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-rw-r--r--content/know/concept/archimedes-principle/index.pdc11
-rw-r--r--content/know/concept/metacentric-height/index.pdc185
-rw-r--r--content/know/concept/metacentric-height/sketch.pngbin0 -> 171361 bytes
-rw-r--r--content/know/concept/runge-kutta-method/index.pdc267
4 files changed, 457 insertions, 6 deletions
diff --git a/content/know/concept/archimedes-principle/index.pdc b/content/know/concept/archimedes-principle/index.pdc
index 3a063ec..fb91b67 100644
--- a/content/know/concept/archimedes-principle/index.pdc
+++ b/content/know/concept/archimedes-principle/index.pdc
@@ -39,7 +39,7 @@ $$\begin{aligned}
Where $\va{g}$ is the gravitational field,
and $\rho_\mathrm{b}$ is the density of the body.
Meanwhile, the pressure $p$ of the surrounding fluid exerts a force
-on the surface $S$ of $V$:
+on the entire surface $S$ of $V$:
$$\begin{aligned}
\va{F}_p
@@ -75,18 +75,17 @@ and zero on the "non-submerged" side, we find:
$$\begin{aligned}
0
- = \mathrm{g} (\rho_\mathrm{b} - \rho_\mathrm{f}) V
= \mathrm{g} (m_\mathrm{b} - m_\mathrm{f})
\end{aligned}$$
-In other words, the mass $m_\mathrm{b}$ of the submerged portion $V$ of the body,
+In other words, the mass $m_\mathrm{b}$ of the entire body
is equal to the mass $m_\mathrm{f}$ of the fluid it displaces.
This is the best-known version of Archimedes' principle.
-Note that if $\rho_\mathrm{b} > \rho_\mathrm{f}$, then,
+Note that if $\rho_\mathrm{b} > \rho_\mathrm{f}$,
+then the displaced mass $m_\mathrm{f} < m_\mathrm{b}$
even if the entire body is submerged,
-the displaced mass $m_\mathrm{f} < m_\mathrm{b}$,
-and the object will continue to sink.
+and the object will therefore continue to sink.
diff --git a/content/know/concept/metacentric-height/index.pdc b/content/know/concept/metacentric-height/index.pdc
new file mode 100644
index 0000000..1fc6aca
--- /dev/null
+++ b/content/know/concept/metacentric-height/index.pdc
@@ -0,0 +1,185 @@
+---
+title: "Metacentric height"
+firstLetter: "M"
+publishDate: 2022-03-11
+categories:
+- Physics
+- Fluid mechanics
+
+date: 2021-05-08T19:03:36+02:00
+draft: false
+markup: pandoc
+---
+
+# Metacentric height
+
+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%;display:block;margin:auto;">
+</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.
diff --git a/content/know/concept/metacentric-height/sketch.png b/content/know/concept/metacentric-height/sketch.png
new file mode 100644
index 0000000..f33fe36
--- /dev/null
+++ b/content/know/concept/metacentric-height/sketch.png
Binary files differ
diff --git a/content/know/concept/runge-kutta-method/index.pdc b/content/know/concept/runge-kutta-method/index.pdc
new file mode 100644
index 0000000..ac2eabf
--- /dev/null
+++ b/content/know/concept/runge-kutta-method/index.pdc
@@ -0,0 +1,267 @@
+---
+title: "Runge-Kutta method"
+firstLetter: "R"
+publishDate: 2022-03-10
+categories:
+- Mathematics
+- Numerical methods
+
+date: 2022-03-07T14:10:18+01:00
+draft: false
+markup: pandoc
+---
+
+# Runge-Kutta method
+
+A **Runge-Kutta method** (RKM) is a popular approach
+to numerically solving systems of ordinary differential equations.
+Let $\vb{x}(t)$ be the vector we want to find,
+governed by $\vb{f}(t, \vb{x})$:
+
+$$\begin{aligned}
+ \vb{x}'(t)
+ = \vb{f}\big(t, \vb{x}(t)\big)
+\end{aligned}$$
+
+Like in all numerical methods, the $t$-axis is split into discrete steps.
