Doing the Math

Electric Motors and Controllers

Doing the Math

Postby Kingfish » Tue Aug 31, 2010 3:24 pm

EDIT: Appending TOC.
  1. Math Modeling
  2. Axial Flux Design, Halbach, Ironless-Stator, Axial Motor
  3. Axial Flux Design, Non-Halbach, Ironless-Stator, Axial Motor

Math Modeling

Greetings –

I am working through some equations however I need some other eyes to cross-check my math; it’s difficult at times to work solo in cave…

Equation:
Power (watts) = Torque(Nm) x Angular Velocity(rads/sec), or

    P = τ * ω
    Where...
    ω = (2 π *rpm)/60 = rads/s

Example Problem:
Solve for theoretical value of Torque when a 24” wheel is using 2 hp to spin at 30 mph.

Rearranging the equation from above, τ = P/ω

Therefore…

24-inch tire = 2 foot diameter (d).
Circumference of 2-foot tire = π d -> π * 2 = 6.283 feet (per revolution)
If 1 mile = 5280 feet, 30 miles = 30 * 5280 = 158400 feet.

With 1 hour = 60 minutes,
rpm = (158400 feet / 6.283 feet) / 60 = 420.2

ω = (2 π *rpm)/60 -> (2 π * 420.2) / 60 = 44 rads/s.

If 1 hp = 745.7 watts then P = 2 * 746 watts = 1491.4 w

Resolving τ = P/ω -> τ = 1491.4/44 = 33.9 Nm

The problem I have is A) do I have the math correct, and B) that my unit conversions are correct.

Thanks, KF
Last edited by Kingfish on Tue Oct 26, 2010 2:11 pm, edited 1 time in total.
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Re: Doing the Math

Postby Goethe » Tue Aug 31, 2010 4:42 pm

Aaargh, imperial! :wink:

The power equation seems right to me.

Kingfish wrote:Equation:
Power (watts) = Torque(Nm) x Angular Velocity(rads/sec), or

    P = τ * ω
    Where...
    ω = (2 π *rpm)/60 = rads/s



To make it easier for me to solve the torque I translated it to metric and it I got the same end result as you.

Solve for theoretical value of Torque when a 24” wheel is using 2 hp to spin at 30 mph.

30 mph= 48,3 km/h
24-inch tire = 0,61 m diameter (d).
Circumference of 0,61 m tire = π d -> π * 0,61 = 1,916 m (per revolution)
48,3 km/h / 3,6 = 13,42 m/s
rps =13,42 m/s / 1,916 m = 7 rps
ω= 7 rps * 2π = 44 rads/s

Resolving τ = P/ω -> τ = 1491.4/44 = 33.9 Nm

/Goethe
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Re: Doing the Math

Postby Kingfish » Tue Aug 31, 2010 10:37 pm

Imperial: :lol: My apologies, I will try not to do this to you again :D

OK, thank you – I really appreciate the confirmation!

Continuing & rhetorically, if I may…
Restating, mechanical power is given as:

    P = τ * ω
Similarly, electrical power can be given as:

    P = I V
Defining the Motor Constant, this is given as:

    τ = Kt * I = Ke * ω
where Kt (Nm/A) = Ke (Nm/Ω)

Solving for K,

    Kt = τ/ω = 33.9 / 44 = 0.77 Nm/Ω
From above, if we calculate for a 3-Phase system, then we would divide by 3 to solve for a single circuit:

    Kt/3 = 0.77/3 = 0.257 Nm/Ω

Am I good so far?
~KF

PS - Superscript and Subscript are not supported in the message editor although it is very easy to program/add these features. If you want Super- and Subscript I suggest you write to the ES Forum demigods and make the request as mine has fallen on deaf ears.
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Re: Doing the Math

Postby Goethe » Wed Sep 01, 2010 1:14 pm

I´m not with you regarding Ke, but I might be missing something.

