Instruments and their design

The hole that spoils a note

A tone hole shortens the tube. A register hole does the opposite job: it is small enough to shorten nothing and is placed where it will wreck the fundamental's resonance and leave the third harmonic's alone, so the note jumps a twelfth instead of retuning. The place that does both is a pressure node of the harmonic being kept — a third of the way along whatever length is sounding — and the length changes with every fingering while the key does not. One key is at the right place for exactly one note, and the note it is worst for is in the throat of the instrument, which is where players say the instrument is worst.

Assumes: Above a certain note the holes stop working · A tube that skips every other partial

Every hole in a woodwind is there to shorten the tube. Open one and the air column ends earlier; open the next one up and it ends earlier still; and the whole scale of the instrument is a sequence of lengths.

There is one hole that is not doing that, and it is the smallest one on the instrument. It is worth separating from the other exception before starting, because both are small holes that do something other than shorten: a cross-fingering leaves a hole open below a closed one and changes the tube’s effective end in a graded way, which is one hole doing a dozen jobs and is still about length. The register hole is not about length at all. On a clarinet it is worked by the left thumb and it makes the note jump a twelfth. If it were shortening the tube it would raise the pitch by a step or two; instead it moves the sound to a completely different mode of the same tube, and the mechanism is worth working out because it explains a defect the instrument has been famous for since it was invented.

What an open hole does to a mode

A clarinet is a tube closed at the mouthpiece and open at the far end, so its standing waves have a pressure antinode at the reed and a pressure node at the bell. That is why it skips every other partial: only the odd harmonics fit those two conditions.

What a 60 cm tube supports, by how its ends are closedThe first 6 modes of a stopped cylinder and an open 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.stopped cylinder143 Hz fundamental1434291902 cents — a twelfthopen cylinder286 Hz fundamental2865721200 cents — an octave2204408801760hertz, on a logarithmic axisevery mode the tube supports, and the jump from the first to the second
Fig. 1 The modes of a closed-open tube against those of an open one. The closed tube’s resonances are at 1, 3, 5, 7 times the fundamental, which is why overblowing a clarinet gives a twelfth and overblowing a flute gives an octave.

Now put a small open hole somewhere along the side. An open hole is a place where the pressure inside must be near atmospheric — a pressure release — and how much that disturbs a given mode depends on how much pressure that mode was trying to have there.

At a point where a mode has a pressure node, the mode already has no pressure and the hole does nothing to it. At a pressure antinode, the mode is trying to hold the largest pressure it has and the hole destroys it. In between the effect goes as the square of the mode’s pressure at that point, which is the standard first-order perturbation result and is all the model here needs.

So a hole can be placed to spoil one mode and leave another. That is not a trick; it is a consequence of the two modes having their nodes in different places.

Where each family's tone-hole lattice stops reflecting. The cutoff frequency of an open tone-hole lattice, from Benade's formula, for four woodwind geometries: clarinet 1824 Hz, oboe 2990 Hz, flute 1690 Hz, bassoon 506 Hz. Below its cutoff a note's wave turns round at the first open hole and the instrument is a tube of that length; above it the wave passes through the whole lattice and radiates from the far end, so the upper part of every note's spectrum leaves the instrument from the same place whichever note is fingered. That is what gives a family one recognisable voice across its range.
Fig. 2 For contrast, what an ordinary tone hole is measured by. A row of open holes behaves as a filter with a cutoff frequency, and above that frequency the holes stop working — the wave runs past them and out of the bell. That is a property of the lattice and it is what limits the top of the instrument. A register hole is not part of the lattice and none of it applies.

Where the place is

For a closed-open tube of length L the pressure of mode m is cos((2m−1)πx / 2L), measured from the closed end. The first mode has no node before the open end. The second mode — the third harmonic, the one the clarinet jumps to — has a node at x = L/3.

At a third of the way along, therefore:

  • the third harmonic’s pressure is zero, so a hole there does nothing to it;
  • the fundamental’s pressure is cos(π/6) = 0.866, so the square is 0.75 and a hole there wrecks three quarters of what an antinode would.

