Equation of a great circle passing through two points

Question

I've searched everywhere for something to help me with this problem, but I can't find anything. What I want to calculate is the midpoint between two locations (latitude and longitude) on a sphere. The midpoint must lie on the shortest path between them. And for this, I need the equation of the great circle on this sphere that passes through these two points.

What I tried to do is first start with an arbitrary great circle given by the following parametric equation:

${x=0}$

${y=cos\space \theta}$

${z=sin\space \theta}$

Or:

$\left( \begin{array}{} 0 \\ cos\space \theta \\ sin\space \theta \end{array} \right)$

Then if we multiply this vector by three rotation matrices $R_x, R_y,$ and $ R_z$ about angles $\alpha, \beta, $ and $\gamma$, and then solve the equations for the rotation angles using our known values of the radius and the coordinates of the two points, we should get the exact equations of this circle in 3D space, but the trigonometric equations are extremely messy and I'm not sure if my assumption is correct.

There must be a statement of the equation somewhere but I just can't find it. What is it?

Perhaps a simpler method would be to calculate the midpoint in geographic coordinates and then convert them to cartesian, but I'm not sure how geometry works in geographic coordinates.

Answer 1

In the following, we consider that we are working on a unit sphere ($\|V_1\|=\|V_2\|=1$).

If your longitude and latitude coordinates are $(\theta_k,\lambda_k) \ k=1,2.$ Let

$$V_1=\begin{bmatrix}\cos \theta_1 \cos \lambda_1 \\ \sin \theta_1 \cos \lambda_1 \\ \sin \lambda_1 \end{bmatrix} \ \ \ V_2= \begin{bmatrix}\cos \theta_2 \cos \lambda_2 \\ \sin \theta_2 \cos \lambda_2 \\ \sin \lambda_2 \end{bmatrix}$$

Set $V_3=V_1+V_2$, and $V_4=\dfrac{V_3}{\|V_3\|}$: it is the desired point.

If one desires the spherical coordinates of $V_4$ out of its cartesian coordinates $(a,b,c)$, it suffices to set : $$\lambda = \text{asin}(c) \ \ \ \text{and} \ \ \ \theta = \text{asin}(\dfrac{b}{\cos \lambda})$$

Remark : if one wants a parametrized equation for the great circle passing through $V_1$ and $V_2$, here is a classical way to obtain it (cometimes called "orthonormalization"). It suffices to obtain an orthogonal basis of plane $(P)$ defined by $V_1$ and $V_2$. This is easily done by taking as first vector $V_1$ and as a second vector a linear combination $V=\alpha V_1+\beta V_2$ such that $\|V\|^2=1$ and $V \perp V_1 \iff V \cdot V_1=0$, generating two equations with two unknowns (let us recall that we have assumed that $\|V_1\|=\|V_2\|=1$):

$$\begin{cases}\alpha^2 +\beta^2+ 2 d \alpha \beta &=&1\\\alpha+d \beta=0\end{cases}\tag{1}$$

(where $d$ is defined as the dot product $d:=V_1\cdot V_2$) which is amenable to a quadratic equation with two solutions (in general) because its discriminant is $>0$.

Let $W_1=\alpha V_1+\beta V_2$ be one of these two solutions ; the looked for parametric equation of the great circle passing through $V_1$ and $V_2$ is

$$V = \cos(t) V_1 + \sin(t) W_1$$