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12. In the following table give the correct magnetic bearing of the distant object, and thence the deviation :

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With the deviation as above, give the courses you would steer by the Standard Compass to make the following courses, correct magnetic.

Correct magnetic courses, E.S.E., S.W., N. 80° E.,

W.S.W.

Compass courses,

Supposing you have steered the following courses by the Standard Compass, find the correct magnetic courses made, from the above deviation table.

Compass courses, E. by S., West, S. 30° W., N. 10° E. Magnetic courses,

You have taken the following bearings of two distant objects by your Standard Compass as above; with the Ship's head at S.E., find the bearings, correct magnetic.

Compass bearings, N. E., West, S. 45° E.
Bearings magnetic,

G

SUMNER'S METHOD OF FINDING THE
LATITUDE AND LONGITUDE.

FOR ONLY MATES, FIRST MATES, AND MASTERS.
RULES FOR SUMNER'S METHOD.*

1. With the first Greenwich Mean Time find the Sun's declination and polar distance also the Equation of time.

2. Then correct the first observed Altitude in the usual manner and so obtain the True Altitude.

3. With this True Altitude, the lesser assumed Latitude, and the polar distance, proceed as in "Longitude by Chronometer" to find the hour angle, and from thence deduce a Longitude.

4. Then with the same True Altitude, the greater assumed Latitude, and the same polar distance, find another hour angle and another Longitude.

5. On the Chart lay off the first of these Longitudes on the parallel of the lesser Latitude, and for easy reference call this point A.

6. Then lay off the second Longitude on the parallel of the greater Latitude, and name this point B.

7. Join the points A and B. This gives the "Line of Position," the direction of which is required. To find the direction, place the edge of the parallel rulers on the line A B, and move them to the centre of the compass, the direction is then seen.

* Charts for this problem, price 6d. each, to be had of J. Newton, Sailors' Home, London, E., by Post 7d.

8. From any point in the "Line of Position" lay off the course and distance the Ship has made in the interval of time between the two observations, and through the end of this distance line draw another line with the parallel rulers parallel to the line A B. Name this line you have just drawn C D. 9. At the present time there is no necessity to work out the two longitudes from the second observation as they will be given to the Candidate at the examination. Lay off these two Longitudes on their respective parallels of Latitude and join the points obtained by a straight line. Where this line cuts CD is the Ship's position when the second Altitude was taken.

10.

II.

The Latitude and Longitude of the Ship's position are found in the usual way on the graduated meridian and the graduated parallel, and when found are to be written down.

To find the Sun's bearing at the time of taking the first altitude, draw a line from the Ship's position at right angles or eight points from the Northerly direction of the "Line of Position,"-to the Eastward if the first Altitude was taken A.M., but to the Westward if P.M.

EXAMPLE.

If at sea on November 15th, A.M., 1885, and uncertain of my position when the Chronometer showed 14d 20h 44m 20s G.M.T., the observed Altitude of the Sun's L.L, was 8° 18' 40" and again P.M. the same day, when the Chronometer showed 15d 2h 24m 40s G.M.T., the observed Altitude of the Sun's L.L. was 16° 19' 10" the Ship having made 27 miles on a true S.E. E course in the interval, height of eye 12 ft. Required the line of position when the first Altitude was taken, also the bearing of the Sun by projection, and the Latitude and Longitude of the Ship when the second Altitude was observed. The Ship being supposed to be between the parallels of 48° 12' N. and 48° 42' N.

Note-The results obtained by working out this question are laid down on the accompanying Chart and should be carefully studied by the Candidate.

SOLUTION OF FOREGOING EXAMPLE.

(FIRST OBSERVATION ONLY.)

Greenwich Mean time 14d. 20h. 44m. 20s., or 3h. 15m. 40s. from noon 15th

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14

App. Time at Ship....141
...Eq. Time

........

....

20 14 30.. Mean Time at Ship.... 14-20 17 7 =20 44 20.. Mean time at Greenwich 14-20 44 20

14 2

4)29 50

Long.......7° 27′ 30′′ W.

4)27 13

Long....6° 48′ 15′′ W.

(see chart opPOSITE.)

These two Longitudes are laid off on the Chart in their respective Latitudes, and the line A B is obtained. Then from any point in the line A B lay off the run S.E.‡E. 27 miles as directed in No. 8 Rule, and the line Č D is obtained. By taking the second Greenwich Mean Time and the second altitude from the question and working them out, using the same two Latitudes as above, we can find the other two Longitudes to be 5° 42′ W. and 6° 51' W. It is at present not necessary to work out

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