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Font size v X (gh + g } h v )
B, Z B G - { + B G cos

B G.

We know that the difference between 1 and cos 6 is versin 6 (1
cos = versin). It will therefore be seen that the difference

The Theory and Design of British Shipbuilding. 99

between B G and B G cos tf will be B G versin d, and this being the
amount by which the minus quantity is in excess, we now write :

X (gh +gr, h t )

Dynamical lever = - - B G versin 0,

V

which is the vertical separation of G and B.

v X (gh + g } A,)

Dynamical moment = W I B G versin 8 I

L V J

which is the work done. This is known as Moseley's Formula.

The dynamical Stability up to any angle, or between any two
angles, is equal to the area of the curve of statical stability for the
corresponding distance of inclination, therefore, by calculating the
area of a statical curve up to a few inclinations, and setting the

results up on their
respective ordin-
ates, we can obtain
the points through
which to draw the
curve of dynam-
ical stability,
which is shown by
Fi g- 46 - D S in Fig. 46.
It will be noticed that the dyanmical curve reaches its maximum
when the statical curve, S S, reaches zero. By subtracting one
ordinate from another, such as n m, we can obtain the amount
of work that is exerted in inclining the vessel from m to n, or other-
wise the amount of dynamical stability. Between any infinitesi-
mally small strip, such as a b c d, Fig. 46, the dynamical moment
will equal the statical moment a b multiplied by the circular measure
of the small angle contained in the strip, this also giving the area
of the strip. From this it will be seen that it is necessary to make
the scale of inclination also in circular measure, or at least to regard
it as such in the calculation. Suppose we had given the curve S S,
Fig. 46, of righting levers, the scales being as shown. First of all,
take an ordinate x, and by means of Simpson's rule, or, preferably,
the planimeter, find the area in square inches, of the curve up to
that point. We have next to take into consideration the scales that

100 The Theory and Design of British Shipbuilding,

the curve is drawn to, this being done as follows : Find what 1 in.
is equal to in each scale, and by multiplying these together we then
have a multiplier with which to multiply the area already found,
and we then have the dynamical lever in feet, which, when multiplied
by the vessel's weight, gives the dynamical moment in foot-tons.
Suppose that the area up to x was found to be 10 square in. actual
area of the curve, and the weight of the vessel 100 tons.

Vertical Scale, 2J ins. = F 1 ft, G Z

. . J in. = -1 ft, G Z and 1 in. = -4 ft. G Z.

Inclination Scale, 10 in. == ol~s deg.

therefore 1 sq. in. = = *04 foot-radians.

10 sq. in. Area x '04
= -4 ft. dynamical lever.

4 ft. x 100 tons ==40 foot- tons dynamical moment.

Having in this manner found the dynamical moments up to the
ordinates such as x, y and z in Fig. 46, and then setting the values
off on them, we have the points through which to draw the curve of
dynamical stability D S, which represents the amount of work
exerted during the inclination of the vessel. This is the method
usually employed in constructing a curve of dynamical stability and
is known as graphic integration.

Stability of Sailing Ships. In Fig. 47 the section is shown of
a vessel with some sail set. The centre of gravity of the sail area
is at E, which is known as the centre of effort, through which the
total pressure P of the wind may be said to be acting. Upon the
lee-side we have a pressure of water, p acting in the opposite direc-
tion to the wind pressure. This water pressure acts through the
centre of lateral resistance, which may be taken to be the centre
of gravity of the under- water middle-line vertical plane. When
P and p are equal the vessel makes no leeway. In Fig. 47 the
distance vertically between the centre of effort and centre of lateral
resistance = x feet. The pressure of wind upon the vessel's sails.
5 P tons. Therefore the inclining moment == (P X x) foot-tons.
Now, this moment inclines the vessel over to the angle 9, at which
we have a righting lever G Z, therefore

Bi Xj<z (inclining moment) = W X G Z (righting moment), W being
weight of vessel.

The Theory and Design of Britwh

101

&- r

P, the pressure on the vessel's sails, is effected by sail area and wind
pressure ; x is effected by the distribution of the sails. The area is
that taken from the projection of the sails on to the centre line of
the vessel. It is obvious that as the vessel inclines this projected
area will become less. Notice in the adjoining sketch in Fig. 47
that when m is inclined to angle 9 the vertical distance A becomes

A 1 . Now, suppose
that A represented
the sail area in
square feet when
upright, then A }
will equal the pro-
jected area when
inclined to 9, being
equal to A x cos
&. Let x be the
distance from the
centre of effort to
centre of lateral
resistance when
upright, then x X
cos 9 will be the
vertical distance
Fig. 47. when inclined to 9.

