Satellite Range

"Terry Pinnell" <terrypin@dial.pipex.com> wrote in message
news:9um290pd7put9p28sn4hh5lhltp9p7v7l2@4ax.com...
"Philip Newman" <nojunkmail@ntlworld.com> wrote:

well, I used these equations to work them out:

Px = (R+r)Cos(Lambda) - rCos(phi)Cos(Theta)
Py = (R+r)Sin(Lambda) - rCos(phi)Sin(Theta)
Pz = -rSin(phi)

R is the radius of the earth 6378km, r is the geostationary altitude,
35786km, ? is the satellite's longitude, 30W, ? is the observer's
longitude,
5.17W and ? is the observer's latitude 50.05N

That last paragraph doesn't make sense to me. What do the three '?'
mean? Are they meant to be Lambda, Theta and phi respectively? If so,
what units are expected, degrees or radians?

--
Terry Pinnell
Hobbyist, West Sussex, UK
Click to expand...
well, thats what i wrote anyway!! And they are in degrees. 50.05(deg) etc

Sorry, copy and paste gone wrong me thinks...

In respect to Active8's reply, I checked out some of the links that were on
the site you sent me to, and there are extra calculations that might need to
be done in order to get the range, but that is surely far too complicated
for a supposedly easy calculation? I'm talking about Orbital Coordinate
Systems, Part II link.

Phil
 
"Philip Newman" <nojunkmail@ntlworld.com> wrote in message
news:eYdkc.192$xt3.78@newsfe1-gui.server.ntli.net...
"Terry Pinnell" <terrypin@dial.pipex.com> wrote in message
news:9um290pd7put9p28sn4hh5lhltp9p7v7l2@4ax.com...
"Philip Newman" <nojunkmail@ntlworld.com> wrote:

well, I used these equations to work them out:

Px = (R+r)Cos(Lambda) - rCos(phi)Cos(Theta)
Py = (R+r)Sin(Lambda) - rCos(phi)Sin(Theta)
Pz = -rSin(phi)

R is the radius of the earth 6378km, r is the geostationary altitude,
35786km, ? is the satellite's longitude, 30W, ? is the observer's
longitude,
5.17W and ? is the observer's latitude 50.05N

That last paragraph doesn't make sense to me. What do the three '?'
mean? Are they meant to be Lambda, Theta and phi respectively? If so,
what units are expected, degrees or radians?

--
Terry Pinnell
Hobbyist, West Sussex, UK


well, thats what i wrote anyway!! And they are in degrees. 50.05(deg)
etc

Sorry, copy and paste gone wrong me thinks...

In respect to Active8's reply, I checked out some of the links that were
on
the site you sent me to, and there are extra calculations that might need
to
be done in order to get the range, but that is surely far too complicated
for a supposedly easy calculation? I'm talking about Orbital Coordinate
Systems, Part II link.

Phil

It's so easy that no-one in their right minds *doesn't* use a spreadsheet or
Click to expand...
calculator these days if you're doing more than one for anything other than
as a demonstration. Go search on the keywords I gave previously - there's
several example programs out there.

Ken
 
"Ken Taylor" <ken123@xtra.co.nz> wrote in message
news:c6rub2$fs3gv$1@ID-76636.news.uni-berlin.de...
"Philip Newman" <nojunkmail@ntlworld.com> wrote in message
news:eYdkc.192$xt3.78@newsfe1-gui.server.ntli.net...

"Terry Pinnell" <terrypin@dial.pipex.com> wrote in message
news:9um290pd7put9p28sn4hh5lhltp9p7v7l2@4ax.com...
"Philip Newman" <nojunkmail@ntlworld.com> wrote:

well, I used these equations to work them out:

Px = (R+r)Cos(Lambda) - rCos(phi)Cos(Theta)
Py = (R+r)Sin(Lambda) - rCos(phi)Sin(Theta)
Pz = -rSin(phi)

R is the radius of the earth 6378km, r is the geostationary altitude,
35786km, ? is the satellite's longitude, 30W, ? is the observer's
longitude,
5.17W and ? is the observer's latitude 50.05N

That last paragraph doesn't make sense to me. What do the three '?'
mean? Are they meant to be Lambda, Theta and phi respectively? If so,
what units are expected, degrees or radians?

--
Terry Pinnell
Hobbyist, West Sussex, UK


well, thats what i wrote anyway!! And they are in degrees. 50.05(deg)
etc

Sorry, copy and paste gone wrong me thinks...

In respect to Active8's reply, I checked out some of the links that were
on
the site you sent me to, and there are extra calculations that might
need
to
be done in order to get the range, but that is surely far too
complicated
for a supposedly easy calculation? I'm talking about Orbital Coordinate
Systems, Part II link.

Phil

It's so easy that no-one in their right minds *doesn't* use a spreadsheet
or
calculator these days if you're doing more than one for anything other
than
as a demonstration. Go search on the keywords I gave previously - there's
several example programs out there.

Ken
Click to expand...
I did manage to find a suitable software/calculator that gave me an answer
of 38123km, which sounds correct. However, I would like to work it out
manually, just as a personal issue now. It seems that I don't need to know
this really, but its niggling me now.

Phil
 
Philip Newman <nojunkmail@ntlworld.com> wrote:
"Ken Taylor" <ken123@xtra.co.nz> wrote in message
news:c6rub2$fs3gv$1@ID-76636.news.uni-berlin.de...
"Philip Newman" <nojunkmail@ntlworld.com> wrote in message
news:eYdkc.192$xt3.78@newsfe1-gui.server.ntli.net...

