## 20200420

### 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.

### Pragmatics

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

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

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 $f$:

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

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:

$$L^2 \approx \frac{1420}{f}\sqrt{D_o^2 + D_i^2}$$

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)
165 249
175241
185235
196228
208221
220215
233209
247203
262197
277192
294186
311181
330176

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.