Heeling mement when upright =
Tons per sq. ft. wind pressure X A x x = M.
Heeling moment when inclined =

Tons per sq. ft. wind pressure X A cos <9 X x cos 9 = M cos 2 9.
The effect of a sudden gust of wind upon the sails of a ship will,
as a rule, heel her over to an extreme angle of heel of about twice
the steady angle at which the same constant pressure of wind would
keep her. The following method is one by means of which we can
determine the angle of extreme roll and the ultimate angle of stead v
heel of a ship when struck by a squall of known force. In Fig. 48
we have a curve of statical stability represented by S.S. Upon
the same diagram, and to the same scales, proceed to construct
the curve of varying wind pressures as follows : First of all find
the area of the sails and the position of the centre of effort relative
to the centre of lateral resistance. Next assume the wind pressure
of, say, 2 Ibs. per sq. ft., which is a very common figure to take for

102 Tfie -Thfon, and Design of British Shipbuilding.

these calculations, with all canvas set, at which time the vessel
should possess a reserve and the angle of lurch should not exceed
the angle of maximum G Z.

Multiplying the area in square feet by 2 Ibs. and dividing by
2,240, we obtain the total number of tons pressure on the vessel's
sails when upright

A x 2

- tons wind pressure.
2,240

A x 2

this, multiplied by x in feet == - - tons X x feet =

2,240

foot- tons heeling moment when upright. Now, we have already seen
that the heeling moment or wind pressure moment varies as the
cos 2 h during inclination , therefore, by multiplying the above result
by cos 2 P for the various angles, we obtain the curve of " varying
wind pressure moments," as shown by W P M. Suppose the squall
suddenly to strike the vessel, causing her to heel away from the
upright towards leeward. When the angle of inclination a is reached,
it will be noticed that the righting moment and the heeling moment
are equal ; this must, therefore, be at the angle of steady heel at
which the vessel would sail if the wind pressure were steadily applied.
The wind pressure moment up to a is represented by c d e a, part
of which is counter-balanced by the area c e a of the righting moment
curve S S. We therefore have left the portion of the wind pressure
moment c d e, which yet remains to be counter-balanced by Irighting
moment ; therefore, on this account, the vessel continues the in-
clination until a point is reached where the excess c d e is counter-
balanced by an area of the S S curve lying above the W T P M curve,
as is shown by the area e f g. In Fig. 48 this point is the angle of
inclination b, which is, therefore, the angle of extreme roll, up to
which the wind force has absorbed an equal amount of the vessel's
righting force, being equal to the work done., After reaching the
angle b the vessel returns to the angle a, at which she settles down
until the force is relaxed. Now, suppose that the curve of statical
stability had taken the shape as shown by S , S , ; it will be noticed
that the portion of the righting moment curve area above the W P M
curve would not be sufficient to counter- balance the excess c d e of
wind pressure moment, therefore the vessel would capsize because

The Theory and Design of British Shipbuilding.

103

Fig. 48

the total force exerted upon her could not be counter-balanced by
her righting force. A large area of the curve of statical stability
is, therefore, seen to be most valuable in the case of sailing vessels.
Of course it must be remembered that in the above cases no account
is taken of fluid resistance upon the immersed portion of the hull
or air resistance of the portions above water, the deductions by the
above method being purely theoretical. The effect of the resistances,
however, may be regarded as a margin of safety.

Effect of Dropping a Weight Overboard. If a weight be
suddenly dropped from off one side of a vessel such as, for instance,
a weight falling owing to the breaking of a rope suspended from a
derrick we have an effect similar to the above suddenly applied
force. Say a weight of 10 tons, suspended at 30 ft. from the centre
line, were to fall in such a manner, a force of 10 X 30 = 300 foot-
tons would be suddenly withdrawn from one side of the vessel. To
find the angles of extreme roll and steady heel, the line x y is drawn
in Fig. 48 at the corresponding moment of 300 foot- tons. At the
angle z the inclining moment represented by the area c x y z is counter
balanced by the area c m z of the righting moment curve, z therefore
being the angle of theoretical extreme roll. At w it will be seen
that the inclining moment and righting moment are equal, this
therefore being the angle of steady heel.

104 The Theory and Design of British Shipbuilding.

CHAPTER X.