"Terry Pinnell" <terrypin@dial.pipex.com> wrote in message
news:9um290pd7put9p28sn4hh5lhltp9p7v7l2@4ax.com...
"Philip Newman" <nojunkmail@ntlworld.com> wrote:

well, I used these equations to work them out:

Px = (R+r)Cos(Lambda) - rCos(phi)Cos(Theta)
Py = (R+r)Sin(Lambda) - rCos(phi)Sin(Theta)
Pz = -rSin(phi)

R is the radius of the earth 6378km, r is the geostationary altitude,
35786km, ? is the satellite's longitude, 30W, ? is the observer's
longitude,
5.17W and ? is the observer's latitude 50.05N

That last paragraph doesn't make sense to me. What do the three '?'
mean? Are they meant to be Lambda, Theta and phi respectively? If so,
what units are expected, degrees or radians?

--
Terry Pinnell
Hobbyist, West Sussex, UK


well, thats what i wrote anyway!! And they are in degrees. 50.05(deg)
etc

Sorry, copy and paste gone wrong me thinks...

In respect to Active8's reply, I checked out some of the links that were
on
the site you sent me to, and there are extra calculations that might
need
to
be done in order to get the range, but that is surely far too
complicated
for a supposedly easy calculation? I'm talking about Orbital Coordinate
Systems, Part II link.

Phil

It's so easy that no-one in their right minds *doesn't* use a spreadsheet
or
calculator these days if you're doing more than one for anything other
than
as a demonstration. Go search on the keywords I gave previously - there's
several example programs out there.

Ken



I did manage to find a suitable software/calculator that gave me an answer
of 38123km, which sounds correct. However, I would like to work it out
manually, just as a personal issue now. It seems that I don't need to know
this really, but its niggling me now.
Click to expand...
Work out the length of the other side of the triangle that goes from the
center of the earth to the surface station, and the satellite to the center
of the earth.
You know both, one is radius of earth, the other is height of GEO.
 
"Philip Newman" <nojunkmail@ntlworld.com> wrote in message
news:yxykc.4154$N13.3048@newsfe1-win...
"Ken Taylor" <ken123@xtra.co.nz> wrote in message
news:c6rub2$fs3gv$1@ID-76636.news.uni-berlin.de...
"Philip Newman" <nojunkmail@ntlworld.com> wrote in message
news:eYdkc.192$xt3.78@newsfe1-gui.server.ntli.net...

"Terry Pinnell" <terrypin@dial.pipex.com> wrote in message
news:9um290pd7put9p28sn4hh5lhltp9p7v7l2@4ax.com...
"Philip Newman" <nojunkmail@ntlworld.com> wrote:

well, I used these equations to work them out:

Px = (R+r)Cos(Lambda) - rCos(phi)Cos(Theta)
Py = (R+r)Sin(Lambda) - rCos(phi)Sin(Theta)
Pz = -rSin(phi)

R is the radius of the earth 6378km, r is the geostationary
altitude,
35786km, ? is the satellite's longitude, 30W, ? is the observer's
longitude,
5.17W and ? is the observer's latitude 50.05N

That last paragraph doesn't make sense to me. What do the three '?'
mean? Are they meant to be Lambda, Theta and phi respectively? If
so,
what units are expected, degrees or radians?

--
Terry Pinnell
Hobbyist, West Sussex, UK


well, thats what i wrote anyway!! And they are in degrees.
50.05(deg)
etc

Sorry, copy and paste gone wrong me thinks...

In respect to Active8's reply, I checked out some of the links that
were
on
the site you sent me to, and there are extra calculations that might
need
to
be done in order to get the range, but that is surely far too
complicated
for a supposedly easy calculation? I'm talking about Orbital
Coordinate
Systems, Part II link.

Phil

It's so easy that no-one in their right minds *doesn't* use a
spreadsheet
or
calculator these days if you're doing more than one for anything other
than
as a demonstration. Go search on the keywords I gave previously -
there's
several example programs out there.

Ken



I did manage to find a suitable software/calculator that gave me an answer
of 38123km, which sounds correct. However, I would like to work it out
manually, just as a personal issue now. It seems that I don't need to
know
this really, but its niggling me now.

Phil


Clearly I wouldn't do this, but there are several sites out there with
Click to expand...
calculators, as you've found. When you have the page do a 'View Source'.
Most enlightening. More intellectually satisfying is getting hold of one of
the texts and looking at how it's all derived. See:
"Satellite Communications", Dennis Roddy, McGraw-Hill (ISBN 0-07-053370-9)
"Satellite Communications Systems", M. Richharia. McGraw-Hill. (ISBN
0-07-052374-6)

Ken
 
On Fri, 30 Apr 2004 21:10:01 GMT, Ian Stirling wrote:

Philip Newman <nojunkmail@ntlworld.com> wrote:

"Ken Taylor" <ken123@xtra.co.nz> wrote in message
news:c6rub2$fs3gv$1@ID-76636.news.uni-berlin.de...
"Philip Newman" <nojunkmail@ntlworld.com> wrote in message
news:eYdkc.192$xt3.78@newsfe1-gui.server.ntli.net...

"Terry Pinnell" <terrypin@dial.pipex.com> wrote in message
news:9um290pd7put9p28sn4hh5lhltp9p7v7l2@4ax.com...
"Philip Newman" <nojunkmail@ntlworld.com> wrote:

well, I used these equations to work them out:

Px = (R+r)Cos(Lambda) - rCos(phi)Cos(Theta)
Py = (R+r)Sin(Lambda) - rCos(phi)Sin(Theta)
Pz = -rSin(phi)

R is the radius of the earth 6378km, r is the geostationary altitude,
35786km, ? is the satellite's longitude, 30W, ? is the observer's
longitude,
5.17W and ? is the observer's latitude 50.05N

That last paragraph doesn't make sense to me. What do the three '?'
mean? Are they meant to be Lambda, Theta and phi respectively? If so,
what units are expected, degrees or radians?

