## Doing the Math

Electric Motors and Controllers

### 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
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### 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

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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
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The hands acquire shakes, the shakes become a warning.
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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

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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.
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### Re: Doing the Math

The math is fairly straightforward, no argument there.

I don't think I understand why you seem to be using 3" as the radius for the hubmotor, though. My brain is too fried at the moment to look further up to see if there's an explanation.
Eric

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### Re: Doing the Math

rhitee05 wrote:The math is fairly straightforward, no argument there.

I don't think I understand why you seem to be using 3" as the radius for the hubmotor, though. My brain is too fried at the moment to look further up to see if there's an explanation.

I'm writing it up now; patience my master
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### Re: Doing the Math

Calculate Poles and Windings
The new radius value of 75 mm was selected because it is the midpoint of a wide magnetic pole. It has to be wide to accommodate the extra wire length. We are bound to that radius by a couple of simple rules I set up early in the thread:

A hub equivalent in size to a 9C 2806: Having a magnetic diameter ~ 200 mm, and not wider than about 50 mm; less if possible.

If we look at Figure 4, the outer blue circle denotes the limit of the copper windings along the periphery; that is the 200 mm dimension Ã¢â‚¬â€œ and we need to work inwards from that. If we were building a Radial flux motor the magnet ring would be in the inside of the blue ring.

• The distance allowed for the winding to loop back is 10 mm. This leaves us with a magnetic OD of 180 mm. If the magnets are 30 mm long, the midpoint is 150 mm diameter, or r = 75 mm.
• It is possible to have magnets longer than 30 mm, however as we do the average radius moves inward as well and that makes the wire conductor even longer still.
• We also have to consider the ratio of number of poles and the number of teeth; the more poles Ã¢â‚¬â€œ the shorter our wire.

Plan-A
We have to start someplace, so Plan-A is:
10 pole-pairs/ 10:1, with 21 teeth
21 Teeth / 3-Phase = 7 Teeth/Phase
Earlier we calculated that the Tire rotation is 7 rps, therefore 7 * 10 p = 70 Hz
And the Current (I) per Tooth was given as
Itooth = Itotal / 10 / Ã¢Ë†Å¡3 => 44 / 10 / Ã¢Ë†Å¡3 = 2.54 A
Solve for length of wire per phase per tooth (Lpp):
Length of Wire (Lw) = 32.3 m = 2d
Lpp = Lw / 3 / 7 => 32.3 / 3 / 7 = 1.53 m which is also equal to 2d,
1.53 / 2 = 768.4 mm as a single diameter.
Using CAD, I measured the perimeter of the ID of a single quasi-racetrack-shaped turn of copper that could effectively over the 30 mm long magnetic pole with a little tolerance for misalignment, and that value is 86.9 mm. If we convert that to a true circle the calculation is
Perimeter (Pc) = 2 Ãâ‚¬ r => r = Pc / (2 Ãâ‚¬) = 86.9 / (2 * Ãâ‚¬) = 13.83 mm
We want diameter; 2 * r = d => 2 * 13.83 = 27.66 mm
A crude estimate of the number of turns per tooth would be 768.4 / 27.66 = 27.8
ThatÃ¢â‚¬â„¢s a lot of windings! We have however an advantage in that the Amps / Phase / Cycle are low, and we do not have to rely on 11 AWG, or even 16 AWG wiring. With Itooth = 2.54 A we could safely drop down to 20 AWG and still keep a clamp on Resistance. Using 4:1 Flat Wire, 20 AWG = 0.39 High x 1.56 wide; the Archimedes Winding Calculator suggests that 27.8 turns with an ID of 14 mm = 2.2 m of conductor, and a 36 mm OD. This is completely achievable. If we were really worried about heat there is plenty of room to emplace a second winding in parallel.

The question that I have is with a racetrack shape, is the diameter calculated from the perimeter, or is it along the axis parallel to the length of the magnet? That distance is 36.7 mm; what are the turns if we divvy it up? 768.4 / 36.7 = 21.

I spent some time this evening trying to find more information about winding shapes; for the types of motors we are discussing the trapezoidal shape is typical. Still, I think more research is warranted.

What happens if we make the magnets longer? Plan-B next.
~KF
Last edited by Kingfish on Fri Mar 04, 2011 2:58 pm, edited 1 time in total.
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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

Kingfish wrote:I spent some time this evening trying to find more information about winding shapes; for the types of motors we are discussing the trapezoidal shape is typical. Still, I think more research is warranted.

