1.5 turn strategy
I haven't worked this out, but it is my feeling that there should be a "better" way to distribute the x-overs and pack more wire onto the armature.
The way I wound the armature resulted in all of the x-overs being in one slot and none of them in the other slot. Here's the 1st three windings, the way I did it:
1a) start at commutator, put wire in slot #1
1b) lay wire over top of armature, to the right of the shaft
1c) put wire into slot #6
1d) lay wire over bottom of armature, to the right of the shaft
2a) put wire in slot #1
2b) lay wire over top of arm. to the left of the shaft
2c) put wire into slot #6
2d) lay wire over bottom of arm. to the left of the shaft
3a) put wire in slot #1
3b) lay wire over top of arm. to the right of the shaft
3c) put wire in slot #6
3d) lay wire over bottom of arm. to the right of the shaft
...
If you try this, you will find that the cross-overs happen at (2a) and (3a), always in slot #1.
The original goal was to make sure that the wire was evenly balanced on each side of the shaft.
I think that it should be possible to do something like:
1a) start at commutator, put wire in slot #1
1b) lay wire over top of armature, to the right of the shaft
1c) put wire into slot #6
1d) lay wire over bottom of armature, to the right of the shaft
1e) put wire into slot #1
1f) lay wire over top of armature, to the right of the shaft
then repeat 1a-1f on the left side of the shaft. 3 rights, then 3 lefts.
I haven't tried this, but I think it will result in an even number of strands on each side of the shaft, plus it will have the same number of x-overs in slot #1 and slot #6.
I think that the best way to test my theory is to try it. If it doesn't work, maybe the solution will still be obvious.
At the moment, I don't have an empty armature to try it myself, but, if you have trouble I'll unwind one and we can try to suss it out together.
If you try it, let us know.
pt
Originally posted by FRC
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The way I wound the armature resulted in all of the x-overs being in one slot and none of them in the other slot. Here's the 1st three windings, the way I did it:
1a) start at commutator, put wire in slot #1
1b) lay wire over top of armature, to the right of the shaft
1c) put wire into slot #6
1d) lay wire over bottom of armature, to the right of the shaft
2a) put wire in slot #1
2b) lay wire over top of arm. to the left of the shaft
2c) put wire into slot #6
2d) lay wire over bottom of arm. to the left of the shaft
3a) put wire in slot #1
3b) lay wire over top of arm. to the right of the shaft
3c) put wire in slot #6
3d) lay wire over bottom of arm. to the right of the shaft
...
If you try this, you will find that the cross-overs happen at (2a) and (3a), always in slot #1.
The original goal was to make sure that the wire was evenly balanced on each side of the shaft.
I think that it should be possible to do something like:
1a) start at commutator, put wire in slot #1
1b) lay wire over top of armature, to the right of the shaft
1c) put wire into slot #6
1d) lay wire over bottom of armature, to the right of the shaft
1e) put wire into slot #1
1f) lay wire over top of armature, to the right of the shaft
then repeat 1a-1f on the left side of the shaft. 3 rights, then 3 lefts.
I haven't tried this, but I think it will result in an even number of strands on each side of the shaft, plus it will have the same number of x-overs in slot #1 and slot #6.
I think that the best way to test my theory is to try it. If it doesn't work, maybe the solution will still be obvious.
At the moment, I don't have an empty armature to try it myself, but, if you have trouble I'll unwind one and we can try to suss it out together.
If you try it, let us know.
pt
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