That is the register hole. It kills the fundamental’s resonance, leaves the third harmonic’s intact, and the reed then oscillates at the only frequency the tube still supports.

The pressure inside each tube, for the first three modes. Pressure along the bore for the first three modes of an open cylinder, a stopped cylinder and a cone. A closed end forces a pressure antinode and an open end forces a node, so the stopped cylinder fits an odd number of quarter-wavelengths and cannot fit an even one. The cone's apex is closed and yet its modes are the complete series, because the spherical wave inside a cone falls as one over the distance from the apex and vanishes wherever a plane wave in an open tube would.
Fig. 3 Where the place comes from, which is the pressure inside the tube rather than anything about holes. For a closed-open tube the pressure of mode m is cos((2m−1)πx / 2L) from the closed end: the first mode has no node before the open end, and the second — the third harmonic, the one a clarinet jumps to — has one at exactly x = L/3. At that point the third harmonic’s pressure is zero, so a small hole does nothing to it, and the fundamental’s is 0.866, whose square is 0.75. The hole wrecks three quarters of the fundamental’s resonance and leaves the twelfth intact, and the reed then oscillates at the only frequency the tube still supports. Nothing in that mentions the hole’s size, the reed, or how hard the player blows.

It is worth noticing what this does not require. Nothing about the argument mentions the size of the hole beyond “small”, nothing mentions the reed, and nothing mentions how hard the player blows. The position is fixed by the mode shapes alone, and the mode shapes are fixed by the boundary conditions — a closed end and an open one, and the fact that the open one is not quite where the tube stops.

And the tube keeps changing length

Here is the problem, and it is forced rather than a matter of craftsmanship.

The register key is drilled once, at a fixed distance from the mouthpiece. The sounding length is whatever the fingering makes it, and on a clarinet’s lowest register that runs from about 58 centimetres down to about 21.

A hole at a fixed 12.8 centimetres is a third of the way along a 38-centimetre tube. On a 58-centimetre tube it is 22 per cent of the way along; on a 21-centimetre tube it is 62 per cent.

One key, nineteen notes, one right answerA register hole disturbs a mode in proportion to the square of that mode's pressure at the hole, so the place that spoils the fundamental and leaves the third harmonic alone is the third harmonic's own pressure node — a third of the way down whatever length is sounding. The length changes with every fingering and the key does not move, so the two curves cross at one note. At A3 the key sits almost exactly on the node and the twelfth speaks cleanly; at A♭4 it is at 62 per cent of the tube and disturbs the third harmonic by 96 per cent of what an antinode would, which is the throat of the instrument and is exactly where players say the notes are worst.D3F3A♭3B3D4F4A♭400.20.40.60.81the note being overblownfraction of the maximumthe fundamental,spoiled — wantedthe third harmonic,disturbed — not wantedthe key, as afraction of the tubeone key at 22 per cent of the longest tube
Fig. 4 One key at 22 per cent of the longest tube, scored against every note it has to serve. At the bottom of the register it spoils 88 per cent of the fundamental and disturbs 26 per cent of the third harmonic. At A3, where it lands exactly on the node, it spoils 76 per cent and disturbs nothing. At the top of the register it spoils only 31 per cent of the fundamental and disturbs 96 per cent of the third harmonic — which is the worst possible combination of the two.

Read the last row again, because it is the finding. At the top of the chalumeau register the key has almost stopped doing its job — it barely disturbs the fundamental — and it has begun actively damaging the note it is trying to produce.

Those notes are the throat tones, and they are the notes every clarinettist is taught to work on. The instrument’s reputation for a weak, stuffy band around written G to B♭ is usually attributed to the tube being short and the tone holes being few, and both of those are true; this adds a third reason and it is the one the register key is directly responsible for.

The written throat notes — G, A, B♭, B and C at the top of the clarinet’s lower register — are the five the compromise falls on, and they are the five every clarinettist is taught to compensate for.

