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2.74 22.34 22.36 11.88 23.42 16.00 57. 48 5.50

Jeta.' square of the distance between the two axes will equal the moment eat the about ordinate 5.

I equalplane, and consequently at the B.M. without the labor of the middle = foregoing calculation by multiplying the Length by the Breadth by

à coefficient, which coefficient will be determined by a and Apps selected from the table given on page 48. By referring to this e are & table, we find for a (value .694) that the coefficient “ į" (inertia

coefficient) is equal to .0414, whence we get I=LX B3 Xi = 100
x 123 x .0414=1154 moment of inertia, which is sufficiently close
for all purposes, and :-

7154
B.M. = = 2.76.

2583.7
By transposing and taking the calculated I, we find

7191
i=

=.0416.

100 X 123
Longitudinal Metacentre (L.M.C.)
From the definition given for the transverse metacentre it will
be seen that if the ship be inclined longitudinally, instead of, as in
the former case, transversely, through a small angle that the point
in which the vertical through the altered C.B. intersects the
original one will also give a metacentre known as the longitudinal,
or L.M.C. Its principal use and value are in the determination of
the moment to alter trim and the pitching qualities of the vessel,
or longitudinal stability. It will be obvious that the moment of
inertia of the water plane must be taken through an axis at right

angles to the previous case, viz., at right angles to the centre line 8.00 : through the centre of gravity of water plane, which will be where

the original and new water planes cross one another in a longitu-
dinal view.

11 of Water Plane about its C.G..
L.M.C. above C.B. -

Volume of Displacement
Therefore, to calculate the M I1, we must figure the moment of
inertia with, say, ordinate 5 (or any other one) as an axis when
the moment about a parallel axis through the centre of gravity
plus the product of the area of water plane multiplied by the

The moment of inertia about the midship ordinate we shall call
I, and the distance of the centre of gravity from this station = x.
The moment of inertia about the centre of gravity of plane = 11.
We then have I=llt Ax?, or I1=I-Ax2. A clearer conception
of this will be obtained from the tabulated arrangement.

1.00

5.11

1

6.6 7.07

1.38

3.70

Je to

ded br

1

ne, we water

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head specti pivota

ship's

produ

there height

be cal

Area of water plane = 62.41 x (x 10) X 2.

=832.14 square feet. Distance of centre of flotation abaft ordinate 5 (67.57 — 52.41) 10

= 2.42 feet.

62.41 Moment of inertia of water plane about ordinate 5

= 324.13 X (3 10) X 102 2 = 432,172 =1. Moment of inertia of water plane about axis through its centre of flotation.

= 432,172 — (832.14 x 2.422) = 427,304 = 11. Longitudinal metacentre above C.B.

I 427,304
V

= 165 feet = Longitudinal B.M.
2583.7

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RODUCTS

FOR IOMENTS

OF VERTIA

An excellent approximate formula for the longitudinal B.M. is given by J. A. Normand in the 1882 transactions of the I.N.A. Taking the symbols we have been using :

A2 X I

L.B.M. = .0735

B XV Applying this formula to the vessel with which we are dealing, we find :

832.142 x 100 L.B.M. = .0735

= 164.12 feet.

12 X 2583.7 which is a very close appróxrimation to the calculated result of 165 feet.

We may also use the aphr-eximate formula which we applied in the case of the transverse B.M. altered to suit the new axis with a modified coefficient, as :

L.B.M. = 13 x B xii.

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Moment to Change Trim (M1).
As the centre of gravity of the displacement (or centre of buoy-
oncy), either in the vertical or the longitudinal direction may be
an entirely different locus from the ship's centre of gravity, it is
obvious that unless the moment of the weights of the ship and
engines, with all equipment weights, balances about the centre of
buoyancy we shall have a preponderating moment deflecting the
head or stern, as the moment is forward or aft of the C.B., re-
spectively, until the vessel shall have reached a trim in which the
pivotal point or C.B. is in the same vertical line as the completed
ship's centre of gravity. To determine the moment necessary to
produce a change of trim (Mi) in a given ship, it is necessary to know
the vertical position of the centre of gravity of the vessel and the
height of the longitudinal metacentre (L.M.C.). The former may
be calculated in detail or preferably proportioned from a similar
type ship whose centre of gravity has been found by experiment;
although great accuracy in the location of this centre in calculat-
ing the moment is not as important as in the case of G. M. for
initial stability, as small variations in its position can only affect
the final result infinitesimally. To investigate the moment affect-
ing the trim, let us move a weight P already on board of the 100-
foot steamer whose calculations are being figured.

D= Weight of ship including weight P = 73.82 tons.
BM= 165 feet.

P = 5 Tons.
GM = 160 feet.

1 = 50 feet (distance moved). L = 100 feet (length of vessel).

entre

In the figure we have the centre of gravity G to Gı, and the centre of buoyancy from B to B1, due to the shifting of the weight P forward for a distance represented by I, giving a moment

PX1
DX GG1 = P xl, and GG1

D

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WW1

The new water line is at WiLi and BiG1 are in the same verti cal and at right angles to it, and the point of intersection of the original and new water line at 0equal to the centre of gravity (flotation) of water plane, therefore the triangles GMG1, WOW, and LOL1, are of equal angle, so that GG1

LLI WW1 + LLI.
GM WO LO WO + LO
But WW1+LL, is the change of trim, and WO+LO is the length
of the vessel = 1, then

change of trim WW1 + LL
L

WO +LO but we have seen that

GM X change of trim ΡΧί
GG1=
L

D

PXL XL
Then
Change of trim

feet.

D X GM
Substituting the values, we get :-
PXL XL 5 x 50' X 100

= 2.116 feet = 24] inches.
D X GM 73.82 X 160

Calling this change of trim 24 inches, and assuming that the point of intersection “0” is at the centre of the length, we should have

the stem immersed 12 inches and the stern raised 12 inches from the original water line, the sum of these figures equalling the total change.

Moment to Alter Trim One Inch (M”). From the foregoing it will be seen that the total change of trim being known for a given moment, inversely we may get the amount necessary to alter the trim for one inch only, this being a convenient unit with which to calculate changes of trim when a complexity of varying conditions are being dealt with. As we have seen Px1=Mthe moment to change trim, and

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In designing preliminary arrangements of vessels, it is necessary that we should know fairly accurately the moment which it will take to alter the trim one inch (M") to enable us to arrange the principal weights in the ship, and the varying effects on the trim consequent on their alteration in position or removal. For this purpose a close approximation to this moment (M'') is desirable and may be calculated from Normand's formula as follows: M" .0001726, or

on2 x 30.9 B

B Where A2=the square of the water plane area, and B=the greatest breadth of water plane. Applying this approximate formula to the foregoing example, we have:

832.142

x .0001725 = 9.95 foot-tons,

12 as against 9.84 foot-tons found by actual calculations, a difference too insignificant to affect noticeably the change in trim.

This inoment is useful to have for various draughts, and consequently should be calculated for light and load conditions, and for one or two intermediate spots and a curve of M“ run on the usual sheet of “ Curves of Elements."

M" =

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