From: "Peter Cone" <brinc@...>

Date: Thu, 17 Jan 2002 10:47:16 -0500

Date: Thu, 17 Jan 2002 10:47:16 -0500

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----- Original Message -----
From: "Ashley Butterworth" <macbse@...>
To: "FutureBASIC List" <futurebasic@...>
Sent: Friday, January 18, 2002 12:00 AM
Subject: Re: [FB] A Pointy Problem definition?
> >Ashley , can you clarify which of the following corresponds to your
problem?
> >
> >[1] As used by Laurent, a finite teepee bounded by its base?
> > In this case do you have to worry about the other finite teepee on
the
> >other side of the peak being included?
> >
<snip>
> >Thanks,
> >
> >Cheers, Peter
>
> 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).
>
> I hope that makes it clearer without confusing it anymore,
>
> Thanks for the help,
>
> Regards,
>
> Ashley ~)~
>
> =============================================================
> Ashley Butterworth
> Email: macbse@...
> =============================================================
I think I have it - assuming the satellite illuminates the whole side of the
earth facing it - I.E., the satellite's beam is *not* focussed to a smaller
cone.
If this is the case, then can't you just compare the distances AT and AM?
Where A is the satellite, M is the receiver on earth, and T is the
tangential contact of the cone with the earth?
If AM < AT then M has to be visible
If AM > AT them M is not visible
Cheers, Peter
PS I'm a little puzzled by a cone being perpendicular to tangents -
perhaps this is a way of saying the axis of the cone goes through the
confluence of the perpendiculars to all the tangents?
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