NavList:

Samuel,

In the "Sight Reduction Procedures" section of the Nautical Almanac, formulae are given for finding a DR position by calculation:

DR Longitude = Lf+t(V/60)sinT/cosBf

DR Latitude = Bf+t(V/60)cosT

...where

• Lf = Longitude of the last fix
• Bf = Latitude of the last fix
• t = Time elapsed
• V = Speed
• T = Course

This assumes a spherical Earth and is only good for relatively short distances, e.g. calculating the DR of a ship from day to day.

Another favorite of mine is the use of traverse tables. While similarly not-so-precise, they are good enough to find a DR from which to pick an AP to use with a celestial observation. I've even memorised the extremely short "Simplified Traverse Table" from the "Emergency Navigation" section in Bowditch. This table can be used with nothing more than pencil and paper to find a reasonably good DR:

Angle: 0___18___31___41___49___56___63___69___75___81___87___90

Factor: _1___.9____.8___.7___.6____.5___.4____.3___.2___.1____0

To use this table: First, reckon the course angle from North or South to East or West. (So a true course of 135° would be [from S] 45° [to E]. A true course of 300° would be [from N] 60° [to W].) Next, enter the table with the course angle and multiply the distance run (in NM) by the factor obtained. The result will be the difference in latitude in NM. Simply add or subtract this to the latitude of the starting position. Then, enter the table with the compliment of the course angle and multiply the distance run (again, in NM) by the factor obtained. The result will be the difference in longitude in NM. Next, enter the table a third time with the mid-latitude and divide the difference in longitude in NM by the factor obtained. The result will be the difference in longitude in minutes. Finally, add or subtract this to the longitude of the starting position.

Sounds complicated, I know. But with practice it becomes quite easy. Here's an example:

Let's say we're starting from the mouth of the Chesapeake Bay (37°N, 76°W) and travelling at 7kts. on a course of 120°. What would our position be after 12 hours?

Solution:

Our distance run will be: 7kts. x 12hrs. = 84NM; and our course angle will be [from S] 60° [to E]. We enter the table with 60° and get a factor of 0.5; 84NM x 0.5 = 42NM = 42' of latitude. Since we're traveling SE, we subtract this from our north latitude: 37°N - 0°42' = 36°18'N. Next, we enter the table with the compliment of our course angle (90° - 60° = 30°) and get a factor of 0.9; 84NM x 0.9 = 75.6NM. Then, we enter the table with our mid-latitude (we'll use 37°) and get a factor of 0.8; 75.6NM / 0.8 = 94.5' (1°34.5'). Since we're traveling SE, we subtract this from our west longitude: 76°W - 1°34.5' = 74°25.5'W. So, our DR will be 36°18.0'N, 74°25.5'W. A quick check with the N.A. formulae yields: 37°+12(7/60)cos(120) = 36°18.0', -76°+12(7/60)sin(120)/cos(37) = -74°28.9'; a difference of 3.4' of longitude and no difference in latitude. Not bad for a pencil and paper calculation, IMHO.

Regards,

Sean C.