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(1) Multiply 92 by 43, by common logs.
(2) Divide 5900 by 347, by common logs.
(3)

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1 S. S.W. 3 6 W.

14. A point of land in 2 3 6

lat. 29° 2' S. 3 3 4

long. 168° 2' E. 4 3 9

bearing by compass 5 S.E. by E. 4 2 N. E. by E. 13 W. 6 4 4

Dist. 15 miles. 7

4 0 8

4 4 9 S. by E. 4 1 E. by S. 14 10

3 7

3 6 12

3 6 1 N.W. by N. 3 4 N.E. by N. 13 Variation 14 pts. E. 2

3 2 3

3 0 4

3 0 5 N. by W. 2 7 W. by N. 18 A current set the 6 2 9

ship by compass 7 2 8

S. by W. W. 8

2 6 9 E.by N.N. 2 9 W. by N. 0 Dist. 10 miles from 10 3 1

the time the depar11 3 7

ture was taken to 12 4 3

the end of the day. (4) 1870, September 21st, in long. 111° 10' E., the observed

meridian altitude of the Sun's 1.1. was 57° 10' 10", bearing S., index error 1' 20', height of the eye 20 feet. Find the latitude,

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(5) In latitude 71° 34', the departure made good was 36 miles.

Required the d.long. by parallel sailing. (6) Required the course and distance from A to B, by

calculation on Mercator's principle.

A-Lat. 10° 17' S. A-Long. 3° 5' W.
B-Lat. 1 15 N. B—Long. 9 1 E.

ADDITIONAL FOR ONLY MATE.

(7) 1870, May 1st. Find the A. M. and P. M. tides at Greenock,

Leith, and Aberdeen. (8) 1870, December 31st, at 4h. 40m. A. M. apparent time at

ship, lat. 38° 50'S., long. 145o E., the Sun's magnetic amplitude was S.E. by E. { E. Find the true amplitude

and variation. (9) 1870, January 22nd, A. M. at ship, in lat. 30° 14' N., the

observed altitude of the Sun's 1. 1. was 21° 31' 45”, index error +3' 15", height of the eye 12 feet, time by chronometer 22d. lh. 56m. 49s., which was fast for mean noon at Greenwich 9m. 378., on October 17th, 1869, and losing daily 1s5. Find the longitude by chronometer.

ADDITIONAL FOR FIRST MATE.

(10) 1870, April 1st, at 8h. 10m. A. M. mean time at ship, in

lat. 28° N., long. 120° W., the observed altitude of the Sun's 1. 1. was 30° 33' 30", height of the eye 16 feet,

bearing by compass S. 96° 54' E. Find the variation. (11) 1870, May 30th, A. M. at ship, in latitude by account

48° 17' N., long. 168o W., the observed altitude of the Sun's l. l., South of the observer, was 63° 4' 18", height of the eye 14 feet, time by watch 3h. 40m., which had been found to be 4h. 9m. 31s. fast on apparent time at ship, the difference of longitude made to the East was 20 miles since the error was determined. Find the latitude by reduction to the meridian.

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ADDITIONAL FOR MASTER.

(12) 1870, March 29th, the observed altitude of the star

Antares, was 66° 16' 47", bearing N., index error
height of the eye 17 feet. Find the latitude.

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(1) Multiply 1601 by 4, by common logs. (2) Divide 1090 by 547, by common logs. (3)

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1 N.E. À E. 9 8 N by WW A point of land in
2
9 6

Îat. 43° 37' S.
9

long. 146° 49' E. 4 8 4

bearing by compass 5 N. 1 W. 8 1 WNW 1 W 11 W.S.W. À S. 6 7

Dist. 8 miles. 7

7 5 8

7 9 9 W., N. 8 0 SSW. W. 1 Variation 11 pt. W. 10

8 4 11

8 9 12

7 4 1 S by W. W 8 6 W. I S. 1 2

9 5 3

9 8 4

9 1 5 E.S.E. 8 7 S.

A current set the 6 7 4

ship by compass 7 6 7

N.W. ; w. 8

8 3 9 S. 9

E.S.E. 11 Dist. 24 miles from 10 6 5

the time the depar11 7 6

ture was taken to 12 9 0

the end of the day. (4) 1870, December 1st, in long. 160° 30' E., the observed meridian altitude of the Sun's

1. l. was 12° 47' 30", bearing S., index error 1' 10", height of the eye 18 feet. Find the latitude.

