# Measuring the Distance to the Sun: Mathematical Details

To determine the point of intersection of the two lines of view we take use of vector calculus. We transform all of the spherical vectors in rectangular vectors with respect to the coordinate system the x-axis and z-axis of which are pointing toward the vernal equinox and the celestial north pole, respectively. The y-axis is oriented so that the system forms a "right hand system". The direction of the y-axis is therefore (alpha=6h, delta=0deg).
1. 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!)

2. In order to calculate the intersection of the two lines of view we need the equatorial positions of the two observers:
1. 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".)

2. 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.
3. There are two possibilities to determine local sideral time:
1. 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 positive).

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.

2. Today you can easily get a computer algorithm for calculating local sideral time (see our program, for instance).
3. 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).

4. Now we are able to determine the point of intersection of the two lines of view:

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!

5. 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), respectively

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:

6. The geocentric distance dp of the asteroid can now be calculated easily: