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To find the Cubic Capacity of a Circular Tank 6 feet in diameter and 15 feet

deep, also the Number of Gallons it would contain, and the Weight Area of circular section equals diameter squared multiplied by 7854.




6 X 6 X 7854 28:2744 ft. area required.
28.2744 x 15 = 424•1 ft. cubic capacity required,
424:1 X 6.25 2650-6 gallons

= 11.8 tons nearly.



35 cubic feet of salt water, and 36 cubic feet of fresh water, weigh i ton. Hence if the tank were full of salt water of specific gravity 1.025, divide by 35.

To find the Draught to which a Vessel may be loaded in Water which differs

from Salt Water whose Specific Gravity is 1.025

Find the specific gravity (by hydrometer) of the water in which the ship is loading, and take the difference between this and the specific gravity of salt water and form a fraction ; multiply the allowance for fresh water by the fraction just found, and the result is the amount the vessel's draught may be increased.

Example.--A vessel is loading in water whose specific gravity is 1•015. How much may the salt water loadline be immersed, the allowance for fresh water being 6 inches?

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The salt water loadline may, therefore, be immersed 2

inches. On


going into salt water she would rise 2- inches.


The above method, although not absolutely correct, is sufficiently correct for all practical purposes.


PLANE SAILING Plane Sailing is the art of navigating a ship upon principles deduced from the supposition of the earth being an extended plane. On this supposition the meridians are considered as being all parallel to each other, the parallels of latitude at right angles to the meridians, and the length of a degree on the meridian, equator, and parallels of latitude everywhere equal. In this sailing there are four principal parts, viz., the course, distance, difference of latitude, and departure. Difference of longitude is not considered. The methods of calculation by plane sailing are sufficiently accurate in low latitudes, but only partially accurate in the higher latitudes, unless the distance sailed be small and the course near a meridian.

The Course is the angle which a ship’s track or path makes with the terrestrial meridian, and is hence the true course, always reckoned from the north or south point of the horizon, towards east or west, either in points of the compass, or in degrees.

DISTANCE is the number of nautical miles between any two places reckoned on the arc of the rhumb curve; or it is the length of a ship's track when she sails on a direct course, in a given time, from one place to another.

Rhumb Line.—Looking from the centre of the compass towards any point on the horizon, the direction is the rhumb line as indicated by the compass; but the continued rhumb line which a ship traces out when sailing from E to A (see Fig.), without altering the course, is the rhumb curve, which may be defined as the shortest line which can join two points on the globe, cutting all the meridians which it crosses at the same angle. The length of any portion of the rhumb curve is the nautical distance.

This line, also known as the loxodromic curve, is in its continuity an unending spiral, winding round the pole, but never reaching it.

DIFFERENCE OF LATITUDE is the distance which a ship makes in a north or south direction from one place to another, and is reckoned on a meridian.

DEPARTURE is the true east or west distance (in nautical miles) that a ship makes when sailing on an oblique course.

It receives its name from being taken to (approximately) measure how far a ship has departed from the meridian ; it must not be confounded with difference of longitude, although it is the basis whence longitude by dead reckoning is derived.

If a ship sail due north or south, she sails on a meridian, makes no departure, and her distance and difference of latitude are the same.

If a ship sail due east or west, she runs on a parallel of latitude, makes no difference of latitude, and her departure and distance are the same.

But when a ship sails in any other direction than on a meridian or on a parallel, she makes both difference of latitude and departure, and these, with the distance, form a right-angled triangle, the hypotenuse of which is the distance sailed, the perpendicular is the difference of latitude, the base is

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the departure, the angle opposite to the base is the course, and the angle opposite to the perpendicular is the complement of the course ; hence, any two of these parts being given, the rest may be found by plane trigonometry.

When a ship's course is 4 points, or 45 degrees, the difference of latitude and departure are equal.

When the course is less than 4 points, or 45 degrees, the difference of latitude is greater than the departure.

When the course is more than 4 points, or 45 degrees, the departure is greater than the difference of latitude.

Given the True Course and Distance, to find the Difference of Latitude and

Departure Example.-A ship from lat. 48° 40' N., sails N.E. by N. 296 miles : required her present latitude, and the departure made good.


