- The measured topocentric positions (alphai, deltai) give us the directions
of the lines of view. We transform them in normalized rectangular vectors
(see the picture in the main paper):
(You can easily prove that these vectors are of unit length!)
- In order to calculate the intersection of the two lines of view
we need the equatorial positions of the two observers:
- The picture below shows an observer and the vernal equinox just laying in
the same plane.
This picture shows you: The declination of an observer is equal to his
geographical latitude. (For good results we will have to take into account
that the earth's body is not a perfect sphere but an ellipsoid with
excentricity e = 0.08182 (see, for instance O. Montenbruck et al: Astronomy
on the personal computer, Springer, New York 1994). Our program does it in its procedure "observersCoord".)
- At the moment of the above picture the right ascension of the observer is
obviously alpha=0h. The vernal equinox is just culminating in the observer's
horizontal system. Therefore, his local sideral time is 0h. Due to the
earth's daily revolution the right ascension of the observer will increase by
a rate of 24h in 23h56min (the earth's period of rotation) with exactly the
same rate as his local sideral time! That means: The right ascension of
an observer is equal to his local sideral time.
- There are two possibilities to determine local sideral time:
- From an astronomical almanach you can get the sideral time st0(Gr)
of the day for 0h UT at Greenwich longitude. At this time the local sideral
time st0(Obs) of an observer is
(l = geographical longitude of the observer, east of Greenwich taken as
At the time t the observer's local sideral time is
The factor in this equation is due to the fact that sideral time increases by
24h in 23h56min approximately.
- Today you can easily get a computer algorithm for calculating local
sideral time (see our program, for instance).
- The sperical coordinates (alphaOi, deltaOi) of the observers can be
transformed in rectangular coordinates as the asteroid's coordinates:
(rhoi is approximately the earth's equatorial radius but, more exactly, the
geocentric distance of the observer expressed as a multiple of the
earth's equatorial radius).
- Now we are able to determine the point of intersection of the two lines
These are three equations with only two unknown quantities. Unfortunately, due
to measuring errors the two lines will not intersect usually and the above
system will not have any solution!
- For this reason, we are forced to calculate the closest distance of the
respective lines instead of their point of intersection. That means we are
looking for two Points P1 and P2 (or two vectors r1 and r2),
so that the vector connecting these points is perpendicular to the respective
directions of view:
This is a system of two linear equations with two unknown quantities.
Therefore, the problem has a unique solution.
Try to find the following solution by yourself:
Now it is easy to calculate lambda and mu:
- The geocentric distance dp of the asteroid can now be calculated