Fly By Night

Dimensional Analysis

Pendulum Period

Simple Dimensional Analysis

Time \(T\) have the dimension of time \([T] = T\). If we want ot express time in terms of length \(l\) and acceleration \(a\), with \([l] = L\) and \([a] = L/T^2\). Thus the period must be:

\[ T \sim \sqrt{l/a} \]

The exact equation should be \(T = 2 \pi \sqrt{l/a}\). The \(2 \pi\) can be determined by experiment.

Or if we consider angular speed and with energy conservation equation, we can get the direct equation for \(\omega\) and \(T\):

\[ E = \frac{1}{2} m \dot{\theta}^2 + m g l (1 - \cos \theta) \simeq \frac{1}{2} m \dot \theta^2 + \frac{1}{2} m g l \theta^2 \]

The \(\simeq\) was obtained from the small angle approximation (with taylor series). Plug in \(\theta = \theta_0 \cos{\omega t}\), we get:

\[ E = \frac{1}{2} m \omega^2 \theta_0^2 \sin^2{\omega t} + \frac{1}{2} m g l \theta_0^2 \cos^2{\omega t} \]

In order for \(E\) to be a constant, \(m \omega^2 \theta_0^2 = m g l \theta_0^2\). Thus \(\omega = \sqrt{g/l}\), and \(T = 2 \pi / \omega = 2 \pi \sqrt{l/g}\).

Crossing the River

With this example, we can learn the significance of guessing the functions with the limit conditions. Usually in physics, the conditions are \(0\) or \(1\). By \(0\), we are meaning the usual \(0\) and the usual limit \(1/\infty\).

(Spent 1 hours drawing this with SVG, so complicated haha)

A person wants to cross the river starting from \(O\) to reach \(F\), crossing the river at point \(P\). She is moving with speed \(v_l\) on land and \(v_w\) in the water. Since we are talking about a human, we assume:

\[ \gamma \equiv \frac{v_l}{v_w} > 1 \]

So where should \(P\) be for here to reach \(F\) ASAP?

The Fly by Night Method

With some dimensional analysis, we can say that \(x\) should be something like this:

\[ x = w f(\gamma) \]

With \(\gamma \rightarrow 1\), the person (lets say Beque), should walk and swim at the same speed. Which make it no difference to swim or walk directly to the finish. So \(x \rightarrow \infty\) as \(\gamma \rightarrow 1\).

With \(\gamma \rightarrow \infty\), Beque should walk as fast as possible. She should spend the shortest amount of time swimming. So \(x \rightarrow 0\) as \(\gamma \rightarrow \infty\).

With these two limits, we can guess that \(f(\gamma) = \frac{1}{\gamma (\gamma - 1)}\) or \(\frac{1}{\gamma^2 - 1}\). Also, we restricted \(\gamma > 1\), the usual thing to do is to limit it with imaginary parts. So I find both \(f(\gamma) = \frac{1}{\sqrt{\gamma (\gamma - 1)}}\) and \(f(\gamma) = \frac{1}{\sqrt{\gamma^2 - 1}}\) reasonable.

The Proper Analysis

First, the distance from \(C\) to \(F\) is not important. So the time from \(P\) to \(F\), is

\[ \frac{L_{cf} - x}{v_l} = \frac{L_{cf}}{v_l} - \frac{x}{v_l} = T - \frac{x}{v_l} \]

The time from \(O\) to \(F\) is irrelevant in the choice of \(x\), and will disappear after differentiation. So the time from \(O\) to \(P\) is:

\[ t = T - \frac{x}{v_l} + \frac{L_{po}}{v_w} = T - \frac{x}{v_l} + \frac{\sqrt{w^2 + x^2}}{v_w} \]

Taking the derivative of \(t\) with respect to \(x\) and set it to \(0\):

\[ \frac{d t}{d x} = - \frac{1}{v_l} + \frac{x}{v_w \sqrt{x^2 + w^2}} = 0 \]

Solve for \(x\) and we get:

\[ x = \frac{w}{\sqrt{\gamma^2 - 1}} \]

This fits our guess perfectly! How Amazing!