--
Terry Pinnell
Hobbyist, West Sussex, UK


well, thats what i wrote anyway!! And they are in degrees. 50.05(deg)
etc

Sorry, copy and paste gone wrong me thinks...

In respect to Active8's reply, I checked out some of the links that were
on
the site you sent me to, and there are extra calculations that might
need
to
be done in order to get the range, but that is surely far too
complicated
for a supposedly easy calculation? I'm talking about Orbital Coordinate
Systems, Part II link.

Phil

It's so easy that no-one in their right minds *doesn't* use a spreadsheet
or
calculator these days if you're doing more than one for anything other
than
as a demonstration. Go search on the keywords I gave previously - there's
several example programs out there.

Ken



I did manage to find a suitable software/calculator that gave me an answer
of 38123km, which sounds correct. However, I would like to work it out
manually, just as a personal issue now. It seems that I don't need to know
this really, but its niggling me now.

Work out the length of the other side of the triangle that goes from the
center of the earth to the surface station, and the satellite to the center
of the earth.
You know both, one is radius of earth, the other is height of GEO.
Click to expand...
The OP used the given eqs (per link) to get the vector from the
earth station to the bird and the range ain't right when you plug
the vector into pythagorus. I'll hold off on deriving those eqs as a
check until I go to use them unless I get bored :)
--
Best Regards,
Mike
 
"Active8" <reply2group@ndbbm.net> wrote in message
news:1fhsfalbu6u7m.dlg@news.individual.net...
On Fri, 30 Apr 2004 21:10:01 GMT, Ian Stirling wrote:

Philip Newman <nojunkmail@ntlworld.com> wrote:

"Ken Taylor" <ken123@xtra.co.nz> wrote in message
news:c6rub2$fs3gv$1@ID-76636.news.uni-berlin.de...
"Philip Newman" <nojunkmail@ntlworld.com> wrote in message
news:eYdkc.192$xt3.78@newsfe1-gui.server.ntli.net...

"Terry Pinnell" <terrypin@dial.pipex.com> wrote in message
news:9um290pd7put9p28sn4hh5lhltp9p7v7l2@4ax.com...
"Philip Newman" <nojunkmail@ntlworld.com> wrote:

well, I used these equations to work them out:

Px = (R+r)Cos(Lambda) - rCos(phi)Cos(Theta)
Py = (R+r)Sin(Lambda) - rCos(phi)Sin(Theta)
Pz = -rSin(phi)

R is the radius of the earth 6378km, r is the geostationary
altitude,
35786km, ? is the satellite's longitude, 30W, ? is the observer's
longitude,
5.17W and ? is the observer's latitude 50.05N

That last paragraph doesn't make sense to me. What do the three
'?'
mean? Are they meant to be Lambda, Theta and phi respectively? If
so,
what units are expected, degrees or radians?

--
Terry Pinnell
Hobbyist, West Sussex, UK


well, thats what i wrote anyway!! And they are in degrees.
50.05(deg)
etc

Sorry, copy and paste gone wrong me thinks...

In respect to Active8's reply, I checked out some of the links that
were
on
the site you sent me to, and there are extra calculations that might
need
to
be done in order to get the range, but that is surely far too
complicated
for a supposedly easy calculation? I'm talking about Orbital
Coordinate
Systems, Part II link.

Phil

It's so easy that no-one in their right minds *doesn't* use a
spreadsheet
or
calculator these days if you're doing more than one for anything other
than
as a demonstration. Go search on the keywords I gave previously -
there's
several example programs out there.

Ken



I did manage to find a suitable software/calculator that gave me an
answer
of 38123km, which sounds correct. However, I would like to work it out
manually, just as a personal issue now. It seems that I don't need to
know
this really, but its niggling me now.

Work out the length of the other side of the triangle that goes from the
center of the earth to the surface station, and the satellite to the
center
of the earth.
You know both, one is radius of earth, the other is height of GEO.

The OP used the given eqs (per link) to get the vector from the
earth station to the bird and the range ain't right when you plug
the vector into pythagorus. I'll hold off on deriving those eqs as a
check until I go to use them unless I get bored :)
--
Best Regards,
Mike
Click to expand...
I'm not going to post the equations as I have no easy way of doing the
characters and I'm not going to breach people's copyright. I suggest the OP
go and buy the bleedin' text - the formulae posted thus far are *wrong*.
We're talking spheres here!

Ken
 
"Ken Taylor" <ken@home.nz> wrote in message
news:F%2lc.1048$8J.36504@news.xtra.co.nz...
"Active8" <reply2group@ndbbm.net> wrote in message
news:1fhsfalbu6u7m.dlg@news.individual.net...
On Fri, 30 Apr 2004 21:10:01 GMT, Ian Stirling wrote:

Philip Newman <nojunkmail@ntlworld.com> wrote:

"Ken Taylor" <ken123@xtra.co.nz> wrote in message
news:c6rub2$fs3gv$1@ID-76636.news.uni-berlin.de...
"Philip Newman" <nojunkmail@ntlworld.com> wrote in message
news:eYdkc.192$xt3.78@newsfe1-gui.server.ntli.net...

