NOAA KLM User's Guide

Appendix I.3

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Appendix I.3: Defining the Satellite Subpoint


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 Unit vector P points in the opposite direction from the Satellite position vector. So, if the satellites's position vector in earth-centered-inertial coordinates is

Satellite position vector P sub sat composed of componants X sub sat, Y sub sat, and Z sub sat in the earth-centered-coordinate 
frame


then

The position vector P hat is composed of componants X sub p, Y sub p and Z sub p and is equal to the componants minus X sub sat 
divided by the absolute magnitude of the vector P sub sat, minus Y sub sat divided by the absolute magnitude of the vector P sub sat and  minus Z sub sat divided by the absolute magnitude of the vector P sub sat in the earth-centered-inertial coordinate frame


where The absolute value of the vector P sub sat is equal to the square root of X sub sat squared plus Y sub sat squared + Z sub sat 
squared .

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

Satellite position vector P sub sat composed of componants X sub sat, Y sub sat, and Z sub sat in the earth-centered-coordinate 
frame .

The magnitude of the component of this vector that is in the equatorial plane is

DIST sub sateq is equal to the square foot of X sub sat squared plus Y sub sat squared.

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)).

Tan of phi sub satgc equals

quantity r sub p squared plus h sub sat times r sub p times the square root of the quantity r sub e divided by r sub p squared 
times the cos squared of phi sub satgd + sin squared of phi sub satgd divided by the quantity quantity r sub e squared plus h sub sat times r sub p times the square root 
of the quantity r sub e divided by r sub p squared times the cos squared of phi sub satgd + sin squared of phi sub satgd times tan of phi sub satgd

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

tan of phi sub satgc equal to  z sub sat divided by DIST sub sateq

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

 The Vector P sub sat2gd is equal to the cpmponants x sub x minus X sub sat, y sub s minus Y sub sat and z sub s minus Z sat 
with reference to the earth-centered-inertial frame

and its magnitude will be

The absolure value of the vector P sub sat2gd is equal to the square root of x sub x - X sub sat squared plus y sub s minus 
Y sub sat squared plus z sub s - Z sub sat squared .

The unit vector pointing from the satellite toward its geodetic subpoint will be, then,

The vector P hat has componants X sub p, Y sub p and Z sub p with respect to the earth-centered-inertial frame and is 
equal to the quantity x sub s minus X sub sat divided by the absolute value of the vector P sub sat2gd, quantity y sub s minus Y sub sat divided by the absolute value 
of the vector P sub sat2gd and quantity z sub s minus Z sub sat divided by the absolute value of the vector P sub sat2gd

Cross section in the plane of 0 degrees E and 180 degrees E of the earth ellipsoid showing the geocentric and geodetic 
latitudes of a feature.

Amended May 8, 2006


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