Air distance

In the last post a proof of the spherical law of cosines was made.

For the figure below, following equality holds:

cos(a)cos(b) + sin(a)sin(b)cos(u) = cos(c)   (***)

The earth is more like an ellipsoid but for computing the air distance with enough accuracy it is fine to consider it a sphere. Imagine that the earth is the sphere from the figure, A' has the GPS coordinates (x_A, y_A) and B' has GPS coordinates (x_B, y_B). All the coordinate are expressed in radians. Note that the angle u is the difference between longitudes because NA'' and NB'' are meridians: u=y_B-y_A

The longitude of A' will be x_A=\frac{\pi}{2}-a and the B' will be x_B=\frac{\pi}{2}-b.

It follows that we can substitute those values in (***) to find c:

cos(c)=cos(a) \cdot cos(b) + sin(a) \cdot sin(b) \cdot cos(u) = cos\left(\frac{\pi}{2}-x_A\right) \cdot cos\left(\frac{\pi}{2}-x_B\right) + sin\left(\frac{\pi}{2}-x_A\right) \cdot sin\left(\frac{\pi}{2}-x_B\right) \cdot cos(y_B-y_A) = sin(x_A) \cdot sin(x_B) + cos(x_A) \cdot cos(x_B) \cdot cos(y_B-y_A)

Having the c angle, the length of the arc A'B' can be computed by multiplying with the earth radius.

s=R \cdot arccos(sin(x_A) \cdot sin(x_B) + cos(x_A) \cdot cos(x_B) \cdot cos(y_B-y_A))

Note that this formula holds for sure if both A' and B' are in the north hemisphere and if the difference between longitudes is smaller than \pi. For other situations it may need some \pm adjustments.

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