Miles
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### Re: Doing the Math

Okay, that explains the design choice. I'm a little confused by the details however. You say that the magnet OD is 180 mm and the magnets are 30 mm long, which would imply that the ID is 150 mm and the average diameter is 165 mm. But you say that the average (midpoint) diameter is 150 mm?

I'll remind you again that the equation and derivation I posted are only valid for the circular case. You'd have to start back at the integral form and make appropriate substitutions for C and dl if you want to analyze a different geometry.

For the geometry you posted (fig 4), I actually think we can make a big hand-waving simplification and say that the effective length of each coil is 2x the magnet length. Why you ask? Especially for the coreless case, we can assume with a reasonable degree of accuracy that the magnetic field is uniform across the face of each magnet and zero outside the face. There will be some fringing fields, edge effects, etc, but this assumption makes things very easy and is fairly accurate. Since the only portion of the coil of interest is the portion within the magnetic field, and due to the coil shape these sections are entirely in the radial direction, we can simply say that the effective length is 2x the magnetic length (one for each side of the coil). Simple, eh?

Another suggestion I'll toss out is to consider using overlapping windings. That will generally achieve higher copper fill (more power density), but at the cost of having more stator poles and thus higher electrical frequency. But, that doesn't seem like it should be a problem here.
Eric

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### Re: Doing the Math

Another comment about the number of windings. You may already know this, but I'll add it for the benefit of others who might be reading along.

The total number of wires per slot determines the torque/power potential of the motor. Using the terms "slot" or "tooth" is a little imprecise when talking about a coreless motor here, but I think everyone will understand the intended meaning. How those wires are connected then determines the Kv/Kt of the motor. For example, if you made each winding a single coil of 28 turns in series, you would have a motor with very low Kv and high Kt. If you did the other extreme and placed 28 single coils in parallel, you would have a motor with high Kv and low Kt, with a whole spectrum of choices in between.
Eric

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rhitee05
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### Re: Doing the Math

rhitee05 wrote:Okay, that explains the design choice. I'm a little confused by the details however. You say that the magnet OD is 180 mm and the magnets are 30 mm long, which would imply that the ID is 150 mm and the average diameter is 165 mm. But you say that the average (midpoint) diameter is 150 mm?

• Magnets in Plan-A are 30 mm in length.
• Let's start with the 180 mm OD and convert that to r = d/2 = 90 mm.
• Subtract 30 from 90 = 60; 60 mm r = 120 mm ID.
• 1/2 of 30 mm is 15 mm; 60 + 15 = 75; midpoint = 75 mm r or 150 mm diameter.
Make sense?

For the geometry you posted (fig 4), I actually think we can make a big hand-waving simplification and say that the effective length of each coil is 2x the magnet length. Why you ask? Especially for the coreless case, we can assume with a reasonable degree of accuracy that the magnetic field is uniform across the face of each magnet and zero outside the face. There will be some fringing fields, edge effects, etc, but this assumption makes things very easy and is fairly accurate. Since the only portion of the coil of interest is the portion within the magnetic field, and due to the coil shape these sections are entirely in the radial direction, we can simply say that the effective length is 2x the magnetic length (one for each side of the coil). Simple, eh?

Works for me

Another suggestion I'll toss out is to consider using overlapping windings. That will generally achieve higher copper fill (more power density), but at the cost of having more stator poles and thus higher electrical frequency. But, that doesn't seem like it should be a problem here.

Considered overlapping however the Book AF Machines makes a rather persuasive case for non-overlap windings, and above that, non-overlaps are easier to assemble
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The hands acquire shakes, the shakes become a warning.
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### Re: Doing the Math

rhitee05 wrote:Another comment about the number of windings. You may already know this, but I'll add it for the benefit of others who might be reading along.

The total number of wires per slot determines the torque/power potential of the motor. Using the terms "slot" or "tooth" is a little imprecise when talking about a coreless motor here, but I think everyone will understand the intended meaning.

Normally I am good with using slots over teeth; the issue I face is that winding can be confused with windings and that means exactly the same to me as well. We could propose to reserve winding(s) as refering to the turns of copper. A tooth or slot is the same to me in a coreless design, though since there are no teeth or slots we are again left to consider the proper term.

