spherical polar geometry change in elevation angle

Question

how to calculate change in elevation angle if you know coordinates of two point on surface of sphere. let us say assume that a point move on the surface of sphere from [x1 y1 z1 ] = [0.1 0.1 0.9899] to [x2 y2 z2] = [-0.1 -0.1 0.9899].

Answer 1

There are several ways to define a spherical coordinate system. The term elevation angle is used with the definition $$\bbox{ \vec{p} = \left [ \begin{matrix} x \\ y \\ z \end{matrix} \right ] = \left [ \begin{matrix} r \cos\theta \cos\varphi \\ r \cos\theta \sin\varphi \\ r \sin\theta \end{matrix} \right ] \tag{1}\label{NA1} }$$ where $\varphi$ is the angle on the $xy$ plane, with $0\text{°}$ being on the positive $x$ axis, and $90\text{°}$ on the positive $y$ axis; and $\theta$ being the angle to the $xy$ plane, with $0\text{°}$ referring to the $xy$ plane ($z = 0$), positive angles up to $90\text{°}$ referring to the hemisphere with positive $z$, and negative angles dow to $-90\text{°}$ referring to the hemisphere with negative $z$. In other words, $\theta$ is the elevation angle.

The inverse of $\eqref{NA1}$ is $$\bbox{\begin{cases} r = \sqrt{x^2 + y^2 + z^2} \\ \varphi = \arctan(y, x) \\ \theta = \arcsin(z / r) \end{cases} \tag{2}\label{NA2}}$$

Thus, if you have a vector $(x, y, z)$, its elevation angle is $$\bbox{ \theta = \arcsin\left( \frac{z}{\sqrt{x^2 + y^2 + z^2}} \right) \tag{3}\label{NA3}}$$

If we compare two points, $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, the change in the elevation angle is $$\bbox{ \theta_2 - \theta_1 = \arcsin\left( \frac{z_2}{\sqrt{x_2^2 + y_2^2 + z_2^2}} \right) - \arcsin\left( \frac{z_1}{\sqrt{x_1^2 + y_1^2 + z_1^2}} \right) }\tag{4}\label{NA4}$$ Unfortunately, I do not see any way of simplifying that expression, unless we place some restrictions on the coordinates.

For example, when $z_1 = z_2$, and $x_1^2 + y_1^2 = x_2^2 + y_2^2$, the elevation angle does not change. (Essentially, the points are then on the same latitude.)

Therefore, in OP's case, the elevation angle does not change between $(0.1, 0.1, 0.9899)$ and $(-0.1, -0.1, 0.9899)$; it is $81.869\text{°}$, or $1.42889$ radians in both cases.