looking at the relatively thick steel plate. Aluminum in the 70-series has similar strength properties to steel alloy given slight dimensionality adjustments, however the weight is significantly less, perhaps as much as 40% less. I like not having iron in my motor, and I wouldnÃ¢â‚¬â„¢t have any at all Ã¢â‚¬â€œ except for sayÃ¢â‚¬Â¦ for the bearings. This is the motor I wish to build.

Teeth and Poles:
The Windmill magnets commonly available would allow for Qty-16, or 8 pole-pairs/8:1 gearing. This likely would translate to a low-voltage high-current fast-spinning rotor. In contrast, my 9C 2806 has 51 Teeth, & 46 magnets/23 pole-pairs/23:1 gearing. This motor will consume about 2 hp to push me up a steep hill at 30 mph/48,3 km/h. I want to craft an AF design that re-creates that experience Ã¢â‚¬â€œ and then improve upon it by making it go faster with the same amount of power.

Observation: The most optimum Teeth/Pole-Pair ratio follows this patternÃ¢â‚¬Â¦
• Teeth/Pole-Pair Ã¢â€°Ë† 1:1 where
• Teeth/Pole-Pair Ã¢â€°Â  1, and
• Teeth/Pole-Pair Ã¢â€°Â  integer, and
• Pole-Pair/Teeth Ã¢â€°Â  integer, and
• Teeth = (Pole-Pair +/- 1 or 2)
Here is a wiring diagram that estimates beyond 18 Teeth & 20 Poles given previously for up to 36x36:

Wiring Diagram for optimum Teeth/Poles up to 36x36

My Plan is to shoot for 20 Poles and 21 Teeth, 3-Phase. This is not set in stone; we can change it Ã¢â‚¬â€œ though we have to start someplace, yes?

Questions?
~KF
Last edited by Kingfish on Fri Mar 04, 2011 2:55 pm, edited 1 time in total.
* My 2WD Garden Wall
* Current ride: 2WD Disc EBikeKit (9C 2806-equivalent) / Dual Lyen 12FET / 15S6P LiPo when commuting.
* Going to California: 2011: Trip completed
* Club Member: 40-mph & 101. 10k-Club: 10,478 miles-to-date, 4,988 as 2WD.

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The hands acquire shakes, the shakes become a warning.
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Kingfish
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### Re: Doing the Math

KF,
If your going to attempt ironless, i see no reason not to follow the leaders in the feild. they have been very generous in documenting some of their "secret sauce" in their version of dual Halbach arrays in axial motors
http://www.launchpnt.com/Documents/halb ... deshow.pdf
still spinning fater than your goal from my observations...but that may be addressable....to a degree.

i did a simpler take-off of single halbauch aray & went with 16 poles (before you posted your ideal senario) it looks a little like this:

notice i have the pole askew on the intermediats....this may or may not have any real effect.....perhaps FEMM would tell us more....need to search for that.

I contacted KJ magnetics for a quote but that info is long gone...it was over my budget by 10x & I don't recall the exact #s.(something like 2k\$ for custom wedge molds & multi set ups for biased magnifcation & min orders)

Also, where is Madact's input in this exchange? he is on the same path:

I can resubmit a RFQ & see where the #s come in....tell me a diameter & thickness you desire, I will model it & send it through for quote. & copy the info here. we'll see if it a reasonable enough request for board sponcership.
Thud
get some......

All information & advice provided by Thud are "Open Source" & free for personal use & distribution under the following agreement linked below.

Thud
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### Re: Doing the Math

You could play with the gizmo here to give you an idea of magnet costs.

Miles
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### Re: Doing the Math

Thanks Miles,
that a nice site & a lovely way to order special magnets....I ran a quote for some mid range magnets to do the 20 single halbauch....
178mm od x 50.8mm id. 9deg x 40 segments (I asume aposing magnet arrays so X2)

The conversion comes to roughly \$1338.00 USD before shipping costs. The lead time is reasonable also. if i were totaly commited to building an ironless motor with a halbach array....i may be able to justify that cost. but not in the forseable future. it would be fun to build a motor utilizing the configuration though.
get some......

All information & advice provided by Thud are "Open Source" & free for personal use & distribution under the following agreement linked below.

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### Re: Doing the Math

This seems like an opportune time to point out that you could use other magnet shapes to build a (non-Halbach) high pole-count rotor. I've seen a few designs using circular puck magnets, I think rectangular magnets would work fine as well. Choose size to suit your desired pole count and diameter. A flux ring would be required.

