Amateur League

The results are in

As promised in the last post, we have now calculated the lengths of aluminium tubing (of a certain grade) required to produce a set of tubular chimes - and even that is much too grandiose a term for the rather toy-like set of factory-cut hollow rods purchased, at the very reasonable price of about thirty quid (including delivery), for the experiment.

The material arrived the next day, and I measured the lengths principally because I needed accurate data for my calculations below - and only secondarily to check that they were as ordered (which they pretty much were, and certainly within the vendor's advertised tolerance). I also weighed them, principally to check that the density matched the theoretical density used in the frequency calculations - which of course it did, to within about a half of a percent. As I have no machinery to check its Young's modulus of elasticity, I continue to rely on its wikipedially reported value.

Measuring their frequencies was the most difficult task, and involved suspending the tubes with a hot-glued thread connecting the top of each tube to a shelf, and recording them - with my trusty H2N Zoom recorder - being lightly struck by a similar aluminium tube. Here's a composite of the noises made by this 'instrument'.

Thirteen Tings

Do not be fooled by the levels. Although the samples in this recording are untreated (apart from being chopped up into segments), those small cylinders cannot shift very much air and the sounds are quite quiet. It's the H2N being held (by hand) up close.

The samples were brought into Cockos Reaper whereupon the use of its built-in band-pass filter, ReaEQ, (to cut out the highly audible harmonics) and frequency counter (ReaTune) plugins proved adequate to the task. Here are all the measurements, plus a picture of the set of tubes.

lengthweightfrequency$\kappa$13 Pipelings
25110.6351432.4 Thirteen Chimes

As expected, the frequencies were not even close to the design targets (respectively 165, 175, 185, 196, 208, 220, 233, 247, 262, 277, 294, 311, 330) and were - on average - 3.14 times higher (to the three significant figures employed throughout this exercise). The final column in the above table is $\kappa$, the diffusivity mentioned in the previous post, and is calculated as $L^2 \times f$ for each row ($L$ being the metre value). The average value for $\kappa$ is 32.2, the same multiple (3.14, weirdly close to $\pi$) of the value of 10.2 expected.

But - and also as expected - the instrument does indeed play a very serviceable chromatic scale. It may not start on the 165 Hertz concert pitch E (as requested) but instead a rather higher 514 Hertz non-concert pitch flat C, not quite a quarter tone below the 'proper' one. That's quite a way out, but our measured value of $\kappa$ could now be used to calculate a new series of $L$ values from $L=\sqrt{\kappa/f}$ - provided of course we were to use exactly the same material. All lengths would, accordingly, be multiplied by a factor of 1.77 (the square root of 3.14).

For this particular kind of aluminium alloy the previous post shows that $\kappa = 1420 \sqrt{D_o^2 + D_i^2}$, where $D_o = 0.006 m$ (i.e. 6mm) and $D_i = 0.004 m$ (i.e. 4mm). It's perfectly reasonable to incorporate the 'blame' for the (fudge) factor of 3.14 into the 'material constant' of 1420 and claim that in practice this constant should be 4450. This leaves us free to use larger inner and outer diameters as long as it's the same alloy to allow us to shift more air and make louder instruments. For example, if we have our heart set on a more robust 165 Hertz concert pitch E, we choose a larger diameter tubing (say half an inch with a 6.2mm bore). This would give us a new value for $\kappa$ - based on $D_o = 0.0127\,m$ and $D_i=0.0062\,m$ - of $4450 \times 0.01413 = 63$. From this we'd calculate a length of 618 mm.

And so on


So why was the calculation incorrect? Why would we need to use 4450 rather than 1420? As the 1420 was itself calculated - as $(22.4/8\pi) \sqrt{E/\rho}$ - from easily verifiable physical properties of a particular alloy of aluminium, the only places where the discrepancy could occur are in the value of 22.4 and from the way that $I$ was defined (in terms of $D_o$ and $D_i$). Perhaps an engineer can let me know where I went wrong?

