Text-book of theoretical naval architecture by Attwood Edward L. (Edward Lewis) 1871-

Author:Attwood, Edward L. (Edward Lewis), 1871-
Language: eng
Format: epub
Tags: Naval architecture
Publisher: London, New York [etc.] Longmans, Green, and co.
Published: 1899-03-25T05:00:00+00:00

Fig. 66.

collision bulkhead 6 feet from the forward end, and floating at the level water-line WL. It is well to do this problem in two stages—

1. Determine the amount of mean sinkage due to the loss of buoyancy.

2. Determine the change of trim caused.

I. The lighter, due to the damage, loses an amount of buoyancy which is represented by the shaded part GB, and if we assume that she sinks down parallel, she will settle down at a water-line wl such that volume wQ = volume GB. This will determine the distance x between wl and WL.

For the volume wQt = «/H x 40 feet x x and the volume GB = GL X 40 feet x 3 feet

. ^ = 4o_x6X3^ ^^^^

94 X 40 = 2\ inches nearly

2. We now deal with the change of trim caused.

The volume of displacement = 100 x 40 X 3 cubic feet

The weight of the lighter = --^'*- ^ = 2^ tons

and this weight acts down through G, the centre of gravity, which is at 50 feet from either end.

But we have lost the buoyancy due to the part forward of bulkhead £F, and the centre of buoyancy has now shifted back to B' such that the distance of B' from the after end is 47 feet. Therefore we have W, the weight of lighter, acting down through G, and W, the upward force of buoyancy, acting through B'. These form a couple of magnitude—

W X 3 feet = ^^ X 3 = ^^^ foot-tons

tending to trim the ship forward.

To find the amount of this trim, we must find the moment to change trim i inch—

_ W X G M 12 X L

using the ordinary notation.

Now, GM very nearly equals BM;

2400

.'. moment to change trim i inch = 7 x bm

^ 12 X 100

= f X BM

lo

BM = ^

where I© = the moment of inertia of the intact water-plane about

a transverse axis through its centre of gravity; and V = volume of displacement in cubic feet.

I = iV(94 X 40) X (94)' V = 12,000

., BM = '^^^(?_4)!

144000

and moment to alter trim i inch = ' ^,^° ^ <94f

7 X 144000

= 66 foot-tons nearly

.-. the change of trim = ^P- -r- 66

= 15^ inches

The new water-line W'L' will pass through the centre of gravity of the water-line wl at K, and the change of trim aft and forward must be in the ratio 47 : 53; or—

Decrease of draught aft = -^ x 15^ = 7^ inches Increase of draught forward = |^ X 15 i = H inches

therefore the new draught aft is given by—

and the new draught forward by—

3' o" + 2i" + 8i' = 3' lol"

The same result would be obtained by considering the weight of water in the compartment GB acting downwards, and taking its moment about the centre of flotation K of the intact part of the water-line wl. This gives a moment forward of—

/ 6 X 40 X 3 \ foot-tons = ^^ foot-tons

as obtained above.

It will

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