Ahh well, my
Cabernets are crushed, but not my spirit! So – 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 o’ 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. I’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 – 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 on…

Double-Sided Rotor & Stator arrangements
Simplistically we’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.
B
m0 = Br[1 - exp(-βh
M)]((sin(À/n
M))/(À/n
M)) [3.42]
Solve for B
m0, where
hM = 6.35mm (1/4-inch)
nM = 4
β = 2À/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 I’m a little bit lost is
βx and
βz, however I have space between the discs (z) as 10 mm (again for the sake of discussion).
I’m going to hold on the Winding Calculation until we sort out the Flux Density.
Have I lost anyone?
~
KF