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I just stared at a globe, and it’s possible, but not from a geostationary orbit. Think about a hemisphere projected onto a 2d circle - you can see all points o

I just stared at a globe, and it’s possible, but not from a geostationary orbit. Think about a hemisphere projected onto a 2d circle - you can see all points on the surface of the hemisphere on the projection. Logically it must be possible to select a hemisphere, such that any two random points on a sphere lie on that hemisphere, and would thus be visible in the projection. The only edge cases is two points that are diametrically opposite - i.e north and south pole, which would only both be visible at infinite distance. Points nearly diametrically opposed would only be visible from extreme distance. This doesn’t take into account inconvenient mountains, but they can easily be moved.

caymanjim 6y ago
In this case I was considering three points, but yeah, they're all in the northern hemisphere, so your scenario applies.
zaarn 6y ago
If the two points you select on the sphere are poles of eachother (ie, their distance through the sphere is equal to the sphere's diameter, then you need to have an infinite distance from the sphere. What you're looking for is a triangle between the two points and a third point forming your observer. The two points are visible at the same time if the derivative of the sphere's surface and the side of the triangle between a point form a positive angle (the derivative intersects the triangle at a point other than the points). So the question then becomes "if the two points are poles, what does that mean"; easy, if the two points are poles and they're only visible if the surface derivative is within the triangle, the angle between the observer, the point and the other point must be 90°. This must hold for both points as the solution is symmetric. However, there is no well defined triangle that has two angles of 90° inside. So the solution is "doesn't work" (or possibly infinite distance).

caymanjim 6y ago

In this case I was considering three points, but yeah, they're all in the northern hemisphere, so your scenario applies.

zaarn 6y ago

If the two points you select on the sphere are poles of eachother (ie, their distance through the sphere is equal to the sphere's diameter, then you need to have an infinite distance from the sphere. What you're looking for is a triangle between the two points and a third point forming your observer. The two points are visible at the same time if the derivative of the sphere's surface and the side of the triangle between a point form a positive angle (the derivative intersects the triangle at a point other than the points). So the question then becomes "if the two points are poles, what does that mean"; easy, if the two points are poles and they're only visible if the surface derivative is within the triangle, the angle between the observer, the point and the other point must be 90°. This must hold for both points as the solution is symmetric. However, there is no well defined triangle that has two angles of 90° inside. So the solution is "doesn't work" (or possibly infinite distance).