NOAA KLM User's Guide
There area at least two ways to define the satellite subpoint. The first way is to call it the intersection with the earth ellipsoid surface of a line from the satellite to the earth ellipsoid's center - call this the geocentric subpoint. The second way is to call it the intersection with the earth ellipsoid's surface of a line from the satellite perpendicular to the ellipsoid - call this the geodetic subpoint. (See Figure I.3-1 and imaging that the feature being located is the satellite.) When the satellite is over the North or South Pole or over the Equator, geodetic and geocentric subpoints will be colocated; when it is over 45° North or South, the distance between the geocentric and geodetic subpoints will be about 2.5 kilometers for NOAA satellites. Documents describing the NOAA K, L and M Attitude Detection and Control System (ADACS) define the subpoint to be geodetic.
Case 1 - The satellite subpoint is defined to be the geocentric subpoint.
In this case, the unit vector points in the opposite direction from the Satellite position vector. So, if the satellites's position vector in earth-centered-inertial coordinates is
Case 2 - The satellite subpoint is defined to be the geodetic subpoint.
In this case, the earth-centered-inertial coordinates of the satellite will be known. The problem will be to use them to find the direction cosines from the satellite toward the geodetic subpoint. Again, the position vector of the satellite will be
The magnitude of the component of this vector that is in the equatorial plane is
If DISTsateq = 0 and Zsat > 0.0 or if DISTsateq = 0 and Zsat < 0.0, or if Zsat = 0.0, or if Zsat =0.0 the satellite is over one of the earth's poles or the earth's equator and the geodetic subpoint is the same as the geocentric subpoint.
If DISTsateq = 0 0 and Zsat latitude (See "I.4 CONVERSION BETWEEN GEODETIC AND GEOCENTRIC LATITUDE",equation (I-36)).
where hsat, the height of the satellite above the ellipsoid, has been substituted for hf, the height of the feature of interest, the satellite's geocentric latitude, φsatgc, has been substituted for φgcf, the geocentric latitude of the feature, and the satellite's geodetic latitude, φsatgd, has been substituted for φgdf, the geodetic latitude of the feature. The tangent of the satellite’s geocentric latitude is given by
Use it in the above equation, and solve for φsatgd by the procedures given in section, I.4, B "Conversion from Geodetic Latitude", Case 2. The right ascension of the satellite is Θsat = arctan(Ysat/Xsat). Find the earth-centered-inertial position vector of the geodetic satellite subpoint, (xs,ys,zs), by using equations (I-31), (I-32) and (I-33) and replacing φgdf with φsatgd and θf with Θsat. The vector from the satellite to its geodetic subpoint will be
and its magnitude will be
The unit vector pointing from the satellite toward its geodetic subpoint will be, then,
Amended May 8, 2006
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