"Terry Pinnell" <terrypin@dial.pipex.com> wrote in message
news:9um290pd7put9p28sn4hh5lhltp9p7v7l2@4ax.com...
"Philip Newman" <nojunkmail@ntlworld.com> wrote:

well, I used these equations to work them out:

Px = (R+r)Cos(Lambda) - rCos(phi)Cos(Theta)
Py = (R+r)Sin(Lambda) - rCos(phi)Sin(Theta)
Pz = -rSin(phi)

R is the radius of the earth 6378km, r is the geostationary
altitude,
35786km, ? is the satellite's longitude, 30W, ? is the
observer's
longitude,
5.17W and ? is the observer's latitude 50.05N

That last paragraph doesn't make sense to me. What do the three
'?'
mean? Are they meant to be Lambda, Theta and phi respectively?
If
so,
what units are expected, degrees or radians?

--
Terry Pinnell
Hobbyist, West Sussex, UK


well, thats what i wrote anyway!! And they are in degrees.
50.05(deg)
etc

Sorry, copy and paste gone wrong me thinks...

In respect to Active8's reply, I checked out some of the links
that
were
on
the site you sent me to, and there are extra calculations that
might
need
to
be done in order to get the range, but that is surely far too
complicated
for a supposedly easy calculation? I'm talking about Orbital
Coordinate
Systems, Part II link.

Phil

It's so easy that no-one in their right minds *doesn't* use a
spreadsheet
or
calculator these days if you're doing more than one for anything
other
than
as a demonstration. Go search on the keywords I gave previously -
there's
several example programs out there.

Ken



I did manage to find a suitable software/calculator that gave me an
answer
of 38123km, which sounds correct. However, I would like to work it
out
manually, just as a personal issue now. It seems that I don't need
to
know
this really, but its niggling me now.

Work out the length of the other side of the triangle that goes from
the
center of the earth to the surface station, and the satellite to the
center
of the earth.
You know both, one is radius of earth, the other is height of GEO.

The OP used the given eqs (per link) to get the vector from the
earth station to the bird and the range ain't right when you plug
the vector into pythagorus. I'll hold off on deriving those eqs as a
check until I go to use them unless I get bored :)
--
Best Regards,
Mike

I'm not going to post the equations as I have no easy way of doing the
characters and I'm not going to breach people's copyright. I suggest the
OP
go and buy the bleedin' text - the formulae posted thus far are *wrong*.
We're talking spheres here!

Ken
Click to expand...
I'm sure there is a copy in the library. Sorry, i was being lazy :)

I looked at the source code for the calculator on the web i found, but still
can't get it to work!!

Not to matter, its only a point of interest that's all.

Thanks for the help

Phil
 
"Ken Taylor" <ken@home.nz> wrote in message
news:Q6Ekc.372$8J.20835@news.xtra.co.nz...
"Philip Newman" <nojunkmail@ntlworld.com> wrote in message
news:yxykc.4154$N13.3048@newsfe1-win...

"Ken Taylor" <ken123@xtra.co.nz> wrote in message
news:c6rub2$fs3gv$1@ID-76636.news.uni-berlin.de...
"Philip Newman" <nojunkmail@ntlworld.com> wrote in message
news:eYdkc.192$xt3.78@newsfe1-gui.server.ntli.net...

"Terry Pinnell" <terrypin@dial.pipex.com> wrote in message
news:9um290pd7put9p28sn4hh5lhltp9p7v7l2@4ax.com...
"Philip Newman" <nojunkmail@ntlworld.com> wrote:

well, I used these equations to work them out:

Px = (R+r)Cos(Lambda) - rCos(phi)Cos(Theta)
Py = (R+r)Sin(Lambda) - rCos(phi)Sin(Theta)
Pz = -rSin(phi)

R is the radius of the earth 6378km, r is the geostationary
altitude,
35786km, ? is the satellite's longitude, 30W, ? is the observer's
longitude,
5.17W and ? is the observer's latitude 50.05N

That last paragraph doesn't make sense to me. What do the three
'?'
mean? Are they meant to be Lambda, Theta and phi respectively? If
so,
what units are expected, degrees or radians?

--
Terry Pinnell
Hobbyist, West Sussex, UK


well, thats what i wrote anyway!! And they are in degrees.
50.05(deg)
etc

Sorry, copy and paste gone wrong me thinks...

In respect to Active8's reply, I checked out some of the links that
were
on
the site you sent me to, and there are extra calculations that might
need
to
be done in order to get the range, but that is surely far too
complicated
for a supposedly easy calculation? I'm talking about Orbital
Coordinate
Systems, Part II link.

Phil

It's so easy that no-one in their right minds *doesn't* use a
spreadsheet
or
calculator these days if you're doing more than one for anything other
than
as a demonstration. Go search on the keywords I gave previously -
there's
several example programs out there.

Ken



I did manage to find a suitable software/calculator that gave me an
answer
of 38123km, which sounds correct. However, I would like to work it out
manually, just as a personal issue now. It seems that I don't need to
know
this really, but its niggling me now.

Phil


Clearly I wouldn't do this, but there are several sites out there with
calculators, as you've found. When you have the page do a 'View Source'.
Most enlightening. More intellectually satisfying is getting hold of one
of
the texts and looking at how it's all derived. See:
"Satellite Communications", Dennis Roddy, McGraw-Hill (ISBN 0-07-053370-9)
"Satellite Communications Systems", M. Richharia. McGraw-Hill. (ISBN
0-07-052374-6)

Ken
Click to expand...
I looked at the source code for the calculator on the web i found, but still
can't get it to work!!

Not to matter, its only a point of interest that's all.

Thanks for the help

Phil
 
Active8 <reply2group@ndbbm.net> wrote:
On Fri, 30 Apr 2004 21:10:01 GMT, Ian Stirling wrote:

Philip Newman <nojunkmail@ntlworld.com> wrote:
snip
I did manage to find a suitable software/calculator that gave me an answer
of 38123km, which sounds correct. However, I would like to work it out
manually, just as a personal issue now. It seems that I don't need to know
this really, but its niggling me now.

