# Re: 2.2' of arc discrepancy between CalSKY and SkyMap ISS tracks

From: Thomas Fly (tfly@alumni.caltech.edu)
Date: Mon Jun 28 2004 - 01:28:54 EDT

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```As always, the Devil's in the details, but conceptually at least, computing the
ISS / Venus ground track only involves 3 "simple" steps:

1) compute the precise apparent (i.e., light-delayed) position of Venus;
2) compute the precise location of the ISS; and
3) compute the intersection of the line connecting the two with the Earth's
surface.

Computing the sky track of the ISS relative to Venus likewise boils down to
those 3 steps.

I re-ran my WorldView program, using Tomás' location (to within about 10
meters), which produced the following result (Tomás would not have received the
information relating to Venus, because I normally report only sunlit planetary
encounters, when the Sun is at least 8° below the horizon):

name: ISS Venus
latitude: 48.2579 N
longitude: 17.0272 E
elevation: 208 m
time zone: using UTC

A - travel distance (kilometers) and direction
B - date
C - time
D - elevation angle of the ISS
E - azimuth angle of the ISS ( + is East from North; - is W from N)
F - range (kilometers)
G - latitude for observing the transit
H - longitude
I - how far (kilometers) can I be from the centerline?

For other than solar transits:
J - lunar transits: is space station sunlit?
planetary encounters: 1=Mercury; 2=Venus; 4=Mars; 5=Jupiter; 6=Saturn
K - sun elevation angle
L - sun/moon or sun/planet separation angle

A------- B----- C-----  D--- E----- F--- G------- H-------- I---- J K---- L----
8.7 SW  8 Jun 100916  63.3  157.8  410  48.2109   16.9334   2.1
1.9 S   8 Jun 100917  63.3  158.0  410  48.2413   17.0222   2.1
1.6 SE            17.16                 48.2461   17.0361   2.1
6.4 E   8 Jun 100918  63.3  158.3  410  48.2717   17.1111   2.1

7.4 SW  8 Jun 100916  63.2  158.2  412  48.2256   16.9392   2.1 2  63.5   0.2
0.3 SE  8 Jun 100917  63.2  158.4  412  48.2560   17.0281   2.1 2  63.5   0.2
7.4 NE  8 Jun 100918  63.1  158.6  412  48.2863   17.1170   2.1 2  63.5   0.2

I subtracted 100° of longitude from these locations (moving Tomás and his family
to Canada), and plotted the tracks using Microsoft Streets & Trips 2001.  This
put him essentially 218 meters NNW of the 10:09:17 UTC point (my "travel
distance" computation is slightly off, apparently), meaning that my MCC track
was about 153 meters SSE of the actual track -> or 1.18 Venus diameters ->
68.3" -> 1.14' of arc.

One thing that's a bit worrisome is the computation of the ECI position of the
observer at a given instant.  The idea behind deltaT is that the earth's
rotational period is not constant, and "leap seconds" have been added to keep
UTC within a second of UT.  In the computation of the ECI position of the
observer, should one use UT rather than UTC then?  At 45° of latitude, the earth
rotates about 328 meters in one second, so a fraction of a second worth of the
earth's rotation is not an insignificant error!

I'll post some intermediate results from my computations sometime this week.
Last year, I'd posted some results of SGP4 computations:
http://satobs.org/seesat/Apr-2003/0259.html

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