Satellite Range

P

Philip Newman

Guest
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?

I have had a look at some literature, but nothing seems to be what I need.
What equations do I need?

Phil
 
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?
Click to expand...
40000Km+-10%.
Or do you need it very accurately?
 
"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?
Click to expand...
I would prefer some workings! How did you come to that value?? Understanding
is important for me...

Phil
 
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...
Click to expand...
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.
 
"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...
And here was me converting the long/lat to planar and doing geometry on it
to work out the range.

hmm, thanks

Phil
 
Philip Newman wrote:

"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.


And here was me converting the long/lat to planar and doing geometry on it
to work out the range.
Click to expand...
For figuring satellite TV signal strength the "rule of
thumb" is fine, but for GPS you need to do the math. What
accuracy do you really need?

Mark L. Fergerson
 
Mark Fergerson <nunya@biz.ness> wrote:
Philip Newman wrote:

"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.


And here was me converting the long/lat to planar and doing geometry on it
to work out the range.

For figuring satellite TV signal strength the "rule of
thumb" is fine, but for GPS you need to do the math. What
accuracy do you really need?
Click to expand...
You'll have really big problems if you'r doing sums on GPS assuming the
birds are in geostationary, or even geosynchronus orbits.
 
Trigonometric.

BTW, your problem statment doesn't say so, but I assume the "Atlantic Ocean
Satellite" is in geosynchronous orbit.

"Philip Newman" <nojunkmail@ntlworld.com> wrote in message
news:0uRjc.80$RI.53@newsfe1-win...
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?

I have had a look at some literature, but nothing seems to be what I need.
What equations do I need?

Phil
Click to expand...
 
"Philip Newman" <nojunkmail@ntlworld.com> wrote in message
news:K9Tjc.193$273.172@newsfe1-gui.server.ntli.net...
"Ian Stirling" <root@mauve.demon.co.uk> wrote in message
news:WaTjc.37105$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.

And here was me converting the long/lat to planar and doing geometry on it
to work out the range.

hmm, thanks

Phil


Urk! You need to use great circle geometry. If you want a text, buy
Click to expand...
"Satellite Communications" by Roddy (it's on Amazon). There are others of
course, but I'm familiar with that one. If you just want the equations on
the web, take a peek at:
http://www.amsat.org/amsat/articles/g3ruh/111.txt (comes with ref's) or
http://www.celestrak.com/columns/v04n09/

Ken
 
On Wed, 28 Apr 2004 12:09:50 -0700, Mark Fergerson wrote:

Philip Newman wrote:

"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.

And here was me converting the long/lat to planar and doing geometry on it
to work out the range.

For figuring satellite TV signal strength the "rule of
thumb" is fine, but for GPS you need to do the math. What
accuracy do you really need?

Mark L. Fergerson
Click to expand...
Does this mean you know how to do hyperbolic trilateralization?
--
Best Regards,
Mike
 
"Active8" <reply2group@ndbbm.net> wrote in message
news:rff7zj92pncq$.dlg@news.individual.net...
On Wed, 28 Apr 2004 12:09:50 -0700, Mark Fergerson wrote:

Philip Newman wrote:

"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.

And here was me converting the long/lat to planar and doing geometry on
it
to work out the range.

For figuring satellite TV signal strength the "rule of
thumb" is fine, but for GPS you need to do the math. What
accuracy do you really need?

Mark L. Fergerson

Does this mean you know how to do hyperbolic trilateralization?
--
Best Regards,
Mike
Click to expand...
All the formulae are in the texts - the rest is indeed hyperbole.

Sorry, couldn't resist. :)

Ken
 
Ken Taylor wrote:
"Active8" <reply2group@ndbbm.net> wrote in message
news:rff7zj92pncq$.dlg@news.individual.net...

On Wed, 28 Apr 2004 12:09:50 -0700, Mark Fergerson wrote:


Philip Newman wrote:


"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.

And here was me converting the long/lat to planar and doing geometry on

it

to work out the range.

For figuring satellite TV signal strength the "rule of
thumb" is fine, but for GPS you need to do the math. What
accuracy do you really need?
^^^^^^^^
Click to expand...

Shoulda been "precision". Duh.

Does this mean you know how to do hyperbolic trilateralization?
Click to expand...
No.

Best Regards,
Mike


All the formulae are in the texts - the rest is indeed hyperbole.

Sorry, couldn't resist. :)
Click to expand...
Aww, you beat me to it!

Mark L. Fergerson
 
Ian Stirling wrote:

Mark Fergerson <nunya@biz.ness> wrote:

Philip Newman wrote:


"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.


And here was me converting the long/lat to planar and doing geometry on it
to work out the range.

For figuring satellite TV signal strength the "rule of
thumb" is fine, but for GPS you need to do the math. What
accuracy do you really need?


You'll have really big problems if you'r doing sums on GPS assuming the
birds are in geostationary, or even geosynchronus orbits.
Click to expand...
OK, GPS was an extreme example, and no, they aren't. I
was just wondering how close the OP had to get.

Mark L. Fergerson
 
On Thu, 29 Apr 2004 02:15:47 -0700, Mark Fergerson wrote:

Ken Taylor wrote:
"Active8" <reply2group@ndbbm.net> wrote in message
news:rff7zj92pncq$.dlg@news.individual.net...

On Wed, 28 Apr 2004 12:09:50 -0700, Mark Fergerson wrote:


Philip Newman wrote:


"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.

And here was me converting the long/lat to planar and doing geometry on

it

to work out the range.

