Satellite Distance

Thinsat arrays will not always be directly above the ground antenna, but will be at an elevation angle e above the horizon. If the array orbits at a radius A \bullet R_E , what is the distance D \bullet R_E between the arrayand the ground antenna? Note that D = 1 when elevation angle e is \pi / 2 or 90°.

The sin law states that:

{ { sin( d ) } \over D } = { { sin( a ) } \over A } = { { sin( c ) } \over 1 }

sin( a ) = sin( e + { \pi \over 2 } ) = \cos( e )

sin( c ) = { { sin( a ) } \over A } = { { cos( e ) } \over A }

cos( c ) = \sqrt{ 1 - \left( { cos( e )^2 } \over A \right)^2 }

The sum of the corners of a triangle is \pi , so

\pi = a + c + d = e + { \pi \over 2 } + c + d

{ \pi \over 2 } = e + c + d

d = { \pi \over 2 } - ( e + c )

sin( d ) = sin( { \pi \over 2 } - ( e + c ) ) = cos( e + c ) = cos( e ) cos( c ) - sin( e ) sin( c )

D = sin( d ) / sin( c ) = cos( e ) cos( c ) / sin( c ) - sin( e ) sin( c ) / sin( c )

D = cos( e ) cos( c ) / ( cos( e ) / A ) - sin( e ) = A cos( c ) - sin( e )

D = A \sqrt{ 1 - \left( { cos( e )^2 } \over A \right)^2 } - sin( e )

D = \sqrt{ A^2 - cos( e )^2 } - sin( e )

D = \sqrt{ ( A^2 - 1 ) + sin( e )^2 } - sin( e )