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N 43°W

-N 59°W

-N 62°W

EXAMPLE 4.-The true course from
Saugatuck to Port Washington is N 59° W.
the variation is 3° E, and the deviation
according to Table III. Find the course
to steer by compass.
SOLUTION. - See Fig. 17.

True course = N 59° W
Variation

3o E
Magnetic course = N 62° W

Deviation 19° W

Compass course N 43° W. Ans. FIG. 17 On examining Table III, it will be seen that the deviation for courses between N 53° W and N 61o W is constant; hence, we apply a deviation of 19° W to the magnetic course.

EXAMPLES FOR PRACTICE 1. The compass course is north, deviation 224° W, variation 111°E. leeway 1 point, wind W NW. Find the true course.

Ans. True course = north 2. The compass course is N W by W, variation 8° E, deviation according to Table III. Required the true course.

Ans. True course = - N 651° W. 3. The true course is S 75o W, variation 33° 10' W, deviation 1° 50 W. Find the compass course.

Ans. Comp. course N 70° W. 4. The compass course is S 45° E, wind S W, leeway 19°, variation 11° W, deviation by Table III. Find the true course.

Ans. True course = S 61° E. 5. The true course is N 27° 30' W, variation 18° 30' E, deviation 3° 15' W. Find what course to steer. Ans. Comp. course = N 42° 45' W

6. The true course, according to the chart, from the east point of Manitou Island to Battle Island is N 1° E, mean variation 2° 30' E Use deviation for the nearest magnetic course found in Table III and determine what course to steer by compass.

Ans. Comp. course = N 91° E. 7. The true course from Port Huron to Point Clark is N 25° E, variation 3° W. Use deviation for the nearest magnetic course and find the compass course.

Ans. Comp. course = N 32° E.

37. As previously stated, when the deviations are large and changing rapidly from point to point (which very seldom happens nowadays when compasses are so carefully adjusted), the deviation for a certain point should be found by proportion to a reasonable degree of accuracy.

For instance, if, in example 3, in the Examples for Practice, instead of a small deviation the deviation in Table III had been used, the solution would be as follows:

True course = S 75° 0 W

Variation 33° 10' W
Magnetic course S 108° 10' W

Or = N 71° 50' W
Deviation (by proportion) = 17° 30' W
Compass course = N 54° 20' W

Or = N W Å W, nearly. 38. Cautionary Remarks. - Learn to master the subject of course corrections thoroughly, as it is very important. A mistake in the application of variation or deviation may at times prove very costly. Not very long ago the captain of a large steamer, in shaping his compass course from the true course as indicated by the chart, applied the deviation the wrong way.

After a few hours' run, the weather being misty, the steamer was grounded and wedged in between solid rocks that afterwards crushed through her plates, causing a loss of thousands of dollars to her owners and the reputation and position of her master; and all this simply because of the mistake made in applying westerly deviation to the left instead of to the right of the magnetic course. It is better to apply no deviation at all than to apply it the wrong way.

MEASUREMENT OF SPEED AND DEPTH

THE LOG

39. Being particular in noting the true direction in which a ship is advancing, it is of equal importance to know the distance traversed on a certain course or combination of courses. This is done by measuring the speed of the vessel. To determine the speed and thence the distance covered, an instrument called the log is used.

40. Principles of the Log. – The principle on which this instrument is founded is as follows: Some light, floating

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object is thrown overboard; as soon as it strikes the water it ceases to partake of the ship's onward motion and becomes stationary. The distance of this stationary object from the ship is then measured after a certain interval of time has passed, and from this measurement the approximate rate of sailing is ascertained.

41. The log consists of three parts, viz., the log chip, the log line, and the log glass.

The log chip is the floating object thrown overboard, the log line measures the distance, and the log glass defines the interval of time.

42. The log chip is a triangular piece of light wood c, Fig. 18, about 5 inches in diameter, the lower edge of which

t

Log Line

FIG. 18

This peg

is rounded and weighted with a strip of lead sufficiently heavy to make it float in an upright position, as shown in the figure.

In each corner is a hole, the log line being knotted to the one pointing upwards; in the lower holes is fastened a sling, or bridle, at the bight of which is a peg that fits snugly into a wooden sockett, commonly called the toggle. can be released from the toggle by a jerk on the log line, thus allowing the chip to be pulled in with little resistance. The inboard end of the line is attached to a reel r, around which it is wound.

43. The Log Line. – The first 15 to 20 fathoms of the log line from the chip is called the stray line, and is usually marked by a small piece of white or red bunting; the purpose of the stray line is to allow the chip to get clear of the vessel's eddy, or wake, before the measuring commences. The rest of the line, or the log line proper, is divided into parts of equal length, called knots, by pieces of cord fastened between the strands of the line. Each piece of cord carries the requisite number of knots according to its order from the stray-line mark. The length of each knot, when the nautical mile is taken as the unit, is 47 feet 3 inches.

44. The log glass is a sand glass of the same shape and construction as the old hour glass. A vessel usually carries two log glasses, one of which runs out in 28 seconds and the other in 14 seconds; the latter is used when the vessel is going at a high rate of speed, when the number of knots of the line run out is doubled.

45. Relation Between the Knot and the Nautical Mile. – In order to determine the speed of a ship per hour, the length of each knot must bear the same ratio to the nautical mile (6,080 feet) as the time of the log glass does to the hour. Hence, the following proportion:

As the number of seconds in an hour is to the number of feet in a mile, so is the number of seconds in the log glass to the number of feet in the knot; or,

3,600 seconds : 6,080 feet = 28 seconds : x; whence, the length of the knot represented by x is 6,080 X 28

47.29 feet, or 47 feet 3 inches. 3,600

EXAMPLE. – What should be the length of the knot, using the statute mile (5,280 feet) as a unit, and a glass of 30 and 28 seconds, respectively?

SOLUTION. - The length of the knot in the former case would be 5,280 x 30 = 44 feet, and in the latter case

5,280 X 28

41.06 ft. Ans. 3,600

3,600

46. Method of Heaving the Log. - Generally, an officer and two men constitute the logging party, one man holding the reel and the other attending to the glass. When the stray-line mark crosses the rail, the officer says "turn," at which the assistant holding the glass turns it instantly, watches the sand running down, and says "stop" when the last grain passes through the opening. The officer then checks the line and reads off the number of knots run out. The chip should be hove out well to the leeward of the stern, and the reel should be held in such a position as to allow the line to run out freely.

47. Errors of the Chip Log. - In order to guard against error caused by an incorrect length of the knot, the log line should be examined quite frequently by comparing each knot with its proper length, as determined by permanent marks fastened to the deck. There may also be errors attached to the log glass; the condition of the sand or the size of the hole through which it runs may be affected by the change in temperature. Hence, it is important to know how to allow for any of these errors.

It is evident that the longer time it takes the sand to run out, the more line will pass over the rail; and the longer the knot, the smaller will be the number of knots passed out in a given interval of time. Therefore, the speed shown is directly proportional to the length, or time, of the glass, and inversely proportional to the length of the knot. In other words, when an erroneous glass is used, the distance shown by the log is to the correct distance as the number of seconds of the erroneous glass is to 28 seconds; when the line is wrong, the distance shown by the log is to the correct distance as 47.3 feet is to the length of the knot used.

48. Correction of Errors. – Let

x = correct distance;
d = distance shown by log;
g = erroneous time of glass;

1 = incorrect length of knot. We then have the following proportions:

28 d (Glass wrong) d : x = g : 28; whence, x =

d 1 (Knot wrong) d : x = 47.3:6; whence, x =

47.3

(1.)

(2.)

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