[jamessb]

But the correct statement is "no more than" not "at least".

Consider a right-angled triangle with base length d and height 550, corresponding to transmission from a base-station to a satellite. The hypotenuse has length sqrt(d^2 + 550^2), so the difference in length between the hypotenuse and base is sqrt(d^2 + 550^2) - d.

This has a maximum of 550 when d=0 (i.e., shooting straight up), and decreases as d increases: https://www.wolframalpha.com/input/?i=plot+sqrt%28d%5E2+%2B+...

Alternatively, consider the triangle inequality: the sum of the lengths of any two sides must be greater than or equal to the length of the remaining side. This directly implies that the difference in length between the hypotenuse and base is less than or equal the height [base + height >= hypotenuse implies height >= hypotenuse - base].

[sliken]

Er, no, "the difference in length between the hypotenuse and base is sqrt(d^2 + 550^2) - d".

The hypotenuse is cos(angle)*base.

If you think about it at a minute if a sat is 500 miles up directly overhead that's the closest it ever will be, as it flies off the hypotenuse gets longer, not shorter.

So ideally you bounce off a sat overhead, (distance of 1100), any single hop will be longer, and to get across an ocean you'll likely need more than one hop.

Basically the sin(beam path) will will never be less than 550 and the length of the beam will never be less than 550.

[jamessb]

~~D'oh - yes, that formula is true only if the triangle is right-angled, which is true for only a single base length.~~

Edit: Actually, this is always true: we are considering a right-angled triangle where the base is the horizontal distance from the ground station to point under the satellite, the vertical part is the 550 miles between the point under the satellite and the satellite, and the hypotenuse is the line joining the satellite and ground station.

> if a sat is 500 miles up directly overhead that's the closest it ever will be, as it flies off the hypotenuse gets longer

Yes: as the horizontal distance d increases, then the length of the hypotenuse (sqrt(d^2 + 550^2)) increases.

However, the difference between this and the horizontal distance (sqrt(d^2 + 550^2) - d) decreases.

-----------------------------------------------

If the angle from the horizontal to the line between the satellite and base-station is theta, then:

sin(theta) = 550/hypotenuse => hypotenuse = 550/sin(theta)

tan(theta) = 550/base-length => base-length = 550/tan(theta)

difference in length = 550/sin(theta) - 550/tan(theta)

[which simplifies to 550 tan(theta/2)]

We are interested in angles between 0 degrees (horizontal - corresponding to the limiting case of infinite horizontal distance between the satellite and base station) and 90 degrees or pi/2 radians (straight up): https://www.wolframalpha.com/input/?i=plot+550%2Fsin%28x%29+...

This is always between 0 and 550. The triangle inequality holds: for a single hop from base-station to satellite, the increase in length is never more than 550.

But as you point out, there may also be multiple hops.

> So ideally you bounce off a sat overhead, (distance of 1100),

This is the shortest total ground-satellite-ground distance, but as you cover 0 horizontal distance it is the worst case: the difference between the ground-satellite-ground distance and the length of the direct ground-ground line is maximised.

[imoverclocked]

Are all base stations directly underneath a satellite?

I think this is an over-simplification if we are chasing pedantics; There are cases where it will be more and others less so the slightly more precise wording might actually be "about 1100km."

To the larger picture: it seems we often lose that order of length on the ground due to existing network topologies and geographical limitations.

[jamessb]

Yes, this is an oversimplification: the original statement seemed to be based on getting a fact about trigonometry backwards, and I was just trying to resolve the underlying confusion.