[Getdp] FW: how to get coordinates of potential vector
michael.asam at infineon.com
michael.asam at infineon.com
Fri May 17 08:30:56 CEST 2013
Hi Frederic,
the function X[] accepts no arguments. It just delivers the actual x-coordinate
during the integration.
But GetDP can handle axi-symmetry directly by applying VolAxi and/or SurAxi as
Jacobian-type. See for example:
https://geuz.org/trac/getdp/wiki/TorusCurrent
or
https://geuz.org/trac/getdp/wiki/RadiativeHeatTransfer
Cheers
Michael
-----Original Message-----
From: getdp-bounces at ace20.montefiore.ulg.ac.be [mailto:getdp-bounces at ace20.montefiore.ulg.ac.be] On Behalf Of Frederic Trillaud
Sent: Thursday, May 16, 2013 4:39 PM
To: getdp at geuz.org
Subject: [Getdp] how to get coordinates of potential vector
Hi,
I am trying to compute the magnetic energy using the axi-symmetry of the
problem and the following formula: integralOverConductor of (pi*X*A*J),
where X is the coordinate of A and the radius in the axi-symmetric
problem (Y corresponds to the axis of symmetry), A potential vector, and
J the density of current in the conductor. I need to get the coordinate
X of A. Here is what I tried to do:
{Name EMagAJ; Value{Integral {[(2.*Pi/2.)*X[{A}]*{A}*{Je}]; In
magneticSolutionDomain; Jacobian jacobianVolumeRegion; Integration
basicIntegration;}}}
which gives the following error message:
Error : './methodForMagnetostaticProblem.pro', line 190 : Wrong number
of arguments for Function 'X' (1 instead of 0)
Best,
Frederic
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