+If a step has size $h$, then as long as $h$ is small enough,
+we can make the following approximation:
+
+$$\begin{aligned}
+ \vb{x}'(t) + a h \vb{x}''(t)
+ &\approx \vb{x}'(t \!+\! a h)
+ \\
+ &\approx \vb{f}\big(t \!+\! a h,\, \vb{x}(t \!+\! a h)\big)
+ \\
+ &\approx \vb{f}\big(t \!+\! a h,\, \vb{x}(t) \!+\! a h \vb{x}'(t) \big)
+\end{aligned}$$
+
+For sufficiently small $h$,
+higher-order derivates can also be included,
+albeit still at $t \!+\! a h$:
+
+$$\begin{aligned}
+ \vb{x}'(t) + a h \vb{x}''(t) + b h^2 \vb{x}'''(t)
+ &\approx \vb{f}\big(t \!+\! a h,\, \vb{x}(t) \!+\! a h \vb{x}'(t) \!+\! b h^2 \vb{x}''(t) \big)
+\end{aligned}$$
+
+Although these approximations might seem innocent,
+they actually make it quite complicated to determine the error order of a given RKM.
+
+Now, consider a Taylor expansion around the current $t$,
+truncated at a chosen order $n$:
+
+$$\begin{aligned}
+ \vb{x}(t \!+\! h)
+ &= \vb{x}(t) + h \vb{x}'(t) + \frac{h^2}{2} \vb{x}''(t) + \frac{h^3}{6} \vb{x}'''(t) + \:...\, + \frac{h^n}{n!} \vb{x}^{(n)}(t)
+ \\
+ &= \vb{x}(t) + h \bigg[ \vb{x}'(t) + \frac{h}{2} \vb{x}''(t) + \frac{h^2}{6} \vb{x}'''(t) + \:...\, + \frac{h^{n-1}}{n!} \vb{x}^{(n)}(t) \bigg]
+\end{aligned}$$
+
+We are free to split the terms as follows,
+choosing real factors $\omega_{mj}$ subject to $\sum_{j} \omega_{mj} = 1$:
+
+$$\begin{aligned}
+ \vb{x}(t \!+\! h)
+ &= \vb{x} + h \bigg[ \sum_{j = 1}^{N_1} \omega_{1j} \, \vb{x}'
+ + \frac{h}{2} \sum_{j = 1}^{N_2} \omega_{2j} \, \vb{x}''
+ + \:...\, + \frac{h^{n-1}}{n!} \sum_{j = 1}^{N_n} \omega_{nj} \, \vb{x}^{(n)} \bigg]
+\end{aligned}$$
+
+Where the integers $N_1,...,N_n$ are also free to choose,
+but for reasons that will become clear later,
+the most general choice for an RKM is $N_1 = n$, $N_n = 1$, and:
+
+$$\begin{aligned}
+ N_{n-1}
+ = N_n \!+\! 2
+ ,\quad
+ \cdots
+ ,\quad
+ N_{n-m}
+ = N_{n-m+1} \!+\! m \!+\! 1
+ ,\quad
+ \cdots
+ ,\quad
+ N_{2}
+ = N_3 \!+\! n \!-\! 1
+\end{aligned}$$
+
+In other words, $N_{n-m}$ is the $m$th triangular number.
+This is not so important,
+since this is not a practical way to describe RKMs,
+but it is helpful to understand how they work.
+
+
+## Example derivation
+
+For example, let us truncate at $n = 3$,
+such that $N_1 = 3$, $N_2 = 3$ and $N_3 = 1$.
+The following derivation is very general,
+except it requires all $\alpha_j \neq 0$.
