Calculating great circle course and distance using HO 249

Calculating a great circle route can either be done graphically using charts specifically designed for the task, by use of a calculator, or the sight reduction tables. The use of the sight reduction tables, to my mind, is the most interesting method. By using either HO 249 or HO 229 we can calculate the great circle distance and true course from a departure point to our destination. Since this method requires that we use the destination latitude as our declination we are better off using HO 229 where the declinations go beyond the 30° in HO 249. For the sake of this example though, we will be using HO 249 and will calculate a great circle true course and distance from New York to Dakar in Senegal. Follow me along and see just how this procedure works. Basically the departure position is considered to be the assumed position and the destination the geographical position. Instead of solving for our distance from the celestial geographical position we are solving for our distance and direction to a physical destination.

Position of departure in New York is 42° 30’ N by 72° 30’ W. The position of our destination in Dakar is 14° 34’ N by 17° 29’ W.

Enter departure latitude as latitude in HO 249.
Latitude of destination is entered as declination.
To find LHA subtract destination latitude from departure latitude.
Extract Hc, d, and Z from tables. Factor in d correction to minutes of declination.
Subtract the Hc from 90° and convert to nm.
Z will give the true course to initially follow.

There are rules for finding the true course. They are as follows:
In the northern hemisphere going east Z = great circle course. If proceeding in a westerly direction the great circle course is = 360°-Z.

In the southern hemisphere, if proceeding east then add 180° to Z to get great circle course. If proceeding in a westerly direction subtract Z from 180° to get great circle course. It may sound complicated, but if you reflect on the direction that you are going you will be able to follow the directions without too much trouble. Let’s do the problem:

1. Departure latitude 42° 30’ N, enter that in HO 249 as latitude.
2. Latitude of destination 14° 34’ W is declination.
3. Difference in departure and destination latitudes is LHA.
42° 30’
-14° 34’
LHA= 27° 56’ or 28°
4. Enter HO 249 Vol 3 latitude page 42° declination 0-14° Same name as Latitude (page 20).
5. Under column of 14° declination and LHA 28° Hc 53.00 d +48 Z 131°.
6. d correction in Table 5 for 34’ of declination (+48) is +27’.
7. Subtract HC from 90° to get ZD.
8. 90°.00’
ZD 36°33’ converted to nautical miles = great circle distance of 2,193 nm on an initial course of 131°.

Since the great circle course changes during a passage, the navigator would take regular sights in order to get a fix and once the fix is recorded it can be used as the new departure point in the exercise using HO 249. The destination, of course will remain the same.

By Ocean Navigator