Difference between revisions of "Tutorial:Satellite Coverage"

Revision as of 16:04, 3 June 2013

This tutorial is about how to calculate the coverage of a celestial body. This is for example useful if using Tutorial:RemoteTech.

General coverage calculations

To determine what altitude is needed to cover a complete body only the radius of the orbited body and the number of used satellites is important.

When all satellites in orbit have the same orbital period, the most efficient way to place them is, is in the corners of a equiangular and equilateral convex polygon. The interior angle of each point is then expressed with following formula:

${\displaystyle \alpha ={\frac {n-2}{n}}\cdot 180^{\circ }}$

Each satellite is placed on the same circle with following distance:

${\displaystyle \delta ={\frac {360^{\circ }}{n}}}$

The satellites with an angle of ${\displaystyle \delta }$ ahead and behind need to be at least ${\displaystyle \alpha }$ degrees away. The celestial body with a radius of ${\displaystyle r}$ has at an altitude of ${\displaystyle a}$ above mean sea level this angular size:

${\displaystyle \beta =2\cdot \sin ^{-1}\left({\frac {r}{r+a}}\right)}$

To have the satellites in sight the celestial body's size ${\displaystyle \beta }$ has to be lower than ${\displaystyle \alpha }$:

${\displaystyle {\frac {n-2}{n}}\cdot 180^{\circ }=\left(1-{\frac {2}{n}}\right)\cdot 180^{\circ }=\alpha \geq \beta =2\cdot \sin ^{-1}\left({\frac {r}{r+a}}\right)}$

The lowest altitude is then, when ${\displaystyle \alpha =\beta }$:

${\displaystyle {\begin{aligned}\left(1-{\frac {2}{n}}\right)\cdot 180^{\circ }&=2\cdot \sin ^{-1}\left({\frac {r}{r+a}}\right)\\\sin \left(\left(1-{\frac {2}{n}}\right)\cdot {\frac {180^{\circ }}{2}}\right)&={\frac {r}{r+a}}\\{\frac {1}{\sin \left(\left(1-{\frac {2}{n}}\right)\cdot 90^{\circ }\right)}}&={\frac {r+a}{r}}=1+{\frac {a}{r}}\\\left({\frac {1}{\sin \left(\left(1-{\frac {2}{n}}\right)\cdot 90^{\circ }\right)}}-1\right)\cdot r&=a\\\end{aligned}}}$

This determines the height needed for n satellites, but can be reversed to get how many satellites are needed for a specific altitude:

${\displaystyle {\begin{aligned}\left(1-{\frac {2}{n}}\right)\cdot 180^{\circ }&=2\cdot \sin ^{-1}\left({\frac {r}{r+a}}\right)\\1-{\frac {2}{n}}&={\frac {2}{180^{\circ }}}\cdot \sin ^{-1}\left({\frac {r}{r+a}}\right)\\-{\frac {2}{n}}&={\frac {1}{90^{\circ }}}\cdot \sin ^{-1}\left({\frac {r}{r+a}}\right)-1\\{\frac {1}{n}}&=-{\frac {{\frac {1}{90^{\circ }}}\cdot \sin ^{-1}\left({\frac {r}{r+a}}\right)-1}{2}}\\n&=-{\frac {2}{{\frac {1}{90^{\circ }}}\cdot \sin ^{-1}\left({\frac {r}{r+a}}\right)-1}}\\\end{aligned}}}$

These calculations work only with at least three satellites.