Tossing another wrench into things, have you considered that axial flux machines can be "stacked" for higher power? Instead of your R-S-R configuration, you could add another rotor and stator to make it R-S-R-S-R (extend as far as you like). I think that would pretty much double the power/torque. Since the two stators could be wired in either series or parallel, you could even put in a switch to get two different Kv's.
Eric

Trek FX 7.3 hybrid - Cyclone 650W setup in process! Still...

rhitee05
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### Re: Doing the Math

Thud:
Yeah, I expect the 10:1 ratio to be too fast; I just need to make a stab at it, run through the calcs, and build a worse-case scenario. I have seen that LaunchPoint presentation before in the hunt for information. Thanks for putting that back in front of me; itÃ¢â‚¬â„¢s good for review.

In reference to that doc, I am going to pass on the 45*F flux pieces; that doubles again the part count. Did you note how they were cooling it with the baffled separator between the rotors? I had a similar design drawn up weeks before seeing theirs. Also they are using the adjacent coils where I do not plan to do so; the power requirements are lower for my design; no need to go higher than 5 hp on race day unless we introduce LN, right?

The Plan is to have only two orientations for wedge-shaped, or one orientation for bar, then flipping as prescribed. I have a good impression of the bar cost but the wedge is more interesting. The OD will probably be similar to the 8Ã¢â‚¬Â Windmill, though I may opt for higher ID to keep the mass down. LetÃ¢â‚¬â„¢s evaluate for moment the 9C 2806 magnets:

3 x 13.5 x 27 mm. I am guessing they are N45. LFP said someone on ES had a Gauss meter and they could measure. LetÃ¢â‚¬â„¢s just said for conversation that IÃ¢â‚¬â„¢ll use the same. The dimensions will be 6 mm thick and slightly longer. With two rotors weÃ¢â‚¬â„¢ll have twice as many poles - though our radius arm will be a little less because we are axial not radial. Though comparing apples and oranges, I still expect the torque to be greater and the motor to be more efficient.

Budget: Man if someone quoted me \$2k IÃ¢â‚¬â„¢d get put-off too. We have to do better.

Thanks for sharing Madact's thread; it was very interesting; lots of user-experience there worth noting. The FEMM software I downloaded had a Trojan attached; it was trapped and deloused, but I havenÃ¢â‚¬â„¢t got back to it.

Miles:
Wow what a neat wizard Ã¢â‚¬â€œ and with instant quotes! Thanks man, IÃ¢â‚¬â„¢ve added that to the Magnetic Wire & Nd Magnets Reference.

Eric:
I am committed to doing Halbach, however I think it would be prudent, prudent (fingers pointing), to have a Plan-B with the traditional bar magnets in parallel.

Multiple Rotor-Stators: Oh yes, that was conceived for the next motor such as a single gear rear hub. If you get rid of the disc brake you can fit two stators in the front hub. What are the distances between mountings for endure-class motorcycle? IÃ¢â‚¬â„¢d think weÃ¢â‚¬â„¢d be able to do two stators in front and three in the rear. ItÃ¢â‚¬â„¢s got to be modular construction though; Kingfish idiot-proof assembly.

~KF
* My 2WD Garden Wall
* Current ride: 2WD Disc EBikeKit (9C 2806-equivalent) / Dual Lyen 12FET / 15S6P LiPo when commuting.
* Going to California: 2011: Trip completed
* Club Member: 40-mph & 101. 10k-Club: 10,478 miles-to-date, 4,988 as 2WD.

It is by caffeine alone I set my mind in motion.
It is by the beans of Java that thoughts acquire speed.
The hands acquire shakes, the shakes become a warning.
It is by caffeine alone I set my mind in motion.

Kingfish
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### Thread Summary & General Review: Monday 9/20

This is a brief summation of what was learned to date, with some corrections.

