Curve of Buoyancy

Figure 2.2: Curve of buoyancy for a floating body.

Consider a floating body in vertical force balance that is slowly rotated about a horizontal axis normal to one of its vertical symmetry planes. Let us take the center of gravity, $G$ , which necessarily lies in this plane, as the origin of a coordinate system that is fixed with respect to the body. As illustrated in Figure 2.2, as the body rotates, the locus of its center of buoyancy, $H$ , as seen in the fixed reference frame, appears to traces out a curve, $AB$ , in the plane of symmetry. This curve is known as the curve of buoyancy. Let $r$ represent the radial distance from the origin, $G$ , to some point, $H$ , on the curve of buoyancy. Note that the tangent to the curve of buoyancy is always orientated horizontally. This follows because, as was shown in the previous section, small rotations of a floating body in vertical force balance cause its center of buoyancy to shift horizontally, rather than vertically, in the plane perpendicular to the axis of rotation. Thus, the difference in vertical height, $z_{GH}$ , between the center of gravity and the center of buoyancy is equal to the perpendicular distance, $p$ , between $G$ and the tangent to the curve of buoyancy at $H$ . An equilibrium configuration therefore corresponds to a maximum or a minimum of $p$ as point $H$ moves along the curve of buoyancy. However, the equilibrium is only stable if $p$ is minimized. If $R$ is the radius of curvature of the curve of buoyancy then, according to a standard result in differential calculus (Lamb 1928),

$$R = r\,\frac{dr}{dp}.$$ (2.50)

Writing this result in the form

$$\delta r = \frac{R}{r}\,\delta p,$$ (2.51)

it can be seen that maxima and minima of $\delta p$ , which are the points on the curve of buoyancy where $\delta p =0$ , correspond to the points where $\delta r= 0$ , and are, thus, coincident with maxima and minima of $r$ . In other words, an equilibrium configuration corresponds to a point of maximum or minimum $r$ on the curve of buoyancy: that is, a point at which $GH$ meets the curve at right-angles. At such a point, $r=p$ , and the potential energy consequently takes the value $W\,r$ .

Let $H_0$ be a point on the curve of buoyancy, and let $r_0$ , $p_0$ , and $R_0$ be the corresponding values of $r$ , $p$ , and $R$ . For neighboring points on the curve, we can write

$$r-r_0=\left. \frac{dr}{dp}\right\vert _{H_0}\,(p-p_0),$$ (2.52)

or

$$r -r_0 = \frac{R_0}{r_0}\,(p-p_0).$$ (2.53)

It follows that $p-p_0$ has the same sign as $r-r_0$ (because $R_0$ and $r_0$ are both positive). [The fact that $R_0$ is positive (i.e., $dr/dp>0$ ) follows from the previously established result that the metacenter, which is the center of curvature of the curve of buoyancy, always lies above the center of buoyancy, implying that the curve of buoyancy is necessarily concave upwards.] Hence, the minima and maxima of $r$ occur simultaneously with those of $p$ . Consequently, a stable equilibrium configuration corresponds to a point of minimum $r$ on the curve of buoyancy: that is, a minimum in the distance $GH$ between the center of gravity and the center of buoyancy.

We can use the previous result to determine the stable equilibrium configurations for a beam of square cross-section, and uniform specific gravity $s$ , that floats with its length horizontal. In order to achieve this goal, we must calculate the distance $GH$ for all possible configurations of the beam that are in vertical force balance. However, we need only consider cases where $s<1/2$ , because, according to the analysis of Section 2.6, for every stable equilibrium configuration with $s=s_0<1/2$ there is a corresponding stable inverted configuration with $s = 1-s_0>1/2$ , and vice versa.

Figure 2.3: Beam of square cross-section floating with two corners immersed.

