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Re: [femm] about Mixed boundary condition
- To: femm@xxxxxxxxxxxxxxx
- Subject: Re: [femm] about Mixed boundary condition
- From: David Meeker <dmeeker@xxxxxxxx>
- Date: Thu, 30 Jan 2003 23:29:43 -0500
- References: <1043409122.1030.21610.m12@yahoogroups.com> <3e33a8565adda@wp.pl>
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Jan Zimon wrote:
HI !
Could anybody help me find some information about Mixed
boundary condition in Axi-symmetrical problems (i'm beginner).
Where can i find any informations about (*.pdf or webside)?
Is it true condition
A(r,z)= SIGMA[m=1..oo]((a_m/r^m)cos(mz+alpha_m)?
Thank you for any help.
Jan
I guess that what you are really asking for is a reference on the
Axisymmetric case of the Asymptotic Boundary Condition. The form of the
general solution for vector potential A isn't the same in the
axisymmetric case as it is in the planar case, but everything comes out
in the wash such that you enter in the same 1/(R*mu_o) value in the "c0
coefficient" box of the dialog, where R is the radius of a circular (or
half circle, for axisymmetric) domain.
A reference for this would be:
S. Gratkowski, L. Pichon, and H. Gajan, "Asymptotic boundary conditions
for open boundaries of axisymmetric magnetostatic finite-element
models," IEEE Transactions on Magnetics, 38(2):469-472, March 2002.
They derive the boundary condition as dA/dR + 2 A/R = 0, where R is
distance from the origin (i.e. working in spherical coordinates). If I
remember correctly, there was sort of a subtle trick to converting it to
the boundary condition form that is implemented in FEMM. The boundary
condition that the program implements is (1/u)Grad(A).n + c0 A + c1 = 0.
In the axisymmetric case, Grad(A) = A/R + dA/dR, so that to get the
dA/dR + 2 A/R = 0 form, you have to pick c0=1/(u*R) and c1=0. Every
time I run across that paper, I think "where did the 2 go?" and end up
running through this same thought process again....
Dave.
--
David Meeker
dmeeker@xxxxxxxx
http://femm.berlios.de/dmeeker