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+---
+title: "Metacentric height"
+date: 2022-03-11
+categories:
+- Physics
+- Fluid mechanics
+layout: "concept"
+---
+
+Consider an object with center of mass $G$,
+floating in a large body of liquid whose surface is flat at $z = 0$.
+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:
+
+$$\begin{aligned}
+    (x_0, y_0)
+    \equiv \frac{1}{A_{wl}} \iint_{wl} (x, y) \dd{A}
+\end{aligned}$$
+
+Where $A_{wl}$ is the cross-sectional area
+enclosed by the "waterline" around the "boat".
+Note that the boat's center of mass $G$
+does not coincide with the origin in general,
+as is illustrated in the following sketch
+of our choice of coordinate system:
+
+<a href="sketch.png">
+<img src="sketch.png" style="width:67%">
+</a>
+
+Here, $B$ is the **center of buoyancy**, equal to
+the center of mass of the volume of water displaced by the boat
+as per [Archimedes' principle](/know/concept/archimedes-principle/).
+At equilibrium, the forces of buoyancy $\vb{F}_B$ and gravity $\vb{F}_G$
+have equal magnitudes in opposite directions,
+and $B$ is directly above or below $G$,
+or in other words, $x_B = x_G$ and $y_B = y_G$,
+which are calculated as follows:
+
+$$\begin{aligned}
+    (x_G, y_G, z_G)
+    &\equiv \frac{1}{V_{boat}} \iiint_{boat} (x, y, z) \dd{V}
+    \\
+    (x_B, y_B, z_B)
+    &\equiv \frac{1}{V_{disp}} \iiint_{disp} (x, y, z) \dd{V}
+\end{aligned}$$
+
+Where $V_{boat}$ is the volume of the whole boat,
+and $V_{disp}$ 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 $\theta$ around the $x$-axis,
+in which case the old waterline, previously in the $z = 0$ plane,
+gets shifted to a new plane, namely:
+
+$$\begin{aligned}
+    z
+    = \sin(\theta) \: y
+    \approx \theta y
+\end{aligned}$$
+
+Then $V_{disp}$ changes by $\Delta V_{disp}$, which is estimated below.
+If a point of the old waterline is raised by $z$,
+then the displaced liquid underneath it is reduced proportionally,
+hence the sign:
+
+$$\begin{aligned}
+    \Delta V_{disp}
+    \approx - \iint_{wl} z \dd{A}
+    \approx - \theta \iint_{wl} y \dd{A}
+    = 0
+\end{aligned}$$
+
+So $V_{disp}$ is unchanged, at least to first order in $\theta$.
+However, the *shape* of the displaced volume may have changed significantly.
+Therefore, the shift of the position of the buoyancy center from $B$ to $B'$
+involves a correction $\Delta y_B$ in addition to the rotation by $\theta$:
+
+$$\begin{aligned}
+    y_B'
+    = y_B - \theta z_B + \Delta y_B
+\end{aligned}$$
+
+We find $\Delta y_B$ 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:
+
+$$\begin{aligned}
+    \Delta y_B
+    = - \frac{1}{V_{disp}} \iint_{wl} y z \dd{A}
+    \approx - \frac{\theta}{V_{disp}} \iint_{wl} y^2 \dd{A}
+    = - \frac{\theta I}{V_{disp}}
+\end{aligned}$$
+
+Where we have defined the so-called **area moment** $I$ of the waterline as follows:
+
+$$\begin{aligned}
+    \boxed{
+        I
+        \equiv \iint_{wl} y^2 \dd{A}
+    }
+\end{aligned}$$
+
+Now that we have an expression for $\Delta y_B$,
+the new center's position $y_B'$ is found to be:
+
+$$\begin{aligned}
+    y_B'
+    = y_B - \theta \Big( z_B + \frac{I}{V_{disp}} \Big)
+    \approx y_B - \sin(\theta) \: \Big( z_B + \frac{I}{V_{disp}} \Big)
+\end{aligned}$$
+
+This looks like a rotation by $\theta$ around a so-called **metacenter** $M$,
+with a height $z_M$ known as the **metacentric height**, defined as:
+
+$$\begin{aligned}
+    \boxed{
+        z_M
+        \equiv z_B + \frac{I}{V_{disp}}
+    }
+\end{aligned}$$
+
+Meanwhile, the position of $M$ is defined such that it lies
+on the line between the old centers $G$ and $B$.
+Our calculation of $y_B'$ has shown that the new $B'$ always lies below $M$.
+
+After the rotation, the boat is not in equilibrium anymore,
+because the new $G'$ is not directly above or below $B'$.
+The force of gravity then causes a torque $\vb{T}$ given by:
+
+$$\begin{aligned}
+    \vb{T}
+    = (\vb{r}_G' - \vb{r}_B') \cross m \vb{g}
+\end{aligned}$$
+
+Where $\vb{g}$ points downwards.
+Since the rotation was around the $x$-axis,
+we are only interested in the $x$-component $T_x$, which becomes:
+
+$$\begin{aligned}
+    T_x
+    = - (y_G' - y_B') m \mathrm{g}
+    = - \big((y_G - \theta z_G) - (y_B - \theta z_M)\big) m \mathrm{g}
+\end{aligned}$$
+
+With $y_G' = y_G - \theta z_G$ being a simple rotation of $G$.
+At the initial equilibrium $y_G = y_B$, so:
+
+$$\begin{aligned}
+    T_x
+    = \theta (z_G - z_M) m \mathrm{g}
+\end{aligned}$$
+
+If $z_M < z_G$, then $T_x$ has the same sign as $\theta$,
+so $\vb{T}$ further destabilizes the boat.
+But if $z_M > z_G$, then $\vb{T}$ counteracts the rotation,
+and the boat returns to the original equilibrium,
+leading us to the following stability condition:
+
+$$\begin{aligned}
+    \boxed{
+        z_M > z_G
+    }
+\end{aligned}$$
+
+In other words, for a given boat design (or general shape)
+$z_G$ and $z_M$ 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).
+
+
+
+## References
+1.  B. Lautrup,
+    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+    CRC Press.
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