Determination of Metacentric Height

Suppose that the floating body considered in the previous section is in an equilibrium state. Let $A_0$ be the cross-sectional area at the waterline: that is, in the plane $z=0$ . Because the body is assumed to be symmetric with respect to the $x=0$ and $y=0$ planes, we have

$$\int_{A_0}x\,dx\,dy = \int_{A_0}y\,dx\,dy=\int_{A_0}x\,y\,dx\,dy=0,$$ (2.22)

where the integrals are taken over the whole cross-section at $z=0$ . Let $\delta(x,y)$ be the body's draft: that is, the vertical distance between the surface of the water and the body's lower boundary. It follows, from symmetry, that $\delta(-x,y)=\delta(x,y)$ and $\delta(x,-y)=\delta(x,y)$ . Moreover, the submerged volume is

$$V_0 =\int_{A_0}\int_0^\delta dx\,dy\,dz= \int_{A_0}\,\delta (x,y)\,dx\,dy.$$ (2.23)

It also follows from symmetry that

$$\int_{A_0} x\, \delta(x,y)\,dx\,dy= \int_{A_0}y\,\delta(x,y)\,dx\,dy = 0.$$ (2.24)

The depth of the unperturbed center of buoyancy below the surface of the water is

$$h = \frac{\int_{A_0}\int_0^\delta z\,dx\,dy\,dz}{V_0}=\frac{1}{2\,V_0}\,\int_{A_0} \delta^{\,2}(x,y)\,dx\,dy=\frac{\delta_0^{\,2}\,A_0}{2\,V_0},$$ (2.25)

where

$$\delta_0 = \left[\frac{\int_{A_0}\delta^{\,2}(x,y)\,dx\,dy}{A_0}\right]^{1/2}.$$ (2.26)

Finally, from symmetry, the unperturbed center of buoyancy lies at $x=y=0$ .

Suppose that the body now turns through a small angle $\theta$ about the $x$ -axis. As is easily demonstrated, the body's new draft becomes $\delta'(x,y)\simeq \delta(x,y) - \theta\,y$ . Hence, the new submerged volume is

$$V_0'=\int_{A_0} [\delta(x,y) -\theta\,y]\,dx\,dy = V_0 + \theta\int_{A_0} y\,dx\,dy = V_0,$$ (2.27)

where use has been made of Equations (2.22) and (2.23). Thus, the submerged volume is unchanged, as should be the case for a purely angular displacement. The new depth of the center of buoyancy is

$$h' = \frac{\int_{A_0}\int_0^{\delta'} z\,dx\,dy\,dz}{V_0}=\frac{1... ...\,2}(x,y) -2\,\theta\,y\,\delta(x,y) + {\cal O}(\theta^{\,2})\right]dx\,dy = h,$$ (2.28)

where use has been made of Equations (2.24) and (2.25). Thus, the depth of the center of buoyancy is also unchanged. Moreover, from symmetry, it is clear that the center of buoyancy still lies at $x=0$ . Finally, the new $y$ -coordinate of the center of buoyancy is

$$\frac{\int_{A_0}\int_0^{\delta'}y\,dx\,dy\,dz}{V_0}=\frac{\int_{A... ...[\delta(x,y)-\theta\,y]\,dx\,dy}{V_0}=-\theta\,\frac{\kappa_x^{\,2}\,A_0}{V_0},$$ (2.29)

where use has been made of Equation (2.24). Here,

$$\kappa_x = \left(\frac{\int_{A_0}y^2\,dx\,dy}{A_0}\right)^{1/2},$$ (2.30)

is the radius of gyration of area $A_0$ about the $x$ -axis.

It follows, from the previous analysis, that if the floating body under consideration turns through a small angle $\theta$ about the $x$ -axis then its center of buoyancy shifts horizontally a distance $\theta\,\kappa_x^{\,2}\,A_0/V_0$ in the plane perpendicular to the axis of rotation. In other words, the distance $HH'$ in Figure 2.1 is $\theta\,\kappa_x^{\,2}\,A_0/V_0$ . Simple trigonometry reveals that $\theta \simeq HH'/MH'$ (assuming that $\theta$ is small). Hence, $MH' = HH'/\theta=\kappa_x^{\,2}\,A_0/V_0$ . Now, $MH'$ is the height of the metacenter relative to the center of buoyancy. However, the center of buoyancy lies a depth $h$ below the surface of the water (which corresponds to the plane $z=0$ ). Hence, the $z$ -coordinate of the metacenter is $z_M=\kappa_x^{\,2}\,A_0/V_0- h$ . Finally, if $z_G$ and $z_H=-h$ are the $z$ -coordinates of the unperturbed centers of gravity and buoyancy, respectively, then