To me is Ke the Voltage constant given in volts/radians/seconds or volts/krpm (Volts per Thousand rpm).
Kt = Ke when given in volts/radians/seconds.

The ohmic relationship seems wrong because if you change the strands in parallel you will change the resistance but not the BEMF.

/Goethe
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Re: Doing the Math

Postby Kingfish » Wed Sep 01, 2010 2:31 pm

You bring up a good point. I have I think the Ke units incorrect; double-checking with another source, Ke units should be V/rads/s – just as you have stated: Thank you for putting me straight :)

For clarity,

    Kt = Torque constant
    Ke = back EMF constant
Re-Solving for Ke

    τ = Ke * ω -> Ke = τ / ω = 33.9 / 44 = 0.77 V/rads/s
Am I good now? :)

The relationship between Power – Mechanical and Power – Electrical is perhaps better scribed as

    P = τ * ω -> Pm = τ * ωm, and
    P = I V -> Pe = eb I, where

      Pm = Power, Mechanical
      Pe = Power, Electrical
      Ï„ = Torque (Nm)
      ωm = Angular Velocity – Motor, mechanical
      eb = back EMF
      I = Current (A)

We can further extrapolate the relationship of back EMF as:

    Pe = ebI = BLvI = τωm, where
    B = Flux Density in Telsa (T)
    L = Length of wire (not to be confused with Inductance) in meters
    v = velocity in m/s
EDIT: corrected formula

However – I didn’t want to leap off this diving board just yet; if we just step back and examine the big picture, I’d like to return to Square One.
If Kt = Ke when expressed in the same units, could we calculate current (I)?
Solve for I:

    I = Ï„ / Kt -> 33.9 / 0.77 = 44A

So far so good? :)
KF
Last edited by Kingfish on Wed Sep 01, 2010 4:41 pm, edited 1 time in total.
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Re: Doing the Math

Postby Goethe » Wed Sep 01, 2010 4:24 pm

Looks good to me!

Could be good thing to clarify that the V in BLVI is the velocity in m/s and not Voltage. Like you did with the B and the L.

These formulas was slowly slipping of my mind, thanks for bringing it back! :D


/Goethe
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Re: Doing the Math

Postby Goethe » Wed Sep 01, 2010 5:03 pm

Edit: Forget about what I wrote below! That only counts if you use the tooth length in BLv.

Sorry for being picky! :oops:

But it's actually ∆B because magnetic field that induces current changes direction. Eg -1T to +1T, so the ∆B is 2T.


/Goethe
Last edited by Goethe on Wed Sep 01, 2010 5:25 pm, edited 1 time in total.
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Re: Doing the Math

Postby Kingfish » Wed Sep 01, 2010 5:20 pm

:D I used to perform a lot of physics calcs in my youth, but not of these types; it’s all new to me. Thanks for spot-checking me… I’ve been at this for days with spreadsheets and no one to talk to.

OK, I edited/corrected the formula notation above. On we go…

Calculate Steady State:

Previously stated:

Given P = I * V, solve for V:

    V = P / I -> 1491.4 / 44 = 33.9
This is amusing. From the relationship of P = I * V we can play around and manipulate the factors until we arrive at suitable products for the system.

The part that bugs me is that the Motor Torque Constant (Kt) is relative and not constant at all unless we choose it to be so for control. In other words, if I am going 30 mph (er, sorry 48.3 kph) for hours on level ground and my battery drops voltage then Kt must change. Similarly, if I have a current limit set and along my journey I encounter wind or an incline, Kt must change. Perhaps it’s a moot point but it has had me twisted up – at least until I could explain it: K is an ideal theoretical value only. Do you agree? :)

Meat & Potatoes:
The whole point of this exercise is to mathematically anticipate the requirements for a steady-state wheel in motion given three values: diameter, velocity, and power consumed. I think we’ve flogged the observed system enough, therefore I wish to analyze the internal workings and construct a model of the electromotive forces to complete the energy balance equation.

