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I have a meshgrid of spherical coordinates. $\theta$ and $\phi$ only (unit sphere). I have data on the grid that looks something like this. Farfield

This is an antenna far-field. The 'big-picture' goal is to find the half-power beam width around the peak in two orthogonal great circles. It's a trivial problem when the peak is centered around $\theta_0 = \phi_0 = 0^{\circ}$. The two great circles would be $\phi=0^{\circ}, -90^{\circ}<\theta<90^{\circ}$ and $\phi=90^{\circ}, -90^{\circ}<\theta<90^{\circ}$.

When the maximum is at non-zero $\theta, \phi$, the first is chosen (to be simple), $\phi = 45^{\circ}, -90^{\circ}<\theta<90^{\circ}$. I'm not sure how the extract the $\theta/\phi$ coordinates that describe the second circle and corresponding data.

I've sketched out candidate great circles that are useful, with the one I'm describing being the black lines. Hopefully with the information given, I can derive the white lines.

Great Circles

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  • $\begingroup$ It is not clear: what do you mean for example with "that describe the second circle"? Which second circle, on which figure ? $\endgroup$ – Jean Marie Jan 30 '17 at 22:44
  • $\begingroup$ Do you really think in terms of spherical coordinates ? If yes, are they necessary as the second drawing is a plane drawing. $\endgroup$ – Jean Marie Jan 30 '17 at 22:46
  • $\begingroup$ @JeanMarie the 'second circle' would be the black line that's orthogonal to the black line along $\phi=45$. The first drawing is the data plotted as the meshgrid, the second drawing would be the projection where $\phi$ is marked around the circle and $\theta$ are the dashed circles. Spherical coordinates are convenient since it's a meshgrid of spherical coordinates and the math used to generate the data is convenient in spherical coordinates. $\endgroup$ – Jason Jan 31 '17 at 0:03

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