Minimum orbital altitude for a three satellite full coverage communication network

[Invader Myk]

I realize that in the real universe we would need to account for either the altitude of the layers of the ionosphere or the deflection and/or energy absorption they cause. However we can ignore that for game purposes as no body has more than a notional magnetosphere.

For the purposes of this problem we will use the following definitions:

r₁ is the radius of the orbited body at its highest point

r₂ is the radius of the orbited body at sea-level

o is the orbital altitude above sea-level (asl)

h is the hypotenuse

Three satellites with line of sight to each other would form a continuous communication network.

These satellites would form three equilateral triangles with a common point at the center of the orbited body and the other two points being two of the satellites.

The line of each triangle from one satellite to another would be tangent to r₁ for any the satellites to have line of sight with each other.

Taking a line from the center of the orbited body to the tangent point of any triangle would give us a right triangle with the hypotenuse being o plus r₂ , the angle at the center of the orbited body being 30^o, and the opposite being r₁.

This gives us the equation:

sin30^o = 0.5 = r₁/h

Simplifying this to find the hypotenuse with a known r₁ we receive the equation:

h = r₁/0.5

To figure for o of any given satellite we would then subtract r₂ giving us the equation:

o = h-r₂ = (r₁/0.5)-r₂

Practical example:

Kerbin has a radius of 300km at sea-level and a highest point of 6.767km. To create a three satellite communication network around Kerbin each satellite would need a minimum orbital altitude of 313.534km asl.

((300km+6.767km)/0.5)-300km = 313.534km asl

Refute me. Preferably with proof.

[Iskierka]

While there's a couple other errors I see, the main one is that you assume full coverage of a planet is possible with three satellites. Without ionospheric effects and similar, it's not - there will always be points at the poles that are just out of their line of sight. The minimum number of satellites for full planetary coverage is four, in a very irregular arrangement known as Draim's Tetrahedron. The minimum practical number of satellites in circular orbits, to provide consistent view resolution/communications delay/etc. is six.

[sgt_flyer]

There was once a challenge in ksp+remotetech about draim's tetrahedral constellation

It shows quite nicely how complicated it is

http://forum.kerbalspaceprogram.com/threads/33711-Scenario-Tetrahedral-Satellite-Configuration-4-sat-full-surface-coverage

[Ralathon]

I realize that in the real universe we would need to account for either the altitude of the layers of the ionosphere or the deflection and/or energy absorption they cause. However we can ignore that for game purposes as no body has more than a notional magnetosphere.

For the purposes of this problem we will use the following definitions:

r₁ is the radius of the orbited body at its highest point

r₂ is the radius of the orbited body at sea-level

o is the orbital altitude above sea-level (asl)

h is the hypotenuse

Three satellites with line of sight to each other would form a continuous communication network.

These satellites would form three equilateral triangles with a common point at the center of the orbited body and the other two points being two of the satellites.

The line of each triangle from one satellite to another would be tangent to r₁ for any the satellites to have line of sight with each other.

Taking a line from the center of the orbited body to the tangent point of any triangle would give us a right triangle with the hypotenuse being o plus r₂ , the angle at the center of the orbited body being 30^o, and the opposite being r₁.

This gives us the equation:

sin30^o = 0.5 = r₁/h

Simplifying this to find the hypotenuse with a known r₁ we receive the equation:

h = r₁/0.5

To figure for o of any given satellite we would then subtract r₂ giving us the equation:

o = h-r₂ = (r₁/0.5)-r₂

Practical example:

Kerbin has a radius of 300km at sea-level and a highest point of 6.767km. To create a three satellite communication network around Kerbin each satellite would need a minimum orbital altitude of 313.534km asl.

((300km+6.767km)/0.5)-300km = 313.534km asl

Refute me. Preferably with proof.

Nope, you are completely correct in the case of a 2D planet. You won't have polar coverage though. Best way to fix that is to add another 3 satellites in a polar orbit.

This is a well known relation for equilateral triangles. The distance from the center to one of the corners is twice the radius of an inscribed circle.

[Invader Myk]

While there's a couple other errors I see, the main one is that you assume full coverage of a planet is possible with three satellites. Without ionospheric effects and similar, it's not - there will always be points at the poles that are just out of their line of sight. The minimum number of satellites for full planetary coverage is four, in a very irregular arrangement known as Draim's Tetrahedron. The minimum practical number of satellites in circular orbits, to provide consistent view resolution/communications delay/etc. is six.

What are the other errors?

And since you're right on the polar coverage, lets simplify this to work towards "full equatorial coverage"

There was once a challenge in ksp+remotetech about draim's tetrahedral constellation

It shows quite nicely how complicated it is

This is exactly what I was aiming to figure out. Thanks.

Now to find Draim's equations so I can figure out Dunian and Joolian constellation.