A single straight wire passing through a uniform magnetic field produces a force, (formula derived from the Lorentz force equation F = qv x B):

    F = IL x B, where

      F = Force in Newtons (N)
      I = Current in Amps (A)
      L = Length of wire in Meters (m)
      B = Flux density in Telsa (T) or Gauss (g x 10,000)
Given as Torque:

    Ï„ = r x F, where

      r = radius in meters
In a single electric circuit (one loop, turn, or winding), we have two sides pulling in opposite directions along a common axis, therefore:

    Ï„ = 2rF = 2rBIL
For the sake of discussion let us assume that r = ½ d of the original wheel. The value for Flux density (B) is picked arbitrarily (we can discuss calcs a bit later).

    r = 12 inches / 0.3048 meters
    B = 0.5 T
    I = 44 A
    Ï„ = 33.9 Nm
Solve for L:

    L = Ï„ / 2rBI -> 33.9 / (2 * 0.3048 * 0.5 * 44) = 2.53 m
Observations:
  • This is a single-phase solution.
  • The frequency would be 420.2 rpm or 7 Hz.
  • The conductor would have to be 11 AWG to carry 44 A safely.
The heat generated by the system would be as follows:

    R = 4.1328 ohms/km for 11 AWG -> (4.1328 * 2.53) / 1000 = 0.01046 ohms
    R = V / I
    P = I^2 * R = V^2 / R
    P = 44^2 * 0.01046 = 20.2 watts
Calculate the Efficiency:

    Pe = (Pi – Pr) / Pi -> (1491.4 – 20.2) / 1491.4 = 98.6%, where

      Pe = Power, Efficiency
      Pi = Power, inital or imperical
      Pr = Power, resistance
This is probably the best this theoretical system will ever see. To overcome the loss of efficiency the system would need to provide > 1.4% more power.

Summation:
In the first post of the thread I calculated the Angular Velocity (ω) and Torque (τ) given wheel diameter (d), velocity (v), and power (P) used. The second half of this thread develops a model which predicts the electromotive forces for a single circuit. Thus we can represent the elementary Energy Balance Equation as:

    Efinal = Ein - Eout, where

      Ein = (1512w @ 44A x 34.4V) – (20.5w required to overcome electrical resistance**); Work on the system, constant current
      Eout = 1491.4w; Work by the system
      Efinal = 0; balanced :)
**Calculation: 1491.4w / 98.6% = 1512w; 1512 - 1491.4 = 20.5. If current (I) is constant then V must rise to 34.4V.

EDIT: Corrected EBE arithmetic.

Are you still with me? Are my calculations correct? :?: :) hehe

Thanks for checking, KF
Last edited by Kingfish on Fri Sep 03, 2010 2:22 am, edited 1 time in total.
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Re: Doing the Math

Postby Goethe » Wed Sep 01, 2010 6:59 pm

Damn, you got me using my brain at 2 am!

I have to go to bed now but I will get right back to this when I get to work. :mrgreen:

There was one thing about the Kt that I did not agree on. You can give me an internet slap if understood you wrong?!

Kingfish wrote:The part that bugs me is that the Motor Torque Constant (Kt) is relative and not constant at all unless we choose it to be so for control. In other words, if I am going 30 mph (er, sorry 48.3 kph) for hours on level ground and my battery drops voltage then Kt must change. Similarly, if I have a current limit set and along my journey I encounter wind or an incline, Kt must change. Perhaps it’s a moot point but it has had me twisted up – at least until I could explain it: K is an ideal theoretical value only. Do you agree?


If the voltage drops then the motor speed also needs to drop to a level of BEMF where the battery voltage over come the BEMF and it can push current again through the motor. The force to over come will be lower and also the current but the Kt will stay the same.

