# 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 array and the ground antenna? Note that D = 1 when elevation angle e is \pi / 2 or 90°. Signals between satellite and ground are attenuated by distance, though at 70 GHz attenuation due a longer slant path through the atmosphere is more important. For the M288 orbit, where A ≈ 2, D = \sqrt{ 3 } if e = 0 , attenuating signals by -10 log 3 or -4.8 dB. The atmospheric attenuation is far greater. Unless there are multiple ground antennas pointed at different parts of the sky, a flat antenna pointed at zenith talking to a satellite at 20° elevation will get -10 log( sin( 20°)) or -4.7 dB less signal due to smaller projected signal capture area, while the distance attenuation is -3.1 dB . See PathAttenuation for the sum of all effects.

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 ) / ( cos( e ) / A ) - sin( e ) sin( c ) / sin( c ) \;\;\; = A cos( c ) - sin( e )

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

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

SatDistance (last edited 2014-05-20 21:03:42 by KeithLofstrom)