LONGITUDINAL STABILITY : METACENTRE AND CALCULATION. TRIM
AND MOMENT TO CHANGE TRIM ONE INCH. ESTIMATIMG THE TRIM.

Longitudinal Metaeentre and method of calculation. Fig. 49
shows a vessel in two conditions of trim.

B = Centre of Buoyancy \ Corresponding to the upright water-

G = Centre of Gravity j line W.L.

B, == Centre of Buoyancy j Corresponding to the inclined water-

G, == Centre of Gravity i W, L,.

CF = The centre of Flotation, being the point of intersection of

the two water-lines,
g and g, = Centres of Gravity of emerged and immersed wedges

respectively.

M 7 = Longitudinal Metaeentre.
w = A weight on board which is moved aft, causing the vessel

to change trim stern wards.
d = distance w is moved.

Let V = the volume of displacement, and v the volume of either
wedge, then

^ X 9 <3\

- B B,, and
V

B B, = B M^ X circular measure of the angle of inclination.
(Let be the angle of inclination, being equal to either a, b or c.)

Therefore B M 7 X circ. m =
v X gg t

and B M 7 =

V X circ. m

At A take an infinitesimally small transverse slice of the wedge,
assuming d x to be its breadth in a fore and aft direction, and the
length across to be y. The depth of the slice will be x circ. m H
and its volume will be y X dx x x X circ. m 0. The moment of
this small volume about C F will be

(y X dx X x X circ. m tf) x x = x 2 X y X dx x circ. m 0.

The Theory and Design of British Shipbuilding.

105

If all such moments throughout the length of the wedges were
summed up, we would obtain the total moment of transference
viz., v X g </,, which is therefore equal to

x 2 X y X dx X circ. m 9.
This can now be substituted for v X g g , .

* M ' - v 7 ' * ^ a = /^-V dx * "T"**'

V x circ. -m* \ x oirc. m fl-

x 2 X ij X dx
Circ. m cancelling out, we have B M 7 = /

7 V

CALCULATION FOR LONGITUDINAL METACENTRE.

No. of
Ordi-
nates.

Lengths
of Half-
Ordinates.

S.
M

Func-
tions.

Levers.

1
i 3
Products for >

Moments.

! J

Products for Mo-
ment of Inertia.

aft
1
'2
3

1
4
2
4

4
3

2

1

4

Moments aft 3
2

1

4

2

sum = 297-28 aft

00 -1 <3i O(

4
2
4
1

1
2
3
4

1
Moments forward 1 2
3

1 4

Sum of

Functions = 367-18 <-
x i longitl.

interval = 12-5

Sum

282-56 ford;
= 297-28 aft I

Sum = 1272-92
longitl.

interval 12-5

4589-75
For both

sides x 2

Waterplane

area - 9179-50 s. ft

Excess - 14-72 aft
x Longitl.

interval 37*5

, 15911-5

Longitl.
interval
squared 37 '5 8

> 367-18)552-00 22375551

Centre of notation aft of amidships 1-5 ft. [For both x 2

Correction for moment of
through the centre of flota-
tion = area of waterplane x
distance of C F from 'mid-
ships squared

9179-5 x 1-52 - 20654 to
be deducted in all cases). <

44751102
-> Correction 20654

-4- By volume of displacement )44730448(MI)

Longitudinal Metacentre above CB = 487-8 ft.
Centre of buoyancy above base = 8-52 ft.

Longitudinal Metacentre above

base = 496-32 ft,

106

The Theory and Design of British Shipbuilding.

It will be seen that the numerator is the algebraic way of expressing
moment of inertia, because we have the small areas y x d x multi-
plied by the distance squared that their centres are from the axis.

In the calculation the area and centre of notation of the water -
plane is first found in the usual way, and then the products for
moments are again multiplied by the levers, the products for moments
of inertia being thereby obtained since the functions of the half-
ordinates are now multiplied by their levers squared. The sum of
the products for M I is multiplied by J of the common longitudinal
interval, so as to complete Simpson's Rule, and then by the interval
squared, the latter being necessary to convert the leverages into
actual distances which are squared. Multiplying by two for both
sides, we have then the longitudinal M I about the 'midship ordinate,
but as the centre of notation is the correct axis, a correction is next

Trim. By " trim " is meant the amount of change of draught
aft plus the change of draught forward. In Fig. 49, W, W -j- L L,
= Trim. The tipping centre is taken as being the centre of flotation.
Change of trim is caused in the following ways : first, by moving a
weight, already on board, in a fore and aft direction ; second, by

)i.v 40k &

g

^ S *

Fig. 49.

