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Q: About satellite geometry ( Answered 4 out of 5 stars,   0 Comments )
Subject: About satellite geometry
Category: Science > Earth Sciences
Asked by: timnan-ga
List Price: $50.00
Posted: 27 Sep 2002 02:14 PDT
Expires: 27 Oct 2002 01:14 PST
Question ID: 69640
For using satellites data, when I know the location (in Lon, Lat) and
the altitude of a satellite (from two-line elements), how do I
calculate the lon-lat value of a viewing point (on earth) with a
specified nadir angle and azimuth angle?

Request for Question Clarification by livioflores-ga on 27 Sep 2002 09:29 PDT
Can you clarify me about the altitude data, you know the altitude in
meters (for example) of the satellite from the location point (like
the altitude of a plane over the Earth surface) or you know the
altitude in degrees?

Clarification of Question by timnan-ga on 27 Sep 2002 21:55 PDT
Dear livioflores:

 for example: the NOAA polor orbiting satellites is 800km above the earth.


Request for Question Clarification by livioflores-ga on 28 Sep 2002 00:27 PDT
I can solve this using basic geometry, polar coordinates and
trigonometry in order to obtain a very aproximate lat-lon value,
without correction for the geodesic form of the Earth, considering the
Earth surface as a plane and suposing, in the calculations, the
satellite position at a given time, with no prediction to a later
I can give you a page with very complex calculations where you can see
the position and prediction calculations did for the NOAA Polar
Do you consider this a good answer?

Request for Question Clarification by livioflores-ga on 28 Sep 2002 00:37 PDT
Clarification of my request:
When I say "considering the Earth surface as a plane", I am
considering it locally, because a small amount of Earth surface is
If this kind of answer donŽt satisfy you please let me know the degree
complexity that you require.
Thank you.

Clarification of Question by timnan-ga on 28 Sep 2002 05:14 PDT
Yes, your answer already meet my requirement. Answer without geodesic
correction (regarding earth as a ellipsoid) is good enough for me, coz
I am now dealing with the oceanic data, and trying to do some
first-order correction.


Clarification of Question by timnan-ga on 28 Sep 2002 06:09 PDT
Also, if there is any reference, please list me one.

Subject: Re: About satellite geometry
Answered By: livioflores-ga on 28 Sep 2002 10:17 PDT
Rated:4 out of 5 stars
Well dear timnan, I will try to get an answer:

First of all some definitions:
-Azimut angle (Aa): The azimuth angle of a satellite is the angle on
the horizon that begins to be measure from the North cardinal point
towards to the east to the vertical of the object.
-Nadir Point (Np): The point on the earth directly below the satellite
at any given time during its orbit.
-Nadir Angle (Na): The angle between the Satellite-Nadir point line
(S-Np) and Satellite-Visualization point line (S-Vp).
-Visualization Point (Vp): the point on the Earth where the observer

Now I can start the calculations:
Considering the local Earth surface as a plane, we have a right
triangle formed by the Satellite, the Nadir point and the
Visualization point. And we know the Nadir angle and the lenght of the
adjacent side of this angle, such is the altitude of the satellite

If D is the distance between the Nadir point and the Visualization

tan(Na) = D/H, then D = H x tan(Na)

Actually we know the Azimuth angle and the distance D from the
Visualization point to the Nadir point. If the Np is taken as origin
of coordinates, we have the polar coordinates of the Vp. We need to
convert it to a (x,y) pair coordinate system, where x express the
distance from origin in the East-West direction (EW) and y the same
but in the North-South direction (NS).

It is not a hard task, but we must consider four situations:

0 =< Aa < 90
cos(Aa) = NS/D, then NS = D x cos(Aa) to the South.
sen(Aa) = EW/D, then EW = D x sen(Aa) to the West.

90 =< Aa < 180
cos(180-Aa) = NS/D, then NS = D x cos(180-Aa) to the North.
sen(180-Aa) = EW/D, then EW = D x sen(180-Aa) to the West.

180 =< Aa < 270
cos(Aa-180) = NS/D, then NS = D x cos(Aa-180) to the North.
sen(Aa-180) = EW/D, then EW = D x sen(Aa-180) to the East.

270 =< Aa < 360
cos(360-Aa) = NS/D, then NS = D x cos(360-Aa) to the South.
sen(360-Aa) = EW/D, then EW = D x sen(360-Aa) to the East.

If your calculator not support the calculation of trigonometric
functions with the angle expressed in degrees, you must divide the
angles by 57.2958 in order to convert it to radians.

Now we need to convert the distances in each direction to degrees.
We know the lat-lon values for the Nadir point.
 If R is the radius of the Earth and O is its center (considering the
Earth as a sphere), to specify the latitude of some point P on the
surface, we must draw the radius OP to that point. Then the elevation
angle of that point above or below the equator is its latitude. For
each latitude angle we can draw a circle of radius r(lat) = R x
Then the circunference of that circle is:
2 x pi x R x cos(lat); 
and a degree is a 360th part of this circunference (for this given
latitude only, in this case the latitude of the Nadir point), then we
1șEW = 2 x pi x R x cos(lat)/360 = 3.1415 x 6378.137 x cos(lat)/180

If we consider the semi-circle that describes any meridian, we can
divide it in 180 parts that each one is equal to one degree variation
in latitude.
1șNW = 3.1415 x 6378.137 /180 [km] 
We can convert the EW and NS distances to EW and NS degrees by a
simple division, for example:
EW degrees = EW/1șEW

Now you only need to add or substract this calculated degrees to the
lat-lon values of the Nadir point.

You will found some interesting (but very complicated) calculations
and related info at:
"National Oceanic and Atmospheric Administration (NOAA)" website


"APPENDIX I.2: Calculating the Earth's Coordinates" article

Some ideas I got from:
"Latitude and Longitude" from NASA website

I hope this satisfy you as an answer, if you need some clarification
and/or more info, please feel free to post a request for it.

Best Regards
timnan-ga rated this answer:4 out of 5 stars
Thanks, this helps a lot.

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