With the current limit set and you encounter an increase in force to over come, the result will only be a lower speed to equal out the added force. The torque produced by the motor stays the same also the current and there by the Kt.

But your right in a way if we take in count that the efficiency will change over the speed range and regarding current we should only calculate with motor current and not battery current.


/Goethe
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Re: Doing the Math

Postby Kingfish » Wed Sep 01, 2010 10:46 pm

I thought about this on my bike ride this evening (with blood flowing mightily through the brain and all) and I can see how my statement is conflicted. Let me see if I can make it less murky:

When I am climbing a hill I pull about 2 hp however it is due in part to an artificial current limit so that I can maximize my battery capacity. The wheel in the example could go much faster; I just picked 2 hp and 30 mph because that’s pretty much what I duplicate when I'm on a ride. However I concede your good point about this being a function of battery current and not motor current. I stand corrected :)

Goethe, thanks for your help; I shall hold off on more calcs until you check me. Also, I know I should format the vectors in non-Italic boldface; I’ll try to clean that bit up going forward: I have four open sources in front of me with each author/team having their own style of expressing equations. They all approach the modeling differently, and yet I feel compelled to knit together a cohesive explanation. Surely other people are struggling to understand this stuff :o

Having fun, KF
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Re: Doing the Math

Postby Goethe » Thu Sep 02, 2010 4:26 pm

You are doing a great job sorting these equations out!

I did the same a while ago but when it came to dealing with the different flux densities in the motor I took a break...a looong brake. :roll:

Nothing to argue about the equations! There was a typo "Efinal = Ein + Eout" Should be a minus right?

The single electric circuit equations does not tell so much about how a "our" motors perform, but I expect that you will deliver the maths for that soon. :D


Keep it up!

/Goethe
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Re: Doing the Math

Postby Kingfish » Fri Sep 03, 2010 2:56 am

<humble bow> Salute and thank you! Hmmm, so you noticed I snuck in the Royal “we” did ya? :lol:
All the world’s indeed a stage and we are merely players…

Energy Balance Equation (EBE): Yes – good eyes, you are correct; I have made the edit :D
Single Electric Circuit: Funny you should mention that…

The Story to Date:
As observers of science we have developed a model for evaluating the electromotive conditions which theoretically fit a wheel in constant motion: For a 24-inch/ 0,61 m diameter wheel spinning at 30 mph/ 48,3 km/h providing 2 hp, we calculated that the required current would be 44A at 34.4V along a single 11AWG conductor that is 2.53 m in length, with an ideal efficiency of 98.6%.

Let us now depart from the passive role of observers, and overtly craft a theoretical motor from the inside out that could deliver at least the minimum aspects of the observed wheel.

For the sake of discussion we are going to exclude road- and wind resistances, and bearing friction.

Reviewing the solution for the Single Electric Circuit, simplistically - the magnetic field used in the example would need to be as long as the calculated length of the circuit. In the real world we know this to be different :wink:

Firstly, we need to reduce the area exposed to the magnetic field to a practical level and then fit our 2.53 m wire within those boundaries. We can do this by rolling up the wire into a flat Archimedes spiral one-wire deep in thickness (also called “multi-layer single-row”) so that the entire disc of copper faces perpendicular to the magnetic field. An American engineer Harold Wheeler developed a neat little formula that helps us with this calculation for the Flat "pancake" coil:

    L (μH) = A^2 * n^2 / (8 * A + 11 * w)

      Where
      L = Inductance, measured in micro-Henry’s (μH)
      A = Average radius of the coil*, m
        Given as
        A = [((OD – ID)/2) + ID] /2
        Where
        OD = Outside Diameter
        ID = Inside Diameter
      n = number of turns
      w = diameter of the wire

But wait…
The problem with this formula is that this gives us the Inductance (L) and not the length, and we have to tell it what the expected OD is which is lame. However we can still use part of the formula for extrapolation.

