Categories: Fluid mechanics, Physics.

Metacentric height

Consider an object with center of mass GG, floating in a large body of liquid whose surface is flat at z=0z = 0. For our purposes, it is easiest to use a coordinate system whose origin is at the area centroid of the object’s cross-section through the liquid’s surface, namely:

(x0,y0)1Awlwl(x,y)dA\begin{aligned} (x_0, y_0) \equiv \frac{1}{A_{wl}} \iint_{wl} (x, y) \dd{A} \end{aligned}

Where AwlA_{wl} is the cross-sectional area enclosed by the “waterline” around the “boat”. Note that the boat’s center of mass GG does not coincide with the origin in general, as is illustrated in the following sketch of our choice of coordinate system:

Boat's coordinate system

Here, BB is the center of buoyancy, equal to the center of mass of the volume of water displaced by the boat as per Archimedes’ principle. At equilibrium, the forces of buoyancy FB\vb{F}_B and gravity FG\vb{F}_G have equal magnitudes in opposite directions, and BB is directly above or below GG, or in other words, xB=xGx_B = x_G and yB=yGy_B = y_G, which are calculated as follows:

(xG,yG,zG)1Vboatboat(x,y,z)dV(xB,yB,zB)1Vdispdisp(x,y,z)dV\begin{aligned} (x_G, y_G, z_G) &\equiv \frac{1}{V_{boat}} \iiint_{boat} (x, y, z) \dd{V} \\ (x_B, y_B, z_B) &\equiv \frac{1}{V_{disp}} \iiint_{disp} (x, y, z) \dd{V} \end{aligned}

Where VboatV_{boat} is the volume of the whole boat, and VdispV_{disp} is the volume of liquid it displaces.

Whether a given equilibrium is stable is more complicated. Suppose the ship is tilted by a small angle θ\theta around the xx-axis, in which case the old waterline, previously in the z=0z = 0 plane, gets shifted to a new plane, namely:

z=sin(θ)yθy\begin{aligned} z = \sin(\theta) \: y \approx \theta y \end{aligned}

Then VdispV_{disp} changes by ΔVdisp\Delta V_{disp}, which is estimated below. If a point of the old waterline is raised by zz, then the displaced liquid underneath it is reduced proportionally, hence the sign:

ΔVdispwlzdAθwlydA=0\begin{aligned} \Delta V_{disp} \approx - \iint_{wl} z \dd{A} \approx - \theta \iint_{wl} y \dd{A} = 0 \end{aligned}

So VdispV_{disp} is unchanged, at least to first order in θ\theta. However, the shape of the displaced volume may have changed significantly. Therefore, the shift of the position of the buoyancy center from BB to BB' involves a correction ΔyB\Delta y_B in addition to the rotation by θ\theta:

yB=yBθzB+ΔyB\begin{aligned} y_B' = y_B - \theta z_B + \Delta y_B \end{aligned}

We find ΔyB\Delta y_B by calculating the virtual buoyancy center of the shape difference: on the side of the boat that has been lifted by the rotation, the center of buoyancy is “pushed” away due to the reduced displacement there, and vice versa on the other side. Consequently:

ΔyB=1VdispwlyzdAθVdispwly2dA=θIVdisp\begin{aligned} \Delta y_B = - \frac{1}{V_{disp}} \iint_{wl} y z \dd{A} \approx - \frac{\theta}{V_{disp}} \iint_{wl} y^2 \dd{A} = - \frac{\theta I}{V_{disp}} \end{aligned}

Where we have defined the so-called area moment II of the waterline as follows:

Iwly2dA\begin{aligned} \boxed{ I \equiv \iint_{wl} y^2 \dd{A} } \end{aligned}

Now that we have an expression for ΔyB\Delta y_B, the new center’s position yBy_B' is found to be:

yB=yBθ(zB+IVdisp)yBsin(θ)(zB+IVdisp)\begin{aligned} y_B' = y_B - \theta \Big( z_B + \frac{I}{V_{disp}} \Big) \approx y_B - \sin(\theta) \: \Big( z_B + \frac{I}{V_{disp}} \Big) \end{aligned}

This looks like a rotation by θ\theta around a so-called metacenter MM, with a height zMz_M known as the metacentric height, defined as:

zMzB+IVdisp\begin{aligned} \boxed{ z_M \equiv z_B + \frac{I}{V_{disp}} } \end{aligned}

Meanwhile, the position of MM is defined such that it lies on the line between the old centers GG and BB. Our calculation of yBy_B' has shown that the new BB' always lies below MM.

After the rotation, the boat is not in equilibrium anymore, because the new GG' is not directly above or below BB'. The force of gravity then causes a torque T\vb{T} given by:

T=(rGrB)×mg\begin{aligned} \vb{T} = (\vb{r}_G' - \vb{r}_B') \cross m \vb{g} \end{aligned}

Where g\vb{g} points downwards. Since the rotation was around the xx-axis, we are only interested in the xx-component TxT_x, which becomes:

Tx=(yGyB)mg=((yGθzG)(yBθzM))mg\begin{aligned} T_x = - (y_G' - y_B') m \mathrm{g} = - \big((y_G - \theta z_G) - (y_B - \theta z_M)\big) m \mathrm{g} \end{aligned}

With yG=yGθzGy_G' = y_G - \theta z_G being a simple rotation of GG. At the initial equilibrium yG=yBy_G = y_B, so:

Tx=θ(zGzM)mg\begin{aligned} T_x = \theta (z_G - z_M) m \mathrm{g} \end{aligned}

If zM<zGz_M < z_G, then TxT_x has the same sign as θ\theta, so T\vb{T} further destabilizes the boat. But if zM>zGz_M > z_G, then T\vb{T} counteracts the rotation, and the boat returns to the original equilibrium, leading us to the following stability condition:

zM>zG\begin{aligned} \boxed{ z_M > z_G } \end{aligned}

In other words, for a given boat design (or general shape) zGz_G and zMz_M can be calculated, and as long as they satisfy the above inequality, it will float stably in water (or any other fluid, although the buoyancy depends significantly on the density).


  1. B. Lautrup, Physics of continuous matter: exotic and everyday phenomena in the macroscopic world, 2nd edition, CRC Press.