$$z_M = \frac{\kappa_x^{\,2}\,A_0}{V_0}+ z_H,$$ (2.31)

and the metacentric height, $\lambda= z_M-z_G$ , becomes

$$\lambda = \frac{\kappa_x^{\,2}\,A_0}{V_0}-z_{GH},$$ (2.32)

where $z_{GH} = z_G-z_H$ . Note that, because $\kappa_x^{\,2}\,A_0/V_0>0$ , the metacenter always lies above the center of buoyancy.

A simple extension of the previous argument reveals that if the body turns through a small angle $\theta$ about the $y$ -axis then the metacentric height is

$$\lambda = \frac{\kappa_y^{\,2}\,A_0}{V_0}-z_{GH},$$ (2.33)

where

$$\kappa_y = \left(\frac{\int_{A_0}x^2\,dx\,dy}{A_0}\right)^{1/2},$$ (2.34)

is radius of gyration of area $A_0$ about the $y$ -axis. Finally, as is easily demonstrated, if the body rotates about a horizontal axis which subtends an angle $\alpha$ with the $x$ -axis then

$$\lambda = \frac{\kappa^{\,2}\,A_0}{V_0} -z_{GH},$$ (2.35)

where

$$\kappa^{\,2} = \kappa_x^{\,2}\,\cos^2\alpha+ \kappa_y^{\,2}\,\sin^2\alpha.$$ (2.36)

Thus, the minimum value of $\kappa^{\,2}$ is the lesser of $\kappa_x^{\,2}$ and $\kappa_y^{\,2}$ . It follows that the equilibrium state in question is unconditionally stable provided it is stable to small amplitude angular displacements about horizontal axes normal to its two vertical symmetry planes (i.e., the $x=0$ and $y=0$ planes).

As an example, consider a uniform rectangular block of specific gravity $s$ floating such that its sides of length $a$ , $b$ , and $c$ are parallel to the $x$ -, $y$ -, and $z$ -axes, respectively. Such a block can be thought of as a very crude model of a ship. The volume of the block is $V=a\,b\,c$ . Hence, the submerged volume is $V_0=s\,V = s\,a\,b\,c$ . The cross-sectional area of the block at the waterline ($z=0$ ) is $A_0=a\,b$ . It is easily demonstrated that $\delta(x,y)=\delta_0=V_0/A_0=s\,c$ . Thus, the center of buoyancy lies a depth $h=\delta_0^{\,2}\,A_0/2\,V_0=s\,c/2$ below the surface of the water. [See Equation (2.25).] Moreover, by symmetry, the center of gravity is a height $c/2$ above the bottom surface of the block, which is located a depth $s\,c$ below the surface of the water. Hence, $z_H = -h=-s\,c/2$ , $z_G= c/2-s\,c$ , and $z_{GH} = c\,(1-s)/2$ . Consider the stability of the block to small amplitude angular displacements about the $x$ -axis. We have

$$\kappa_x^{\,2} = \frac{\int_{-a/2}^{a/2}\int_{-b/2}^{b/2} y^{\,2}\,dx\,dy}{a\,b} = \frac{b^{\,2}}{12}.$$ (2.37)

Hence, from Equation (2.32), the metacentric height is

$$\lambda = \frac{b^{\,2}}{12\,s\,c} - \frac{c}{2}\,(1-s).$$ (2.38)

The stability criterion $\lambda>0$ yields

$$\frac{b^{\,2}}{6 \,c^{\,2}} - s\,(1-s)> 0.$$ (2.39)

Because the maximum value that $s\,(1-s)$ can take is $1/4$ , it follows that the block is stable for all specific gravities when

$$c < c_0 = \sqrt{\frac{2}{3}}\,b.$$ (2.40)