Work out the length of the other side of the triangle that goes from the
center of the earth to the surface station, and the satellite to the center
of the earth.
You know both, one is radius of earth, the other is height of GEO.

The OP used the given eqs (per link) to get the vector from the
earth station to the bird and the range ain't right when you plug
the vector into pythagorus. I'll hold off on deriving those eqs as a
check until I go to use them unless I get bored :)
Click to expand...
Pythagorus works just fine.
Though not really applicable, as this is not a right-angled triangle.
You need to be able to work out the resultant angle from two components,
and you can use simple geometry.
The above triangle is in a plane that intersects the center of the earth,
the earth station, and the satellite.
The satellite is above the equator at geosynchronus altitude.
It's above the point with the given longitude at the equator.
Work out the angle between the line going from the above point to the center
of the earth, and the one going from the earth station to the center of
the earth.
This gives you one angle for the above triangle.
The other is given by the angle that's required to move the distance
(direct, not along the earth) between the point on the equator and the
earth station in the distance out to the satellite, and the other
by the angles inside the triangle summing to 180 degrees.
Now, you know three angles, and two lengths for the inside of the triangle,
and can solve for the third length.
 
On Sun, 02 May 2004 13:19:59 GMT, Ian Stirling wrote:

Active8 <reply2group@ndbbm.net> wrote:
On Fri, 30 Apr 2004 21:10:01 GMT, Ian Stirling wrote:

Philip Newman <nojunkmail@ntlworld.com> wrote:
snip
I did manage to find a suitable software/calculator that gave me an answer
of 38123km, which sounds correct. However, I would like to work it out
manually, just as a personal issue now. It seems that I don't need to know
this really, but its niggling me now.

Work out the length of the other side of the triangle that goes from the
center of the earth to the surface station, and the satellite to the center
of the earth.
You know both, one is radius of earth, the other is height of GEO.

The OP used the given eqs (per link) to get the vector from the
earth station to the bird and the range ain't right when you plug
the vector into pythagorus. I'll hold off on deriving those eqs as a
check until I go to use them unless I get bored :)

Pythagorus works just fine.
Though not really applicable,
Click to expand...
why not? the range vector (were it correct) has x, y, and z
components.

as this is not a right-angled triangle.
Click to expand...
see above.

You need to be able to work out the resultant angle from two components,
and you can use simple geometry.
The above triangle is in a plane that intersects the center of the earth,
the earth station, and the satellite.
The satellite is above the equator at geosynchronus altitude.
It's above the point with the given longitude at the equator.
Work out the angle between the line going from the above point to the center
of the earth, and the one going from the earth station to the center of
the earth.
This gives you one angle for the above triangle.
The other is given by the angle that's required to move the distance
(direct, not along the earth) between the point on the equator and the
earth station in the distance out to the satellite, and the other
by the angles inside the triangle summing to 180 degrees.
Now, you know three angles, and two lengths for the inside of the triangle,
and can solve for the third length.
Click to expand...

--
Best Regards,
Mike
 
Active8 <reply2group@ndbbm.net> wrote:
On Sun, 02 May 2004 13:19:59 GMT, Ian Stirling wrote:

Active8 <reply2group@ndbbm.net> wrote:
On Fri, 30 Apr 2004 21:10:01 GMT, Ian Stirling wrote:

Philip Newman <nojunkmail@ntlworld.com> wrote:
snip
I did manage to find a suitable software/calculator that gave me an answer
of 38123km, which sounds correct. However, I would like to work it out
manually, just as a personal issue now. It seems that I don't need to know
this really, but its niggling me now.

Work out the length of the other side of the triangle that goes from the
center of the earth to the surface station, and the satellite to the center
of the earth.
You know both, one is radius of earth, the other is height of GEO.

The OP used the given eqs (per link) to get the vector from the
earth station to the bird and the range ain't right when you plug
the vector into pythagorus. I'll hold off on deriving those eqs as a
check until I go to use them unless I get bored :)

Pythagorus works just fine.
Though not really applicable,

why not? the range vector (were it correct) has x, y, and z
components.
Click to expand...
Well, of course.
 
"Ian Stirling" <root@mauve.demon.co.uk> wrote in message
news:3A6lc.36674$h44.5514287@stones.force9.net...
Active8 <reply2group@ndbbm.net> wrote:
On Fri, 30 Apr 2004 21:10:01 GMT, Ian Stirling wrote:

Philip Newman <nojunkmail@ntlworld.com> wrote:
snip
I did manage to find a suitable software/calculator that gave me an
answer
of 38123km, which sounds correct. However, I would like to work it
out
manually, just as a personal issue now. It seems that I don't need to
know
this really, but its niggling me now.

Work out the length of the other side of the triangle that goes from
the
center of the earth to the surface station, and the satellite to the
center
of the earth.
You know both, one is radius of earth, the other is height of GEO.

The OP used the given eqs (per link) to get the vector from the
earth station to the bird and the range ain't right when you plug
the vector into pythagorus. I'll hold off on deriving those eqs as a
check until I go to use them unless I get bored :)

Pythagorus works just fine.
Though not really applicable, as this is not a right-angled triangle.
You need to be able to work out the resultant angle from two components,
and you can use simple geometry.
The above triangle is in a plane that intersects the center of the earth,
the earth station, and the satellite.
The satellite is above the equator at geosynchronus altitude.
It's above the point with the given longitude at the equator.
Work out the angle between the line going from the above point to the
center
of the earth, and the one going from the earth station to the center of
the earth.
This gives you one angle for the above triangle.
The other is given by the angle that's required to move the distance
(direct, not along the earth) between the point on the equator and the
earth station in the distance out to the satellite, and the other
by the angles inside the triangle summing to 180 degrees.
Now, you know three angles, and two lengths for the inside of the
triangle,
and can solve for the third length.
Click to expand...
As the good doctor (as we used to call him) was fond of saying: "and the
rest is conceptually trivial!" The tricky bits are those angles and
distances around the sphere. If it was all two dimensional we could do it in
our heads. Mind you, if it was two-dimensional we wouldn't need a
satellite.....