+Renaming $\omega_{mj}$, we start from:
+
+$$\begin{aligned}
+ \vb{x}(t \!+\! h)
+ &= \vb{x} + h \bigg[ (\alpha_1 + \alpha_2 + \alpha_3) \, \vb{x}'
+ + \frac{h}{2} (\beta_2 + \beta_{31} + \beta_{32}) \, \vb{x}''
+ + \frac{h^2}{6} \gamma_3 \, \vb{x}''' \bigg]
+ \\
+ &= \vb{x} + h \bigg[ \alpha_1 \vb{x}'
+ + \Big( \alpha_2 \vb{x}' + \frac{h}{2} \beta_2 \vb{x}'' \Big)
+ + \Big( \alpha_3 \vb{x}' + \frac{h}{2} (\beta_{31} + \beta_{32}) \vb{x}'' + \frac{h^2}{6} \gamma_3 \vb{x}''' \Big) \bigg]
+\end{aligned}$$
+
+As discussed earlier, the parenthesized expressions
+can be approximately rewritten with $\vb{f}$:
+
+$$\begin{aligned}
+ \vb{x}(t \!+\! h)
+ = \vb{x} + h &\bigg[ \alpha_1 \vb{f}(t, \vb{x})
+ + \alpha_2 \vb{f}\Big( t \!+\! \frac{h \beta_2}{2 \alpha_2}, \;
+ \vb{x} \!+\! \frac{h \beta_2}{2 \alpha_2} \vb{x}' \Big)
+ \\
+ & + \alpha_3 \vb{f}\Big( t \!+\! \frac{h (\beta_{31} \!\!+\!\! \beta_{32})}{2 \alpha_3}, \;
+ \vb{x} \!+\! \frac{h \beta_{31}}{2 \alpha_3} \vb{x}' \!+\! \frac{h \beta_{32}}{2 \alpha_3} \vb{x}'
+ \!+\! \frac{h^2 \gamma_3}{6 \alpha_3} \vb{x}'' \Big) \bigg]
+ \\
+ = \vb{x} + h &\bigg[ \alpha_1 \vb{k}_1
+ + \alpha_2 \vb{f}\Big( t \!+\! \frac{h \beta_2}{2 \alpha_2}, \;
+ \vb{x} \!+\! \frac{h \beta_2}{2 \alpha_2} \vb{k}_1 \!\Big)
+ \\
+ & + \alpha_3 \vb{f}\Big( t \!+\! \frac{h (\beta_{31} \!\!+\!\! \beta_{32})}{2 \alpha_3}, \;
+ \vb{x} \!+\! \frac{h \beta_{31}}{2 \alpha_3} \vb{k}_1 \!+\! \frac{h \beta_{32}}{2 \alpha_3}
+ \vb{f}\Big( t \!+\! \frac{h \gamma_3}{3 \beta_{32}}, \;
+ \vb{x} \!+\! \frac{h \gamma_3}{3 \beta_{32}} \vb{k}_1 \!\Big) \!\Big) \bigg]
+\end{aligned}$$
+
+Here, we can see an opportunity to save some computational time
+by reusing an evaluation of $\vb{f}$.
+Technically, this is optional, but it would be madness not to,
+so we choose:
+
+$$\begin{aligned}
+ \frac{\beta_2}{2 \alpha_2}
+ = \frac{\gamma_3}{3 \beta_{32}}
+\end{aligned}$$
+
+Such that the next step of $\vb{x}$'s numerical solution is as follows,
+recalling that $\sum_{j} \alpha_j = 1$:
+
+$$\begin{aligned}
+ \boxed{
+ \vb{x}(t \!+\! h)
+ = \vb{x}(t) + h \Big( \alpha_1 \vb{k}_1 + \alpha_2 \vb{k}_2 + \alpha_3 \vb{k}_3 \Big)
+ }
+\end{aligned}$$
+
+Where $\vb{k}_1$, $\vb{k}_2$ and $\vb{k}_3$ are different estimates
+of the average slope $\vb{x}'$ between $t$ and $t \!+\! h$,
+whose weighted average is used to make the $t$-step.
