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 magnetsMagnetic 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.
= Br[1 - exp(-ÃŽÂ²hM
Solve for Bm0
hM = 6.35mm (1/4-inch)
nM = 4
ÃŽÂ² = 2Ãâ‚¬/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
, 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?