Restate the Basics:
Given:
24-inch tire = 0,61 m diameter (d), traveling at
Velocity (v) = 30 mph / 48,3 km/h linear speed, using
Power (P) = 2 hp
We Calculated:
tire spins at 7 rps
Angular Velocity (Ãâ€°) = 7 rps * 2Ãâ‚¬ = 44 rads/s
Current (I) = 44 Amps
Torque (Ãâ€ž) = P/Ãâ€° -> Ãâ€ž = 1491.4/44 = 33.9 Nm

Windings and Effective Length:
Given
F = IL x B
Ãâ€ž = 2rF = 2rBIL
B = 0.5 T
We Calculated:
Wire Length (L) = Ãâ€ž / 2rBI -> 33.9 / (2 * 0.3048 * 0.5 * 44) = 2.53 m
However, as noted - our winding is circular, and the effective length is Â½ the circumference, we therefore correct as follows:
L = d; actual Lw = Circumference (C) / 2
Circumference (C) = 2rÃâ‚¬ = dÃâ‚¬;
1/2C = dÃâ‚¬/2; Lw = (2.53 m * Ãâ‚¬) / 2 = 3.974 m

Problem: The winding will likely not be circular; it could be trapezoidal, rectangular, or oval. Therefore the actual effective length of Â½ the winding will need to factor in the shape and adjust for losses where corners do not contribute.

Example: Rectangle with full-radius ends (racetrack shape). The most effective part of the coil is the straight length that is tangent to the radius-arm. Therefore a relationship between the effective length and the overall length should be determined in advance.

The quickest method is through CAD: Develop the desired shape, measure the effective length, and contrast with measurement the parameter; this provides a rough correction ratio. If the effective length is not entirely tangential then more rigorous calculations are in order.

Correct Efficiency and Voltage:
Let us presume then that the effective length of our winding is 3.974 m however it is divided up.
Recalculating Resistance:
R = 4.1328 ohms/km for 11 AWG -> (4.1328 * 3.974) / 1000 = 0.01642 ohms
P = I^2 * R = V^2 / R
P = 44^2 * 0.01642 = 31.8 watts
Calculate the Efficiency:
Pe = (Pi Ã¢â‚¬â€œ Pr) / Pi -> (1491.4 Ã¢â‚¬â€œ 31.8 ) / 1491.4 = 97.9%
Calculate the Volts:
1491.4w / 97.9% = 1524w; 1524 - 1491.4 = 32.5w (the amount of energy required to overcome the inefficiency, in theory).
Reverse to get to R:
Given P = I * V, solve for V:
V = P / I -> 1524 / 44 = 34.6V

Halbach and Simple Magnet Arrays:
This conversation is about Axial Flux machines, having a Rotor-Stator-Rotor arrangement. A Simple magnet array is constructed as a series of magnets in a circular arrangement with alternating Up-Down/North-South faces. Assemblage would require a backing plate to constrain the potential flux leakage. This is considered the less-expensive solution.

A Halbach array consists of at least twice as many magnets which are configured by alternating the orientation N-L-S-R and so on that effectively doubles the flux on one side through redirection, thus saving the potential weight of the backing plate. However this assembly is more difficult to achieve. In addition, if a shape other than arc is used then high-inductance/low-reluctance filler material would enhance the flux field rather than retard if the air gap remained.

In a Simple array we want a small air gap between the magnets; in a Halbach array we want the magnets to touch.

A coreless stator does not have iron or ferrous material, thus we eliminate cogging, which increases the efficiency of the motor. However our torque value drops because the coreless design cannot focus and concentrate the flux in the same manner than an iron stator. Therefore the closer the two rotor faces are to each other the greater the flux density in the air gap between; the goal is to have high flux density.

Windings (Teeth) to Poles (Magnets) Ratio:
Though not exclusive, the more optimum ratios of Stator Teeth (Windings) to Magnet Poles are close to but not exactly 1:1. There is a direct relationship between the increase in the amount of torque per rotation and the amount of magnetic poles. Increasing the count of pole-pairs also directly increases the frequency of the signal pulse, given as
ÃŽâ€ Magnet Pole pairs (p) = ÃŽâ€ Signal Frequency (f)
If we had 10 pole-pairs in a motor, each pair would provide 1/10th of the total toque, and by the same token require 1/10th of the amount of current per pair.

