Metacentre and laboratory determination of height

Metacentre

When a floating body is tilted by a small angle, the centre of buoyancy $B$ shifts to a new position $B'$ because the shape of the displaced volume changes. The vertical line through $B'$ meets the original vertical axis of the body at a point $M$, the metacentre. The distance $GM$ is the metacentric height.

        upright               heeled
         |  M                  M
         |                    /|
         G                   / G
         B                  B'  W
   weight W down, buoyancy W up through B'

• If $M$ is above $G$ ($GM>0$): a righting couple $W\cdot GM\sin\theta$ acts, so the body is stable.
• If $M$ is below $G$: an overturning couple acts, so the body is unstable.
• If $M$ coincides with $G$: neutral equilibrium.

$$GM=BM-BG=\frac{I}{V}-BG$$

where $I$ is the second moment of the waterline area about the axis of tilt and $V$ is the displaced volume. A larger $GM$ gives greater stability but a shorter rolling period.

Laboratory determination of $GM$ of a rectangular vessel (pontoon)

Apparatus: a rectangular pontoon, a water tank, a movable weight $w$ on a cross-bar, a plumb line and a graduated scale to read the tilt angle.

Steps:

1. Weigh the pontoon with its fittings; its total weight is $W$.
2. Float it in the water in the tank and check that it floats level, with the plumb line at zero on the scale.
3. Place the movable weight $w$ at the centre of the cross-bar, then shift it a measured distance $x$ to one side.
4. The pontoon heels; read the angle $\theta$ from the plumb line on the scale (or find $\tan\theta$ from the deflection).
5. Repeat for several positions $x$ on both sides, and for different values of $w$.
6. Compute $GM$ for each reading, then take the average:

$$w\,x=W\,GM\tan\theta\ \Rightarrow\ GM=\frac{w\,x}{W\tan\theta}$$

7. As a check, calculate the theoretical value $GM=\dfrac{I}{V}-BG$ with $I=\dfrac{LB^{3}}{12}$ for the rectangular waterline, and compare it with the experimental value.