+They are given by:
+
+$$\begin{aligned}
+ \boxed{
+ \begin{aligned}
+ \vb{k}_1
+ &\equiv \vb{f}(t, \vb{x})
+ \\
+ \vb{k}_2
+ &\equiv \vb{f}\bigg( t + \frac{h \beta_2}{2 \alpha_2}, \;
+ \vb{x} + \frac{h \beta_2}{2 \alpha_2} \vb{k}_1 \bigg)
+ \\
+ \vb{k}_3
+ &\equiv \vb{f}\bigg( t + \frac{h (\beta_{31} \!\!+\!\! \beta_{32})}{2 \alpha_3}, \;
+ \vb{x} + \frac{h \beta_{31}}{2 \alpha_3} \vb{k}_1 + \frac{h \beta_{32}}{2 \alpha_3} \vb{k}_2 \bigg)
+ \end{aligned}
+ }
+\end{aligned}$$
+
+Despite the contraints on $\alpha_j$ and $\beta_j$,
+there is an enormous freedom of choice here,
+all leading to valid RKMs, although not necessarily good ones.
+
+
+## General form
+
+A more practical description goes as follows:
+in an $s$-stage RKM, a weighted average is taken
+of up to $s$ slope estimates $\vb{k}_j$ with weights $b_j$.
+Let $\sum_{j} b_j = 1$, then:
+
+$$\begin{aligned}
+ \boxed{
+ \vb{x}(t \!+\! h)
+ = \vb{x}(t) + h \sum_{j = 1}^{s} b_j \vb{k}_j
+ }
+\end{aligned}$$
+
+Where the estimates $\vb{k}_1, ..., \vb{k}_s$
+depend on each other, and are calculated one by one as:
+
+$$\begin{aligned}
+ \boxed{
+ \vb{k}_m
+ = \vb{f}\bigg( t + h c_m,\; \vb{x} + h \sum_{j = 1}^{m - 1} a_{mj} \vb{k}_j \bigg)
+ }
+\end{aligned}$$
+
+With $c_1 = 1$ and $\sum_{j = 1} a_{mj} = c_m$.
+Writing this out for the first few $m$, the pattern is clear:
+
+$$\begin{aligned}
+ \vb{k}_1
+ &= \vb{f}(t, \vb{x})
+ \\
+ \vb{k}_2
+ &= \vb{f}\big( t + h c_2,\; \vb{x} + h a_{21} \vb{k}_1 \big)
+ \\
+ \vb{k}_3
+ &= \vb{f}\big( t + h c_3,\; \vb{x} + h (a_{31} \vb{k}_1 + a_{32} \vb{k}_2) \big)
+ \\
+ \vb{k}_4
+ &= \:...
+\end{aligned}$$
+
+The coefficients of a given RKM are usually
+compactly represented in a **Butcher tableau**:
+
+$$\begin{aligned}
+ \begin{array}{c|ccc}
+ 0 \\
+ c_2 & a_{21} \\
+ c_3 & a_{31} & a_{32} \\
+ \vdots & \vdots & \vdots & \ddots \\
+ c_s & a_{s1} & a_{s2} & \cdots & a_{s,s-1} \\
+ \hline
+ & b_1 & b_2 & \cdots & b_{s-1} & b_s
+ \end{array}
+\end{aligned}$$
+
+Each RKM has an **order** $p$,
+such that the global truncation error is $\mathcal{O}(h^p)$,
+i.e. the accumulated difference between the numerical
+and the exact solutions is proportional to $h^p$.
+
+The surprise is that $p$ need not be equal to the Taylor expansion order $n$,
+nor the stage count $s$.
+Typically, $s = n$ for computational efficiency, but $s \ge n$ is possible in theory.
+
+The order $p$ of a given RKM is determined by
+a complicated set of equations on the coefficients,
+and the lowest possible $s$ for a desired $p$
+is in fact only partially known.
+For $p \le 4$ the bound is $s \ge p$,
+whereas for $p \ge 5$ the only proven bound is $s \ge p \!+\! 1$,
+but for $p \ge 7$ no such efficient methods have been found so far.
+
+If you need an RKM with a certain order, look it up.
+There exist many efficient methods for $p \le 4$ where $s = p$,
+and although less popular, higher $p$ are also available.
+
+
+
+## References
+1. J.C. Butcher,
+ *Numerical methods for ordinary differential equations*, 3rd edition,
+ Wiley.