On the other hand, if $c>c_0$ then the block is unstable for intermediate specific gravities such that $s_-< s< s_+$ , where

$$s_\pm = \frac{1\pm \sqrt{1-c_0^{\,2}/c^{\,2}}}{2},$$ (2.41)

and is stable otherwise. Assuming that the block is stable, its angular equation of motion is written

$$I\,\frac{d^{\,2}\theta}{dt^{\,2}} = - W\,\lambda\,\sin\theta\simeq - W\,\lambda\,\theta,$$ (2.42)

where

$$I = \frac{W}{g\,V}\int_{-a/2}^{a/2}\int_{-b/2}^{b/2}\int_{-s\,c}^... ...,dx\,dy\,dz= \frac{W}{12\,g}\left(b^{\,2}+ 4\,[(1-s)^3+s^{\,3}]\,c^{\,2}\right)$$ (2.43)

is the moment of inertia of the block about the $x$ -axis. Thus, we obtain the the simple harmonic equation

$$\frac{d^{\,2}\theta}{dt^{\,2}} = -\omega^{\,2}\,\theta,$$ (2.44)

where

$$\omega^{\,2} = \frac{W\,\lambda}{I}= \frac{g}{s\,c}\,\frac{c_0^{\,2}-4\,s\,(1-s)\,c^{\,2}}{c_0^{\,2}+(8/3)\,[(1-s)^3+s^{\,3}]\,c^{\,2}}.$$ (2.45)

We conclude that the block executes small amplitude angular oscillations about the $x$ -axis at the angular frequency $\omega$ . For the case of rotation about the $y$ -axis, the previous analysis is unchanged except that $a\leftrightarrow b$ . The previous analysis again neglects the phenomenon of added mass, and, therefore, underestimates the effective inertia of the block. (See Sections 5.9 and 7.10.)

The metacentric height of a conventional ship whose length greatly exceeds its width is typically much less for rolling (i.e., rotation about a horizontal axis running along the ship's length) than for pitching (i.e., rotation about a horizontal axis perpendicular to the ship's length), because the radius of gyration for pitching greatly exceeds that for rolling. As is clear from Equation (2.45), a ship with a relatively small metacentric height (for rolling) has a relatively long roll period, and vice versa. An excessively low metacentric height increases the chances of a ship capsizing if the weather is rough, if its cargo/ballast shifts, or if the ship is damaged and partially flooded. For this reason, maritime regulatory agencies, such as the International Maritime Organization, specify minimum metacentric heights for various different types of sea-going vessel. A relatively large metacentric height, on the other hand, generally renders a ship uncomfortable for passengers and crew, because the ship executes short period rolls, resulting in large g-forces. Such forces also increase the risk that cargo may break loose or shift.

We saw earlier, in Section 2.4, that if a body of specific gravity $s$ floats in vertical equilibrium in a certain position then a body of the same shape, but of specific gravity $1-s$ , can float in vertical equilibrium in the inverted position. We shall now demonstrate that these positions are either both stable, or both unstable, provided the body is of uniform density. Let $V_1$ and $V_2$ be the volumes that are above and below the waterline, respectively, in the first position. Let $H_1$ and $H_2$ be the mean centers of these two volumes, and $H$ that of the whole volume. It follows that $H_2$ is the center of buoyancy in the first position, $H_1$ the center of buoyancy in the second (inverted) position, and $H$ the center of gravity in both positions. Moreover,

$$V_1\,\,H_1G = V_2\,\,H_2G = \left(\frac{V_1\,V_2}{V_1+V_2}\right)H_1H_2,$$ (2.46)

where $H_1G$ is the distance between points $H_1$ and $G$ , et cetera. The metacentric heights in the first and second positions are

$$\lambda_1 = \frac{\kappa^{\,2}\,A}{V_1} - H_1G =\frac{1}{V_1}\left[A\,\kappa^{\,2}- \left(\frac{V_1\,V_2}{V_1+V_2}\right) H_1H_2\right],$$ (2.47)

$$\lambda_2 = \frac{\kappa^{\,2}\,A}{V_2} - H_2G =\frac{1}{V_2}\left[A\,\kappa^{\,2}- \left(\frac{V_1\,V_2}{V_1+V_2}\right)H_1H_2\right],$$ (2.48)

respectively, where $A$ and $\kappa$ are the area and radius of gyration of the common waterline section, respectively. Thus,

$$\lambda_1\,\lambda_2 = \frac{1}{V_1\,V_2}\left[A\,\kappa^{\,2}- \left(\frac{V_1\,V_2}{V_1+V_2}\right) H_1H_2\right]^{\,2}\geq 0,$$ (2.49)

which implies that $\lambda_1\gtreqqless 0$ as $\lambda_2\gtreqqless 0$ , and vice versa. It follows that the first and second positions are either both stable, both marginally stable, or both unstable.