Consider a great circle between $[lat_1, lon_1]$ and $[ lat_2, lon_2]$, on a perfectly spherical earth.
Consider a second one : between $[lat_1, lon_1 +b]$ and $[ lat_2, lon_2+b]$. For a very small constant $b$, if the great circles themselves are small, they can be considered parallel to each other.
Both are segments of circles and thus does not go completely around the world. I want to project one great circle segment upon the other to find out if these segments "overlap" each other.
Overlap :
Overlap happens in between the two green dotted lines. If I move the second great circle by $-b$, then the "overlap" is the common sub-segment of both segments of great circles.
Edit after @michaelseifert 's comment :
The green lines should be normals dropped from one segment to another. However, I also do not have a proper algorithm / formula to do that. After the projection on the cylinder, I was painstakingly evaluating each endpoint to drop these normals. Even tho, it is done on a computer, I think there should be better ways of doing it.
Edit 2 :
Once I project the Great Circles on a cylinder (see below in attempt to solution), the arcs of latitude and longitude become straight lines. For very short lengths of great circle segments, I can approximate them as straight lines. Then I can drop normals.
The outcome will be the same, if I could draw the shortest cross track great circle segments between the two given great circle segments defined by $[lat_1, lon_1]$ , $[ lat_2, lon_2]$, and $[lat_1, lon_1 +b]$ , $[ lat_2, lon_2+b]$ - from the points that delimit the "overlapping". I realize that overlapping and the normals are defined w.r.t. each other, and that bad. But the plan is to extract parallel GPS tracks, and trim them if one exceeds a certain length beyond overlapping.
Attempt to solution
I tried to convert the polar coordinates to cartesian, butthey are in 3D.
In order to reduce them to 2D i am projecting them on a cylinder like mercator projection and unwrapping it.
Problem with this approach
At higher latitudes the surface of the cylinder is very far away from the surface of the sphere leading to large errors.
Question
How can I project one great circle segment on another without first having to project them on a cylinder or other surface?
Thank you
Comment... trying to understand your question..i.e,
By their definition, two geodesics/great circles can never be parallel. They intersect at two points.
For parallels at least one should be a small circle.
Every small circle is non-geodesic, that is, the normal to the surface at a given point P and normal to the small circle at P makes the same angle between them.
I.e., the $\phi=5^{\circ}$ parallel circle on a sphere has this angle constant all along the small circle length.
Two given points and sphere center determine a great circle. With respect to a great circle can define the relative latitude for a small parallel circle $\phi$ thus:
For a great circle Clairaut circle radius $ is a constant (blue). For a small circle it varies as shown (red). Using Liouville theorem of geodesic curvature we can establish that
$$\tan \phi= \dfrac{d (r \sin \psi)}{dz}=\dfrac{d (r_{Clairaut})}{dz}; $$ It vanishes for a great circle but is a constant for small circle, $\psi$ being angle between meridian and the small circle.