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What altitude would you have to be at to have line of site to Tokyo and London from Chicago? They're probably aimed somewhere above and in between https://en.w

What altitude would you have to be at to have line of site to Tokyo and London from Chicago? They're probably aimed somewhere above and in between https://en.wikipedia.org/wiki/Skywave https://en.wikipedia.org/wiki/Skywave

caymanjim 6y ago
You'd need to be in orbit. Edit: actually it might not be possible at all from a single point anywhere in the universe.
- filmfact 6y ago
  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).
jsherer 6y ago
You will not usually have nor do you need line of sight for HF radio. The radio waves are reflected by the ionosphere at various angles and heights, allowing the signal to be received at many different distances at once. Radio waves at the correct frequency hit the ionosphere like stones skipping along the surface of water.
tzs 6y ago
Assuming the same tower height at both ends, and ignoring refraction due to the atmosphere, I get that you would need 1500 mile tall towers for Chicago to Tokyo. (I did miles rather than meters because my Google for the distance between them came back in miles and I was too lazy to convert). For Chicago to London I get 560 mile tall towers.
- robocat 6y ago
  From Chicago ORD to Tokyo NRT is 6274 miles on great circle. Distance around world is 24901 miles. Two towers add 1120 miles (and ignoring extra length of straight line transmission compared to great arc) then a fibre optic cable on great arc would be faster than straight line if light signal within fibre travels at > 85% of speed of light (also ignoring slowdown by atmosphere for radio).

caymanjim 6y ago

You'd need to be in orbit. Edit: actually it might not be possible at all from a single point anywhere in the universe.

filmfact 6y ago
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).

jsherer 6y ago

You will not usually have nor do you need line of sight for HF radio. The radio waves are reflected by the ionosphere at various angles and heights, allowing the signal to be received at many different distances at once. Radio waves at the correct frequency hit the ionosphere like stones skipping along the surface of water.

tzs 6y ago

Assuming the same tower height at both ends, and ignoring refraction due to the atmosphere, I get that you would need 1500 mile tall towers for Chicago to Tokyo. (I did miles rather than meters because my Google for the distance between them came back in miles and I was too lazy to convert). For Chicago to London I get 560 mile tall towers.

robocat 6y ago
From Chicago ORD to Tokyo NRT is 6274 miles on great circle. Distance around world is 24901 miles. Two towers add 1120 miles (and ignoring extra length of straight line transmission compared to great arc) then a fibre optic cable on great arc would be faster than straight line if light signal within fibre travels at > 85% of speed of light (also ignoring slowdown by atmosphere for radio).