Today a friend gave me this Sparta bicycle motor and the intention is to turn it into a windmill generator.
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If you count the number of magnets you get 20, so there are ten north poles and ten south poles. If you count the number of coils you get twentyfour. So the ratio is 24/20 = 6/5. These different numbers are chosen to prevent a high peak in the cogging torque but I doubt if this is effective enough to use this motor as a generator for a wind turbine. The angle in between the magnet poles of the armature is 360/20 = 18°. The angle in between the stator poles is 360/24 = 15°. So the difference is 3°. This means that you will get 360/3 = 120 preference positions per revolution. This isn't a high number and therefore the preference positions will be rather strong. These rather strong preference positions won't be a problem for a bicycle motor but it may give your wind turbine a rather high starting wind speed.
You must know very well what you are doing if you want to use this motor as a generator and if you use the original winding or if you make a new one. The winding must be a 3-phase winding. This means that there are 8 coils of phase U, 8 coils of phase V and 8 coils of phase W. I assume that the coil sequence is: U1, U2, V1, V2, W1, W2, U3, U4, V3, V4, W3, W4, U5, U6, V5, V6, W5, W6, U7, U8, V7, V8, W7, W8 right hand. So every phase has four coil bundles of each two coils. I assume that the sequence of the magnets is: N1, S1, N2, S2, N3, S3, N4, S4, N5, S5, N6, S6, N7, S7, N8, S8 right hand.
Next you should make a picture of the armature and the stator and in this picture you should take coil U1 opposite to magnet N1. If coil U1 is opposite to magnet N1, coil U2 is about opposite to magnet S1 (with 3° shift). This means that the winding direction of the two coils of one coil bundle must be opposite to make that the voltage generated in coil U1 is strengthened by the voltage generated in coil U2. Assume that coil U1 is wound right hand. So coil U2 must be wound left hand.
In the picture you can see that if magnet N1 is opposite to coil U1, magnet N6 is opposite to coil U5 and that magnet S6 is about opposite to coil U6. So coil U5 has to be wound right hand and coils U6 has to be wound left hand. So the winding direction of two opposite coil bundles is the same!
In the picture you can see that if magnet N1 is opposite to coil U1, magnet S3 is opposite to coil U3 and that magnet N4 is about opposite to coil U4. So now a south pole is opposite to the left coil of a coil bundle. This means that coil U3 must be wound left hand and that coil U4 must be wound right hand. The same counts for the opposite coils U7 and U8. So the winding direction of these four coils of phase U is just opposite as the winding direction of the first four coils of phase U! So it is very important to make no mistake in the winding direction of the eight coils of one phase.
The eight coils of phase V and the eight coils of phase W have the same pattern for the winding direction. It is easy to prove that there is a phase angle of 120° in between the coils of phase U, V and W. However, the coil patterns have to be positioned such with respect to each other that there is an angle of 12° in between a coil of phase V and a north pole and an angle of 24° in between a coil of phase W and a north pole. This is the case for coil V3 and pole N4 and for coil W5 and pole N7. So these two coils must be wound right hand just as coil U1. The mechanical angle in between coils U1, V3 and W5 is 120°. Assume right hand is R and left hand is L. So the sequence of the coil number and the winding direction is: U1R, U2L, V1L, V2R, W1R, W2L, U3L, U4R, V3R, V4L, W3L, W4R, U5R, U6L, V5L, V6R, W5R, W6L, U7L, U8R, V7R, V8L, W7L, W8R.
The armature has the same magnetic position with respect to the stator if it has rotated 36°. So a rotational angle beta = 36° corresponds to a phase angle alpha = 360°. So a rotational angle beta = 1° corresponds to a phase angle alpha = 10°. So a rotational angle beta = 3° corresponds to a phase angle alpha = 30°. This means that there is a phase angle of 30° in between the voltages generated in coil U1 and coil U2. Adding two identical sinusoidal voltages which are out of phase results in a new sinusoidal voltage which has a maximum if the magnet N1 has rotated 1.5° right hand from the drawn position in the figure. Beta = 1.5° corresponds to alpha = 15°. This results in a summerised maximum voltage of 2 * sin 75° * Vmax = 1.932 * Vmax. If both voltages would be in phase, the maximum voltage would be 2 * Vmax if Vmax is the maximum voltage of one coil. The fact that the voltages are 30° out of phase, results therefore in a reduction of the voltage by a factor 1.932 / 2 = 0.966 which is certainly acceptable.
If a new winding is laid, the number of turns per coil and the wire thickness can be found by try and error for a certain wind turbine rotor. However, I see no good reason why you should rewind the stator if the winding direction of the coils is as I predicted.