When you rotate the south pole (i.e. the point where theta=180) to the center of the screen, this point is always set in the program with fixed texturing coordinates (u=0.5, v=1). This means that it no longer has any defined azimuthal component - all phi (azimuth) values converge there.
Rotation around the Y and Z axes essentially only changes the azimuthal component (direction to the sphere), but for the poles this component is irrelevant because it is ignored and a fixed value is taken. Therefore, if you have the south pole exactly in the center, any rotation around the Y or Z axis, or the X and Z axes, will have the same effect - the point will remain the same because its determination is not based on azimuth.
In short:
For the poles, the azimuth angle is not taken into account due to the fixed setting of the texturing coordinates (to avoid seam problems).
Rotations around the Y and Z axes and X and Z axes primarily affect this angle.
Since there is no difference in the azimuth value at the South Pole, the visual effect of rotation around the Y axis is identical to the effect of rotation around the Z axis.
Maybe another solution could be used. But instead of additional mathematics, I would use glSphere, if it can be used in your program. It would be the fastest (and proven) solution. OpenGL sphere is one in quadrics here: https://qb64phoenix.com/forum/showthread...579&page=3
Thank you, Pete
Rotation around the Y and Z axes essentially only changes the azimuthal component (direction to the sphere), but for the poles this component is irrelevant because it is ignored and a fixed value is taken. Therefore, if you have the south pole exactly in the center, any rotation around the Y or Z axis, or the X and Z axes, will have the same effect - the point will remain the same because its determination is not based on azimuth.
In short:
For the poles, the azimuth angle is not taken into account due to the fixed setting of the texturing coordinates (to avoid seam problems).
Rotations around the Y and Z axes and X and Z axes primarily affect this angle.
Since there is no difference in the azimuth value at the South Pole, the visual effect of rotation around the Y axis is identical to the effect of rotation around the Z axis.
Maybe another solution could be used. But instead of additional mathematics, I would use glSphere, if it can be used in your program. It would be the fastest (and proven) solution. OpenGL sphere is one in quadrics here: https://qb64phoenix.com/forum/showthread...579&page=3
Thank you, Pete