We can reduce Current (I) further by increasing the number of phases per cycle. There is a direct relationship to the number of phases to the number of winding teeth, and the ratio for the amount of current per pulse is specified as the square root of the number of phases; therefore the current pulse of a 3-phase motor is given as:
I1p = I3p / Ã¢Ë†Å¡3 Ã¢â€°Ë† I/1.732
Thus 44 A / 1.732 = 25.4 A per phase.
If we had a 3-phase motor with 10 pole-pairs (p) and 21 teeth, calculate the amount of current per phase.
21 Teeth / 3-Phase = 7 windings per phase, wound as:
AaAaAaABbBbBbBCcCcCcC
Tire rotation is 7 rps, therefore 7 * 10 p = 70 Hz
Itooth = Itotal / 10 / Ã¢Ë†Å¡3 => 44 / 10 / Ã¢Ë†Å¡3 = 2.54 A
=> 2.54 A @ 70 Hz Ã¢â‚¬â€œ at least theoretically

If there are no questions I should like very much to move onward and recalculate the Force as we move it originally from the Wheel circumference towards the smaller Hub working diameter.
Thanks, KF
* My 2WD Garden Wall
* Current ride: 2WD Disc EBikeKit (9C 2806-equivalent) / Dual Lyen 12FET / 15S6P LiPo when commuting.
* Going to California: 2011: Trip completed
* Club Member: 40-mph & 101. 10k-Club: 10,478 miles-to-date, 4,988 as 2WD.

It is by caffeine alone I set my mind in motion.
It is by the beans of Java that thoughts acquire speed.
The hands acquire shakes, the shakes become a warning.
It is by caffeine alone I set my mind in motion.

Kingfish
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Posts: 3864
Joined: Wed Feb 03, 2010 11:23 am
Location: Redmond, WA, USA, Earth, Sol, Orion–Cygnus Arm, Milky Way. Age: > yesterday < tomorrow

### Re: Doing the Math

Kingfish wrote:Problem: The winding will likely not be circular; it could be trapezoidal, rectangular, or oval. Therefore the actual effective length of Â½ the winding will need to factor in the shape and adjust for losses where corners do not contribute.

Example: Rectangle with full-radius ends (racetrack shape). The most effective part of the coil is the straight length that is tangent to the radius-arm. Therefore a relationship between the effective length and the overall length should be determined in advance.

The quickest method is through CAD: Develop the desired shape, measure the effective length, and contrast with measurement the parameter; this provides a rough correction ratio. If the effective length is not entirely tangential then more rigorous calculations are in order.

The effective length for a circular coil shape is actually a ratio of 2/pi. You can skip the fancy math explanation and just say that the effective length is the vertical (radial) dimension of the coil x2, for both the upward and downward sides. For a circle of diameter 1 that would give an effective length of 2, given the total circumference of pi, the ratio is 2/pi - tada!

That makes it fairly easy to calculate the effective length of any given structure. A racetrack shape is probably a good choice.
Eric

Trek FX 7.3 hybrid - Cyclone 650W setup in process! Still...

rhitee05
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### Re: Doing the Math

OK, now I am completely confuseÃ¢â‚¬â„¢ed:

I mean Ã¢â‚¬â€œ I get that the Diameter (d) x 2 (both sides) of the circle affects torque production. What I am trying to do is mathematically link
F = IL x B ultimately to the number of turns
Let me scribble in the sandbox here and see if I can grasp the concept:
• If I have a straight conductor then the Length (L) is 2.53 m
• If I have a circular conductor then we have to account for Â½ the circumference. That came out to 3.974 m.
However Ã¢â‚¬â€œ if what you say is true, then all we really care about is 2d, and the number of them.

Leaping off the edge, I have 7 teeth per phase. If I want one winding per tooth, then I take the straight measurement and divide by (7 * 2d) or
d = 2.53 m / (7 * 2) = 0.18 m/7.1 inches.
Or is it that I take the 3.974 m (1/2 circumference) measurement =>
d = 3.974 m / (7 * 2) = 0.28 m/11.2 inches.
In either case, d is very large. The best that we can do is use the Archimedes Winding and recalculate with a small inside diameter (ID) of 15 mm and figure the amount of turns.

Choosing the worst case of d = 0.28 m/280 mm, calculate turns per tooth:
280 mm / 15 mm = 18.7
If we used 11 AWG the OD becomes pretty huge, therefore I recalculated using 16 AWG Flat Wire at 4:1, where Width = 2.42 mm x 0.61 mm High, and changed the ID from 15 mm to 12 mm which resulted in 19.05 Turns and an OD of 35.9 which is < 36 mm and enough for an air gap between the copper windings for cooling. Granted the resistance will be higher, and depending on the Current (I) this may be a factor.

Regardless, if the width of the winding is true then we can narrow the gap between the rotors and increase the Flux Density significantly, which in turn reduces the length of copper conductor; win-win.

But I canÃ¢â‚¬â„¢t be sure until I understand precisely the mathematical expression that takes us from F = IL x B, to L = 2d. I require a bit of hand-holding: Can you display the proof please?