Ken
 
Ken Taylor <ken123@xtra.co.nz> wrote:
"Ian Stirling" <root@mauve.demon.co.uk> wrote in message
news:3A6lc.36674$h44.5514287@stones.force9.net...
Active8 <reply2group@ndbbm.net> wrote:
On Fri, 30 Apr 2004 21:10:01 GMT, Ian Stirling wrote:

Philip Newman <nojunkmail@ntlworld.com> wrote:
snip
I did manage to find a suitable software/calculator that gave me an
answer
of 38123km, which sounds correct. However, I would like to work it
out
manually, just as a personal issue now. It seems that I don't need to
know
this really, but its niggling me now.

Work out the length of the other side of the triangle that goes from
the
center of the earth to the surface station, and the satellite to the
center
of the earth.
You know both, one is radius of earth, the other is height of GEO.

The OP used the given eqs (per link) to get the vector from the
earth station to the bird and the range ain't right when you plug
the vector into pythagorus. I'll hold off on deriving those eqs as a
check until I go to use them unless I get bored :)

Pythagorus works just fine.
Though not really applicable, as this is not a right-angled triangle.
You need to be able to work out the resultant angle from two components,
and you can use simple geometry.
The above triangle is in a plane that intersects the center of the earth,
the earth station, and the satellite.
The satellite is above the equator at geosynchronus altitude.
It's above the point with the given longitude at the equator.
Work out the angle between the line going from the above point to the
center
of the earth, and the one going from the earth station to the center of
the earth.
This gives you one angle for the above triangle.
The other is given by the angle that's required to move the distance
(direct, not along the earth) between the point on the equator and the
earth station in the distance out to the satellite, and the other
by the angles inside the triangle summing to 180 degrees.
Now, you know three angles, and two lengths for the inside of the
triangle,
and can solve for the third length.

As the good doctor (as we used to call him) was fond of saying: "and the
rest is conceptually trivial!" The tricky bits are those angles and
distances around the sphere. If it was all two dimensional we could do it in
our heads. Mind you, if it was two-dimensional we wouldn't need a
satellite.....
Click to expand...
It is two-dimensional, if you look at it right.

Ok, some numbers.
Say the satellite is at 28.2 degrees east.
Say the earth station is at 2 degrees west, and 56N latitude.
Ok, what's the as-the-crow flies distance between this and the spot
on the equator.
Well, let's first state some knowns.
The earth has a radius of 6000Km (for easy calculation).
The earth is spherical (nearly is anyway).
GEO is at 40000Km (round numbers)
Drawing an isocelese triangle with two arms of 6000Km from the earth station,
directly south to the equator and to the center of the earth this
results in a triangle with an angle at the center of the earth of 56
degrees.
The other side has a length of 5632Km.
Now, draw another isocelese triangle from the center of the earth to
2 degrees west and to 28.2 east.
The angle at the center of the earth is 30.2 degrees.
And the angle of the other side is 3126Km.

Now, we have a right-angled triangle with three points, all touching
the earths surface, from the earth station directly south to the equator,
west to the point under the satellite.
The direct distance from the earth station to the point under the
satellite is simply sqrt(5632^2+3126^2), or 6441Km.
The angle between the earth station and the point under the
satellite as seen from the earths center is 64.9 degrees.
So, this lets us construct a triangle from the earths center to the
earth station, to the satellite and back.

The angle next to the earths core is 64.9 degrees, the distance to the
earth station from the core is 6000Km, the distance from the core
of the earth to the satellite is 46000Km.
Now, you have a triangle with one known angle, and two known sides.
This gives you the length of the other side after you look up
the appropriate formula.
 
"Ian Stirling" <root@mauve.demon.co.uk> wrote in message
news:r9hlc.36799$h44.5553994@stones.force9.net...
Ken Taylor <ken123@xtra.co.nz> wrote:
"Ian Stirling" <root@mauve.demon.co.uk> wrote in message
news:3A6lc.36674$h44.5514287@stones.force9.net...
Active8 <reply2group@ndbbm.net> wrote:
On Fri, 30 Apr 2004 21:10:01 GMT, Ian Stirling wrote:

Philip Newman <nojunkmail@ntlworld.com> wrote:
snip
I did manage to find a suitable software/calculator that gave me an
answer
of 38123km, which sounds correct. However, I would like to work it
out
manually, just as a personal issue now. It seems that I don't need
to
know
this really, but its niggling me now.

Work out the length of the other side of the triangle that goes from
the
center of the earth to the surface station, and the satellite to the
center
of the earth.
You know both, one is radius of earth, the other is height of GEO.

The OP used the given eqs (per link) to get the vector from the
earth station to the bird and the range ain't right when you plug
the vector into pythagorus. I'll hold off on deriving those eqs as a
check until I go to use them unless I get bored :)

Pythagorus works just fine.
Though not really applicable, as this is not a right-angled triangle.
You need to be able to work out the resultant angle from two
components,
and you can use simple geometry.
The above triangle is in a plane that intersects the center of the
earth,
the earth station, and the satellite.
The satellite is above the equator at geosynchronus altitude.
It's above the point with the given longitude at the equator.
Work out the angle between the line going from the above point to the
center
of the earth, and the one going from the earth station to the center of
the earth.
This gives you one angle for the above triangle.
The other is given by the angle that's required to move the distance
(direct, not along the earth) between the point on the equator and the
earth station in the distance out to the satellite, and the other
by the angles inside the triangle summing to 180 degrees.
Now, you know three angles, and two lengths for the inside of the
triangle,
and can solve for the third length.

