Acoustics & Music Theory
why it sounds like that
Series

Air column — the series

13 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. What a 60 cm tube supports, by how its ends are closed. The first 6 modes of an open cylinder and a stopped cylinder, all of the same acoustic length. An open tube supports every whole multiple of the fundamental and reaches the second mode an octave up. A cylinder stopped at one end supports only the odd multiples and reaches its second mode 1902 cents up, which is a twelfth. Its fundamental is also an octave below the others, because it fits a quarter of a wavelength where they fit a half.

    A tube that skips every other partial

    Stop one end of a cylinder and half its modes vanish. That single fact about where the pressure has to be decides that a clarinet sounds hollow, that it plays an octave below its length suggests, and that it must cover nineteen semitones with fingers before it can overblow — while every other woodwind covers twelve.

    part 1 · instruments
  2. What a 60 cm tube supports, by how its ends are closed. The first 6 modes of a stopped cylinder and a cone, all of the same acoustic length. A cone supports every whole multiple of the fundamental and reaches the second mode an octave up. A cylinder stopped at one end supports only the odd multiples and reaches its second mode 1902 cents up, which is a twelfth. Its fundamental is also an octave below the others, because it fits a quarter of a wavelength where they fit a half.

    A cone is not a cylinder

    A saxophone has a reed at a closed end, exactly as a clarinet does, and it overblows at the octave rather than the twelfth. If the previous essay's argument were about reeds that would refute it. It is about geometry, and a cone closed at its apex has the complete harmonic series for a reason that takes one line of algebra and is genuinely surprising.

    part 2 · instruments
  3. The end correction, for a bore of radius 7.5 mm. How flat a tube sounds against what its physical length alone would predict, because the wave carries on past the opening before it turns round. The correction is 4.6 mm at every note — Levine and Schwinger's 0.6133 times the radius for an unflanged end — and the error it causes is 13 cents on a 60 cm sounding length and 52 cents on 15 cm. It is the same millimetres in both cases.

    The tube ends after it ends

    A wave does not turn round at the opening. It carries on into the room for about six-tenths of the bore radius and reflects there, so every tube is acoustically longer than it is. The correction is a fixed number of millimetres against a wavelength that halves every octave — a rounding error at the bottom of an instrument's range and most of a semitone at the top.

    part 3 · instruments
  4. A reed that shuts at 5000 pascals, and the air it lets through. Volume flow through the reed channel against the pressure across it, in the quasi-static model: Bernoulli flow through an opening that closes linearly with pressure. The flow peaks at 1667 pascals — exactly a third of the closing pressure, for any reed, because that is where the two effects balance — and it is 0.18 litres a second there. Everything to the right of that peak is the argument: the flow falls as the player blows harder, from 0.18 to 0.05 litres a second by 4500 pascals, and a resistance that behaves that way supplies energy instead of taking it. There is no reed inertia in this model, so it cannot squeak.

    The reed is a valve, not a vibrator

    Three essays here have said that a clarinet's reed does not choose the note, and none of them said what it does instead. It chops a steady stream of air, and past a third of the pressure that closes it the flow falls as the player blows harder — a resistance with the wrong sign, which is the only thing in the instrument capable of putting energy into an oscillation that is otherwise losing it.

    part 4 · instruments
  5. A wind instrument is a thermometer, and a string is not. Cents from the pitch at 20 degrees, against the temperature of the air inside a wind instrument and of a steel string, computed from the speed of sound as 343.2 metres a second times the square root of absolute temperature, and from a string's tension falling by Young's modulus times the expansion coefficient per degree. The wind slope is 2.95 cents a degree at 20 degrees, so 0.0 cents at 20, 11.7 cents at 24, 23.3 cents at 28, 34.7 cents at 32. The Pythagorean comma is reached at 28.1 degrees — 8.1 degrees of warming, which a wind instrument does from breath alone within a few minutes. The string goes the other way, 36 cents flat at 34 degrees, so the gap between the two sections opens at 5.4 cents a degree.