Given the length 2.53 m of 11 AWG copper wire, solve for the number of turns of a multi-layer single row spiral if the ID is 15mm.


Caveat: I searched the web to find a short easy solution but I couldn’t find a single one where I could take my length of copper wire and figure out how many turns it would make. Giant Black Hole Mystery. Closest I did find on the subject was here:
http://mathworld.wolfram.com/ArchimedesSpiral.html

I would definitely like to revisit this with a resolution. <Hereby Flagged> :!:

OK, so now we have one big huge flat 11 AWG wire spiral pulling 44 A sitting at 90* on the perimeter of the tire that’s going like 7 Hz. This isn’t very useful, and what we need to do is make a few adjustments and redefine our Wheel by placing a hub at the center with magnetics on the rotor, and locate our spiral conductor onto the stator.

Doing so we must understand that:
  1. For torque (Ï„) to remain the same, the net Force (F) must increase in direct proportion to the decrease of the radius (r).
  2. The result is that our spiral coil will need to grow in size to compensate.
  3. We also need to divide the spiral up into smaller pieces linked in series to better utilize & distribute the applied force around the hub.

Note: At this time we are simply rearranging the geometries to facilitate our further study, and we are not introducing new materials such as Iron and issues such as cogging and eddy currents. K.I.S.S.

Are you with me so far? :)

~KF
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Re: Doing the Math

Postby Kingfish » Thu Sep 09, 2010 1:14 am

Hmmm, have I scared you away? :(

The thing about figuring out how many copper turns was bugging me, and I didn’t much like the awkward method of trying to figure that out given only length and wire diameter. I searched and searched and found nothing but a bunch of complex math.

This has had me busy in my tiny cave, scratching and scribbling away, tinkering on this and that. It’s a wee bit crude, but I have an alpha version that think is good enough for an unveiling…

<drum roll please...>
Announcing my 1st ES-Application

The Archimedes-Winding Calculator!

Instead of relying on a bunch of fancy-schmacy calcs – I used graphic kinematics and iteratively looped around the axis until the length was all used up :mrgreen:

As I said, the application is a bit crude. It needs validation, a little buff & polish, and a few nice features added. I made it with CSharp and Silverlight. It’s a standalone app, and you can use it out-of-browser by right-clicking on the interface and installing it onto your desktop. It should update auto-magically :wink:

Anyway – Walla! There it is. :D

Now that I am unblocked, I shall continue onward and through the fog.
Enjoy! KF
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Re: Doing the Math

Postby Kingfish » Thu Sep 09, 2010 4:45 pm

I have released a new spiffed-up BETA version that also fixes a pesky graphics bug, and is 10X more accurate on calculations. Also we now accept parameters in mm, meters, inches, and feet.

If you downloaded and installed the ALPHA version, please remove it:
  1. Click on the icon to start the application.
  2. Right-mouse-click and select “Remove this application…”
  3. Now click on the hyperlink below to install the BETA version. The auto-update feature should work now :wink:

Archimedes-Winding Calculator BETA

OK, back to the whiteboard…
~KF
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Re: Doing the Math

Postby Goethe » Thu Sep 09, 2010 8:03 pm

Kingfish wrote:Hmmm, have I scared you away? :(


No absolutely not! :D

This is really interesting and I find very educating for me. It's just that I got many projects running and ideas spinning around in my head right now so I haven't got the mental rest to focus at this properly. I will get in the game soon! :D

/Göran
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Re: Doing the Math

Postby Goethe » Fri Sep 10, 2010 12:07 pm

I've tried your calculator and it worked fine!

I think it would be wise to start calculating with the 2.53 m as only the part of the wire that is peripendicular to the magnetic field.

/Göran
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Re: Doing the Math

Postby Kingfish » Fri Sep 10, 2010 7:59 pm

Gotcha. :D Let me state that I have set up a support page in the Technical Reference section the Archimedes-Winding Calculator.

