From: tedd <tedd@...>

Date: Thu, 17 Jan 2002 09:54:05 -0500

Date: Thu, 17 Jan 2002 09:54:05 -0500

>It is [1]. Since I am dealing with a satellite and what it can see >on the ground (known as it's footprint) it depends upon what side of >the earth it and the satellite are on, ie it's no good if the >satellite is over the artic and it tells me that the station in the >antarctic can recieve the signal, just because it happens to be >directly opposite. > > Here is a clearer explanation, I hope. > > The satallite, located at (sX#,sY#,sZ#) in the XYZ (3D) Plane, >centered at the center of the earth such that the X axis comes out >at the equator in the 0th degree of longitude and the Y axis comes >out at the equator 90 degrees further around and the Z axis emerges >from the north pole. A groundstation located at (lX#,lY#,lZ#) wants >to know if it will be in range to recieve data from the satellite >that can transmit in what appears as a cone, ie it would appear as >like and upside down ice-cream cone where the tip of the cone is at >the satellite and the rim of the cone sits on the surface of the >earth such that the central axis of the cone is perpendicular to the >tangents at the surface of the sphere of earth ie if you connected >the tip of the cone with the center of it's base then the line would >extend through the origin (0,0,0). > > If that needs some explaining try thinking about a basketball >where the valve is the Z axis and one of the seams forms the 0th >degree of longitude (hopefully there is one for the equator as well) >and then think of an ice-cream cone sitting on the ball such that it >would be perpendicular to the tangent at the surface or altenativley >make a cone with a segment of a circle of paper to play around with. >Then suppose there is a dot on the surface of the ball that you want >to determine if it's in the cone (mathematically not visually). > > Ashley ~)~ Another way to look at it. So what you have is a circle on the Earth in which you can receive transmissions -- outside that circle, you can't. And, the circle moves as the satellite moves -- correct? Now, knowing where the satellite is, you want to know where the reception circle is, correct? In your current analogy, you are using a cone/sphere intersection, but you can also think about it as a sphere/sphere intersection. One sphere is, of course, the Earth, but the other is the transmission from the satellite -- which radiates outward like a sphere (even if focused). So, you could look at this like the intersection of two spheres. The circle of reception would be the locus of all points common to the surfaces of both spheres. This may make for a simpler problem. HTH tedd -- http://sperling.com