Let us define fixed rectangular axes, $x$ and $y$ , passing through the center of the middle section of the beam, and running parallel to its sides. Let us start with the case where the waterline $PQ$ is parallel to a side. (See Figure 2.3.) If the length of a side is $2\,a$ then Equation (2.16) yields

$$AP=BQ=2\,a\,s.$$ (2.54)

Suppose that the beam is turned through an angle $\theta>0$ such that the waterline assumes the position $P'Q'$ in Figure 2.3, but still intersects two opposite sides. The lengths $AP'$ and $BQ'$ satisfy

$$BQ' - AP' = 2\,a\,\tan\theta.$$ (2.55)

Moreover, the area of the trapezium $P'ABQ'$ must match that of the rectangle $PABQ$ in order to ensure that the submerged volume remains invariant (otherwise, the beam would not remain in vertical force balance): that is,

$$(AP' + BQ')\,a = 4\,a^{\,2}\,s.$$ (2.56)

It follows that

$$AP' = a\,(2\,s-\tan\theta),$$ (2.57)

$$BQ' = a\,(2\,s+\tan\theta).$$ (2.58)

The constraint that the waterline intersect two opposite sides of the beam implies that $AP'>0$ , and, hence, that

$$\tan\theta < 2\,s.$$ (2.59)

The coordinates of the center of buoyancy, $H$ , which is the mean center of the trapezium $P'ABQ'$ , are

$$\bar{x} = \frac{\int_{-a}^a\int_{a-h(x)}^a x\,dx\,dy}{\int_{-a}^a\int_{a-... ... = \frac{(2/3)\,a^{\,3}\,\tan\theta}{4\,a^{\,2}\,s}=\frac{a}{6\,s}\,\tan\theta,$$ (2.60)

$$\bar{y} =\frac{\int_{-a}^a\int_{a-h(x)}^a y\,dx\,dy}{\int_{-a}^a\int_{a-h(x)}^a dx\,dy}$$

$$= \frac{4\,a^{\,3}\,s\,(1-s)-(1/3)\,a^{\,3}\,\tan^2\theta}{4\,a^{\,2}\,s}=(1-s)\,a - \frac{a}{12\,s}\,\tan^2\theta,$$ (2.61)

where

$$h(x) = 2\,a\,s+x\,\tan\theta.$$ (2.62)

Thus, if $u=r^{\,2}/a^{\,2}=(\bar{x}^{\,2}+ \bar{y}^{\,2})/a^{\,2}$ then

$$u = \frac{t^{\,2}}{36\,s^{\,2}} + \left[(1-s)-\frac{t^{\,2}}{12\,s}\right]^{\,2},$$ (2.63)

where $t=\tan\theta$ . A stable equilibrium state corresponds to a minimum of $r$ with respect to $\theta$ , and, hence, of $u$ with respect to $t$ . However,

$$\frac{du}{dt} = \frac{t}{36\,s^{\,2}}\left[t^{\,2}-12\,s\,(1-s)+2\right],$$ (2.64)

$$\frac{d^{\,2}u}{dt^{\,2}} = \frac{1}{36\,s^{\,2}}\left[3\,t^{\,2}-12\,s\,(1-s)+2\right].$$ (2.65)

The minima and maxima of $u$ occur when $du/dt=0$ , $d^{\,2}u/dt^{\,2}>0$ and $du/dt=0$ , $d^{\,2}u/dt^{\,2}<0$ , respectively. It follows that the symmetrical position, $t=0$ , in which the sides of the beam are either parallel or perpendicular to the waterline, is always an equilibrium, but is only stable when

$$s^{\,2}-s+\frac{1}{6}>0:$$ (2.66)

that is, when $s<1/2-1/\sqrt{12}= 0.2113$ . It is also possible to obtain equilibria in asymmetric positions such that $t$ is the root of

$$t^{\,2}= 12\,s\,(1-s)-2.$$ (2.67)

Such equilibria only exist for $s>0.2113$ , and are stable. Finally, in order to satisfy the constraint (2.59), we must have $t<2\,s$ , which, in combination with the previous equation, implies that

$$8\,s^{\,2}-6\,s +1>0,$$ (2.68)

or $s<0.25$ .