The compromise a maker actually makes

The figure above can be dragged, and dragging it is the argument: there is no position at which the third harmonic is undisturbed across the register. Move the key down the tube and the low notes improve and the throat notes get worse; move it up and the reverse.

A maker therefore chooses the least bad point, and the choice is not made at the middle of the range. It is made toward the top, because the low notes have margin — at the bottom of the register the key spoils 88 per cent of the fundamental even at a bad fraction, which is plenty — and the throat notes have none.

That is why the register key on a real clarinet sits nearer the mouthpiece than a naive reading of “a third of the way along” would put it, and it is why the same key is asked to do a second job.

Where the least bad point actually is

“Least bad” is an optimisation and it can be run rather than gestured at. Fix the key at a distance d from the mouthpiece, score it against all nineteen sounding lengths of the register, and take the largest damage to the third harmonic anywhere in that set; then choose the d that makes that largest damage as small as possible. It is a minimax, it has no free parameters beyond the ones already in the model, and it takes a sweep.

The answer is 10.2 centimetres, which is 17 per cent of the longest tube, and at that position the worst damage anywhere in the register is 47 per cent.

Two things follow, and the first is about the figure above. The key drawn at 22 per cent is not the compromise; it is a choice that leaves the worst case at 96 per cent when 47 was available. Moving the key two and a half centimetres nearer the mouthpiece halves the damage at the worst note, and it costs the bottom of the register almost nothing — D3’s spoil rises from 89 per cent to 93 while its damage rises from 26 to 47, which is the trade the top of the register needs and the bottom can afford. The direction the essay guessed is right and the magnitude is larger than it guessed.

The second is that 47 per cent is the best that one hole can do, and it is a bad number. At the note where the compromise bites hardest, the mode the key exists to preserve is being destroyed to nearly half of the maximum possible. There is no position that fixes this, and the sweep is the proof: the minimum over every position of the maximum over every note is 47 per cent, and a single key cannot get under it.

That is also the answer to why a real clarinet’s key sits at 22 rather than 17, and the essay has already given it. The hole is doing a second job. A vent for the written B♭ has to be where a tone hole for that note would be, and that is further down the tube than the register optimum; the instrument splits the difference and the throat B♭ is the note that pays. The minimax above is the position of a key that only had one job.

How many keys the geometry asks for

The same optimisation answers the question the essay raised and left as an observation — that counting register keys is counting how badly the compromise bit.

Cut the register into K contiguous spans, give each span its own key, and choose both the cuts and the positions to minimise the worst damage anywhere. That is a small dynamic program over the nineteen lengths.

keys worst damage anywhere positions, as a fraction of the longest tube
1 47% 17%
2 15% 26%, 15%
3 7% 28%, 19%, 13%
4 3% 29%, 22%, 17%, 13%

The second key is the one that matters. It takes the worst case from 47 per cent to 15 — a factor of three — and every key after it buys about half of what is left. That shape is the usual one for a covering problem, and it says something specific about the instruments: an oboe with two or three octave keys is not being fussy, it is taking the large improvement and then most of the small one, while a clarinet with a single key is sitting at the one point on this curve where the compromise is still severe.

It also puts a number on what the clarinet gives up. A second register key would cost a mechanism and a thumb, and it would buy the throat register a reduction from 96 per cent damage to 15. That the instrument has never acquired one, in three centuries of otherwise relentless keywork, is a fact about players and mechanisms rather than about acoustics — the acoustics have been asking for it the whole time.

What a 21 cm tube supports, by how its ends are closedThe first 6 modes of a stopped cylinder, all of the same acoustic length. 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 an octave below that of an open tube of the same length, because it fits a quarter of a wavelength where an open tube fits a half.stopped cylinder143 Hz fundamental1434291902 cents — a twelfth220440880hertz, on a logarithmic axisevery mode the tube supports, and the jump from the first to the second
Fig. 5 The shortest sounding length in the register, which is where the margin runs out. At 21 centimetres the modes are four and a half times as far apart in hertz as they are on the full tube, and the register key — fixed in place — is now a much larger fraction of the way along it. The low notes have margin and the throat notes have none, which is why a maker’s least-bad point sits toward the top of the range rather than in the middle of it: at the bottom the key spoils 88 per cent of the fundamental even at a bad fraction, which is plenty, and at the top there is nothing to spare.