Resuming where I left off, we have the following:
A 24-inch/0,61 m diameter wheel spinning at 30 mph/48,3 km/h providing 2 hp, and a single circuit carrying 44A at 34.4V along a single 11AWG conductor that is 2.53 m in length.

Using the nifty calculator, we can deduce the following, given:

    w = diameter of the wire = 2.30378 mm
    ID = Inside Diameter = 15 mm
    Ag = AirGap = 0
    l = Length = 2.53 m
Yields…

    OD = Outside Diameter, mm = 87.8634418888889
    n = Number of Turns = 15.3138888888889

Now we can go back and attempt to correctly calculate Inductance (L) using inches with the formula given:

    L (μH) = A^2 * n^2 / (8A + 11 w), or more simply
    L (μH) = (nA)^2 / (8A + 11w)
Solving for A:

    Note – A is in inches, therefore
    OD = 87.86/25.4 = 3.459
    ID = 15/25.4 = 0.591
    w = 2.30378 = 0.0907
    A = [((OD – ID)/2) + ID] /2 =>
    [((3.459 – 0.591)/2) + 0.591] /2 =>
    A = 1.0125
Solve for L:

    L = ((1.0125)^2 * (15.3144)^2) / ((8 * 1.0125) + (11 * 0.0907)) =>
    L = 240.43 / 9.0977 = 26.43 μH
Note that this is different by a factor of 2.5 than what is given in the calculator link below:
And is off by even more with this one:
However it is nearly the exact same with this calculator:
Solving for L in meters, the formula given on Wikipedia is:

    L = (A^2 * n^2) / ((2A + 2.8d) * 10^5 =>
    L = (An)^2 / ((2A + 2.8d) * 10^5
Where d = depth of coil; (OD – ID) / 2

    OD = 87.86/1000 = 0.0879
    ID = 15 = 0.0150
    d = (0.0879 – 0.0150) /2 = 0.03645
    A = [((OD – ID)/2) + ID] /2 =>
    A = (0.03645 + 0.0150) /2 = 0.025725
Solver for L:

    L = (0.025725^2 * 15.3139^2) /[(2*0.025725) + (2.8*0.03645)]*10^5 =>
    L = 0.155197 / 15351 = ridiculously wrong. :roll:
Conclusions:
I am at a bit of a loss for a decent formula for open air coil Inductance. Practical sources say to measure it in situ. Can we presume that we can affect the overall inductance with an inline coil to match our controller?

One step back before we take two steps forward. :?
~KF
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Re: Doing the Math

Postby Kingfish » Sat Sep 11, 2010 3:38 pm

Bypassing the Inductance (L) conundrum for a moment, I would like to focus on how to reduce the current through the conductor with the goal to reduce potential ohmic heating.

Our single circuit presently pulls 44 A. The most straight-forward method I can think of to reduce current is to add more magnetic poles to our design. However in doing so we must also up the A/C signal frequency proportionally to compensate.

I haven’t been able to locate the precise formula, however we can assume that

    Δ Magnet Pole pairs (p) = Δ Signal Frequency (f)
An example of that is displayed in this wiring diagram:

Therefore, if our spinning wheel is already rotating at 7 Hz and pulling 44 A, determine the changes when we apply a 7X reduction of current. Given:

    Ii = 44 A
    pi = 2
    fi = 7 Hz =>
    If = 44/7 = 6.3 A
    pf = 2*7 = 14
    ff = 7*7 = 49 Hz ≈ 50 Hz (European A/C frequency)
Realistically, the frequency can go up quite a bit higher before we would see appreciable losses; however it is doubtful this system will ever see a signal beyond 400 Hz even at WOT.

Another way to reduce the current is to divide the single circuit into 3-phases. Converting the current value from a single-phase to a 3-phase is given as I1p = I3p / √3 ≈ I/1.732

To convert our original 44 A single circuit to 3-phase, each individual phase would run at 25.4 A (a strange coincidence). If we took our newly-minted 7-pole paired motor and split it into 3-phase, the 6.3 A value would drop again to 3.6 A per phase.

