Dimensional analysis from first principles

A playground swing and a weight hanging from a string are the same machine: a mass on a line, pulled aside and let go. Suppose a park had two swings, one hanging from chains four times as long as the other. Such a pair is rare, but it is easy to picture. Pull both back by the same small angle and release them together. Does the long swing take four times as long to come back? Twice as long? The same time? Pick an answer before you read on. Figure 1 checks it.

The question is a small case of a large one. If you make something bigger or smaller without changing its shape, does it behave the same way? Engineers ask this every time they test a scale model of a bridge, a dam or a ship, because the model is cheap and the real thing is not. The answer turns out to depend on which quantities the physics cares about, and there is a method for finding them that needs nothing more than the units of measurement. This article builds that method from the swings up. You need school algebra and the idea that a speed is a length divided by a time.

A swing four times as long#

A real swing with a child on it is complicated: the chains have weight, the seat has size, and the child moves. Strip that away and keep a small heavy bob on a light line that doesn't stretch, swinging without friction. That is an ideal pendulum, and it is what Figure 1 simulates. It solves the pendulum's equation of motion exactly, so the swings in the figure are not drawn by hand.

The left pendulum is four times as long. How long does it take to swing out and back, compared with the right one?

4 times2 timesThe same

Release both

The long pendulum takes twice as long, not four times as long. From the starting angle of 10 degrees, its period, the time to swing out and back, is 4.02 seconds; the short one's is 2.01 seconds. Change the length and the ratio of the periods follows the square root of the ratio of the lengths: nine times as long, three times the period. Raise the release angle and both periods grow, but their ratio stays exactly the square root, as long as both start from the same angle. Make the left bob four times as heavy, and nothing changes at all.

Two things need explaining: why a square root, and why mass doesn't matter.

Try it with a string

Tie a small heavy object, such as a nut or a key, to a thread. Hang it so that 25 cm of thread is free, pull it a little to one side, and time ten full swings with a phone's stopwatch. Then do the same with 1 m of thread. For small swings the ideal pendulum predicts 1.00 s and 2.01 s per swing, so about 10 s and 20 s for ten. Your times will differ a little, because the thread has some weight and the object has some size, but the ratio should come out close to 2.

What the units allow#

The period of an ideal pendulum can depend on only a few things: the length of the line, $L$; the strength of gravity, $g$; the mass of the bob, $m$; and the angle it is released from, $\theta_0$. Each of these is measured in units, and an equation that relates them has to come out in the same units on both sides. A period is a time, so whatever combination of $L$, $g$, $m$ and $\theta_0$ gives the period must be measured in seconds.

Look at the units. $L$ is in meters, $g$ in meters per second squared, $m$ in kilograms, and an angle has no units at all. Only one quantity contains kilograms, so no combination can cancel them: if the mass appeared, the period would be measured in something with kilograms in it, and it isn't. The mass drops out. That leaves $L$ and $g$ to make a time, and there is only one way to do it. Dividing $L$ by $g$ gives seconds squared, so

$$T = \sqrt{\frac{L}{g}}\; F(\theta_0),$$

where $F$ is some function of the release angle alone. The units can't tell us what $F$ is. They do tell us that two pendulums released from the same angle share the same $F$, so their periods are in the ratio $\sqrt{L_2/L_1}$. Four times the length gives $\sqrt 4 = 2$ times the period, without solving any equation of motion.

This is dimensional analysis: requiring every physical equation to have the same units on both sides, and using that to find the form of a relation before knowing the relation itself. The quantity $T\sqrt{g/L}$ has no units, so it is called dimensionless, and the argument says it depends on nothing but $\theta_0$.

The method has a clear limit. To find $F$, you have to solve the pendulum's equation of motion. For small swings, $F$ is close to $2\pi$, which gives the familiar $T = 2\pi\sqrt{L/g}$. At larger angles it grows: a pendulum released from 30 degrees takes 1.7% longer than the small-swing formula says, and one released from 60 degrees about 7% longer. Figure 1 uses the exact value, and the readout T√(g/L) shows it.

