242 MECHANICAL ENGINEERING PRINCIPLES
21.14 The stability of floating bodies
For most ships and boats the centre of buoyancy (B)
of the vessel is usually below the vessels’ centre of
gravity (G), as shown in Figure 21.13(a). When this
vessel is subjected to a small angle of keel (θ), as
shown in Figure 21.13(b), the centre of buoyancy
moves to the positionB′,where
W
W
G
B
W
W
M
G
B B′
(a) (b)
q
Figure 21.13
BM=the centre of curvature of the centre of
buoyancy=
I
V
, (given without proof)
GM=the metacentric height,
M=the position of the metacentre,
I=the second moment of area of the water
plane about its centreline, and
V=displaced volume of the vessel.
The metacentric height GM can be found by
a simple inclining experiment, where a weightP
is moved transversely a distancex, as shown in
Figure 21.14.
G
P
B
GG′
BB′ P
M
x
(a) (b)
Figure 21.14
From rotational equilibrium considerations,
W(GM)tanθ=Px
Therefore, GM=
Px
W
cotθ( 21. 1 )
whereW=the weight of the vessel, and
cotθ=
1
tanθ
Problem 13. A naval architect has carried
out hydrostatic calculations on a yacht,
where he has found the following:
M =mass of yacht=100 tonnes,
KB=vertical distance of the centre of
buoyancy(B)above the keel(K)
= 1 .2 m (see Figure 21.15),
BM=distance of the metacentre(M)above
the centre of buoyancy= 2 .4m.
M
G
B
K
Figure 21.15
He then carries out an inclining experiment,
where he moves a mass of 50 kg through a
transverse distance of 10 m across the
yacht’s deck. In doing this, he finds that the
resulting angle of keelθ= 1 °. What is the
metacentric height (GM) and the position of
the centre of gravity of the yacht above the
keel, namelyKG? Assumeg= 9 .81 m/s^2.
P=50 kg× 9. 81 = 490 .5N,
W=100 tonnes× 1000
kg
tonne
× 9. 81
m
s^2
=981 kN,
x=10 m,
θ= 1 °from which, tanθ= 0 .017455 and
cotθ=
1
tanθ
= 57. 29