These changes allow for a robust copper fill – and we haven’t yet optimized the coil windings and distribution.

Am I good so far? :)
~KF
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Re: Doing the Math

Postby Goethe » Sat Sep 11, 2010 4:42 pm

To increase the efficiency in a motor without total redesign I would start to look at the magnets and airgap.

I have been going through my papers and some stuff I've found on the web and I think we have to restart this whole thing from another angle.
The magnetic flux is built up from many components and to get decent figures these components can´t be left out. For example is the permeability constant about 7000 times higher for an electric steel core compared to the air like in the coil we calculating with now.
Magnet data, Air gap geometry, teeth geometry, stator material and more some stuff are parameters that needs to part of the equations to get decent accuracy.

To do this we need to begin with a known stator and rotor geometry and brake it down to 1 slot/pole calculations. After that scale it back to a complete motor.


/Göran
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Re: Doing the Math

Postby Kingfish » Sat Sep 11, 2010 6:05 pm

Actually, the whole time I have been working with coreless to K.I.S.S. My next motor is in fact – a coreless design, so with regards to laminations and eddy currents, we are still on track.

The magnets are Neodymium, the airgap is 1mm between the magnets and the copper, and the Flux Density is 0.5T. I haven’t mucked with that data yet… soon though.

If we can get through the basic calculation so I understands it from beginning to end, that would help me tremendously. :)

Make sense? KF
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Re: Doing the Math

Postby Goethe » Sat Sep 11, 2010 6:41 pm

Ahaa!

Sorry didn´t mean to get you off track!

So you already have a coreless motor? Or are you designing one?
Is the example with a flat spiral coil moving through a uniform magnetic field close to the real design?
And what is K.I.S.S?

/Göran
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Re: Doing the Math

Postby Kingfish » Sat Sep 11, 2010 7:25 pm

OK, I suppose I should come clean… :roll: :lol:

I am trying to design the simplest motor I could think of. Powerful – YES! But still very simple to assemble if one has the correct tooling.

I already have a Radial Flux re-work/modification design completed: I had been working on it for about 3 months and it was going to use a pair of 9C motors to craft my 2WD ebike. But then :idea: I learned about other ways to employ electromotive forces by means of an Axial Flux design. :idea: In this manner, the magnets are collocated on the sides of the rotor rather than at the circumference. Thus we can theoretically maintain a 1 mm airgap.
In order to assemble the two sides together, a ironless-coreless design is required, which has the added benefit of eliminating cogging. 8)

The magnets under consideration were at one time the very same used in windmills, thus the ideal model would have 16 arc magnets 6.35 mm thick x 100 mm ID x 200 mm OD. The problem with that design is that it requires a steel backing plate to capture the backside flux. I don’t want to go down that path. Instead, I am going to use a Halbach array so that I can stay with a lightweight aluminum rotor.

This whole time you and I have been working with mock-data; the final numbers will get sorted out as we narrow down the issues. The biggest one I have faced is calculating the windings. I haven’t found useful tools for figuring that part out – so I created them. :)

My goal is to automate the design process; take the current winding application and add several layers to it so one could literally walk through a wizard and design their motor in one night. It can only be done with coreless axial design through because every other design uses radial or iron and that – as we know makes the calculations more complicated.

K.I.S.S. means “Keep It Simple Silly”. When I was in the Navy it meant “Keep It Simple Stupid” ‘cuz most of the swabbies had about an 8th-grade level education. :roll: The KISS method works for a lot of things. Sometimes people I work with get way too complicated and it's not necessary to complete the task. We're using KISS here for problem-solving :)

Are you still on board? It’s going to get fun – I can promise you that!