The Theory and Design of British Shipbuilding. 107

placing a weight on board ; third, by taking a weight off a vessel.
In the second and third we have sinkage and rise respectively, as
well as change of trim, to take into account. In the first, we have
a weight w in tons moved a distance d (see Fig. 49). w X d = the
trimming moment in foot-tons. Now, if we know the amount of
moment necessary to alter the trim 1 in (i.e., J in. at each end), we
can find the total amount of change of trim caused by the shifting
of a weight already on board the vessel thus : Trimming moment
(w X d) -f- moment to change trim 1 in. (M C T 1 in.).

Moment to Change Trim 1 in.

w x d (W being the total weight of the

In Fig. 49, - = G G, vessel),

W

and G G, = G M, tan

w X d

or - - = G M; tan 0,

W

w X d
. ' . = tan 6.

W x GMj

The total trim = W W, + L L, = T.

T

L (length of water-line) X tan & = T, or = tan 0,

L
w X d T

W x G M; L

If the change of trim is 1 in., then

w x d 1 (L being in feet, must necessarily be

multiplied by 12).
W X G M 7 L x 12

W x GM 7

. . - = w X d, the trimming moment where the trim

L x 12

W X G M;

is 1 in., so we have M C T 1 in., = W being the

L X 12

108 The Theory and Design of British Shipbuilding.

total displacement. G M, the height of the longitudinal metacentre
above the centre of gravity, and L the length of the water-line.

Dividing the trimming moment by M C T 1 in. we have the amount
of trim that is to be divided at each end according to the position
of the centre of flotation, which is the tipping centre. If the tipping
centre is at 'midships, the amount of trim must be divided equally
over each end of the vessel. When the alteration is small, the
tipping centre may be assumed to be at 'midships ; but when the
change is large, it must be proportionately distributed according
to the position of C F. For instance, suppose a vessel 100 ft. long
with C F at 25 ft. aft of 'midships and the total change of trim to
be 12 in. in a stern ward direction, the relative changes would be

i%V of 12 in. = 3 in. trim aft (increase of draught),
yVo of 12 in. = 9 in. trim forward (decrease of draught).

Adding a Weight. First, take the case of a weight of moderate
amount. If it were desired to place it on board in such a position
so that the trim would remain unaltered, it would have to be placed
so that the downward force of the added weight would be acting in
the same vertical line as the upward force of the added buoyancy
contained in a parallel layer that is, the centre of gravity of the
added weight would be placed over the centre of buoyancy of the
added parallel layer of buoyancy. In this case, the weight being
of a moderate amount, the layer would be a thin one and will have
its centre at a position just about coinciding with the centre of
flotation of the former water-line, so, therefore, we may say that,
for moderate amounts of weight placed vertically over or under the
centre of flotation, we have only sinkage and no change of trim.
The sinkage is obtained by dividing the amount of added weight
by the tons per inch corresponding to the water-line.

We therefore see that the following assumptions are made : First,
the tons per inch immersion alters only very slightly as the draught
increases ; and second, the centre of buoyancy of the added layer is
in the same vertical line as the centre of flotation of the former
water-line. Now, having found the position, as above, in which
to place the weight so as not to alter the trim, we can then proceed
to find the trimming moment by multiplying the added weight by
the distance of its centre from the centre of flotation, and this moment
divided by the M C T 1 in., will give the trim. Now, suppose the

The Theory and Design of British Shipbuilding. 109

weight to be of a large amount. We will have to discard the as-
sumptions made in the previous case, and take the following into
account :

1. To find the new parallel draught from the displacement scale
by adding the amount of the weight to the original displacement,
and then to read off the new mean draught corresponding to the
new total displacement.

2. The correct position of centre of buoyancy of added layer.

3. The correct position of centre of flotation of new waterplane.

4. The vertical position of G will be altered most probably.

5. The increase of draught will alter the position of M/.

The 4th and 5th will affect the M C T 1 in. The position of the
centre of buoyancy of the added layer of displacement is found as
follows : From the curve of centres of flotation read off the position
of C F for a water plane at half -depth of the layer between the old
and the new parallel draught, and this position may be used as the
centre of buoyancy of the layer. The C F of the new parallel water-
plane is also found from this curve. Now find the position of G
after the weight has been added, and the new position of M 7 can be
taken from the curve of longitudinal metacentres, the new meta-
centric height being thereby obtained which enables us to find the
new M C T 1 in. The trimming moment is equal to the added weight
multiplied by the distance of its centre from the C B of the added

Fig. 50.

layer, and this moment, divided by the new M C T 1 in., gives the
amount of trim to be distributed according to the position of the

centre of flotation (tipping centre) of the new waterplane.

