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The Kuester Response REVIEWED
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Phil's brief note, dated 1.28.14, saying he now understands an email from Kuester and that Kuester was right. It looks at the H field in the right half of a 6x1 bar and takes the large b/a limit of the Hy field expression at the bar end, giving a pi/2 result. It then tries a simple model for Hx to estimate Li, which comes out proportional to w/d and grows with bar length, so it overestimates that component. Equations are partly garbled by extraction.
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The Kuester Response PhL 1.28.14
Here after a while I grok what Kuester was telling me in his email. He is right, I was wrong.
It took me a while to understand what he is saying, but now I do understand. Here is my own plot of the right half of a 6x1 bar, rotated to match my PDF drawing
No matter how long the bar is, you will still see strong vertical even half way in from the end.
Now in terms of my usual picture, this is
The Hy field at the right end is this:
Hy = - F(y,x=a,b,a)
Maple gives the F function at that point and y = 0
Write this then as
Hy = - [ -8a tan-1(b/2a) + 2b ln()
Now take the limit for large b/a. The ln term goes to ln(1) = 0, and you get
Hy = - [ -8a tan-1(∞)] = (π/2) =
which is a nice number to know about. Now suppose you model this in the bar as follows:
Hy(x,y) = (x/a). = x
But in terms of the Kuester PDF I sent, this says instead
Hx = y
and this is then a model for the Hx component of the H field in the bar. Then maybe
Li = μ d !Syntax Error, Idy (H/I)2 = 2 μ d !Syntax Error, Idy [y]2
= 2μd (4dw)-2 !Syntax Error, Idy y2 = 2μd (4dw)-2 (1/3) (w/2)3
= 2 (1/16)(1/8) μ d1-2 w-2+3 = (1/64) μ w/d
The ratio is wrong here and the result increases with bar length ! So this overestimates that component.