Ashley ~)~

>  >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.
>
>Sort of. When you bring this circle out to the 2D latitude,longitude
>used for the map (one of the rectangular ones) then it is only a
>true circle if the satellite is directly over the equator, as it
>moves further away from the equator the circle deforms such that
>when it reaches its heighest latitude above about 60š it actually
>looks a little like a sinewave.

There are a couple of problems I see here.

1. Unless the satellite is actually "on the fly" redirecting it's
parabolic dish (which I don't think it does), then the "footprint"
will be close to a "circle" regardless of where the satellite is
located. Now, "circle" is loosely used here because it is actually a
circle projection onto an oblate spheroid (the Earth) which will
never be a prefect circle.

2. This is also a map projection problem. You are trying to project
an image on a map -- a map which is a distorted representation of
surface data. You see, when determining where points are on a map
(i.e., survey), one eventually has to render those points in two
dimensions, but they are actually taken in three dimensions. That's
the reason that maps, as one travels north (or south), show man-made
square rectangles (such as townships, counties, etc.) that are not
actually square, but are narrower toward the end closest to the pole.
In other words, your north-south direction can remain constant in
defining things like townships, but your east-west direction must
shorten to accommodate for approaching a pole. When you take this
phenomena out to a global consideration, then you have things like
Antarctica looking much larger than it actually is.

>I have actually done this part (to generate the footprint) as this
>is how I can tell if the function is producing the correct result. I
>now have to work the other way to determine if a location is within
>this footprint.

My guess is that the actual "footprint" remains just about the same
all over the planet, but changes when projected on current maps. The
only exception to this would be if the satellite changes the altitude
of its orbit -- the higher the orbit, the larger the "footprint".
Also, there should be some minor differences in "footprints" between
polar and equatorial satellites -- the polar having a larger range of
distortion. But, outside of that, the "footprint" should remain
fairly constant -- unless you want to consider atmosphere differences
and how that effects radio transmissions, but I think that's beyond
what you want.

>A cone is used because the dishes used on many satellites and probes
>are parabolic, especially the ones in higher orbits. Yes some of the
>lower orbit satellites use normal antennas which would produce the
>sphere but the function used is generic for sensors and comms which
>facilitates the use of a cone rather than a sphere.

This problem can still be thought of as the intersection of two
spheres -- keep in mind that a cone is nothing more than a 3-D arc of
a larger sphere. Think of it as when you use a flashlight -- you have
the center of the light that is focused by the internal parabolic
reflector, but the light is not confined to just that focus. In fact,
the light spreads out much further than where the beam is directed.

But, in any event, the intersection of a cone with a sphere, or a
sphere with a sphere, should render the exact same result. I seem to
think that finding the intersection of two spheres would be easier
than trying to define the intersection of a cone with a sphere.

>I am open to other suggestions if code can be provided or guided to.

Before developing code, it's usually best to define the problem in
simplest terms. As it is now, from what I've read, I still don't
think the problem has been defined. For example, it's not finding a
point within a cone (no offense to the mosquitoes and tee pee
example), but rather where the cone (or sphere) intersects the Earth
AND how to show that intersection on a map -- isn't that the problem
here?

tedd
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