As the good doctor (as we used to call him) was fond of saying: "and the
rest is conceptually trivial!" The tricky bits are those angles and
distances around the sphere. If it was all two dimensional we could do
it in
our heads. Mind you, if it was two-dimensional we wouldn't need a
satellite.....

It is two-dimensional, if you look at it right.

Ok, some numbers.
Say the satellite is at 28.2 degrees east.
Say the earth station is at 2 degrees west, and 56N latitude.
Ok, what's the as-the-crow flies distance between this and the spot
on the equator.
Well, let's first state some knowns.
The earth has a radius of 6000Km (for easy calculation).
The earth is spherical (nearly is anyway).
GEO is at 40000Km (round numbers)
Drawing an isocelese triangle with two arms of 6000Km from the earth
station,
directly south to the equator and to the center of the earth this
results in a triangle with an angle at the center of the earth of 56
degrees.
The other side has a length of 5632Km.
Now, draw another isocelese triangle from the center of the earth to
2 degrees west and to 28.2 east.
The angle at the center of the earth is 30.2 degrees.
And the angle of the other side is 3126Km.

Now, we have a right-angled triangle with three points, all touching
the earths surface, from the earth station directly south to the equator,
west to the point under the satellite.
The direct distance from the earth station to the point under the
satellite is simply sqrt(5632^2+3126^2), or 6441Km.
The angle between the earth station and the point under the
satellite as seen from the earths center is 64.9 degrees.
So, this lets us construct a triangle from the earths center to the
earth station, to the satellite and back.

The angle next to the earths core is 64.9 degrees, the distance to the
earth station from the core is 6000Km, the distance from the core
of the earth to the satellite is 46000Km.
Now, you have a triangle with one known angle, and two known sides.
This gives you the length of the other side after you look up
the appropriate formula.
Click to expand...
This may be one of my less lucid days, but you lost me when you got two
angles of 64.9 degrees. As a check, I used this technique with the right
values for earth's radius and geo height and the value obtained was about
10% low. ??

Ken
 
On Mon, 03 May 2004 01:22:31 GMT, Ian Stirling wrote:

Ken Taylor <ken123@xtra.co.nz> wrote:
"Ian Stirling" <root@mauve.demon.co.uk> wrote in message
news:3A6lc.36674$h44.5514287@stones.force9.net...
Active8 <reply2group@ndbbm.net> wrote:
On Fri, 30 Apr 2004 21:10:01 GMT, Ian Stirling wrote:

Philip Newman <nojunkmail@ntlworld.com> wrote:
snip
I did manage to find a suitable software/calculator that gave me an
answer
of 38123km, which sounds correct. However, I would like to work it
out
manually, just as a personal issue now. It seems that I don't need to
know
this really, but its niggling me now.

Work out the length of the other side of the triangle that goes from
the
center of the earth to the surface station, and the satellite to the
center
of the earth.
You know both, one is radius of earth, the other is height of GEO.

The OP used the given eqs (per link) to get the vector from the
earth station to the bird and the range ain't right when you plug
the vector into pythagorus. I'll hold off on deriving those eqs as a
check until I go to use them unless I get bored :)

Pythagorus works just fine.
Though not really applicable, as this is not a right-angled triangle.
You need to be able to work out the resultant angle from two components,
and you can use simple geometry.
The above triangle is in a plane that intersects the center of the earth,
the earth station, and the satellite.
The satellite is above the equator at geosynchronus altitude.
It's above the point with the given longitude at the equator.
Work out the angle between the line going from the above point to the
center
of the earth, and the one going from the earth station to the center of
the earth.
This gives you one angle for the above triangle.
The other is given by the angle that's required to move the distance
(direct, not along the earth) between the point on the equator and the
earth station in the distance out to the satellite, and the other
by the angles inside the triangle summing to 180 degrees.
Now, you know three angles, and two lengths for the inside of the
triangle,
and can solve for the third length.

As the good doctor (as we used to call him) was fond of saying: "and the
rest is conceptually trivial!" The tricky bits are those angles and
distances around the sphere. If it was all two dimensional we could do it in
our heads. Mind you, if it was two-dimensional we wouldn't need a
satellite.....

It is two-dimensional, if you look at it right.

Ok, some numbers.
Say the satellite is at 28.2 degrees east.
Say the earth station is at 2 degrees west, and 56N latitude.
Ok, what's the as-the-crow flies distance between this and the spot
on the equator.
Well, let's first state some knowns.
The earth has a radius of 6000Km (for easy calculation).
The earth is spherical (nearly is anyway).
GEO is at 40000Km (round numbers)
Drawing an isocelese triangle with two arms of 6000Km from the earth station,
directly south to the equator and to the center of the earth this
results in a triangle with an angle at the center of the earth of 56
degrees.
The other side has a length of 5632Km.
Now, draw another isocelese triangle from the center of the earth to
2 degrees west and to 28.2 east.
The angle at the center of the earth is 30.2 degrees.
And the angle of the other side is 3126Km.

Now, we have a right-angled triangle with three points, all touching
the earths surface, from the earth station directly south to the equator,
west to the point under the satellite.
The direct distance from the earth station to the point under the
satellite is simply sqrt(5632^2+3126^2), or 6441Km.
The angle between the earth station and the point under the
satellite as seen from the earths center is 64.9 degrees.
So, this lets us construct a triangle from the earths center to the
earth station, to the satellite and back.

