It is possible to know how meshgrid organize the points?
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Hello!
I am creating a sphere of points like it is shown in image 1. The ideia is now, for each point, calculate the value of the electric field. That part is in image 2 and you can consider that the calculations are correct. After this, I want to plot the radiation filed with patternCostum, and for that, I need to have the electric field (E_points) organized in a matrix. As I have 10000 points, when I traverse the array of points to calculate the field the E_points array also have 10000x1 points. So I reshape it and then do the patternCostum(). - image 3.
The correct patternCostum is presented in image 4. It seems that my patternCostum is correct, but some E_points doesn't match with the Phi_points and Theta_points arrays. It is possible somehow know which point correspond to each theta and phi? Or the are some easier option?
Thank you a lot!
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William Rose
2022-7-1
编辑:William Rose
2022-7-1
I think the following code illustrates how the points returned by sphere() are arranged. See comments below.
[xs,ys,zs]=sphere(4);
disp(xs); disp(ys); disp(zs);
sphere(n) returns 3 matrices which are the x, y, and z coordinates of a n+1 by n+1 rectangular grid wrapped around the sphere. Suppose the z axis is the Earth's axis of rotation and +x goes through the Greenwich meridian and +y goes through 90 East. Then row 1 of each array corresponds to the south pole (theta=+pi, where theta is the polar angle, measured from th +z axis, as in the physics convention in the wikipedia article on spherical coordinates). The middle row is the equator (theta=+pi/2). Row n+1 corresponds to the north pole (theta=0). Column 1 corresponds to the international date line (phi=-pi, where phi is the azimuthal angle, measured CCW in the x-y plane, as in the physics convention in the diagram in Wikipedia for spherical coordinates), then the columns go east (phi increasing). The middle column corresponds to the Greenwich meridian (phi=0) and the last column is the date line again (phi=+pi). Good luck with your work.
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