Now another way to look at the whole arrangement is this:

For Plan-A, I intend to use Non-Overlap Windings; that is the term to describe the assembly of copper turns as well as the orientation of the structure. Maybe what we need to do is set the definition:
• Winding: A single unit of copper turns, whether multi-level or multi-row.
• Windings: Plural of Winding, the whole assemblage, or a phase-array thereof.
• Turn: One revolution of copper conduit, whether it be of a single conductor, an assembly of parallel wires, or of Litz wire.
• Turns: Plural of Turn, regardless of quantity or structure or layout.
Definitions can be cool

How those wires are connected then determines the Kv/Kt of the motor. For example, if you made each winding a single coil of 28 turns in series, you would have a motor with very low Kv and high Kt. If you did the other extreme and placed 28 single coils in parallel, you would have a motor with high Kv and low Kt, with a whole spectrum of choices in between.

Oh yes: We are just now getting to the meat and potatoes of the design! Imagine that this thread started on August 31st and I am just now comprehending the ratio of Power to Turns

What I'd like to do next is develop a couple of various design changes and determine the physical results. If the math is proving correct and we can anticipate with confidence the results - then I will be able to programmatically develop an AF Motor wizard to speed our tweaking along, similar to how I have been leveraging the Archimedes Winding Calculator.

Make sense?
~KF
* My 2WD Garden Wall
* Current ride: 2WD Disc EBikeKit (9C 2806-equivalent) / Dual Lyen 12FET / 15S6P LiPo when commuting.
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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

Kingfish wrote:# Magnets in Plan-A are 30 mm in length.
# Let's start with the 180 mm OD and convert that to r = d/2 = 90 mm.
# Subtract 30 from 90 = 60; 60 mm r = 120 mm ID.
# 1/2 of 30 mm is 15 mm; 60 + 15 = 75; midpoint = 75 mm r or 150 mm diameter.

Sigh. I seem to be having problems with math recently. You are, of course, correct. Your use of terminology seems reasonable and fairly standard, so there should be little confusion.
Eric

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rhitee05
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### Re: Doing the Math

rhitee05 wrote:
Kingfish wrote:# Magnets in Plan-A are 30 mm in length.
# Let's start with the 180 mm OD and convert that to r = d/2 = 90 mm.
# Subtract 30 from 90 = 60; 60 mm r = 120 mm ID.
# 1/2 of 30 mm is 15 mm; 60 + 15 = 75; midpoint = 75 mm r or 150 mm diameter.

Sigh. I seem to be having problems with math recently. You are, of course, correct. Your use of terminology seems reasonable and fairly standard, so there should be little confusion.

NO NO!! I make mistakes ...but then I don't mind admitting them

We're in this together, you and I, us and we'all

Let's make some motors! KF
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Kingfish
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### Re: Doing the Math

You may have already seen this, but if not I suspect you will find it very useful:

http://web.mit.edu/scolton/www/SCThG.pdf

This is a M.S. thesis by a fellow named Shane Colton at MIT, who I believe lurks around here periodically, on the topic of BLDC motor design and control. The entire document is interesting, but there is one section in particular I think you will find highly relevant. He goes through, in great detail, the design and modeling process for BLDC motors and his particular test case is a coreless axial flux design. He goes through several models with increasing fidelity and ends up with measured data to compare against. Should be a very useful reference for you as you attempt the same task.

Eric
Eric

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rhitee05
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### Re: Doing the Math

I began following Shane's blog last spring; once I completed the elementary motor resign for Radial Flux I switched over to AF based on his discoveries. Cool stuff, huh?

Plan-B
Similar to Plan-A, the length of the Magnets changes from 30 mm to 40, reducing the Torque Arm length by 5 mm.
r = 0.070 m
F = Ãâ€ž / r = 33.9 / 0.070 m = 484 N.
Solve for L:
L = Fx / [(2/Ãâ‚¬)IBz] => 484 /[(2/Ãâ‚¬) *44 * 0.5] = 484 / 14 = 34.6 m (which is equal to 2d)
Solve for length of wire per phase per tooth (Lpp):
Lpp = Lw / 3 / 7 => 34.6 / 3 / 7 = 1.65 m which is also equal to 2d,
1.65 / 2 = .8236 m / 823.6 mm as a single diameter.
With Perimeter determined by CAD; Calculate ID:
Pc = 105.1 mm = 2 Ãâ‚¬ r => r = Pc / (2 Ãâ‚¬) = 105.1 / (2 * Ãâ‚¬) = 16.7 mm
We want diameter; 2 * r = d => 2 * 16.7 = 33.47 mm
A crude estimate of the number of turns per tooth would be 823.6 / 33.47 = 24.6 Turns

Using 4:1 Flat Wire, 20 AWG = 0.39 High x 1.56 wide; the Archimedes Winding Calculator suggests that 24.6 turns with an ID of 16.7 mm = 2.07 m of conductor, and a 36.3 mm OD.