The angle next to the earths core is 64.9 degrees, the distance to the
earth station from the core is 6000Km, the distance from the core
of the earth to the satellite is 46000Km.
Now, you have a triangle with one known angle, and two known sides.
This gives you the length of the other side after you look up
the appropriate formula.
Click to expand...
All of what you said may or may not be right, but the fact is that
the link previously posted that the OP used presented 2 sets of eqs.
One set gave cartesian (unless I seriously f'd up on reading) coords
for the earth station, and coords for the bird. Assuming those eqs
were correct - I haven't derived them myself (I fail to see any
reason why the sherical trig should present any prob at all, despite
the opinions of others) - the article then shows a simple
subtraction which yields the range vector. This Range vector should
give the actual range via sqrt(x^2 + Y^2 + Z^2) BUT IT DON'T.

My conclusion is that the eqs in the link are wrong and that a few
people are blathering on about something they haven't even looked
at.
--
Best Regards,
Mike
 
"Active8" <reply2group@ndbbm.net> wrote in message
news:14hw9cm26zh39$.dlg@news.individual.net...
On Mon, 03 May 2004 01:22:31 GMT, Ian Stirling wrote:


All of what you said may or may not be right, but the fact is that
the link previously posted that the OP used presented 2 sets of eqs.
One set gave cartesian (unless I seriously f'd up on reading) coords
for the earth station, and coords for the bird. Assuming those eqs
were correct - I haven't derived them myself (I fail to see any
reason why the sherical trig should present any prob at all, despite
the opinions of others) - the article then shows a simple
subtraction which yields the range vector. This Range vector should
give the actual range via sqrt(x^2 + Y^2 + Z^2) BUT IT DON'T.

My conclusion is that the eqs in the link are wrong and that a few
people are blathering on about something they haven't even looked
at.
--
Best Regards,
Mike
Click to expand...
My conclusion is that you think the equations are right, acknowledge that
they don't work and then figure other people don't now what they're talking
about. If I could be bothered figuring an easy way to copy/post the
equations in either of my texts I would, but I really can't be fagged. OP
has acknowledged them and will look if/when he gets sufficient
time/interest, so the job's done from my point of view.

Ian's method seems possible on face value but I doubt it's veracity, and it
appears to be out by at least a few per cent, which at present I'll put down
to being due to an invalid assumption somewhere. However it's probably a
reasonable (and easy) first approximation for the interested.

Ken
 
.. OP
has acknowledged them and will look if/when he gets sufficient
time/interest, so the job's done from my point of view.
Click to expand...
I will, thanks :)

Phil
 
"Ian Stirling" <root@mauve.demon.co.uk> wrote in message
news:WaTjc.37105$Y%6.5073786@wards.force9.net...
Philip Newman <nojunkmail@ntlworld.com> wrote:

"Ian Stirling" <root@mauve.demon.co.uk> wrote in message
news:NdSjc.35687$h44.5291991@stones.force9.net...
Philip Newman <nojunkmail@ntlworld.com> wrote:
This is to do with telecommunications, and hopefully someone should
know
the
answer to this!

Assume a ground-station has a longitude of 5.17degW and a latitude of
50.05degN
There is an Atlantic Ocean Satellite with longitude 30.00degW.

How can I calculate the range of the satellite from the earth
station?

40000Km+-10%.
Or do you need it very accurately?

I would prefer some workings! How did you come to that value??
Understanding
is important for me...

Geosynchronus orbit is by its nature around 42300Km in radius.
The earth radius is about 6378Km.

At most the distance can only vary by around 6370Km.
So, the minimum is 36830 (at the equator at 30W), and the maximum is
43200Km
at the great circle normal to this.
Click to expand...
38929.7313km.

I drew it with Mechanical Desktop 6.

see:
http://home.earthlink.net/~rich_grise/images/SatelliteRange.gif

BTW, the numerals for the dimensions are 1000km tall. :)

Cheers!
Rich
 
"Rich Grise" <null@example.net> wrote in message
news:giVlc.151522$L31.72466@nwrddc01.gnilink.net...
"Ian Stirling" <root@mauve.demon.co.uk> wrote in message
news:WaTjc.37105$Y%6.5073786@wards.force9.net...
Philip Newman <nojunkmail@ntlworld.com> wrote:

"Ian Stirling" <root@mauve.demon.co.uk> wrote in message
news:NdSjc.35687$h44.5291991@stones.force9.net...
Philip Newman <nojunkmail@ntlworld.com> wrote:
This is to do with telecommunications, and hopefully someone should
know
the
answer to this!

Assume a ground-station has a longitude of 5.17degW and a latitude
of
50.05degN
There is an Atlantic Ocean Satellite with longitude 30.00degW.

How can I calculate the range of the satellite from the earth
station?

40000Km+-10%.
Or do you need it very accurately?

I would prefer some workings! How did you come to that value??
Understanding
is important for me...

Geosynchronus orbit is by its nature around 42300Km in radius.
The earth radius is about 6378Km.

At most the distance can only vary by around 6370Km.
So, the minimum is 36830 (at the equator at 30W), and the maximum is
43200Km
at the great circle normal to this.

38929.7313km.

I drew it with Mechanical Desktop 6.

see:
http://home.earthlink.net/~rich_grise/images/SatelliteRange.gif

BTW, the numerals for the dimensions are 1000km tall. :)

Cheers!
Rich
Click to expand...
wow, thats great. I can use that as a great example - cheers

Phil
 
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