Nav Problem Solution: From ‘The Saga of Cimba’ (Page 64)

Cimba Solution

Richard Maury, like many navigators, relied on noon sights to fix latitude. The advantage of the noon sight is that with a little effort, the navigator can find latitude, which, combined with a DR, can give a pretty good idea of position. Nowadays, of course, with dedicated celestial navigation computers, it is relatively simple to do more mathematically challenging planet, moon and star sights without too much trouble.

We are precalculating the time of Local Apparent Noon. If you have a celestial computer, all that is required is to enter the DR latitude and longitude, and the time of LAN is given in GMT. You can�t necessarily rely that the time given will be the exact moment when the sun will cross your meridian, for your DR may either be east or west of your exact position. This will become obvious when you take the noon sight and keep track of the time. If the sun crosses your meridian (Hs is at its highest), prior to the time of your calculations, then you are east of the DR. Likewise, if the greatest Hs is after your calculated time, then you are west of the DR.

Problem:

Based on the DR of 35 degrees N by 64 degrees 25 minutes W on Dec. 12, we go to the daily sun pages in the Nautical Almanac and find that the time for Meridian Passage is at 11:54 LMT. Next, we go to the Arc to Time conversion tables at the rear of the Nautical Almanac and convert our longitude into time. 64 degrees of longitude converts to 4 hours 16 minutes of time, and 25 minutes of longitude converts to 1 minute 40 seconds of time. We add these numbers to the time of the Meridian Passage and get the time for Meridian Passage in GMT at our DR:

11:54:00 (Meridian Passage)

+04:16:00 (65 degrees longitude)

+00:01:40 (25 minutes longitude)

16:11:40 Time of Meridian Passage in GMT at DR

The next thing we have to do is reduce the Hs to Ho. This is a lower-limb shot of the sun.

Hs 31 degrees 37.4 minutes

-dip 3.1 minutes

+IC 3.0 minutes

Ha 31 degrees 37.3 minutes

+3rd cor 14.7 minutes

Ho 31 degrees 52.0 minutes

Once we have the Ho we can use the formula Latitude = 90 degrees – Ho = ZD +/- dec. The declination we find in the daily pages under the time of the sight of 16 hours 11 minutes 40 seconds. The declination at 1600 hours is listed as S 23 degrees 08.0 minutes with a d correction of +0.2 minutes. The correction for 11 minutes 40 seconds of time is 0.0 minutes. Our declination of the sun at the time of the shot is then S 23 degrees 08.0 minutes.

So we have:

Latitude = 90 degrees – 31 degrees 52 minutes = 58 degrees 08 minutes – 23 degrees 08 minutes = 35 degrees latitude.