Text-book on Navigation and Nautical Astronomy

ART. 24. Middle-latitude Sailing. This is a combination of plane and parallel sailing, the dep. being common to both, but the latitude used is the "middle lat.," or mean of the two latitudes, for in this case the ship does not sail on a parallel of latitude.

It is evident that the formulæ for rightangled triangles will solve every case, but this method is only made use of in modern navigation for the purpose of finding the D. long. in the "day's work," the formula employed being

D. long. dep. x sec of middle lat.

The mean is not the true" middle lat.," being too small, but is sufficiently near for short distances.

Examples.

Dist

Dep
M.Lat

D.Long

FIG. 73.

w D. Lat

1. In lat. 46° 45' the Departure made good was 110.5 miles required the Difference of Longitude.

Construction. In Fig. 74, scale

inch = 10 miles.

Draw horizontal line AB = 1·1 inches for departure,

make angle BAC = 46° 45', and draw BC perpendicular to AB: then AC represents D. long.

Measure AC = 1.6 inches or 160 miles.

2. In lat. 53° 14′ the D. long. being 3° 16', required the Meridian Distance or Departure.

In Fig. 75 draw horizontal line AB. Make angle BAC = 53° 14', and AC = 0.98 inch; scale inch = 20 miles.

From

C draw CB perpendicular to AB: then AB represents Dep.
Measure AB : = 0.6 inch or 120 miles.

3. On what parallel of latitude would a Meridian Dis. tance of 245 miles give a difference of longitude of 5° 54' ? In Fig. 76, scale inch 20 miles, draw AB = 1.2 long, at B draw a perpendicular to AB, and with A as centre, radius = 1.77 inch, draw an arc, cutting this perpendicular in C: then BAC represents lat.

Measure BAC = 46°.

D. long. 354
Dep. 245

FIG. 75.

Remark. In the first two cases 10 is rejected from the sum of indexes, and in the third case 10 is added to the upper index, because in the Table of Log Sines, etc., the proper index is always increased by 10.

EXERCISES.

1. In lat. 16° 30' N. the dep. made good was 167.6 miles: what will the difference of longitude be by parallel sailing?

2. The difference of longitude between two places in lat. 63° 19′ 30′′ is 4° 20' required the distance as measured along the parallel.

3. In lat. 11° 56' N. the dep. made good was 365 miles: required the difference of longitude in minutes.

4. In what latitude is 200 miles of dep. equal to 600 minutes of longitude?

5. In lat. 35° 40′ S. the dep. made good was 976: required the difference of longitude measured on the parallel.

6. What is the difference of latitude between two parallels of latitude on different sides of the Equator, when between the same two meridians the dep. on the northern parallel is 250 miles, on the southern 350 miles, and on the Equator 470 miles ?

7. On the parallel of 48° 12′ ran true west 1005 miles: if the longitude left was 9° 30′ W., what was the longitude in?

8. From two ports in lat. 40° 20′ Ñ., distant 300 minutes of longitude, two ships sail N. for 300 miles: how many miles apart will they be, and in what latitude?

9. In lat. 33° 15' N. ran east 10.95 miles: how many minutes of longitude will that represent?

10. If two ships in lat. 51° S. should sail directly north until the departure changes from 400 to 500 minutes, what latitude will they be in?

11. In lat. 72° 59' the dep. was 370-5 miles: convert that into minutes of longitude.

12. In what latitude does 300 miles of dep. correspond to 600 minutes of longitude?

13. The latitude by Mer. alt. was 36° 35′ N.; longitude by afternoon sights, 137° 38′ E.; run since noon due east, 39 miles: find noon Jongitude.

14: A vessel sailed due east along the parallel of 45° 24′ S. for 29 miles from long. 43° 40′ E.: what is her longitude in?

15. What distance must a vessel sail on the parallel 48° 25′ N. to change her longitude 2° 25' ?

16. A vessel sailing due east finds her difference of longitude to be 1o 19', and distance made good 50': what is her latitude?

17. How far on the parallel of 38° S. must a vessel sail from long. 47° 29′ E. to reach long. 49° 27' E.?

18. For every 1° diff. long. I have to sail 45 miles due west: along what parallel am I sailing?

ART. 25. Plane and Traverse Sailings.-For short distances, the surface of the Earth may be regarded as a plane. In plane

Dist.

Course

Dep.

FIG. 77.

D.Lat.

and traverse sailings no notice is taken of the curvature, and all cases may be solved by the formulæ for right-angled triangles.

Thus in sailing from A to B, the hypotenuse represents the distance, the perpendicular the departure, and the base the D. lat., whilst the angle which AB makes with AC (lying true N. and S.) is the course.

In the Traverse Table the values of D. lat. and De. in miles are given for every degree of course

up to 89° and all distances up to 600 miles, being computed from the formula

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When D. lat. and dep. are given, the course and distance made good may be found from the formula

ART. 25. The Day's Work and Traverse Sailing. The purpose of the Day's Work is to find the position of the ship at noon by 'Dead reckoning," that is, from the courses and distances sailed since the preceding noon. The direct course and distance made

good are also determined.

The courses given are by compass as per ship's Log Book, but the true courses must be found for computing a "Traverse Table."

Directions for finding the True Courses.-(a) Allow the leeway to the Right (R.) when the wind blows on the left or Port Šide of the ship, and to the Left (L.) when on the Right or Starboard Side. (b) Allow E. deviation and variation (or total correction) to the Right (R.), and W. to the Left (L.).

NOTE.-The Leeway may be determined by towing the log-ship astern and observing the angle the line makes with the keel, but it is often merely estimated.

The Deviation is found by reference to the Deviation Card, and the Variation from a chart.

When the Departure is taken from a known point of land or light, the opposite course to the bearing of the point or light from the ship is called the "Departure Course," and the distance off is the distance belonging to that course.

The "Set" and "Drift " of the current are found by chart, and are treated as a course and distance made by the ship, because the effect is the same on the Day's Work.

NOTE.-Should the "Set" and "Drift" of the current be required, it is the course and distance calculated by Mercator from the D.R. position found by Traverse to the observation position given in the question.

They may also be found by turning D. long. into Dep.: then with the D. lat. and Dep. find Co. and Dist. in Traverse Table, for Set and Drift required.