    A wind instrument is a thermometer

    Pitch goes as the square root of absolute temperature, so a warming clarinet sharpens by about three cents a degree and passes a Pythagorean comma after eight. The strings beside it go flat as they warm. Nobody chose either number, no temperament addresses either of them, and together they are twice the size of the discrepancy this whole field is named after.

    part 5 · tuning
  6. What gets out of an opening, for 4 openings. The fraction of the wave's energy radiated at an open end against frequency, in the baffled-piston model — the radiation resistance of a circular piston, normalised to the tube's own impedance. Each curve runs from nothing at the bottom, where the opening is far smaller than a wavelength and the wave simply turns round, to everything above ka ≈ 2. Half the energy leaves at 6364 Hz for a flute's embouchure end (radius 10 mm), 2015 Hz for a clarinet's bell (radius 30 mm), 975 Hz for a trumpet's bell (radius 62 mm), 403 Hz for a horn's bell (radius 150 mm). The crossover goes as one over the radius, so the widest and narrowest here are 15.8 times apart in frequency. The same number decides how strongly the tube resonates and how much sound it makes, which is why a bell cannot brighten an instrument without also weakening its own resonances.

    The bell decides what gets out

    A tube resonates because the wave turns round at the open end, and it is audible because some of the wave does not. Those are the same number with opposite signs. One quantity — the size of the opening against a wavelength — decides how loud an instrument is, how bright it is and how directional it is, and a bell moves the boundary rather than removing it.

    part 6 · timbre
  7. Blowing harder is playing sharper. How far the played note is pulled from the bore's own resonance, against blowing pressure, for an air jet crossing a 4 mm embouchure. The jet's preferred frequency goes as the square root of the pressure, so it rises by a factor of 3.16 across the tenfold pressure range drawn, and the bore holds it to 272 cents of that — from -174 at the quietest to +98 at the loudest, about the player's own nominal. The rows below give the same pull for a drive one semitone sharp of the bore, for four valves: a clarinet reed is damped by the lip and pulls 4.9 cents, brass lips are not and pull 23.6.

    Blowing harder is playing sharper

    Every frequency computed so far for an air column is a resonance of the tube, and no wind instrument plays at its bore's resonance. It plays between the bore and whatever is driving it, weighted by how sharply each is tuned — and a flute's driver is an air jet whose own preferred frequency goes as the square root of the blowing pressure. Across a tenfold pressure range the jet's preference rises by a factor of 3.16 and the bore holds it to 112 cents of that. A wind player's dynamics and their intonation are one control, and the size of the coupling between them is the valve's Q.

    part 7 · tuning
  8. Which partials of a natural horn can be lipped into tune. Every partial of the natural series against the nearest note of twelve equal, in cents, with the band the player's lips can actually move it drawn around each one. The band is ±23.1 cents, computed from the Q-weighted mean of a bore at Q 40 and lips at Q 12 rather than chosen. 4 of the first 16 partials fall outside it: 7, 11, 13, 14. The worst is the 11th at -48.7 cents, which would need lips of Q 38 to reach — comparable with the bore's own, at which point the bore has stopped deciding the pitch at all.

    The partial the lips cannot reach

    A wind instrument plays between its valve's preferred frequency and its bore's, which suggested that the natural trumpet's notorious eleventh partial would therefore turn out to be a statement about the lips. It is not. Run the same Q-weighted mean over the whole series and the lips can move that partial twenty-three cents, which is a quarter of what it needs — and the Q that would fix it is the Q at which the bore stops choosing the note.

    part 8 · tuning
  9. What a hand in the bell buys, and what it costs. How far the instrument flattens and how much radiation it loses, against the share of the bell's mouth the hand blocks, for an F horn's 15 centimetre mouth on a 3.7 metre acoustic length. The two curves are the same aperture radius read twice: a narrower mouth adds inertance, which lengthens the tube, and is acoustically smaller, which stops radiating. The mark is the 25.6 cents left owed earlier on the eleventh partial — it needs 62 per cent of the mouth blocked and costs 3.8 decibels of radiated power. Fully stopping is a semitone and nine decibels.

