Eleven partials is one too many

There is a way of finding what is left to say about a subject that has worked more than once on this collection: go through every figure an anchor has drawn and find the number all of them hold at the same value.

For this one the answer is not the threshold, which has been swept; not the tolerance on the common-fate cue, which has been sensitivity-tested; and not the exchange rate, which an entire rung exists to sweep because nobody has it.

It is the number of partials. Every census, every arbitration and every verdict this ladder has produced was computed on exactly ten, in six rungs and about twenty figures, and no placement has ever passed another value.

Eleven #

At ten partials the harmonicity census reports that a perfect harmonic series is perfectly fused: ten of ten, at a fundamental of exactly one. That is what it must say. An ideal string is the canonical single object, and a census that split it would be broken.

At eleven it reports five of eleven, at a fundamental of exactly two.

Nothing about the spectrum has changed. Partials one to ten are where they were and an eleventh has been added exactly where a harmonic series puts it. And having received one more partial that fits the series perfectly, the census concludes that six of them do not belong.

Feed that into the competition and the verdict follows: at twelve partials the grouping competition cuts an ideal string after its second partial, and it does so across almost the whole swept range of exchange rates. The most fusible object this ladder has is two objects.

It is not the arbitration #

The obvious diagnosis is a scale problem in the cost model, and it is worth walking through because it is wrong and the way it is wrong is informative.

The competition charges one unit for every partial a hypothesis groups against a cue’s verdict and one for every partial it separates against one, plus a fixed cost for each extra stream it posits. The cue costs therefore grow with the number of partials and the stream cost does not — so a large enough spectrum should make the second stream nearly free, and any spectrum with a fixed proportion of misfit partials should eventually be split.

That is a real defect in the shape of the model and it is not what is happening here.

Sweep the stream cost from a tenth of its value to five times it — a factor of fifty — and every verdict comes back identical. The figure asserts it rather than reporting it: it recomputes the whole grid at both extremes and refuses to draw if any cell disagrees. The exchange rate does not save it either; the boundary between the two runs sits at the same rate before and after.

The reason it cannot matter is structural rather than numerical, and it is the useful half of the diagnosis. A hypothesis’s cost is a weighted sum over the cues’ verdicts — which partials each cue would take out of the note — and the stream cost, the exchange rate and any other weight decide only how those verdicts are traded against one another. They are all downstream of the verdicts. If the verdict vector itself is wrong, no assignment of weights recovers the right answer, because every hypothesis is being scored against the same wrong list.

That is worth carrying beyond this ladder. A cost model with free parameters invites the assumption that a surprising output is a badly chosen parameter, and the free parameters are exactly the part of such a model that is easiest to sweep and hardest to blame. Sweeping them here took a minute and eliminated the whole layer, which is the only reason the layer below it was looked at.

So the arbitration is behaving exactly as designed. It is being handed a wrong answer by the cue underneath it.

The cap and the count are the same number #

The census searches for the fundamental that makes the most partials fuse, over the partials themselves and their small divisors. Left unconstrained that search is useless, and the rung that built it says why at length: a bell’s partials are 0.5, 1.0, 1.2, 1.5, 2.0 and 2.5, which are all whole multiples of one tenth, so an unconstrained search finds a fundamental of 0.1, declares every partial exact, and reports that a bell fuses perfectly.

The constraint that stops it is a cap on how far below the top partial the fundamental may lie: about ten harmonics. The justification given is a fact about listeners rather than about arithmetic — a fundamental twenty-five harmonics below the top partial is not one anybody could use, because nothing that high is resolved and the low harmonics that would establish it are missing.

And a harmonic series of eleven partials has a top partial eleven harmonics above its own fundamental. So the first thing the cap excludes, at a count of eleven, is the right answer. The search then takes the best fundamental it is still allowed — the octave above, which makes the even partials exact and the odd ones half a step off the grid — and the verdict follows mechanically.

There is exactly one integer of margin in the whole arrangement, and the ladder has been sitting on it for six rungs without saying so.

The piano string breaks slightly earlier and slightly worse. At ten partials it already fuses only nine, because stiffness has carried the tenth off the grid — which is the finding the stiffness rung exists for. At eleven it drops to four of eleven at a fundamental of 2.025, and it stays at four however many more partials it is given, because everything above is being measured against a fundamental an octave too high.