O O OT A CO CT OSA COA OO O O O O O

(5) In latitude 71° 34', the departure made good was 36 miles.

Required the d.long. by parallel sailing. (6) Required the course and distance from A to B, by

calculation on Mercator's principle.

A-Lat. 10° 17' S. A-Long. 3° 5' W.
B-Lat. 1 15 N. B-Long. 9 1 E.

ADDITIONAL FOR ONLY MATE.

(7) 1870, May 1st. Find the A. M. and P. M. tides at Greenock,

Leith, and Aberdeen. (8) 1870, December 31st, at 4h. 40m. A. M. apparent time at

ship, lat. 38° 50'S., long. 145° E., the Sun's magnetic amplitude was S.E. by E. E. Find the true amplitude

and variation. (9)

1870, January 22nd, A. M. at ship, in lat. 30° 14' N., the observed altitude of the Sun's l. 1. was 21° 31'45", index error +3' 15", height of the eye 12 feet, time by chronometer 22d. lh. 56m. 49s., which was fast for mean noon at Greenwich 9m. 37s., on October 17th, 1869, and losing daily ls:5. Find the longitude by chronometer.

ADDITIONAL FOR FIRST MATE.

(10) 1870, April 1st, at 8h. 10m. A. M. mean time at ship, in

lat. 28° N., long. 120° W., the observed altitude of the Sun's l.l. was 30° 33' 30", height of the eye 16 feet,

bearing by compass S. 96° 54' E. Find the variation. (11) 1870, May 30th, A. M. at ship, in latitude by account

48° 17' N., long. 168° W., the observed altitude of the Sun's l. 1., South of the observer, was 63° 4' 18", height of the eye 14 feet, time by watch 3h. 40m., which had been found to be 4h. 9m. 31s. fast on apparent time at ship, the difference of longitude made to the East was 20 miles since the error was determined. Find the latitude by reduction to the meridian.

ADDITIONAL FOR MASTER.

(12) 1870, March 29th, the observed altitude of the star

Antares, was 66° 16' 47", bearing N., index error - 3, height of the eye 17 feet. Find the latitude.

CHART. In looking at a Chart, it will be observed that there are lines which run up and down, and others across, the Chart; those which run up and down are due North and South, North being at the top, and South at the bottom; they are termed meridians. Those which run across are called parallels of latitude, and are due East and West, the right hand being East, and the left hand West.

(1) To find the Course and Distance from one Headland to another Headland.

Lay the edge of your parallel rulers on both the Headlands, and bring the rulers to the compass on the Chart, so that the edge passes through its centre; you will then see the course by the compass. In finding the distance, putone leg of your dividers on one Headland and the other leg on the other Headland, and measure the distance on the graduated scale, which runs North and South on the edge of your

Chart. Care should be used that the distance be taken equally on each side of the middle latitude between the two places or Headlands.

(2) Given the ship’s position, that is, the latitude and longitude. Required the Course and Distance to a known Headland.

Mark the latitude off on the graduated scale running N. and S. on one side of your Chart, measure this on the nearest meridian of the ship's longitude, then bringing your parallel rulers up from the parallel

of latitude, all places along the edge of the ruler will be in the same latitude; allowing your parallel rulers to remain, mark your longitude from the scale at the top or bottom of the Chart from the same meridian, East or West, as you require; running this along the edge of your rulers, will give the position of the ship; then to find the Course and Distance, use Rule 1.

(3) To find the position of the ship by Cross Bearings.

Take the bearings of two known Headlands, differing as near 90° as possible, and not less than 45o; lay the edge of your rulers on the centre of the compass on the Chart to the bearing of the first Headland, and bring them up to it, marking with a pencil on the Chart a line from the Headland along the edge of the parallel rulers ; do the same with the other Headland; and where the two lines intersect, that will be the position of the ship.

In the above three rules, if the magnetic points of the compass are not marked on the Chart, care should be taken to allow the variation to the left when Easterly, and to the right when Westerly, contrary to the usual way.

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