And the part of the mechanism this model leaves out is the reed, whose flow curve decides how much pressure it takes for any of these modes to start speaking at all — a vent that weakens a mode also raises its threshold, and a note that will not speak is a worse fault than a note that speaks flat.

Why the flute does not have this problem in the same way

A flute is open at both ends, so its modes are at 1, 2, 3, 4 times the fundamental and overblowing gives an octave. The mode to be kept is the second harmonic, whose pressure node is at the middle of the tube, and the fundamental’s pressure there is at its maximum.

So the flute’s ideal register hole is at L/2, where the fundamental is entirely destroyed and the octave is entirely untouched — a cleaner separation than the clarinet’s 0.75 against 0. In practice a flute has no register key at all: the player changes the jet instead, and the modern instrument’s few vent holes do a slightly different job on the third register.

What a 60 cm tube supports, by how its ends are closedThe first 6 modes of an open 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. open cylinder286 Hz fundamental2865721200 cents — an octave4408801760hertz, on a logarithmic axisevery mode the tube supports, and the jump from the first to the second
Fig. 6 Why the flute does not have this problem in the same way. An open tube supports every whole multiple of its fundamental, so the mode a register vent has to encourage is the second rather than the third — and the second mode’s pressure node sits at the middle of the tube, which is a far less crowded place to drill than a third of the way along. A flute overblows to the octave and a clarinet to the twelfth, and the whole of this essay’s difficulty follows from that one difference in boundary conditions.

The oboe and the bassoon are conical, which puts their modes at 1, 2, 3, 4 as well, and both use octave keys — the oboe has two or three of them, and the reason is exactly the argument above: one key cannot serve the whole register, so the maker adds keys until each covers a small enough span of lengths.

Counting the register keys on an instrument is therefore counting how badly the compromise bit. A flute has none, a clarinet has one, an oboe has two or three, a bassoon has several. That is a design consequence rather than a tradition, and it can be read off the geometry.

A conical bore behaves like an open cylinder in this respect and like a stopped one at its input, which is why an oboe and a saxophone overblow to the octave with a reed at the narrow end — the cone’s mode spacing is the open tube’s, and the register vent problem is the flute’s rather than the clarinet’s.

And the end correction makes every number above approximate rather than exact: the acoustic length is the physical length plus a fixed few millimetres, so a fraction measured from the mouthpiece is a slightly different fraction of the tube the wave actually sees, and the discrepancy is largest on the shortest sounding lengths — which are the ones the compromise already hurts most.

What a 32 cm tube supports, by how its ends are closedThe first 6 modes of a stopped cylinder, all of the same acoustic length. 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 an octave below that of an open tube of the same length, because it fits a quarter of a wavelength where an open tube fits a half.stopped cylinder268 Hz fundamental2688041902 cents — a twelfth4408801760hertz, on a logarithmic axisevery mode the tube supports, and the jump from the first to the second
Fig. 7 And the middle of the register, for the comparison the optimisation is over. Nineteen sounding lengths run from 21 to 60 centimetres, so a key fixed at one distance from the mouthpiece is at 17 per cent of the longest tube and 48 per cent of the shortest — the same hole, in a completely different place, on the same instrument. The minimax answer is 10.2 centimetres and its worst damage is 47 per cent, which is the best one hole can do; the real key sits at 22 per cent because it is also the tone hole for the written B♭, and the throat B♭ is the note that pays.

The general form of the difficulty is worth stating because it is not confined to woodwind. A fixed geometric feature serving a variable geometry is a compromise whose size is the ratio of the extremes, and this collection has met it before: a guitar’s frets are cut for one scale length and every string has a different stiffness, and a piano’s hammer strikes at one fraction chosen for a middle register it then applies to the whole compass. In each case the maker picks the point that minimises the worst case rather than the point that is right anywhere.

