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* Working through a past example
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Uranus ring data extraction and manipulation

Task: Extract three cuts of the Epsilon ring, and compute the
eccentricity of the outer edge of the ring using radiuses and
longitudes. (roughly)

Steps:

1) Went to DOCUMENT:TUTORIAL.TXT, to see overview of data.
 From 2.1 "Derived Ring Profiles", we see that separate profiles
are given for Uranus in EASYDATA.

2) Went to EASYDATA/DATAINFO.TXT to get information for the naming
convention of the files. The smallest resolution is located in
/KM000_1: so used: EASYDATA/KM000_1/PU1P01EI.TAB (SSI) and
PU1P01EE.TAB (SSE). None of the same resolution exists for beta Per
Ingress, however, so extracted the data for the next available data
in the EASYDATA directory. Use: EASYDATA/KM0001/PU2P01EI.TAB (BPI).

3) Opened files, and plotted the data (I used Mac Kaleidagraph).

4) Extracted the three radiuses at the outer edge. (Mac Kaleidagraph
provides a 'microscope' to zoom into x  and y values on curves. I
zoomed into the midpoint of the outer edge of the ring and
determined the following values.)

r_1 (=SSI)  = 50885.7 km
r_2 (=SSE)  = 50758.9 km
r_3 (=BPI)  = 51006.5 km


5) Extracted the corresponding three longitudes at the outer edge.

Went to GEOMETRY/GEOMINFO.TXT to get the information for the naming
convention of the files. GEOMETRY/PU1G01EI.TAB (SSI) and
PU1G01EE.TAB (SSE), and PU2G01EE.TAB (BPI) provide data for
longitudes in 2000 and 1950 coordinates.

Plotted each and blew up the region around r_{1,2,3} above. Used the
Kaleidagraph microscope tool to extract the longitude (this tool
implicitly interpolates) for the longitudes at the ranges of each of
the Epsilon ring cuts.

theta_1 (=SSI) = 266.5 deg (2000)
theta_2 (=SSE) = 305.5 deg (2000)
theta_3 (=BPI) = 21.1 deg (2000)


5)  Compute longitude of periapse: longitude_0
(3 equations, 3 unknowns)
r_1 = a(1-e^2) / (1+e COS (theta_1 - theta_0) )
r_2 = a(1-e^2) / (1+e COS (theta_2 - theta_0) )
r_3 = a(1-e^2) / (1+e COS (theta_3 - theta_0) )

Solving for longitude_0

theta_1 = theta_1 * !pi / 180.
theta_2 = theta_2 * !pi / 180.
theta_3 = theta_3 * !pi / 180.

numerator = COS(theta_2)*(r_2 r_3 - r_1 r_2) + COS(theta_1)*(r_1 r_2 - r_1 r_3)
+ COS(theta_3)*(r_1 r_3 - r_2 r_3)

denominator = SIN(theta_2)*(r_1 r_2 - r_2 r_3) + SIN(theta_1)*(r_1 
r_3 - r_1 r_2)
+ SIN(theta_3)*(r_2 r_3 - r_1 r_3)

long_0 = ATAN[ numerator / denominator ]  = -51.1 deg
(Uranus Notebook #1,pg. 26 = -47.2179 deg)

5)  Compute eccentricity at the outer edge.

e = (r_1 - r_2) / (r2 COS(theta_2 - theta_0) - r1 COS(theta_1 - theta_0) )
e = 9.7 E-03

(Uranus Notebook #1,pg. 26 = 8.24622E-03,
Elliot&Nicholson, _Planetary Rings_ = 7.94E-03)

So then, same magnitude as 'right answer', but off, due to only
working with a few significant digits, and lower resolution of the
cuts used here, especially BPI.