For figuring satellite TV signal strength the "rule of
thumb" is fine, but for GPS you need to do the math. What
accuracy do you really need?
^^^^^^^^

Shoulda been "precision". Duh.

Does this mean you know how to do hyperbolic trilateralization?

No.

Best Regards,
Mike

All the formulae are in the texts - the rest is indeed hyperbole.

Sorry, couldn't resist. :)

Aww, you beat me to it!

Mark L. Fergerson
Click to expand...
Reason I brougth it up is that is what turned up when I checked out
GPS, but even on alt.math.undergrd I couldn't find any info. I
managed to write eqs for intersecting spheres, but the closest link
I found on hype trilat was a site that just mentioned it and said it
was proprietary - prob the code they wrapped it up in.
--
Best Regards,
Mike
 
yes thats right. So far I have managed to work out the vector co-ordinates
of the satellite with respect to the ground-station, which are:
x = 13629.72km
y = 19011.35km
z = -3224.72km

How can I equate this to an actual range?

I've tried pythagoras and things, but can't get it to work! It's really
annoying me now....

Phil

"Richard Henry" <rphenry@home.com> wrote in message
news:45Vjc.2058$fE.724@fed1read02...
Trigonometric.

BTW, your problem statment doesn't say so, but I assume the "Atlantic
Ocean
Satellite" is in geosynchronous orbit.

"Philip Newman" <nojunkmail@ntlworld.com> wrote in message
news:0uRjc.80$RI.53@newsfe1-win...
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?

I have had a look at some literature, but nothing seems to be what I
need.
What equations do I need?

Phil
Click to expand...
 
On Thu, 29 Apr 2004 16:55:16 +0100, "Philip Newman"
<nojunkmail@ntlworld.com> wrote:

yes thats right. So far I have managed to work out the vector co-ordinates
of the satellite with respect to the ground-station, which are:
x = 13629.72km
y = 19011.35km
z = -3224.72km

How can I equate this to an actual range?

I've tried pythagoras and things, but can't get it to work! It's really
annoying me now....

Phil

[snip]
Click to expand...

Why is it that you think Pythagoras' theorem isn't working? If you
have correctly identified the coordinates in an *orthogonal*
coordinate system it certainly should work.

It could be that you are actually measuring from within a spherical
coordinate system and misapplying the data.

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.
 
On Thu, 29 Apr 2004 09:12:41 -0700, Jim Thompson wrote:

On Thu, 29 Apr 2004 16:55:16 +0100, "Philip Newman"
nojunkmail@ntlworld.com> wrote:

yes thats right. So far I have managed to work out the vector co-ordinates
of the satellite with respect to the ground-station, which are:
x = 13629.72km
y = 19011.35km
z = -3224.72km

How can I equate this to an actual range?

I've tried pythagoras and things, but can't get it to work! It's really
annoying me now....

Phil

[snip]

Why is it that you think Pythagoras' theorem isn't working? If you
have correctly identified the coordinates in an *orthogonal*
coordinate system it certainly should work.

It could be that you are actually measuring from within a spherical
coordinate system and misapplying the data.

...Jim Thompson
Click to expand...
Yeah. That vector can't be right if the bird is at roughly 40,000 km
--
Best Regards,
Mike
 
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



"Active8" <reply2group@ndbbm.net> wrote in message
news:7wtv9cw6afqp$.dlg@news.individual.net...
On Thu, 29 Apr 2004 09:12:41 -0700, Jim Thompson wrote:

On Thu, 29 Apr 2004 16:55:16 +0100, "Philip Newman"
nojunkmail@ntlworld.com> wrote:

yes thats right. So far I have managed to work out the vector
co-ordinates
of the satellite with respect to the ground-station, which are:
x = 13629.72km
y = 19011.35km
z = -3224.72km

How can I equate this to an actual range?

I've tried pythagoras and things, but can't get it to work! It's really
annoying me now....

Phil

[snip]

Why is it that you think Pythagoras' theorem isn't working? If you
have correctly identified the coordinates in an *orthogonal*
coordinate system it certainly should work.

It could be that you are actually measuring from within a spherical
coordinate system and misapplying the data.

...Jim Thompson

Yeah. That vector can't be right if the bird is at roughly 40,000 km
--
Best Regards,
Mike
Click to expand...
 
On Thu, 29 Apr 2004 17:51:47 +0100, Philip Newman 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
Click to expand...
Straight from the link provided earlier. Maybe double check your
calcs - crap... I did the first 2 component sand your number's right
so something is FUBAR.
"Active8" <reply2group@ndbbm.net> wrote in message
news:7wtv9cw6afqp$.dlg@news.individual.net...
On Thu, 29 Apr 2004 09:12:41 -0700, Jim Thompson wrote:

On Thu, 29 Apr 2004 16:55:16 +0100, "Philip Newman"
nojunkmail@ntlworld.com> wrote:

yes thats right. So far I have managed to work out the vector
co-ordinates
of the satellite with respect to the ground-station, which are:
x = 13629.72km
y = 19011.35km
z = -3224.72km

How can I equate this to an actual range?

I've tried pythagoras and things, but can't get it to work! It's really
annoying me now....

Phil

[snip]

Why is it that you think Pythagoras' theorem isn't working? If you
have correctly identified the coordinates in an *orthogonal*
coordinate system it certainly should work.

It could be that you are actually measuring from within a spherical
coordinate system and misapplying the data.

...Jim Thompson

Yeah. That vector can't be right if the bird is at roughly 40,000 km
--
Best Regards,
Mike
Click to expand...

--
Best Regards,
Mike
 
"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
Click to expand...
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
 
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