What the number should have been #

The cap’s justification is a claim about resolution, and this collection has the resolution.

A harmonic is separable when the spacing to its neighbours exceeds the ear’s analysis bandwidth at its frequency. The spacing is the fundamental everywhere and the bandwidth grows with frequency, so a series runs out, and where it runs out is a function of pitch: five harmonics below 71 hertz, six to 102, seven to 181, eight to 866, and nine above that.

It never reaches ten. The received statement that the first eight or ten harmonics are resolved is not a measurement of one stimulus; it is what this criterion gives across the whole musical range.

So the number the census should be given is not a constant. It is the number of partials that are individually available at that spectrum’s own pitch, and for a harmonic spectrum it is between five and nine.

Which means the model has been operating inside its own valid range on every essay this ladder has written, and by accident. Ten is one above the largest count a listener ever supplies and one below the count at which the census stops working. Both boundaries were set independently, one by a cap chosen to stop a bell fusing and one by a habit nobody wrote down, and they happen to leave a gap exactly one integer wide.

That is not a comfortable result. A model that is right for a reason it does not contain is a model that will be wrong the first time somebody changes something innocuous — and the innocuous change here is passing a spectrum with a few more partials in it, which is what a figure about a rich timbre would naturally do.

And the constraint binds only on the spectra that fuse #

The other half of the repair is more interesting than the repair.

A harmonic series has a constant spacing and a widening filter, so it runs out and runs out from the top. An inharmonic spectrum does not behave that way at all. A bar’s modes are at 1, 2.76, 5.40 and 8.93 times its fundamental — gaps that grow faster than any filter widens — so every partial of a bar is individually separable at any pitch it is ever sounded at, and the census could honestly be given all of them.

So the property that makes a spectrum fuse is the property that makes its partials uncountable. Harmonicity is regular spacing, regular spacing is what a widening filter eventually swallows, and a spectrum obeying the strongest grouping cue in the model is exactly a spectrum whose partials a listener cannot separately count. The two are the same fact seen from two ends, and neither the census nor the competition contained it.

It also explains why the cap was needed in the first place, which the third rung asserted and could not derive. The cap says a fundamental more than about ten harmonics below the top partial is not one a listener could use. That is precisely the resolvability statement: at the pitch of any such fundamental, the tenth harmonic and above are arriving several to a filter, so the pattern that would establish the fundamental is not available to be matched. The cap was a resolvability limit written as a constant, and writing it as a constant is what made it collide with a count that was also a constant.

That has a consequence for the hypothesis set. The competition offers, as one of its candidates, a cut after the ninth partial — and on a harmonic spectrum below 866 hertz the ninth partial is not separately available, so no listener could entertain the hypothesis. Trimming the candidates to the partials a listener actually has is one line, and when it is trimmed the pathology above cannot arise, because the count can never reach eleven.

Low down, the inharmonic spectra stop being immune, and how they fail is worth naming because it is a different failure.

A bell’s prime and its tierce are a minor third apart, and a minor third fits inside one auditory filter below about three hundred hertz — so a low bell has an unresolvable pair in the middle of its spectrum while its upper partials remain perfectly separable. The minor third a bell is tuned to is the interval every founder shaves metal to place, and at the bottom of a ring it is inside a critical band.

A kettledrum is worse. Its modes are at 1, 1.5, 1.99, 2.44 and 2.89, spaced by about half a fundamental all the way up rather than spreading, so at 131 hertz four of its five partials share filters with a neighbour. That is a computed reason for something the timpani rung observes and does not explain: a timpano has a definite pitch and no audible partials. There is nothing to hear out.

What moves inside the range a listener could supply #

The pathology above happens at eleven, and a listener never supplies eleven. So the fair question is what the competition does across the counts that are actually available — four to nine — and the answer is that almost nothing moves.

Four of the five spectra give the same verdict at every count in that range. The one that changes is the bar: one stream at four and five partials, and a cut after the fourth from six onward.

It would be satisfying if that were a pitch effect, since the harmonic count is a function of pitch, and it is not. A bar’s modes are all separable at any pitch it is sounded at, so the count for a bar is not set by a listener; it is set by how many modes the table carries. Six is the table’s length rather than a listener’s limit, and the verdict moving between five and six is a statement about the truncation.

