We present here a calculation of the location of a star as a function of its declination, the observer’s latitude and time after it is at its highest position in the sky. We use the fact that stars trace out circular paths about a fixed point in the sky.

The image below shows a celestial sphere centred on an observer at the location . Assume the observer is in the northern hemisphere and the point represents the north celestial pole about which the stars appear to rotate anti-clockwise as the earth rotates on its axis (in the southern hemisphere stars rotate clockwise about the south celestial pole). The red ellipse illustrates the circular path of a star over a 24-hour period – point is its highest location while point is where it will be after a quarter of a day. Also shown in the diagram are axes (pointing east), (pointing north), (pointing directly overhead) and corresponding unit basis vectors , and .

Our aim is to find the location of the star (in 3-dimensional coordinates) on the red circle given:

- – the time (as a proportion of a day) after the star has reached its highest point
- – the angle of declination of the star (-90° to 90°) – the angle between the star and the celestial equator (which is the earth’s equator projected skyward)
- – the latitude of the observer (-90° to 90°)

The point is fixed as the earth rotates and in the north direction with an angle of elevation equal to the location’s latitude. Hence we may write

Next consider the angle between and . If , we would have a northern pole star and . If we would have a star on the celestial equator and . More generally,

The point has an angle of elevation of , hence we have

The circular path of the star can be regarded as a point traced around by a line having fixed angle from the fixed line . This circle has its centre at the point which is the projection of onto the segment . Note that does not lie on the sphere but , and do. We have

and

We parameterise any point on this circular path by expressing it in terms of the basis vectors and . For any point on the path we can write

For example,

- if ,
- if ,
- if ,
- if ,

Since has the same length as and is pointing directly west (i.e. perpendicular to both and ), we have from (5):

We now have all the ingredients we need to find :

In this formula, the angle of elevation is found by setting equal to the -component, or

Let us test out formula (8) in some special cases:

- If we recover the formula , the fixed location of the pole star.
- If we recover the formula . This traces a circle with diameter joining and (two antipodal points), hence it is a great circle on the celestial sphere.
- If (observer on the equator) we obtain , hence the -component remains fixed and a circular path is traced.
- If (observer at the north pole) we obtain , hence the star traces out a circle with constant angle of elevation.

We can also determine the times at which a star rises or sets by setting the -component of position in (8) to 0:

This is the so-called sunrise equation and was derived differently in an earlier blog post in [1].

Next let us look at the case when the star is our sun. The declination changes during the year between -23.4° and +23.4° (earth’s axial tilt) between the winter and summer soltice (for the northern hemisphere) as the earth revolves around the sun. Assume that the centre of the sun is at the origin and that earth’s orbit is the x-y plane having the form (assuming a circular orbit with radius starting at at ). The parameter here represents the fraction of year after the winter solstice. Earth’s axis of rotation points in the direction , being tilted away from the sun in the north when .

From the earth’s point of view the sun is in the direction . Then the angle between the sun’s position and the earth’s axis satisfies

For example if , we have . If , we have the equinoxes and . Finally if then .

Substituting (10) into (8) gives the location of the sun given the time of year and time relative to when it’s at its highest point in the sky (solar noon). Note that the equation is not perfect since it does not take into account atmospheric refraction or the fact that the earth’s orbit is not perfectly circular and therefore the earth is not uniform in speed. For more exercises on the length of days based on latitude and the sun’s angle of declination refer to [2].

Finally, we can determine the bearing of a star when it rises or sets in the following manner. The position of the sun is in the x-y plane and so has the form

Secondly and hence

Hence . For example taking the sun on the summer solstice at Melbourne, Australia we have , and so . This corresponds to a bearing of , around 20 degrees south of east. The answer given in timeanddate.com is , slightly more perhaps because it takes into account refraction and the fact that the sun is not pointlike.

### Reference

[1] *The Shortest Day of the Year*, Chaitanya’s Random Pages – https://ckrao.wordpress.com/2012/06/23/the-shortest-day-of-the-year/

[2] Alan Champneys, *The Length of Days* – https://www.bristol.ac.uk/media-library/sites/engineering/engineering-mathematics/documents/modelling/teacher/daylength_t.pdf