The general method: Buckingham's theorem

Write the units of each quantity as powers of mass (M), length (L) and time (T). The pendulum's five quantities give a table of exponents, one column each:

$$\begin{array}{c|ccccc} & T & L & g & m & \theta_0 \\ \hline \mathrm{M} & 0 & 0 & 0 & 1 & 0 \\ \mathrm{L} & 0 & 1 & 1 & 0 & 0 \\ \mathrm{T} & 1 & 0 & -2 & 0 & 0 \end{array}$$

The table has 5 columns and rank 3. Buckingham (1914) showed that a relation among $n$ quantities whose table has rank $r$ can always be rewritten as a relation among $n - r$ independent dimensionless groups. Here $5 - 3 = 2$: for example $T\sqrt{g/L}$ and $\theta_0$, so $T\sqrt{g/L} = F(\theta_0)$. The groups are not unique (any product of powers of them is also dimensionless), and the theorem says nothing about $F$. It also depends on the list: leave out a quantity that matters, and the groups it produces are wrong.

The exact period

For an ideal pendulum released from rest at $\theta_0$, the period is

$$T = 4\sqrt{\frac{L}{g}}\; K\!\left(\sin\frac{\theta_0}{2}\right),$$

where $K$ is the complete elliptic integral of the first kind. It can be computed quickly as $T = 2\pi\sqrt{L/g}\,/\,\mathrm{AGM}\big(1, \cos(\theta_0/2)\big)$, with AGM the arithmetic-geometric mean, which is how the figure computes it. Compared with $2\pi\sqrt{L/g}$, the period is 1.002 times as long at 10 degrees, 1.017 at 30, 1.073 at 60 and 1.180 at 90.

So the swing problem has one dimensionless group besides the period, and two pendulums with the same release angle match it automatically. That makes the long one a faithful scale model of the short one, with time stretched by the square root of the length ratio. Does a smaller copy of anything behave like the original, then, as long as the shape is the same?

Area and volume don't keep pace#

Take a steel beam and build a model of it at a tenth of the size: every length divided by 10, the same shape, the same steel. What fraction of the original's weight does the model have: a tenth, a hundredth, or a thousandth? Decide before reading on.

It is a thousandth. Weight follows volume, and volume is length times width times height, so each of the three factors of 10 counts. The cross-section, a width times a height, shrinks by a factor of 100. In general, if every length is multiplied by a scale factor $s$, areas are multiplied by $s^2$ and volumes by $s^3$.

Applying the length ratio to everything is a tempting shortcut, and it gives wrong answers here. Researchers who gave word problems about the lengths and areas of similar figures to students aged 12 to 13 and 15 to 16 found a strong tendency to apply the length ratio to the areas as well (De Bock, Verschaffel & Janssens, 1998).

The difference matters for anything that carries its own weight. The weight of a part grows as $s^3$, but the cross-section that holds it up grows only as $s^2$. Their ratio, a stress, grows as $s$. For the simplest case, a column of height $H$ standing on its base, the stress at the bottom is $\rho g H$, where $\rho$ is the density: double the height and you double the stress, whatever the width. A model in the same material carries less stress from its own weight than the original does, in proportion to its size. How much less does it sag, then, and can that be fixed?

The beam that sags too little#

Take a cantilever: a beam fixed at one end and free at the other, like a diving board. Let it be 10 m long, in steel, with a solid square cross-section 0.5 m on a side, carrying nothing but its own weight. It sags at the free end by about 22 mm, a little over two thousandths of its length. Now build the 1:10 model: 1 m long, 50 mm square, the same steel. As a fraction of its length, does the model sag the same, a tenth as much, or a hundredth? Decide, then choose in Figure 2.

The model is the same shape, a tenth of the size, in the same steel. Its sag as a fraction of its length is:

The same1/101/100

The model sags a tenth as much, relative to its length. From the standard beam formula, the tip of a cantilever under a uniform load of $w$ per meter sags by $wL^4/(8EI)$, where $E$ is the stiffness of the material and $I$ measures how the cross-section resists bending. For the square section, that gives

$$\frac{\delta}{L} = \frac{3}{2}\,\frac{\rho g L}{E}\left(\frac{L}{b}\right)^{2}.$$

The shape fixes $L/b$, so the model and the original differ only in the group $\rho g L/E$. It has no units: $\rho g L$ is the stress a column of the material as tall as the beam is long would put on its own base, and $E$ is a stress too. It compares the material's weight with its stiffness, and at a tenth of the length, in the same steel, it is a tenth as large. The stress at the root drops the same way, from 46.2 MPa to 4.62 MPa.

To make the model behave like the original, $\rho g L/E$ has to match. There are two ways to change it at the smaller $L$.

