So I am making a similar rotor with real strong magnets to do real tests.. Here's the math..
3.14 Pi is an irrational number so the only way to make this almost perfect is to use magnets that are divisible by 3.14. But good luck finding magnets 3.14mm, 6.28mm, 9.42mm, etc. So we need to fudge it a little and do what we gotta do. The sinewave will have some minor irregularities, but who cares..
I have N42 bar magnets polarized like this. 13mmx13mmx50mm
[
attachment=150]
Now the last rotor was designed to reciprocate 1 polarity back and forth. This one will do both polarities one after another. Exactly like a generator does. But the motion will still be perpendicular to the induction.
So take the magnet face size and divide by 2. 13/2= 6.5.. We divide by 2 because each magnet should overlap 1/2 with the previous to get the smoothest transition.
Now this 6.5mm needs to complete a half circle. So multiply 6.5 X Pi (3.14). And I get 20.41. This number will be the angle of each magnet.
Now to figure out how many magnets.... take 180 degrees / 20.41. In my example- this equals 8.819 magnets.. <--- this number will never be a whole number unless your magnets were a multiple of 3.14. But whatever...
So rounded, I need 9 magnets to complete the rotor because my magnets are long enough to use both polarities. If you are using flat magnets, you need double this many. 1 sine per half rotation.
Now I am off by 0.18079th of a magnet. So we take that 20.41 number and round it to the closest degree which is divisible by 180. in my case 20 degrees. 180/20 = 9 magnets.
So my magnets will be configured like this.. Except red on 1 side, blue on the other.
[
attachment=151]
This design can not have a center shaft because of the single magnets. But with flat magnets, you use double as many and you get to have a center shaft.
Now keep in mind, the last step to complete the revolution needs to also be the first step. 10 magnets puts be back at the starting point, but I only want 9 so the 9th is overlapping with the first at the same ratio all the others are.
Does it have to be this critical? probably not. But if I am using $50 in magnets and designing, I am going to try to get it right.