Herbertkabi says
here that solder wick will not generate significant eddy currents.
Taking a formula from wikipedia:
w = pi^2 * B * d^2 * f^2 / (12 * ro * D)
w is Watt per kg
B is peak flux in Tesla
d is diameter of wire in meters
f is frequency in Hz (rpm*poles/120)
ro is resistivity in Ohm-m (copper = 1.68 * 10^-8)
D is density in kg/m3
With a 0.5mm wire, at 1000 rpm, flux density 0.6, 36 poles, I get 89 watt per kg.
For a flat sheet, the formula is exactly the same, except d is now thickness, and the divisor "12" is only 6. So, twice the loss for the same "d".
However, as noted by Shane Colton in the earlier referred paper, which I had failed to grasp earlier, if there is leakage between neighbouring mags, then there is also a tangential field component, which makes eddies in the flat wire.
As he says, LEAF had a large gap (17.8mm) and also the mags were pretty close.
In a halbach with a small gap, there won't be as much tangential flux I think.
LEAF's noload current with trap drive was 15A at 1500 rpm, 19.7v = 296 watt.
According to the above formula the eddy losses should have been 22.6 watt. Here is what I put into it:
16AWG 1:4 aspect ratio = 1.31mm^2, 0.57 x 2.29 mm
120 turns, 101.6mm active length of each turn. 4 parallell strands. 0.63 kg copper.
16 poles. 0.5T
I think though, that since these watt are made from the motion of the mags, that there is an additional loss associated with eddy currents in motors, that are not accounted for in the formula. That is, there is a resistance loss from the current that is needed to produce 22.6 mechanical watt in the first place. 22.6 watt at 1500 rpm is about 0.15 Nm, and with Kt=0.13 that is 1.16A, which with 19.7v input doubles the "eddy" loss - 45.4 watt.
Is that a valid way to calculate? If so, how do we model/calculate/estimate the losses in flat endwindings?
Slideshow with illustration of eddy currents.