Deducting a Weight. In this case the method above described
is simply reversed. The depth of deducting layer of displacement
and its C B and also the C F of the new waterplane are found in
the same manner. Multiplying the weight by the distance that
its centre was from the C B of the deducted layer we then have the
trimming moment, which, when divided by the M C T 1 in., as-

110 The Theory and Design of British Shipbuilding.

corrected for the new condition, will then give the total amount of
trim to be distributed according to the position of C F of the new
waterplane.

To obtain the position in which to place a weight so that the draught
aft will not be altered by means of its addition, deduction or gradual
consumption. Such a position is often required in connection with
a vessel's bunkers, so that the propeller immersion will remain
constant during fuel consumption. First find the sinkage if the
weight is placed vertically in line with b, the centre of buoyancy
of the added layer of displacement (see Fig. 50). This sinkage is
x, being the increase of draught at each end, since the layer is parallel.
The draught aft in this condition will be exceeding the requirements
by the amount x, therefore the vessel's trim requires to be now
amended so that the draught aft will be reduced to its former amount.
The vessel may be assumed to tip about 'midships, and the required
alteration being a reduction of x draught aft and a further increase
of x forward, the total change will be 2x in a forward direction. The
total change of trim in inches, when multiplied by the M C T 1 in.
corresponding to the condition, will give the amount of trimming
moment required to produce this change. This trimming moment
being divided by the known weight w will then give the distance d
forward of the C B of the layer, in which to place the weight so as
to obtain the required trim :

(x sinkage in ins. X 2) x M C T 1 in. = trimming moment

. . = d

weight w

The Theory and Design of British Shipbuilding. Ill

CHAPTER XI.

RESISTANCE. WATER RESISTANCE AND STREAM LINE THEORY.
FRICTIONAL RESISTANCE, EDDY-MAKING RESISTANCE, WAVE-MAKING
RESISTANCE AND AIR RESISTANCE. MODEL EXPERIMENTS. HORSE
POWER. EFFECTIVE H.P. INDICATED H.P. PROPULSIVE CO-
EFFICIENTS. LOSSES. COMPOSITION OF I. H.P. NOMINAL H.P.

Tow Rope Resistance. The total resistance of a ship is composed
of Skin Friction (R s ,), Eddy-making (R e ), Wave-making (R w ), and
Air Resistance (RJ. Total Tow Rope Resistance = R s , + R, ;
-f R,,, + R (l = R. If a ship were towed in such a manner so that
the presence of the tug would not affect the resistance of the ship,
then the ship would impart to the rope a strain equal to R. This
resistance is augmented in the case of a vessel that is propelled by
her own means, such as screw propeller, etc. In dealing with resist-
ance, it is usual to disregard this propeller augment, and only work
with the tow rope resistance or " resistance of form," allowance
being afterwards made for this augment when estimating the
horse-power.

Resistance of the Water. Water is not a perfect fluid ; when in
motion its particles rub against each other, causing an amount of
friction which is known as viscosity. When rubbing against a body
such as a ship, an amount of frictional resistance is set up. If a
ship-shaped body, as shown in Fig. 51, is submerged and drawn

Fig. 51.

Fig. 52.

through the water, or, say, fixed in a position and the water allowed
to flow past it (which has exactly the same effect), the particles of
water are diverted in the directions as shown b}' the lines outside
the figure, these lines being termed stream lines. The direction of
the flow of water is shown by the arrow. When nearing the body,
the stream lines widen out, reaching a maximum width at the end

112 The Theory and Design of British Shipbuilding.

A. This widening causes a decrease in the speed of the flow at this
point, and, consequently, an increase of pressure. At 'midships
the distances between the stream lines become narrower, which
now causes an increased speed, giving a reduced pressure. Upon
reaching the aft end B a similar effect is obtained as at the fore end
A. The effect of this change of pressure can be noticed in a vessel
moving upon the surface, where the increase of pressure is seen to
cause the water to increase in height at the ends, while the reduced
pressure at 'midships causes a reduction in height (see Fig. 52), the
tw r o waves being known as the " statical crests." The stream lines
of water flowing past the hull of a vessel obtain complicated motions,
the particles becoming confused with each other, and, rubbing

Online LibraryAmos Lowrey AyreThe theory and design of British shipbuilding → online text (page 8 of 14)