Conclusions:
By lengthening the magnets, though constrained by the OD, the length of the conductor becomes longer, however so does the ID of the winding, and fortunately for us the number of turns is reduced, as is the actual length of copper to create the turns, therefore our resistance likewise is reduced, and overall efficiency is improved.

There is however another benefit yet to be calculated that is shared by both of these Plans. And there are other tweaks we can do to this design.

More in a bit, KF
* My 2WD Garden Wall
* Current ride: 2WD Disc EBikeKit (9C 2806-equivalent) / Dual Lyen 12FET / 15S6P LiPo when commuting.
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### Re: Doing the Math

Magnet Height:
The length of the magnets is not the only dimension that could be changed; increasing the height also helps increase the Tesla value Ã¢â‚¬â€œ but only up to a point. I created a graph - Figure 10, which displays roughly the trend and the point of narrowing gain as Height increases and the Tesla value approaches 100%. The break-even point is about 8 mm +/- 1mm.

Rotor Air Gap:
We can also reduce the air gap along the z-axis between the rotor plate to raise the value of Tesla. In Figure 11 the graph is nearly linear for small distances. By reducing the air gap we increase Tesla. There is however a practical cutoff and that is the width of the windings + the air gap between the winding and the magnet surfaces x 2.

Magnet Mass:
LetÃ¢â‚¬â„¢s consider for a moment that we increase the thickness of the magnets to 8 mm. With Plan-B, the mass for one pole is 105 gm (given to us by the magnet wizard here). Since we are using a Halbach array we will require 40 magnets per side: 40 * 2 * 105 gm = 8400 gm / 8.4 kg / 18.5 lbs. That is a shed-load of rotating mass! And even though it has the potential to raise our Tesla value quite significantly, the cost of manufacturing and the weight factor in pretty large. BTW, Plan-A magnets which are 10 mm shorter and only 6 mm high weigh in at 5.064 kg / 11.2 lbs; that is still quite heavy for the power output in my book and I think we can do much better.

Introspection:
Perhaps we need to re-think this problem. We can change the length and height of the magnets, and distance between the opposing faces to affect Tesla. And weÃ¢â‚¬â„¢re kinda stuck with the radius as well Ã¢â‚¬â€œ at least that is the radius needs to be about on par with the 9C 2806 hub. We actually have a little bit of wiggle room to play here but not too much. For example, we know that as the radius decreases that we need to apply more Force (F), and that in turn leads us down the path of larger more powerful magnets and longer conductors.

What happens though if we go the other way; what if we changed the OD of the magnet ring to 200 mm and reduced the length of the magnets from 30 or 40, down to 20 mm since our torque arm is longer. All things being equal, letÃ¢â‚¬â„¢s do the math and find out...

Plan-C
Identical to Plan-A except for the following:
Magnet length = 20 mm; Â½ length = 10 mm.
r = (200 / 2) Ã¢â‚¬â€œ 10 = 90 mm / 0.090 m.
F = Ãâ€ž / r = 33.9 / 0.090 m = 377 N.
Solve for L:
L = Fx / [(2/Ãâ‚¬)IBz] => 377 /[(2/Ãâ‚¬) *44 * 0.5] = 377 / 14 = 26.9 m (which is equal to 2d)
Solve for length of wire per phase per tooth (Lpp):
Lpp = Lw / 3 / 7 => 26.9 / 3 / 7 = 1.28 m which is also equal to 2d,
1.28 / 2 = .6406 m / 640.6 mm as a single diameter.
With Perimeter determined by CAD; Calculate ID:
Pc = 75.5 mm = 2 Ãâ‚¬ r => r = Pc / (2 Ãâ‚¬) = 75.5 / (2 * Ãâ‚¬) = 12 mm
We want diameter; 2 * r = d => 2 * 12 = 24 mm
A crude estimate of the number of turns per tooth would be 640.6 / 24 = 26.7 Turns

Using 4:1 Flat Wire, 20 AWG = 0.39 High x 1.56 wide; the Archimedes Winding Calculator suggests that 26.7 turns with an ID of 24 mm = 2.92 m of conductor, and a 45.2 mm OD.

