justin
It is a good point that the insulation has a bearing on the problem.
A first pass calculation would be a parallel plate capacitive voltage divider.
Let subscript i be solid insulation and a be air.
e = permability
X = capacitive impedance
d = depth
A = area
Voltage divider
Vair = Vtotal * Xair / Xtotal
Vair = Vtotal * Xair / [Xair + 2*Xinsulation]
Xair = 2*pi*f * (da / [ea*A]
Xinsulation = 2*pi*f * ei A / di
factors of 2*pi and A will not show up in the ratio Xair/Xtotal
Vair = Vtotal * da/ea / [da/ea + 2*di/ei]
Typical values:
da = 0.5", di = 0.25"
typical insulation ei ~ 4* ea
Vair = Vtotal * da/ea / [da/ea + 2*di/ei]
Vair = Vtotal * 0.5"/ea / [0.5"/ea + 2*0.25"/(4*ea)]
= Vtotal * 0.5 / [0.5 + (1/8)]
= Vtotal * 0.5 / [0.5 + 0.125]
= Vtotal * 0.5 / [0.625]
~ 80% Vtotal
For your value ei~6*ea, the air voltage would be a higher percentage of the total.
My approximation was that Vair~Vtotal - all of the voltage drop appears accross the air. For that particular approximation we consider the distance from coil surface to coil surface and distance to copper is not relevant. For fuller solution you are right it is relevant.
There are I'm sure also very important geometry factors to consider (effect of localized stress concentration). Also if we design for an average gap, there will be some above and some below and ideally would like to prevent corona on them all.
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