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will give a common interval of 10 feet. The draught of 5 feet must likewise be subdivided into a certain number of equal intervals, which in this case we will fix at 4, so that
5 ft. draught
= 1.25 ft.
4 interval between water lines. These divisions of water lines must be drawn across the body plan of ten sections, and the half breadths read off with a scale and tabulated as in table, on following page.
It should be stated in connection with the subdivision of the base line that the length taken for displacement is measured by some designers from the after side of body post i.e., ignoring the propeller aperture ; and by others from the fore side of body post to the after side of stem omitting the moulded size of these forgiugs. Both of these methods are inaccurate besides leading to confusion, as, in the first case, the displacement of the propeller with its boss will equal the displacement cut out for aperture not to mention the volume of the rudder, which is rarely, if ever, taken into account. And in the second case the tiny amount of displacement added at the knuckle formed by the bearding line of plating when the length is taken to forward and after sides of stem and stern post respectively, is compensated for by the gudgeons on stern post. Therefore the most correct and also the most convenient length is from after side of rudder post to forward side of stem at load water line.
Where vessels have a very flat floor line a half water line should be taken between base line
1 st W.L. and first water plane, and the keel or bottom half-breadth
Keel given a value proportioned to the rise of floor line as in Fig. 5.
Required the half-breadth x at the keel for the displacement sheet, where 10 feet is the actual scaled length L, 6" the rise of floor, 7" the distance from the rise line to first water line at moulded half-breadth of ship and, of course, 13 inches the water line interval, then :
13: 7":: 10 feet : x.
.04 .01 .04 .01 .04 .01.04 .01 1 .03 03.08 .08.18 .18.43 .43 1.41 1.41
.02 .01 .16 .12 .73 .55 1.78 1.33 3.10 2.32 21.02.04 .92 1.84 2.35 4.70 3.78 7.56 4.81 9.62 1 .02 02 2.13 2.13 4.03 4.03 5.16 5.16 5.56 5.56 2 .02 .04 3.20 6.40 4.98 9.96 5.67 11.34 5.96 11.92 1 .02.02 3.54 3.54 5.20 5.20 5.80 5.80 6.00 6.00 2 .02 .04 3.00 6.00 4.66 9.32 5.34 10.68 5.58 11.16 1 .02 .02 2.00 2.00 3.58 3.58 4.42 4.42 4.87 4.87 2.02.04 1.25 2.50 2.28 4.56 3.04 6.08 3.57 7.14
* .02 .01 .48 .36 1.00 .75 1.50 1.12 1.90 1.42 1 .02 .02
.18 .18.50 .50 .74 .74 .97 .97 1
6 7 8 9 97 10
15 = 10 X1.5
of .30 25.16 43.34 54.68 62.41
Prod. Sum ucts. of
1/ 를 2(1) 1() 2(1) Mul
1 tipli. .15 + 50.32 + 43.34 + 109.36 + 31.20
( W.L. interval) x ( ordinate interval) X 2 (both sides)
=coeff. 35 (cub. ft. of S.W. in a ton) (1.25 X ļ) X (10 x ş) 2
54.68 = 148.49
= 46.77 tons. .15 + 50.32 + 21.62
= 22.70 tons. .15 + 25.16
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 "8", or,
LXBXd The range of this coefficient for various types is given elsewhere in the Table of Element Coefficients.
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 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,
12 12 and the weight of water in the layer
А 1 A
= tons per inch. 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 :
Area of water plane 4.00 335.46 577.86 741.06 832.14 12" X 35 =
420 420 420 420 Tons per inch
.01 .79 1.37 1.76 1.98 S.W.
It is often necessary to estimate the tons per inch approxi. mately, 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
=.694. Length x Breadth 1200 Therefore the tons per inch is equal to
420 Its relation to the other element coefficients is
8 a =
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
35 D, Immersed volume in fresh water = 36 D, 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.