The first thing we notice is that OD of the winding is really tall, and what we want is closer to a square fill. There is plenty of room for a second winding. LetÃ¢â‚¬â„¢s round up the number of turns to the next even value divide by two; 28 / 2 = 14 turns x 2 rows. This comes out to 1.32 m of conductor with an OD of 35.3 which is much better, but we still have a ratio of 11:3 height-to-width which is nearly a factor of 4.

Throw in the Monkey wrench:
This has been all fun and good so far but the truth is I want this motor to handle a good blow, like the windstorm I experience on my Road Trip to California. I also donÃ¢â‚¬â„¢t want it to get hot when running WOT in the desert after I pass all them neckkÃ¢â‚¬â„¢id hippies out at Burning Man running away from the loooong arm of the law. So what if I took the hp rating and added another 50% to cover my worries; thatÃ¢â‚¬â„¢s a pretty good engineering margin of error. We throw that all into the mix and pretty soon this design begins so see some issues beyond the rows of turns.

So~ how do we make this critter more flexible? The answer isÃ¢â‚¬Â¦

... KF
(Don't you love a good mystery?)
Last edited by Kingfish on Fri Mar 04, 2011 2:59 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.
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* 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.
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### Re: Doing the Math

Plan-D

Ã¢â‚¬Â¦We need to optimize this design; slow the wheel down and reduce the amount of current per cycle.

Revision:
32 Poles / 16:1, 30 Teeth / 10 teeth per phase
Repeating calcs we did for Plan-C:
Magnet length = 20 mm; Â½ length = 10 mm.
r = (200 / 2) Ã¢â‚¬â€œ 10 = 90 mm / 0.090 m.
F = Ãâ€ž / r = 33.9 / 0.090 m = 377 N
L = 26.9 m (which is equal to 2d).
Solve for length of wire per phase per tooth (Lpp):
Lpp = Lw / 3 / 7 => 26.9 / 3 / 10 = 0.897 m which is also equal to 2d,
0.897 / 2 = .4483 m / 448.3 mm as a single diameter.
With Perimeter determined by CAD; Calculate ID:
Pc = 64 mm = 2 Ãâ‚¬ r => r = Pc / (2 Ãâ‚¬) = 64 / (2 * Ãâ‚¬) = 10.2 mm
We want diameter; 2 * r = d => 2 * 10.2 = 20.4 mm diameter of the winding (dw).
The number of turns will be Lw / dw => 448.3 / 20.4 = 21.98 Turns => letÃ¢â‚¬â„¢s make it 22 Turns even.

Using 4:1 Flat Wire, 20 AWG = 0.39 High x 1.56 wide; the Archimedes Winding Calculator (AWC) suggests that 22 turns with an ID of 20.4 mm = 2.04 m of conductor, and a 38 mm OD. The actual thickness of the windings is calculated as
(OD Ã¢â‚¬â€œ ID) / 2 => (38 Ã¢â‚¬â€œ 24) / 2 = 8.8 mm

It turns out that a 8.8 mm thickness is too much for that many windings and it will not fit properly. Once again I suggest we take the number of turns and divide by two so that we can have two rows.

22 /2 = 11 turns. Calculate new OD and thickness: AWC suggests that 10 turns with an ID of 20.4 mm = 0.87 m of conductor, and a 29.4 mm OD. The actual thickness is
(OD Ã¢â‚¬â€œ ID) / 2 => (29.4 Ã¢â‚¬â€œ 20.4) / 2 = 4.5 mm
The 4.5 mm thickness fits perfectly. The total width of the windings is 1.56 * 2 = 3.12 mm. The thickness and the width are fractionally close to 1.5:1. It is better to have a 1:1 ratio, but this is pretty good as it is.

If we really needed to squeeze a little more out of this winding, we could reduce the ID by an offset of 1 mm, which yields a new perimeter of 57.7 mm, which gives us an ID of 18.4 mm. 448.3 / 18.4 = 24.4 turns; round up to next even number and divide by two yields 13 turns. The AWC suggests 13 turns takes 0.98 m of conductor, and a 29 mm OD. Actual thickness is (29 Ã¢â‚¬â€œ 18.4) / 2 = 5.3 mm, and that falls inside the perimeter of the original calculation by more than a whisker. LetÃ¢â‚¬â„¢s throw one more turn on it:

The AWC suggests 14 turns takes 1.07 m of conductor, and a 29.7 mm OD. Actual thickness is (29.7 Ã¢â‚¬â€œ 18.4) / 2 = 5.67 mm. When drawn up in CAD, the air gap between the windings is slightly more than 1/16-inch / 1.59 mm which by several sources is the minimum given for airflow. The amount of turns has improved from 11 to 14; 14/11 = 27% improvement which is indeed a nice margin. Furthermore the winding thickness does not exceed our preset 200 mm working diameter for the hub. Another item to consider is that we increased the gearing ratio from 10 to 16.