Which computation produced the numbers

The pressure of mode m at position x in a closed-open tube of length L is cos((2m−1)πx / 2L), which is the exact standing wave for an ideal lossless cylinder. The effect of a small open hole on that mode is taken as proportional to the square of that pressure, which is first-order perturbation theory and is the standard treatment.

“Spoil” is that quantity for the fundamental and “damage” is the same quantity for the third harmonic; both are normalised to one at an antinode, so they are fractions of the maximum possible effect rather than frequencies or decibels. Nothing here computes how far the note is pushed, only how strongly each mode is disturbed, and that is the whole of the model’s ambition.

The tube lengths are quarter-wavelengths of each sounding pitch at the speed of sound this collection uses. They are acoustic lengths rather than bore lengths, which is stated above; using bore lengths instead shifts every fraction by the same small amount.

The chalumeau register is taken as the sounding pitches from D3 to A♭4, which is written E3 to B♭4 on a B♭ clarinet — the standard compass of the lower register.

The node position for keeping harmonic 2m−1 is (2k−1)/(2m−1) of the length, evaluated at k = 1: a third for the third harmonic, a fifth for the fifth, a seventh for the seventh. Those are the positions an instrument with several register keys would want.

The two optimisations are over that same scoring and add nothing to the model. The single-key position is a sweep of the drilled distance from two to thirty-two centimetres in quarter-millimetre steps, taking for each the largest damage over all nineteen sounding lengths and keeping the distance that minimises it. The key-count table is the same objective under a partition: the register is cut into contiguous spans, each span gets its own swept position, and a short dynamic program over the nineteen lengths chooses the cuts that minimise the worst span. Both report the worst case rather than the average deliberately — a register key that is excellent on eighteen notes and useless on the nineteenth is an instrument with a bad note on it, which is the complaint the essay started from.

Neither optimisation knows anything the rest of the essay does not. In particular neither one knows that the hole has a second job as a vent, so the position it recommends is the position of a key that does not have to be a tone hole as well. The gap between the 17 per cent it returns and the 22 per cent a real instrument uses is the size of that second constraint, measured indirectly.

What the picture cannot show

A hole is not a point and it is not small. The perturbation model assumes the hole is much smaller than a wavelength and that it does nothing but release pressure; a real register hole has a chimney with its own mass of air in it, and its size trades the strength of the effect against how much it flattens the note it is venting.

It does not model the reed. What actually decides which mode sounds is a nonlinear oscillator choosing among the tube’s resonances, and the choice depends on lip pressure and breath as much as on the tube. The reed is a valve rather than a vibrator and a full account of overblowing needs it.

It has no losses. A real tube’s modes have finite width, so a resonance that is “spoiled” is weakened rather than removed and the competition between modes is a matter of degree. The figures here are ratios of an idealised strength.

It says nothing about what leaves the instrument. A disturbed mode is still a mode; whether the note that results is weak depends on radiation, and the bell decides what gets out at least as much as the bore decides what is inside. The throat tones are quiet as well as unstable and this model only speaks to the second.

And it treats the register as a set of lengths, when in fact a fingering with several holes open is not a simple shortened tube at all — the holes below the first open one keep contributing, and above the cutoff frequency the whole lattice behaves as one object rather than as an end.

The ladder from here

This rung took the one hole on a woodwind that is not shortening the tube and found a compromise forced by geometry. What the ladder still owes is the same analysis on a conical bore, where the mode spacing is different and the number of octave keys is larger, and a treatment of the automatic octave mechanisms that modern oboes and saxophones use, which are an engineering answer to exactly the problem this essay describes.

Part 3 of 7

One essay in the series on tone holes. The essays either side of this one:

What links here

Essays that reach for this one mid-argument — the half of a link its own author cannot write down.

The objects named here

The third way in, after the field and the series: the things themselves, and every essay that touches each one.

BoreBoundary conditionNodeOverblowingRegisterResonanceStanding waveTone hole