In the clouds, KF
* My 2WD Garden Wall
* Current ride: 2WD Disc EBikeKit (9C 2806-equivalent) / Dual Lyen 12FET / 15S6P LiPo when commuting.
* Going to California: 2011: Trip completed
* Club Member: 40-mph & 101. 10k-Club: 10,478 miles-to-date, 4,988 as 2WD.

It is by caffeine alone I set my mind in motion.
It is by the beans of Java that thoughts acquire speed.
The hands acquire shakes, the shakes become a warning.
It is by caffeine alone I set my mind in motion.

Kingfish
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Posts: 3864
Joined: Wed Feb 03, 2010 11:23 am
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### Re: Doing the Math

Here is a derivation of the Lorentz equation for the circular coil case. I attached a PDF because it's much easier to express the equations in LaTeX.

If you don't understand the geometry description, I can add a picture to illustrate. I assume x is the tangential direction, y is radial, and z is axial with the appropriate right-hand convention.
Attachments
Circular Coil Derivation.pdf
Eric

Trek FX 7.3 hybrid - Cyclone 650W setup in process! Still...

rhitee05
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### Re: Doing the Math

OK, now I see the Proof; that makes sense Ã¢â‚¬â€œ and I like the formula
Fx = (2/Ãâ‚¬)IBzL
Where L = Length (Ã¢â‚¬Ëœcos with this particular font you cainÃ¢â‚¬â„¢t tell the difference between I and little L)

And itÃ¢â‚¬â„¢s even better because I need to reference Force (F) in a moment.
Thanks Eric; much appreciated

Moving right along...

Recalculate Force from Wheel Perimeter to Hub Diameter

The Force Law of Equilibrium implies a balance of opposing forces within a system. LetÃ¢â‚¬â„¢s examine Figure 9 (and for a moment forget that F1 and F2 would turn the wheel in opposite directions). If the tire spins at a constant rate, then the torque (Ãâ€ž) is also constant. The amount of Force (F) then is proportional to the radius (r), given by the formula
Ãâ€ž = r x F
We already know the values of r1 and F1, and we have a good idea what r2 will be, so we need to discover F2 so we can recalculate part of our math for a hub motor. If we solve for the Balanced Beam the equation becomes
r1 x F1 = r2 x F2 => F2 = (r1 x F1) / r2
I donÃ¢â‚¬â„¢t think we ever calculated Force (F), so using
Ãâ€ž = r x F => F = Ãâ€ž / r = 33.9 / 0.3048 = 111.2 N (at a radius of 12 inches / 0.3048 m)
Finishing the solution
Set r2 = 0.075 m / 75 mm / ~3 inches
=> F2 = (0.3048 x 111.2) / 0.075 m = 452 N
Cross-checking using Ãâ€ž = r x F, solve for F
F = Ãâ€ž / r = 33.9 / 0.075 m = 452 N.
From the math we can directly observe that the Force required to maintain the Torque within a typical hub circumference is at least 4X greater than at the edge of the tire. The precise ratio is 452 / 111.2 = 4.06 (4X is close enough). It turns out that our length of wire must increase by 4X as well.

With Fx = (2/Ãâ‚¬)IBzL, solve for L:
L = Fx / [(2/Ãâ‚¬)IBz] => 452 /[(2/Ãâ‚¬) *44 * 0.5] = 452 / 14 = 32.3 m (which is equal to 2d)
Now we can appreciate the advantage of Gearing; adding more Pole-Pairs (p) to the motor!

Did I get this correct?
wag-wag, KF
Last edited by Kingfish on Fri Mar 04, 2011 2:56 pm, edited 1 time in total.
* My 2WD Garden Wall
* Current ride: 2WD Disc EBikeKit (9C 2806-equivalent) / Dual Lyen 12FET / 15S6P LiPo when commuting.
* Going to California: 2011: Trip completed
* Club Member: 40-mph & 101. 10k-Club: 10,478 miles-to-date, 4,988 as 2WD.

It is by caffeine alone I set my mind in motion.
It is by the beans of Java that thoughts acquire speed.
The hands acquire shakes, the shakes become a warning.
It is by caffeine alone I set my mind in motion.

Kingfish
1.21 GW

Posts: 3864
Joined: Wed Feb 03, 2010 11:23 am
Location: Redmond, WA, USA, Earth, Sol, Orion–Cygnus Arm, Milky Way. Age: > yesterday < tomorrow

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