Kogs P56 wire calculation
G'day UFO and Team
Number of turns per Pole
If you wind completely one half of a pole i.e. fill P1 N or P1 S the number of windings/turns is equal to two full poles i.e. P1 N&S and P2 N&S we can call this number FHP (Full Half Pole) and therefore the number of turns for each pole is FHP divide by 2 also the number of turns for each half pole is FHP divided by 4.
If the Armature is to be Bifilar then the wire is to be thinner and the number of turns for one Pole would be equal to FHP divided by 2. This is the same as normal winding as we are winding only one strand per turn but being Bifilar when winding the real Pole we will be winding two strands together and so the real number of turns is halved and therefore the number of bifilar winds per half pole would be FHP divided by 8.
The Ohms per Pole
Normal winding If you measure the Total Ohms for the FHP then this would be say TOFHP and therefore would be the total Ohms for two whole poles so the Ohms for one pole would be TOFHP divided by 2
Bifilar winding as the calculating of a bifilar Ohms is Half (near enough) of the total single wire Ohms so the Ohms for one pole would be TOFHP divided by 8.
Now the winding of MY P1 is as below
Using the AWG 18 (0.1mm diam.) wire The FHP is 68 turns and therefore the Number of turns per half pole is 17
The TOFHP measured with 2 different DMM is
First DMM is 0.7 Ohms minus the DMM 0.2 Ohms = 0.5 Ohms
Second DMM is 0.9 Ohms minus the DMM 0.4 Ohms = 0.5 Ohms
The TOFHP is for 2 poles and therefore the one pole is 0.25 Ohms
Using Bifilar the AWG 22 (0.6mm diam.) wire The FHP is 232 turns and therefore the Number of turns per half pole bifilar is 29
The Second DMM is 5.1 Ohms minus the DMM 0.4 Ohms = 4.7 Ohms
The TOFHP is for 2 poles and therefore the one pole is 4.7 Ohms divided by 8 =0.5875 Ohms
I did not use the first DMM here
The P1 for AWG 18 =0.25 Ohms The P1 for AWG 22 =0.5875 Ohms
I hope this info is understood
Please any input would be appreciated
Kindest Regards
Kogs now has a headache
Originally posted by iankoglin
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Number of turns per Pole
If you wind completely one half of a pole i.e. fill P1 N or P1 S the number of windings/turns is equal to two full poles i.e. P1 N&S and P2 N&S we can call this number FHP (Full Half Pole) and therefore the number of turns for each pole is FHP divide by 2 also the number of turns for each half pole is FHP divided by 4.
If the Armature is to be Bifilar then the wire is to be thinner and the number of turns for one Pole would be equal to FHP divided by 2. This is the same as normal winding as we are winding only one strand per turn but being Bifilar when winding the real Pole we will be winding two strands together and so the real number of turns is halved and therefore the number of bifilar winds per half pole would be FHP divided by 8.
The Ohms per Pole
Normal winding If you measure the Total Ohms for the FHP then this would be say TOFHP and therefore would be the total Ohms for two whole poles so the Ohms for one pole would be TOFHP divided by 2
Bifilar winding as the calculating of a bifilar Ohms is Half (near enough) of the total single wire Ohms so the Ohms for one pole would be TOFHP divided by 8.
Now the winding of MY P1 is as below
Using the AWG 18 (0.1mm diam.) wire The FHP is 68 turns and therefore the Number of turns per half pole is 17
The TOFHP measured with 2 different DMM is
First DMM is 0.7 Ohms minus the DMM 0.2 Ohms = 0.5 Ohms
Second DMM is 0.9 Ohms minus the DMM 0.4 Ohms = 0.5 Ohms
The TOFHP is for 2 poles and therefore the one pole is 0.25 Ohms
Using Bifilar the AWG 22 (0.6mm diam.) wire The FHP is 232 turns and therefore the Number of turns per half pole bifilar is 29
The Second DMM is 5.1 Ohms minus the DMM 0.4 Ohms = 4.7 Ohms
The TOFHP is for 2 poles and therefore the one pole is 4.7 Ohms divided by 8 =0.5875 Ohms
I did not use the first DMM here
The P1 for AWG 18 =0.25 Ohms The P1 for AWG 22 =0.5875 Ohms
I hope this info is understood
Please any input would be appreciated
Kindest Regards
Kogs now has a headache
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