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Draw the line B C to represent the meridian the ship sailed from ; with the chord of 60° in the compasses, and one foot in C, describe the arc N E; from N to E lay off the chord of 90°, and draw the line C E: then will NC E represent the N.E. quarter of the compass. Take the course, three points, in the compasses from the line of rhumbs, which lay off from N towards E, and through the point where it cuts the arc draw the line CA, which make equal to the distance 296, taken from a scale of equal parts; through A draw the line B A parallel to CE (Prob. IV. Geom.); then will B A represent the departure equal to 164 miles, and C B the difference of latitude 246 miles.

Distance 296 m.

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NOTE.—This case is the basis on which the Traverse Tables are constructed, and to which constant reference is made in navigation.

* When the courses are given in points of the compass, the log. sine, cosine, tangent, etc., aro to be taken from the Table of Log. sines, etc., for points and quarter points,

To find the Latitude in
Latitude left ....

48° 40' N.
Diff. of latitude 246 miles, or 4

6 N.
Latitude in

52 46 N.
Ans. Lat. in 52° 46' N., departure 164:4 miles.


Previous to resolving the cases by Inspection, read very carefully the Explanation of the Traverse Tables. If the course is less than 4 points, or 45°, it is taken from the top of the page; if more than 4 points, or 45°, take it from the bottom of the page, being careful to find the d. lat. and dep. in their proper columns. When the larger of the two given numbers is the d. lat., seek it in the column marked “d. Lat.” at the top of the page until the smaller number is found by its side in the column marked“Dep.” at the top; then the course corresponding to that d. lat. and dep. will be found at the top of the page, and the distance will be found on the left of the above numbers, in heavy type, in the column marked “ Dist.” at the top and bottom. If the dep. is more than the d. lat. the course will be at the bottom of the page and the d. lat. and dep. columns are reversed in position as compared with the top.

Enter the Traverse Table, and find the course 3 points (at the top of the page), and in one of the columns marked “ Dist." find the distance 296 ; then opposite to this, in the columns marked “ Lat.” and “Dep.,” will be the difference of latitude 246.1, and the departure 164:4.

Given the Difference of Latitude and Departure, to find the True Course

and Distance

Example.—A ship from Funchal, in Madeira, in lat. 32° 38' N., sails a direct course between the south and west until she is in lat. 31° 13' N. by observation, having made 72 miles of departure : required her true course, and distance run.

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Draw the meridian line CB, and from C describe the quadrant SCW; from C to B lay off the difference of latitude 85; through B draw A B parallel to CW, and equal to the departure 72 ; and join A C; then the course A C B will measure 401°, and the distance A C III miles.


Ditt. of Lat.85 m.

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By Inspection.-Seek in the Traverse Table till the difference of latitude 85, and the departure 72, or the nearest thereto, are found side by side in their respective columns, which in this case will be under the course 40 degrees; and the distance answering thereto will be 111 miles.

Ans. Course S. 40° 16' W., distance III.4 miles.

Obs.--. For all practical purposes these Examples might be computed with sufficient accuracy, using only four places of decimals in the logarithms.

Examples for Practice 1. A ship from lat. 36° 30' N., sails S.W. by W. (true) 420 miles : what is her latitude in, and what departure has she made ?

Ans. Lat. in 32° 37' N., and departure 349:3 miles.

2. A ship from lat. 3° 54' S., has sailed N.W. & W. (true) until she arrives at lat. 2° 14' N. : required her distance run, and departure made good.

Ans. Distance 617.8, and departure 496-2 miles.

3. A ship from St. Helena, in lat. 15° 55' S., sails S.S.E. 1 E. (true) until she has made 115 miles of departure : required her latitude in, and the distance she has run.

Ans. Lat. in 19° 30' S., and distance 244 miles.

4. A ship from lat. 28° 20' N., sails north-easterly 486 miles, and finds by observation that she is in lat. 32° 17' N.: what true course and what departure has she made ?

Ans. Course N. 60° 49' E., or N.E. by E. } E. nearly, and departure 424:3 miles.

5. A ship sails between the north and west 170 leagues from a port, in lat. 38° 42' N., until her departure is 98 leagues : required her true course and latitude in.

Ans. Course N. 35° 12' W., or N.W. N. nearly, and lat. in 45° 39' N.

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