• A different material. It needs ten times the ratio of density to stiffness. Aluminum barely helps: it is lighter than steel but also less stiff, in almost the same proportion. Polyethylene is far more flexible for its weight and matches near 1:25, although a real polymer also creeps over time, which the figure ignores.
• More gravity. Spin the 1:10 model in a centrifuge so that its effective gravity is 10 times Earth's, and its own weight produces the same stresses as in the original. Geotechnical engineers use this to test models of soil structures, whose behavior depends strongly on the stresses from their own weight (Taylor, 1995). In Figure 2, set the gravity to 10 times for the 1:10 model (n times for a 1:n model) and the ratio in the readout returns to 1.

The material and gravity were free to change here. A harder case is one where two dimensionless groups have to match and there is only one thing to adjust.

A model ship has one knob#

A ship 100 m long sails at 10 m/s. Engineers want to measure its resistance before it is built, so they tow a 1:100 model, 1 m long, through a tank of water. Two physical effects shape the resistance.

The ship makes waves, and waves are driven by gravity: the pattern a hull makes depends on its speed compared with the speed of gravity waves as long as the hull. The dimensionless group is the Froude number, $\mathrm{Fr} = U/\sqrt{gL}$, with $U$ the speed and $L$ the length.

The water also drags along the hull through its viscosity. That depends on the Reynolds number, $\mathrm{Re} = UL/\nu$, where $\nu$ is the water's kinematic viscosity, about $1.0\times10^{-6}$ m²/s at room temperature.

To make the same pattern of waves as the ship, should the model run faster than 10 m/s, at the same speed, or slower? Decide, then choose in Figure 3.

To make the same pattern of waves as the ship, the 1:100 model must run:

FasterAt the same speedSlower

Matching the waves takes a slower model. Equal Froude numbers with the same $g$ need $U_m/U_p = \sqrt{L_m/L_p}$, so the 1 m model runs at 1 m/s. At that speed its Reynolds number is a thousandth of the ship's: a tenth for the speed, times a hundredth for the length. Switch the figure to match viscosity instead, and the model has to run at 1,000 m/s, which is absurd and would ruin the waves anyway. In the same water, under the same gravity, no single speed matches both. Matching both at once would need a model fluid with a kinematic viscosity $(L_m/L_p)^{3/2}$ times water's, a thousandth of it at 1:100.

Towing tanks don't give up on models because of this. They run the model at the ship's Froude number, so the waves are right, and handle the viscous part separately. The International Towing Tank Conference's procedure splits the model's measured resistance into a friction part, taken from a formula in the Reynolds number, the 1957 correlation line $C_F = 0.075/(\log_{10}\mathrm{Re} - 2)^2$, and the rest, which is mostly wave-making (ITTC, 2017). The rest carries over to the ship unchanged, while friction is recomputed at the ship's Reynolds number from the same line, with allowances for roughness and other effects. It is a deliberate, standardized correction for a group that can't be matched, not a flaw in the test, and its uncertainty is part of what model testing has to judge.

Back to the swings#

Why, then, did the swings work so cleanly? Return to Figure 1 and check Measure time in units of √(L/g). The two curves, which in seconds have different lengths, now lie on top of each other: in that unit of time, the two pendulums make exactly the same motion. The long swing is a perfect scale model of the short one.

It works because in the ideal pendulum nothing else carries a scale. The only length is $L$, the only acceleration is $g$, and the release angle has no units. Once the angles match, every dimensionless group matches, and the motions are the same up to a change of the unit of time. The beam failed because the material brought its own ratio of stiffness to density, which doesn't shrink with the model. The ship failed because water brings its own viscosity. A real playground swing has more of these, too: the weight of the chains, air drag on the rider, and a rider who pumps. Each adds a group, and a model has to match all the ones that matter.

Where it stops working#

Dimensional analysis is only as good as the list of quantities you start from. Leave out one that matters, such as the viscosity of the water, and the groups it gives you are incomplete; it can't warn you. Choosing the list takes physics, not algebra.

It gives the form of a relation, not the relation. It found $T = \sqrt{L/g}\,F(\theta_0)$ but not $2\pi$, and not how $F$ grows with the angle. Those need an equation of motion, an experiment, or both.

And when the groups can't all be matched, as for the ship, the result of a model test has to be corrected, and the corrections are partly empirical. They come with uncertainty that the method itself does not quantify.

In one sentence#

A smaller copy behaves like the original only when every dimensionless group that the physics depends on has the same value in both, and matching the shape fixes only the ratios of lengths.

References#