Tire rotation is 7 rps, therefore 7 * 16 p = 112 Hz
The Current also drops: Itooth = Itotal / 16 / Ã¢Ë†Å¡3 => 44 / 16 / Ã¢Ë†Å¡3 = 1.6 A.

The 20 AWG Flat Wire spec was left unchanged to help depress the internal resistance.

I like this design very much; weÃ¢â‚¬â„¢re reduced the magnet size, maximized the effective torque arm, reduced the Current per Phase per Winding, and we have reduced the z-distance between the rotors which increases our flux density to the point that I can safely use temperature-tolerant magnets.

Now about that money wrenchÃ¢â‚¬Â¦ If I want to add 50% more torque to this motor, how would I do it?
Are you still with me? 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.
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Kingfish
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### Re: Doing the Math

I'm getting a little confused following the details of your calculations for the various coil dimensions. I think I'm also getting a little confused because your calculations still seem to assume the spiral-type winding, rather than the racetrack/sector shape that I was under the impression was the current choice.

Let me try running through some numbers myself to see if we can get on the same page.

For your plan D, the magnets are 20 mm long and you calculate that we need ~900 mm of wire per coil (per "tooth") for the desired torque.

I want to start from the opposite direction by figuring how much space we have. There are a total of 30 coils (10 per phase, 3 phases) around the stator. That works out to 12 degree sectors for each phase. At the inner radius of 80 mm that's an arc length of 16.6 mm and arc length of 20.8 mm at the outer radius of 100 mm. Combined with the magnet length of 20 mm that gives us an almost-square area in which to fit the desired coil.

Assume we are using the sector-shaped windings. We need 900/20/2 = 22.5 turns to get the desired torque (20mm per side, total of 40 mm per turn). Round up to 23. Using your wire dimensions of 0.39mm x 1.56mm, a stack of 23 pieces of wire is ~9 mm high. So, we can't fit the desired number of turns in a single layer. 12 pieces would be 4.7 mm high. That might be okay, and is ~3.1 mm wide. It's important for there to be an enclosed area in the center of the coil. We only get net torque if the two sides of the coil experience different flux, so if they are tight against each other not much will happen.

Note that the above coil dimensions don't leave any spacing between adjacent coils. Putting 30 coils within a 200 mm diameter seems pretty tight to me. I think reducing the number of coils, making the diameter larger, or using an overlapping winding would make it easier. An overlapping winding would not be too difficult. You can just make 2 stators each with half the number of coils, then stack them together offset by 1/2 coil and wire appropriately.
Eric

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rhitee05
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### Re: Doing the Math

Eric, I knew I should have enclosed a picture. Figure 12 represents both the original and the modified Plan-D Windings. They are still a modified racetrack shape with the full-radius on the inside about as small as I dare; about 1.75 mm R. The skinny one on the right is the original with 11 turns, and the chunky one on the left has 14 turns. This is as tight as I can make it, though I donÃ¢â‚¬â„¢t know enough about winding physics to say if this will work or if it is optimum. I completely understand that some space is required.

OK, so you are correct about the second stator, and in fact you mentioned that as an option early on (forgive me: I was deliberately ignoring the suggestion so we could focus where directed). For the monkey wrench problem where I desire 50% more power to cover my bases Ã¢â‚¬â€œ the only way I could see getting this accomplished is with a second stator design. The direct benefit is that we can split the load so-to-speak between the two stators by connecting the phase coils in series.

In the hub design there is room for more than one stator. In fact Ã¢â‚¬â€œ I have room for four stators! See Figure 13.

In the beginning of the thread I believe I mentioned that I wanted to have the motor width limited to 50 mm max. LetÃ¢â‚¬â„¢s do the math:
Two rows per winding =>
Using 4:1 Flat Wire, 20 AWG = 0.39 High x 1.56 wide; 1.56 * 2 = 3.12 mm wide (theoretically)
If I use Â¼-inch /3.175 mm thick material, probably aluminum 70XX for high-strength/low-deflection, the turns will comfortably fit within the width of that stock material. We'll probably have to burnish off a tiny bit to match the winding.

