This Post is mainly directed at JohnRob since we are both working on doing custom motor winds, so bear with me.
After winding a few stator poles with 4 strands of 22awg @ 5 turns on the 3530, I see this is gonna be tough. It's hard to get it in there and make everything lay neat. Each strand has to behave and lay parallel without crossing any of it's "neighbors".
I think I see now why the person behind the PowerCroco site is going with just one strand. Even though it's stiffer, it should be easier to make lay in the turns and easier to manage.
In researching, I have constantly seen references to copper density, wind resistence, and the effects of not having good values as being "copper losses" Copper Losses result in a motor that runs hotter than normal and one that is less efficient.
To Minimize copper losses, one should have the most amount of copper on each stator pole looking at the Total Cross Sectional Area of the winding. In other words, each stator pole must be as full as possible AND have the least amount of resistence in the wind as possible.
I'm going to try to wind with just one strand to hopefully gain some on this. I'm willing to put out the extra work and effort to do the wind and am not too worried about it "being easy on my fingers" LOL
In order to know what awg of wire to use, first some things must be calculated.
By taking the Cross Sectional Area of a given awg of wire, then multiplying by the number of strands used in an already succesful motor, we should be able to come up with a single strand that is the same as using all those multiple strands.
CSA walks hand in hand with wire resistence also.
Here is a chart that was pulled from the net that shows data for various wire sizes used in motor winding.
http://www.powerstream.com/Wire_Size.htm
30awg Dia. .01(.254mm) Ohms per 1000' 103.2 max. amps .86 Cross Sectional Area = .0000785
28awg Dia. .0113(.287mm) Ohms per 1000' 64.9 max. amps 1.4 CSA = .00010023665 x 20 = .002004733 20 strands @ 5 turns = .010023665 Total CSA
26awg Dia. .0159(.403mm) Ohms per 1000' 40.81 max. amps 2.2
24awg Dia. .0201(.510mm) Ohms per 1000' 25.67 max. amps 3.5
22awg Dia. .0254(.645mm) Ohms per 1000' 16.14 max. amps 7 Cross Sectional Area = .0005064506
20awg Dia. .032(.812mm) Ohms per 1000' 10.15 max. amps 11
18awg Dia. .040(1.02mm) Ohms per 1000' 6.38 max. amps 16
16awg Dia. .050(1.29mm) Ohms per 1000' 4.01 max. amps 22 CSA = .0019625
14awg Dia. .064(1.62mm) Ohms per 1000' 2.52 max. amps 32 CSA = .00321536
12awg Dia. .080(2.05mm) Ohms per 1000' 1.58 max. amps 41
10awg Dia. .101(2.58mm) Ohms per 1000' .99 max. amps 55
Area = (pi) x r2
In English, this formula means "area equals pi times the radius squared." A circle's radius is one half of its diameter, or one half of what you get if you measure all the way across its widest part. "Squaring" something means you multiply it by itself. Pi is a number that roughly equals 3.14159.
A = 3.14 x Radius x Radius Radius = Diameter/2
I have been figuring and adding Cross Sectional Area values to the chart.
The 3530 motor was wound using 20 strands of 28awg wire.
This is a .002004733 Cross Sectional Area of Copper.
14awg wire has a CSA of .00321536
16awg wire has a CSA of .0019625
16awg is ever so slightly less, with 14awg being about 1.5 times that.
I'm going to order some more wire. Copper Density and efficiency can be improved if 14awg will fit. But even if 16awg is used, it will duplicate the density/efficiency scores that are on the market now.