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.

234.37

X.316
Displacement to W.L. 4

= 73.82
.15 + 50.32 + 43.34 + 54.68 = 148.49

X.315
Displacement to W.L. 3

= 46.77 tons. .15 + 50.32 + 21.62

= 72.09

X.315
Displacement to W.L. 2

= 22.70 tons. .15 + 25.16

= 25.31

X.315
Displacement to W.L. 1

7.97 tons. The displacement to the load water line being 73.82 tons it is useful to know what relation that weight bears to the vessel if she were of box section, in other words, the amount that has been cut off the rectangular block formed by the length, breadth, and draught, to fine it to the required form, or the block coefficient or coefficient of displacement represented by the symbol "8". It will be evident that this coefficient may readily be computed by multiplying the length x breadth x draught, and dividing the product, which is the volume of the box in cubic feet, by 35 to get the tons displaced by the rectangular block. The displacement as calculated, divided by this result, will give the block coefficient “g”, or,

V

;

LxBxū=.432 nearly.

The range of this coefficient for various types is given elsewhere in the Table of Element Coefficients.

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The area of any of the water planes in the specimen displacement table will simply be the sum of the products of the particular water plane required, multiplied by š the interval between ordinates. This product doubled will be the total area of both sides.

Tons per Inch of Immersion (¥''). It is useful to know the amount of displacement of the vessel for each inch of immersion at various draughts, as from this data small amounts of cargo taken out or placed on board can be accurately determined without reference to, or scaling from, the regular displacement curve. It will be seen that if A represents the area of water plane, that this surface multiplied by a layer 1 inch in thickness and divided by 12 will equal the volume of water displaced in cubic feet at the particular water plane dealt with, and that this volume divided by 35 will equal the displacement in tons for one inch, or in other words, the tons per inch immersion. Or,

1 A

cubic feet,

12 12 and the weight of water in the layer

А 1 A

Х
12 35

= tons per inch.

420 Tons per inch immersion in salt water,

area of water plane.

420 Tons per inch immersion in fresh water,

area of water plane

(12 X 36) = 432 So that referring to the table we have been working out, we get :

Ах

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Area of water plane 4.00 335.46 577.86 741.06 832.14 12" X 35 =

420

420

420 Tons per inch = .01 .79 1.37 1.76

1.98 S.W.

It is often necessary to estimate the tons per inch approximately, and for this purpose the coefficient of the load line or "a" is used. The method of arriving at this coefficient is explained in the chapter on design when the displacement is known.

It has a range of about .6 in fine vessels to .9 in exceptionally full ones. In the above example it is found to be

832.14

832.14

=.694. Length x Breadth 1200 Therefore the tons per inch is equal to

LXBX.694

: 1.98.

420 Its relation to the other element coefficients is

8 E.B

a =

35 D, 36 D,

Immersion Passing from Salt to Fresh Water. From what has been previously said it will be obvious that the draught of water, or immersion of a vessel, will undergo a change in passing from fresh water into the sea or vice versa, owing to the difference in density of the two liquids. If we take the case of the ship passing from salt water to fresh, the immersed volume will be in each case as follows:

Immersed volume in salt water

Immersed volume in fresh water = where D is the displacement in tons, which in the example we have been investigating equals 73.82 tons. Therefore the volume in cubic feet which the vessel has sunk on entering the fresh water is 36 D-35 D=2657 - 2584 = 73 cubic feet. Let T=tons per inch immersion in fresh water . . area of water plane = 432 T and the extent to which the vessel will sink 73

12 X 73 73 feet =

= 1.02 inches. 432 T 432 T inches

36 T Inversely we have the amount that the vessel emerges in passing out of a river into the ocean. Thickness of the layer which vessel has risen in feet

Difference in volume D

Area of the plane and in inches, Difference in Volume D x 12 12 X 73 73

= 1.05 inches. Area of water plane 420 T 69.3 This immersion and emersion is, of course, the mean amount as the vessel will also slightly change her trim due to the altered position of the centre of gravity of water plane, about which the ship's movements are pivotal.

Area of Midship Section (3 A). The area of this, or any of the other sections on the displacement table, is calculated by taking the half-breadths of the water lines and integrating them as explained for water-line area. The sum of the products thus obtained is multiplied by ķ the distance of water lines apart, and that result by 2 for both sides. Where the vessel has little rise of floor a half water line should be introduced, and the bottom half-breadth proportioned to the rise line, as pointed out in the displacement calculation. In the example with which we are dealing, however, the vessel has considerable rise, so that this subdivision has been omitted.

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.01 + 7.08 + 5.20 + 11.60 + 3.00

= 26.89 distance between water lines

X.83 Half area of midship section to L.W.L.

=22.31 For both sides .

X 2 Midship section area

= 44.62 The coefficient of this area, or B, is a very important element of the design as explained elsewhere, and is obtained by dividing the midship area by the area of the rectangle formed by the molded breadth and the draught, or Mid. area

44.62

=.743 coefficient of mid. area. Breadth x draught 60 Its relation to the midship-section cylinder or prismatic co

8 efficient "p” is , and “p” is equal to the volume of dis

B placement divided by the length x mid. area, thus :

L X B X 2 X 8 p=

= prismatic coefficient, LHA LX BΧdΧβ β

and consequently,

B =

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Centre of Buoyancy (C.B.). The centre of buoyancy of the displaced water is simply its centre of gravity, and its location below the load-water line is greater or less in accordance with the form of the immersed body. This distance may be found by dividing the under-water part into a number of planes parallel to the load line, and multiplying the volumes, lying between these water planes, by their depth below load-water line. These moments divided by the displacement volume will give the location of centre of buoyancy below loadwater plane. So that by taking the functions of the products at each water plane on the sheet we have been working and multiplying them by the number of the water line they represent below L. W.L., and dividing the sum of those products by the sum of the functions referred to, we shall have the number of water-line intervals (or fraction of an interval), which the C.B. is below load-water line. This result, multiplied by the common interval between water lines, will give the required distance in feet.

KEEL. W.L. 1. W.L. 2. W.L. 3. W.L. 4. Functions of products 4 3

2

1 0 .60 +150.96 + 86.68 + 109.36 + 0 = 347.60 347.6 ; 234.37 = 1.49

Х Water lines apart

1.25 Centre of buoyancy below W.L. 4

1.86 ft.

}.15 + 50.32 + 43.34 + 109.36 + 31.20 = 234.37

.

The centre of buoyancy may be determined from the displacement curve by calculating the area enclosed within the figure formed by the vertical line representing the draught of 5 ft., the horizontal line equal to the tons displacement at this draught and the curve itself. This area divided by the length of the horizontal line referred to, will give the depth of C.B. below L.W.L. In the present example we have : area = 138.6 sq. feet, and length of horizontal line (displacement in tons) = 73.82, and

138.6

-1.87 feet,

73.82 distance of C.B. below L. W.L.

A like result may also be obtained by taking the sum of the products of each water line, and dividing them by the sum of Simpson's multipliers. The mean half-breadths of water lines 80 obtained may be then used to draw a mean section of the

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