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since a greater angle corresponds to a shorter distance the actual position of the ship will be further off-shore than the result by the method indicates. Owing to the interference of terrestrial refraction the result by this method should be considered only as approximately correct.

97. To Find the Distance of a Known Object Seen on the Sea Horizon. The solution of this problem depends on the uniform curvature of the sea, by means of which all terrestrial objects appear or disappear from sight at certain distances from the observer. These distances, corresponding to different elevations of the eye and object, are found in the Nautical Tables under the heading "Distances of Objects at Sea," expressed in nautical and statute miles, respectively.

In order to give the student a clear idea of this method let 1, Fig. 26, represent a lighthouse situated on the summit. of a mountain, s the lookout stationed in the upper rigging of an approaching vessel, and h the visible horizon of s.

h

FIG. 26

Then as soon as the top of the lighthouse appears above the horizon it is evident that the distance of the ship from the lighthouse is equal to the sum of the distances s h and h l. These distances, depending on the height of s and 7, are recorded in the tables mentioned. Assume, for instance, that the height of the lighthouse above the level of the sea is 250 feet and that of the lookout is 115 feet; what would be the distance of the ship from 7, expressed in statute miles? Consulting the Table of Distances in statute miles we find that for a height of 250 feet the corresponding distance is 20.82 miles, and for a height of 115 feet the distance of the visible horizon is 14.12 miles. Therefore, the distance to the lighthouse must be equal to the sum of these distances, or equal to 20.82+14.12=34.9 statute miles.

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98. This method is very useful at nights in clear weather, when approaching or getting out of the range of visibility of a light, the height of which is known. be remembered, however, that the visibility of lights, as indicated on charts, is that distance that corresponds to a height of the eye of 15 feet; hence, when the observer is at a greater or lesser height than 15 feet above the water-line of the vessel his actual elevation should be taken into consideration and used in entering the table.

EXAMPLE 1.-The officer of the watch having kept his eyes on a certain light sees it is about to disappear below the horizon. He takes its bearing and notes the time. The height of his eye above the water-line is 25 feet and the height of the light according to the chart is 275 feet. Find the distance in nautical miles.

SOLUTION.-Entering the table, we find that The distance corresponding to 25 feet = 5.75 and the distance corresponding to 275 feet = 19.07

Hence,

distance of ship from light=24.8 naut. mi. Ans.

EXAMPLE 2.-A vessel is running for a certain port. At the time expected to get within sight of the lighthouse at the entrance of the harbor a man is sent aloft. His height above the water-line is 60 feet. After a while he discovers the light, the height of which is 75 feet. What is the distance of the ship from the light in nautical miles? SOLUTION.-Entering table, we find that

8.91

The distance corresponding to 60 feet and the distance corresponding to 75 feet = 9.96

Hence,

distance from light=18.9 naut. mi. Ans.

99. To Find the Distance by the Velocity of Sound. A convenient method, whenever available, is to determine the distance by noting the number of seconds elapsed between seeing the flash and hearing the report of a gun fired. The velocity of sound is 1 nautical mile in 5.6 seconds, or .18' (=1,092 feet) in 1 second. Hence, the following rule:

Divide the number of seconds elapsed by 5.6, or, multiply them by .18. The result is the required distance expressed

in miles.

EXAMPLE. The number of seconds counted in the interval of time between the flash and report of a gun is 14. Required the distance. SOLUTION.-Applying the foregoing rule the required distance

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100. This method may sometimes prove useful during a fog when a vessel is coasting or approaching a harbor or coast where guns are fired as fog signals; especially so at night when the reflection of the flash of the gun is noticeable through the fog. By taking a bearing of the flash and noting the number of seconds elapsed from it to the report, the approximate position of the ship may be obtained.

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A moderate breeze in the direction of the sound causes a variation of about 25 feet a second in the velocity; a strong breeze more. Its effect on the result, however, is hardly worth notice in navigation.

101. To Find the Position of a Ship on the Chart in Misty or Foggy Weather by Means of a Line of Sounding. In misty or foggy weather the following method of determining approximately the position.

of a ship is very useful, and should never be neglected whenever circumstances call for its utilization. Take several casts with the lead at frequent intervals, and note the time, and the course and distance run between each cast; note also at each sounding the character of the bottom as shown by the

arming of the lead. After a sufficient number of casts have been made, rule a line a b, Fig. 27, on a piece of tracing paper to represent the direction of the meridian, and another line cd to represent the course (or courses) steered. On this line cd lay off and mark down, according to scale (see Art. 62), the distance run between each cast, and at each point thus obtained mark the respective soundings and character of bottom, as shown in the figure. The result will be a chain or line of soundings. This done, place the tracing paper down on the chart with the end of the line cd on the place where you conceived your position to be when the first cast was taken. If the soundings on the

line cd agree with those on the chart, as seen through the tracing paper, you will know at once where you are.

they do not agree, move the tracing paper (keeping the line ab parallel to the meridian) until you find a place where they do agree. This is the place of the ship.

THE STATION POINTER

102. Description.-In connection with the use of charts and determining the position of a ship while in sight of known objects on land, many instruments have been introduced to facilitate and simplify such operations. Among them the station pointer is important. This instrument, which does away with the possibility of committing errors in applying compass corrections, besides saving considerable time in fixing a position, is shown in Fig. 28.

Right

FIG. 28

But if

Left

It consists of a graduated circle of metal having three arms attached to its center c; of these, ce is fixed, while the other two are movable. The latter are fitted with clamping screws

s and s', so they can be set and secured to any required angle, the center line of the fixed arm serving as the zero point. The instrument is made in different sizes and more or less perfect in detail; some are, like the sextant, fitted with tangent screws and reading microscopes. As a whole, the instrument is simple, easy to handle and adjust, and has no complicated mechanism to get out of order.

103. How to Use the Station Pointer.-Select three objects the positions of which are known, and which are to be found on the chart. Let these three objects be represented in Fig. 29 by a, b, and c; the horizontal angles between

a

ZeroArm

FIG. 29

a and b, and b and c are then measured with a sextant in quick succession by one observer, or separately at the same instant by two observers (the latter to be preferred). This done, take the station pointer, holding the arms from you, and set the left arm so that the angle subtended by it and the zero arm will be equal to the measured angle between a and b. Similarly set the right arm so the angle between it and the zero arm is equal to the measured angle between b and c. Then place the instrument on the chart so that the center of the fixed arm passes through the object b, while the other arms pass through a and c, respectively. The center of the

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