Amplitude Day Ex Meridian Exercises Pole Star
Naut. Almanac    


Celestial Navigation



Ex Meridian


First Point of Aries

Nautical Astronomy is an exact science, if approximation were to be used then the position of a ship would be grossly in error and may end up with the ship in disastrous situations.

Since the axis of the Earth is tilted to about 23.5˚ the sun is assumed to travel around on the ecliptic. This ecliptic cuts the celestial equator at a point called the ‘First Point of Aries’. Since this is as constant as can be obtained the reference point for most of nautical astronomy is this point.

All the stars are referenced to this point, the Hour Angle West of the ‘First Point of Aries’ is called the Sidereal Hour Angle (SHA) and this is tabulated for each star in the Nautical Almanac.

The Greenwich Hour Angle of a star is thus found out by obtaining the GHA (Aries) from the Almanac and then adding the SHA of that particular star.

So instead of having the GHA of each star tabulated for each hour, it is convenient to use a constant like the GHA (Aries), which is tabulated for every hour, and then adding the SHA to obtain the GHA (star).

Ex Meridian:

If the sun is not available during the time of her passage across the meridian – meridian passage then the following observations and calculations are done to obtain the latitude.

This method is only applicable to bodies close to the meridian and the HA of the body should not exceed one hour. The thumb rule being that the HA in minutes should not exceed the number of degrees in the meridian zenith distance nor should the meridian zenith distance be less than 25° except under most favourable conditions.

To start off the altitude of the body is taken together with the GMT, this to obtain the HATS which should be accurate.


In the above we have the observer at Lat. 43° and the body (sun) with a declination of 22°. The sun is approaching the meridian – it is better to take an ex. Meridian before the MP when conditions are not favourable rather than to take the Ex. Meridian after the MP.

WQE is the equinoctial

WZE is the prime vertical

NZS is the observer’s meridian

dMXd’ the parallel of declination

P the pole, Z the zenith, X the sun at the time of observation

M the position of the sun when it is on the meridian

Angle ZPX is the East HA of the Sun (X)

PZ the co-latitude

ZQ the latitude

PX the polar distance (90°+ decl.)

ZX the zenith distance of the body (Sun)

ZM the zenith distance when the sun is on the meridian that is it is the MZD (Meridian Zenith Distance)

ZQ the latitude = (ZM – QM), that is (MZD – decl.)

We have to find ZM the meridian zenith distance.

The least distance of a body occurs when the body is on the observer’s meridian that is at MP, so the zenith distance at any other time must be greater than the MZD and therefore in the above example ZX is greater than ZM.

Thus the calculation of finding the arc ZM is called ‘Reduction to the Meridian’.

The reduction referred above is a very small quantity, which is the difference between the MZD and the observed zenith distance: arc ZX – arc ZM

But the body is so close to the meridian that PX is practically equal to PM and if

(PM – PZ) is equal to ZM, then (PX – PZ) is assumed to be assumed to be also equal to ZM, the meridian zenith distance which we are out to find in triangle PZX.

Now a DR latitude has to be introduced into the above calculation so that the latitude found out may be approximate to the actual latitude.

If the calculated latitude should differ greatly from the DR latitude the triangle has to be reworked, using the new latitude so as to obtain a nearer approximation.

Many Ex _ Meridian tables are published in the nautical Tables and the ‘Reduction’ may be found and directly applied to the observed zenith distance to get MZD.

Using Nories Tables

Let us work with an LHA (sun): 352°44.2’

            Declination: 21°49.5’S

            Latitude: 42°18.0’ N

Enter Table I with:

Lat. 42°18.0’ N, decl. 21°49.5’S (different names) and get 1.5” (A)

Now enter Table II

LHA 352°44.2’ and (A) as 1” and get  14.05(*1)

LHA 352°44.2’ and (A) as 0.5” and get           7.03 (*2) (here read the value for 5” and then shift the decimal point to get for 0.5”)

Therefore adding for a total of 1.5” would give a result of 21.08, and this is the ‘Reduction’ to be subtracted from the observed zenith distance.

Obs. Zenith Dist.          64°32.00

Reduction                     21.08

Therefore MZD            64°10.92 N

Declination                   21°49.5 S

Latitude at sight time = 42°21.4’ N