# Celestial navigation explained - page 2

## Celestial mechanics - a blueprint

 Let's put an observer on the Earth. The point Z on the celestial sphere, which is directly above the head of the observer, is called his zenith. The distance on the celestial sphere between the zenith Z and the star S is the zenith distance z. The zenith distance is like the distance d from the lighthouse of the previous example. We can draw a circle centred on the star S with a radius z. We are somewhere on the projection of this circle on the Earth. This is our new circle of position. The position of the projection of the star on the Earth (latitude, longitude) and the position of the star S itself on the celestial sphere are identical (same angles, same reference plans: the celestial equator and the Greenwich meridian).
 The position of the star on the celestial sphere is given by its declination delta (90°N to 90°S) and its Greenwich Hour Angle (GHA, 0° to 360°). We can find both values for any given time in the Nautical Almanac. Knowing delta & GHA and measuring the zenith distance z, we can find our circle of position and finally our position. Let’s look more closely now: Take the previous figure and wipe-off the surplus. Note the latitude and co-latitude (90° minus lat), the declination delta and polar distance Delta (90° minus delta). The figure is using the equatorial coordinate system (the reference is the celestial equator). However, what we need is a reference to our local coordinate system: the horizontal coordinate system (the reference is the local apparent horizon). At our position (Observer) on the Earth, we can imagine the plan of our horizon with the four cardinal points. If we move this plan to the centre of the Earth and redraw the figure using this plan as reference, we get the next drawing. 