• Air gap between the winding and the magnet surface is 1 mm, typical.
• The thickness/height of the magnets is 4 mm.
• The distance between the two opposing magnetic faces is 5.12 mm.
• The outer face, the thickness of the exterior rotor is 3/16-inch / 4.7625 mm thick Ã¢â‚¬â€œ at least where the magnets are located.
• The thickness of the material retaining the magnets between the stators is irrelevant because the forces are balanced; the magnets are slightly thicker than the retaining material. Because the magnets are short, the beam is about 1-inch / 25.4 mm and weÃ¢â‚¬â„¢re just not going to see a whole lot of deflection given the material and thickness.
• The total distance from exterior face to exterior face is 50.08 mm wide. I can live with the 0.08 mm.*

Thoughts...
I studied the multi-stator concept early on when I dabbled with the thought of building a motorcycle-class ebike knowing that I would need at least double the horsepower per motor of whatever I came up with for the ebike. The other factor we need to evaluate is that I am building a 2WD. The point is that there is a lot of flexibility with what we can do with an AF motor design. For my needs, I want the motor to be fast, robust, and not get terribly hot.

I am also sensitive to weight, and I came up with a slight modification where the magnets are only 3 mm high*, as well as calculating the Flux Density for a 6 mm air gap between the magnetic faces instead of 5 mm. All variations and permutations still predicted that the Flux Density would remain above the 0.5 Tesla values that we have been using throughout this thread. By reducing the gap, I have been able to tone-down my expectations and plan for a lower-Tesla magnet with higher thermal tolerance (120-140*C).

*BTW Ã¢â‚¬â€œ reducing the magnet height to 3 mm also reduces the width of the hub by 5 mm, therefore the new width could become 45.08 mm.

Does this make it clearer?
On the Cat-Bird Seat, KF

PS - I still have questions!
Last edited by Kingfish on Fri Mar 04, 2011 3:00 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.
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* 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.
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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

I was never any good at juggling, and I think the same principle applies here. I'll focus on one point first until we're in agreement, then we can move on.

I think the winding shape is a good place to start, since the winding geometry defines so many of the other motor characteristics. I think we previously agreed on one basic fact: assuming the modified-racetrack winding shape, the effective length for torque production is 2x the magnet length. For the current 20 mm long magnets, we get 40 mm of useful length per turn in the windings.

Now, regarding the spacing/enclosed area part. Let's use a very simple example. Assume for a second a hairpin-shaped piece of wire, that is a very narrow U shape with long legs. Current flows up one leg and back down the other. Let us further assume that the U is immersed in a uniform and perpendicular magnetic field, such as that of the AF motor. The cross-product of the force equation tells us that the two sides of the U will produce a force in opposing directions. For the uniform magnetic field case, there is no net force. If it were somehow possible to create a +z-directed field on one leg of the U and a -z-directed field on the other leg, that gives us the maximum force case. Reality will be somewhere in between, but we obviously want to be closer to the second example than the first.

I don't know if there is a common design rule-of-thumb or other guidance for the winding shape in this context. Your AF text may have some guidance here. But, it seems to me that based on this common-sense analogy, you would want the spacing between the two sides of the winding to be something like the spacing between adjacent magnetic poles (this would be 2N in your Halbach array). If they're much closer than that I think torque production will start to suffer. Put another way, this is probably the reason most motors usually have a similar number of teeth and magnetic poles. I think if you want a significantly higher number of windings/teeth, then you have to go to a distributed/overlapping design. Note that this is distinct from and not equivalent to a two-stator design. We can discuss that separately.
Eric

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### Re: Doing the Math

Good points

Windings Re-Work:
I am trying to avoid the overlap windings because the complexities multiple over the non-overlap, however I understand that the leverage would be greater because the legs of the windings are farther apart. There needs to be at least 1* of arc between the windings to promote air flow and cooling. Ideally the width and height cross-section should be square, actually round, but round is difficult to control.

Certainly we could get more area to work with if the bounding Radius were allowed to expand, however I think that the 2806 hubs are already ungainly in diameter. The only alternative is to move the magnets to the rim Ã¢â‚¬â€œ and maybe that should be explored a bit later Ã¢â‚¬â€œ though I suspect that is another can oÃ¢â‚¬â„¢worms.

Another scenario is to reduce the size of the conductor yet again. If we dropped down from 20 AWG Flat Wire to 24 AWG, the width goes to 1 mm which would allow 3 rows. LetÃ¢â‚¬â„¢s calculate the results and see where the dust settles.