    The hand that changes the bore

    A conjecture refuted here left a question with a number on it: the natural trumpet's eleventh partial is 48.7 cents flat, the lips can move it 23.1, and 25.6 cents are owed by something that is not an embouchure. There is exactly one thing a player can change about the bore while playing, and putting the hand in the bell buys those 25.6 cents at a cost of 3.8 decibels — because the aperture that tunes the instrument is the aperture that radiates it.

    part 9 · tuning
  10. The hand closing, and the note it is holding. The fourth resonance of a 370 cm horn against how much of the bore the hand occludes, at 97 per cent of the way along where the radius is 36 mm. Nothing much happens for the first ninety per cent. The note then falls to -401 cents — most of a fourth — over the next few, and between 99.5 and 99.90 per cent it jumps to 96 cents SHARP, because the lowest resonance has left the bottom of the range and every mode has taken the place of the one below it. The series' ratios are unchanged across the jump.

    The hand goes in, and the note jumps

    A horn player's hand closes the bell and the pitch falls — 19 cents, then 55, then 132, then four hundred, accelerating the whole way. Then, in the last half per cent of closure, it stops falling and lands a semitone above where it started. An earlier essay guessed the mechanism was the boundary condition changing kind and the series going odd-only. It is not. The series never changes at all.

    part 10 · tuning
  11. The first five peaks, followed as the hand closes. Each line is one member of the series, tracked by its rank rather than by its frequency, and each dot's size is that peak's height. The lowest peak falls from 38 hertz to 34 as the hand closes and then jumps to 45, which is the renumbering computed earlier: past the wall the series is one member shorter at the bottom and every peak has taken the place of the one below it. The dots shrink through the middle of the travel and grow again at the far end, so the transition costs the player support as well as pitch — and the cost is temporary, which is why a fully stopped horn is a usable instrument and a nearly stopped one is not.

    A resonance has a strength as well as a frequency

    What eleven earlier essays drew is a row of frequencies, because the solver behind it has no losses and a lossless resonance has no width. Put the losses in and every one of them acquires a height and a Q — and the hand closing a horn's bell turns out to take away nine and a half per cent of the instrument's total support before giving all of it back, in a window a few per cent wide where the horn is genuinely hard to play.

    part 11 · tuning
  12. How much of each instrument is narrow enough to steepen a wave. For each bore, the quantity that decides how nonlinear it is: the narrowest radius divided by the radius at each station, integrated along the tube. The shading is that integrand, so a bar that stays dark is a tube still doing damage to the wave and a bar that fades is a flare that has thinned it out. Divided by the instrument's own length the integral is a pure number: a plain cylinder 1.00, a tenor trombone 0.88, a trumpet 0.86, an F horn 0.59, a cone of a trumpet's length 0.24. A plain cylinder is 1 by construction, a cone of the same length and mouth is 0.24, and the ordering across the brass family is the one players give when asked which of them can be made to blare.

    The partials the tube makes itself

    Eleven earlier essays compute a passive linear resonator, and none of them ever says so. At a real fortissimo the air in a brass instrument is not linear: a compression outruns a rarefaction, the wave leans forward as it travels, and the fourth partial of a loud trumpet note is seventy decibels louder than a scaled-up quiet one — generated in the tube rather than at the lips. How much of it happens is an integral over the bore, and it is why a flugelhorn cannot be blown into being a trumpet.

    part 12 · instruments
  13. The partials the air makes, on the series the bore actually has. The input impedance of a trumpet, with the partials wave steepening manufactures from its A4 at 428 hertz drawn on it as vertical marks. The steepening is a distortion of one periodic waveform, so it makes its partials at exact integer multiples — 856, 1285, 1713, 2141, 2569 hertz. The bore's own resonances are not at integer multiples of anything: the peaks the manufactured partials aim at sit at 886, 1346, 1812, 2278, 2741. So every manufactured partial lands flat of the peak it might have used, by 59, 81, 98, 108, 112 cents — 2.6, 2.7, 3.0, 4.0, 4.9 half-widths of the peaks in question, which is outside the half-power point of every one of them. The dashed line is where the bell stops reflecting, at 2945 hertz; above it there is no peak to land on or miss.

    Which notes go brassy first

    Wave steepening manufactures partials at exact integer multiples of the note being played. A brass instrument's resonances are not at integer multiples of anything, and twelve earlier essays have measured how far off they are without ever putting the two on one axis. Put them there and a manufactured partial never lands on a peak — never once, at any note, by a miss that is the same number of hertz every time. So the alignment cannot be what decides which notes blare, and the thing that does turns out to be the bell.

    part 13 · instruments

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