Figure 2.4: Beam of square cross-section floating with one corner immersed.

Suppose that the constraint (2.59) is not satisfied, so that the immersed portion of the beam's cross-section is triangular. (See Figure 2.4.) It is clear that

$$\frac{BQ'}{BP'} = \tan\theta.$$ (2.69)

Moreover, the area of the triangle $P'BQ'$ in Figure 2.4 must match that of the rectangle $PABQ$ in Figure 2.3, in order to ensure that the submerged volume remain invariant: that is,

$$\frac{1}{2}\,BP'\,\,BQ'= 4\,a^{\,2}\,s.$$ (2.70)

It follows that

$$BP' = (8\,s/\tan\theta)^{1/2}\,a,$$ (2.71)

$$BQ' = (8\,s\,\tan\theta)^{1/2}\,a,$$ (2.72)

or, writing $z^{\,2}=\tan\theta$ and $\omega^{\,2} = (8/9)\,s$ ,

$$BP' = 3\,\omega\,z^{\,-1}\,a,$$ (2.73)

$$BQ' = 3\,\omega\,z\,a.$$ (2.74)

The coordinates of the center of buoyancy, $H$ , which is the mean center of triangle $P'BQ'$ , are

$$\bar{x} = a- BP'/3= a\,(1-\omega\,z^{\,-1}),$$ (2.75)

$$\bar{y} = a-BQ'/3=a\,(1-\omega\,z),$$ (2.76)

because the perpendicular distance of the mean center of a triangle from one of its sides is one third of the perpendicular distance from the side to the opposite vertex. Thus, if $u=r^{\,2}/a^{\,2}=(\bar{x}^{\,2}+ \bar{y}^{\,2})/a^{\,2}$ then

$$u =(1-\omega\,z^{-1})^2 + (1-\omega\,z)^2,$$ (2.77)

$$\frac{du}{dz} = \frac{2\,\omega^{\,2}\,(z^{\,2}-1)\,(z^{\,2}-\omega^{-1}\,z+1)}{z^{\,3}},$$ (2.78)

$$\frac{d^{\,2}u}{dz^{\,2}} = \frac{2\,\omega^{\,2}\,(z^{\,4}-2\,\omega^{-1}\,z+3)}{z^{\,4}}.$$ (2.79)

Moreover, the constraint (2.59) yields

$$z > \frac{3}{2}\,\omega.$$ (2.80)

The stable and unstable equilibria correspond to $du/dz=0$ , $d^{\,2}u/dz^{\,2}>0$ and $du/dz$ , $d^{\,2}u/dz^{\,2}<0$ , respectively. It follows that the symmetrical position, $z=1$ , in which the diagonals of the beam are either parallel or perpendicular to the waterline is an equilibrium provided $\omega<2/3$ , or $s<1/2$ , but is only stable when $\omega>1/2$ , or $s>9/32=0.28125$ . It is also possible to obtain equilibria in asymmetric positions such that $z$ is the root of

$$z^{\,2}-\omega^{-1}\,z+1=0.$$ (2.81)

Such equilibria only exist for $\sqrt{2}/3<\omega<1/2$ , or $1/4<s<9/32$ , and are stable.

In summary, the stable equilibrium configurations of a beam of square cross-section, floating with its length horizontal, are such that the sides are either parallel or perpendicular to the waterline for $s<0.2113$ , such that two corners are immersed but the sides and diagonals are neither parallel nor perpendicular to the waterline for $0.2113<s<0.25$ , such that only one corner is immersed but the sides and diagonals are neither parallel nor perpendicular to the waterline for $0.25<s<0.28125$ , and such that the diagonals are either parallel or perpendicular to waterline for $0.28125<s<0.5$ . For $s>0.5$ , the stable configurations are the same as those for a beam with the complimentary specific gravity $1-s$ .