Whether we pick 22 or 23 turns, going to 24 turns is easily dividable by 3, therefore letÃ¢â‚¬â„¢s use 8 turns per row with a conductor height of 0.25 mm. I am going to calculate using an ID that accepts 24 mm for each leg which to be longer than the height of the magnet; it only affects the length of the conductor which we need for mass and resistance calculations. WeÃ¢â‚¬â„¢ll need to iterate until we can get the maximum ID for the greatest distanceÃ¢â‚¬Â¦ LetÃ¢â‚¬â„¢s use the initial 20.4 mm ID. We want to end up with a perimeter of about 93 mm, or about 30 mm OD.

The AWC suggests that with 0.57 m of conductor we can achieve 8 turns with an OD of 24.6 mm. OK Ã¢â‚¬â€œ this is a good start, so letÃ¢â‚¬â„¢s move the ID out and try againÃ¢â‚¬Â¦

Saving a bit of time, I massaged the values until I came up with:
ID = 25.7 mm, conductor length = 0.71 m, OD = 29.99 (or 30) mm, and turns = 8.08. The cross-section is (30 Ã¢â‚¬â€œ 25.7) / 2 = 2.15 mm x 1 mm x 3 rows = 2.15 mm high x 3 mm wide. Not optimum but doable.
If you will oblige my fancy, I would like to calculate with an ID of 24 mm which when matched with an OD of 30, we would get the 3 x 3 cross-section; how many turns will that provide?

AWC suggests that with an ID = 24 mm, the resultant OD is given as 29.996 mm, using 0.979 m of conductor and providing 11.5 turns. This means that our start and end turns are on opposite sides. Allow me to reduce that to 11 turns even: ID = 24.25 mm and the conductor length becomes 0.945 m.

Unfortunately the current carrying capacity of 24 AWG is about 1/3 of 20 AWG, and yet Ã¢â‚¬â€œ the value is more than twice what we expect weÃ¢â‚¬â„¢ll need for this design.

Caveats:
Originally the design was for one row per winding, however now we are considering 3 rows. I have a concern about resistance and inductance and would like very much to review those values, and how they could affect the final plans.

Eric, can you live with a layout as shown in Figure 14 where the cross-section is close to ideal, the separation between the windings is enough for airflow, and the distance between the up and down legs are at the maximum? The overall turns advances from the original 22 or 23, to 33 turns by using smaller gauge flat wire. If weÃ¢â‚¬â„¢re good on this I think it would be prudent to calculate resistance, inductance, and changes to Current and voltage.

Best, KF
Last edited by Kingfish on Fri Mar 04, 2011 3:00 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
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### Re: Doing the Math

Kingfish wrote:Eric, can you live with a layout as shown in Figure 14 where the cross-section is close to ideal, the separation between the windings is enough for airflow, and the distance between the up and down legs are at the maximum? The overall turns advances from the original 22 or 23, to 33 turns by using smaller gauge flat wire. If weÃ¢â‚¬â„¢re good on this I think it would be prudent to calculate resistance, inductance, and changes to Current and voltage.

The magnet sectors drawn in red in your FIg 14 - what do they represent? I mean, does each sector represent a "full pole" of the Halbach array (a L-S or R-N combination), or is each sector just a single magnet? The spacing looks like it should be good if those are poles.

The triple layer of 24 AWG wire looks like it would work out well. Depending on what Kv you want to achieve, you could connect the layers in parallel and get approximately the same current capacity of the single 20 AWG, but with 50% more turns.

Kingfish wrote:Certainly we could get more area to work with if the bounding Radius were allowed to expand, however I think that the 2806 hubs are already ungainly in diameter. The only alternative is to move the magnets to the rim Ã¢â‚¬â€œ and maybe that should be explored a bit later Ã¢â‚¬â€œ though I suspect that is another can oÃ¢â‚¬â„¢worms.

Just for kicks, I think that would be an interesting design exercise. You can imagine the opposite of a hub motor - sort of a "rim" motor. Construction would be tricky, but you can imagine a way where the magnets are integrated into the rim of the wheel so you get the maximum torque arm possible. If you could get past the construction details, that would be an excellent way to build a high-torque low-RPM motor with a very high pole count.
Eric

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rhitee05
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### Re: Doing the Math

rhiteeo5 said
You can imagine the opposite of a hub motor - sort of a "rim" motor. If you could get past the construction details, that would be an excellent way to build a high-torque low-RPM motor with a very high pole count.

Arlo1 had the same idea awhile back....
http://endless-sphere.com/forums/viewto ... f=2&t=9966

no need for tourque arms if the stator is mounted to the chain stays
get some......

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