So to re-cap and outline the challenge:
  1. We have a wheel, it spins at a velocity, using x-amount of horsepower.
  2. We determined the length of wire, and the amount of current and voltage to theoretically balance the power equation.
  3. The next step is to predict how to reduce the current required if we add more magnets and switch to 3-Phase. There are still no losses, or wind resistance, or bearing friction: It’s just pure electromotive force that we are attempting to tame into a small compact coreless design.

Make sense? :D
Truly, I appreciate your help! 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

Postby Goethe » Sat Sep 11, 2010 8:08 pm

Ok now it make more sense! :mrgreen:

This is really a cool project!

I will reset my brain with some sleep now, and hopefully I have forgotten all about iron stator and rotors and radial flux tomorrow. :wink:

/Göran
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Re: Doing the Math

Postby rhitee05 » Sat Sep 11, 2010 8:45 pm

You should keep in mind that, for a coreless design, the airgap is the distance separating the two magnet rotors. So, its not just the magnet-coil distance that counts but the thickness of the coil as well.

How do you plan to construct the axial-flux Halbach array? I have not seen any edge-magnetized wedge magnets (other than custom pieces). Are you just going to use rectangular magnets? If so, the pie-shaped gaps will also count towards the air gap. More importantly, the flux density will vary across the radius - highest on the inside where the magnets touch and lower toward the outside as the gap increases. Not that it wouldn't work, but it would probably work better if you filled the gaps in between with a ferrous material.
Eric

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Re: Doing the Math

Postby Kingfish » Sat Sep 11, 2010 10:28 pm

EDIT: Appending TOC.
  1. Math Modeling
  2. Axial Flux Design, Halbach, Ironless-Stator, Axial Motor
  3. Axial Flux Design, Non-Halbach, Ironless-Stator, Axial Motor

Axial Flux Design, Halbach, Ironless-Stator, Axial Motor

Image

AirGap: On a single-rotor and single stator, it is measured from the face of the magnet-array to the face of the copper winding according to the book that I have on the subject. However with a double-rotor we get the added benefit of both sides and the calcs are a little different. The two sides will be locked N-S in attraction which will make assembly challenging, though less so if I had used an iron stator. My plan was to keep the gap between the two magnetic faces <= 10mm.

At the moment I wish to use Litz wire. I have pre-built a reference to sources here: Magnetic Wire & Nd Magnets Reference

The Archimedes-Winding Calculator is also specific to the AF design.

There is to be an airgap between the windings for cooling should it come to that. The plan is to keep the ohmic heating down to the minimum - though common sense says there will be heat regardless. I might as well be prepared to deal with it.

The magnets have an airgap as well for the same reason - to foment cooling. In the image above I had considered briefly to use 1/4-inch cubes, however custom-orders as a rectangle are possible and will make assembly much easier. I haven't inquired if the manufacturer will do weggies - that would be the optimum!

The construction, if I stay with rectangular shapes, is to either build modular units of four, or to have a jig where I lay down 1/4 of the magnets at a time against the rotor surface (probably the latter). I did not consider putting ferrous material in the bonding agent for use between the gaps; nothing I've read has that as an option. Might be a good subject for FEA. I understand that the Flux Density is greatest where they touch (ID). One positive aspect is that as the inside radius becomes larger the gap between the magnet array reduces. The Halbach array increases the Flux Density by ~1.4X; the whole business is funny math and took many afternoon naps for me to understand it :wink:

My hope is that there will be plenty of Torque (LFP has suggested as much) - this way I can focus the design towards light-weight, high-rotation, and efficiency. It's for my next ebike which is structurally stronger and designed to go faster.

Eric, want to help with the maths too? I'll be your friend :)

Best, KF
Last edited by Kingfish on Fri Mar 04, 2011 2:47 pm, edited 2 times 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 8)
* Club Member: 40-mph & 101. 10k-Club: 11,340 miles-to-date, 5850 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.
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