Finally, here's a plot of frequency versus length of the tubes received in the mail. The least squares fit (from LibreCalc) is $f = 33.6 \times L^{-1.97}$, satisfyingly close to the model of $L^2 f = \kappa$ (with my measured average of $\kappa=32.2$).

length frequency plot
Thirteen frequencies of chromatic tubical pieces


Amateur Hour

Overtaken by a sudden urge to manufacture a cheap musical instrument, this blogger was minded to research the solutions to the wave equation in cylindrical coordinates, with a view to constructing a set of tubular chimes ('bell' seeming a little grandiose a term for the small scale anticipated).

At the outset I would urge anyone in search of a real instrument to go to a professional (such as Matt Nolan, but do search around for others) because I have zero confidence that what I will receive will be anywhere near the frequencies I've specified, although I do expect them to be in tune(ish) relative to each other. We'll get to the reasons why soon enough - it's certainly not due to the metal merchant but to me, the utter amateur.


Engineers have already collapsed the mathematics into just the amount necessary for the problem at hand - typically the determination of the resonant frequency, $f$, of a pipe made out of a material of known elasticity and density. Consequently we can start with a formula such as is provided by Ron Frend

\begin{equation} f = \frac{1}{2\pi} \cdot 22.4 \cdot \sqrt{\frac{EI}{\mu L^4}} \end{equation}

where we know things like

  • $E$ is the Young's Modulus of Elasticity
  • $I$ is the 4th Polar Moment of Inertia
  • $\mu$ is the Mass per unit length
The formula requires that we use $\mu$ as mass per unit length. The kind of tubing we can get from suppliers is typically specified with an inner and outer diameter, and so the area of the material in the pipe's cross-section is $\pi \cdot (D_o^2 - D_i^2) / 4$, giving us the required $\mu = \pi\rho \cdot (D_o^2 - D_i^2) / 4$, where $\rho$ is the material's density.

Our source provides a formula, $I=0.049 \times(D_o^4 - D_i^4)$. This is the first place where I encounter a problem since other authorities give $I=\pi(D_o^4 - D_i^4)/32$ which evaluates to twice the value of the one provided here (and by others - he's by no means alone) since $0.049 = \pi / 64$.

It's why I have little confidence that I'm going to get the frequencies I imagine I'm asking for. Regardless, we may reformulate - taking this additional information into account and switching the dependency of $f$ upon $L$ to that of $L$ upon $f$:

\begin{equation} \begin{split} L^2 &= \frac{1}{2\pi} \cdot \frac{22.4}{f} \cdot \sqrt{\frac{\pi E \cdot (D_o^4 - D_i^4) \cdot 4}{\pi\rho \cdot (D_o^2 - D_i^2) \cdot 64}}\\ &= \frac{22.4}{8 \pi f} \cdot \sqrt{\frac{E}{\rho} (D_o^2 + D_i^2)} \end{split} \end{equation}

For a typical grade 6063 Aluminium sold by commercial metal merchants, we use $E = 6.83\times 10^9\,kg\,m^{-1}s^{-2}$ and $\rho = 2690\,kg\,m^{-3}$ and employ the three significant digit formula:

\begin{equation} L^2 \approx \frac{1420}{f}\sqrt{D_o^2 + D_i^2} \end{equation}

Consequently, this adventurer has ordered the following cuts of 4mm inner diameter and 6mm outer diameter with an intent to construct a full chromatic scale

f (Hz)L (mm)
165 E F F# G G# A B♭ B C C# D E♭ E 249

As stated, we don't expect the frequencies to be correct - principally due to some conflicting information about the formulæ found online. The pipe vendor guarantees only a cut accuracy of ±2mm. (The formula indicates a discrepancy of 2% in $f$ for every 1% discrepancy in $L$). However, even if the 'base' frequency is way off, we'd expect the entire system to be mostly in tune with itself since the frequency ratios should still be chromatic if the length ratios are as directed.

When the material arrives, we shall measure the actual lengths and frequencies and see how far off we are. As an experiment (even if dire failure frequency-wise) it should still give us a constant value $\kappa$ - a kind of diffusivity - for this particular material which we can use in a formula $L^2 f = \kappa$ to produce a more accurate set, possibly in an exotic scale of 19 TET with a complement of 20 tubes.

My ineptitude may be amusing. Watch this space.