Which is the honest end of this rung: for a harmonic spectrum the count is now a computed quantity, and for the three spectra this ladder actually argues about it remains a convenience. The competition’s verdict on a bell has been a verdict about eight partials because eight is how many the table holds.

Which computation produced the numbers #

The census, the competition and the five spectra are unchanged from the third, fourth and fifth rungs. The threshold at which a partial is heard out is one per cent and is this ladder’s own; the common-fate tolerance is a factor of two; the exchange rate is swept over three decades and the winner is the hypothesis holding the widest run of it.

The count of resolvable harmonics uses the equivalent-rectangular-bandwidth model with the plain criterion — a harmonic is resolved when the spacing to its neighbours exceeds the bandwidth at its frequency — and the Bark model is drawn beside it because the two disagree by a factor of three below 500 hertz, which is exactly the register this argument lives in.

For an inharmonic spectrum that criterion has to be applied partial by partial, since the spacing is not the fundamental. Doing so on a harmonic series reproduces the count computed the other way, which is the check that the two are one question.

The stream-cost sweep runs from a tenth to five times the value the ladder uses, and the assertion compares whole grids rather than single cells.

Where the model stops #

The cap is still a cap. Replacing it with a resolvability limit fixes the failure demonstrated here and does not make the census right; the underlying search for a best fundamental is a piece of machinery with no listener in it, and the reason a bell has a strike note is a question about pitch extraction that this collection treats elsewhere.

Resolvability is a criterion, not a measurement. The whole argument moves if the criterion moves, and the Bark model gives no resolved harmonics at all below 100 hertz.

And the five spectra are idealisations. A real bell has thirty partials and this ladder’s table has eight; a real bar has more modes than six. Every claim about an inharmonic spectrum running out is a claim about a table that stops before the spectrum does.

What the picture cannot show #

It cannot show a resolved partial being used. Separability says a partial is available, not that anything attends to it, and the dominance region is the standing reminder that availability and influence are different quantities.

Nor can it show a spectrum in a room. Every reflection arrives as a second copy of the whole spectrum, and whether two copies of an unresolved pair are more or less resolvable than one is not a question this criterion can be asked.

It cannot show attention. A listener told to listen for the tierce of a low bell does considerably better than a filter model predicts, and no criterion in this ladder has anything to say about that.

And it cannot show onset. The blown note says the third cue’s magnitude depends on an unmeasured criterion; this rung says the first cue’s input depends on an uncounted one. They are independent problems that happen to have the same shape.

Whose spectra, and when #

The spectra are idealisations and none is a measurement of an instrument.

The one place practice has a view is bell founding, and it is a view about exactly the interval this rung finds inside a filter. A founder tunes five partials by turning metal off the inside of a casting, and the tierce — the minor third above the prime — is the one that gives a European bell its character and is the hardest to place. It has been tuned by ear for six centuries, on bells whose primes run from about 200 hertz for a large ring down to under 60 for a bourdon. This rung says that on the largest of them the tierce and the prime cannot be separated by a listener at all. That is either a reason the bottom of a ring is tuned by comparison with its neighbours rather than by hearing the interval within one bell, or it is a fact about a filter model that a founder would recognise as false; and the founders are the ones with the evidence.

Where this ladder goes next #

Eight rungs. The ear builds objects and sometimes offers a choice; what makes two partials one note; a spectrum’s inharmonicity read as a perceptual count; the same count under the cue that needs time; the two arbitrated with the exchange rate swept; the onset cue that made the exchange rate stop mattering; the blown note that does not put it back; and now the number of partials all seven of those held fixed, which is one below the number at which the machinery breaks and one above the largest number a listener ever has.

What is owed after this is the count for a spectrum that is not a series. For a harmonic one the question is answered — it is between five and nine, it is a function of pitch, and it is computed. For a bell, a bar and a drum the answer is the object’s own table, and every table in this ladder stops for reasons of convenience rather than of physics. A real bell has thirty modes and this ladder censuses eight, so every verdict it has produced about a bell is a verdict about a truncation nobody chose deliberately. Finding where a real casting’s modes actually stop being separable is arithmetic once the modes are known, and the modes of a bell are a shell problem this collection has not solved — so this one needs an instrument, and it is the first debt in this anchor that does.