A stiff string gives the bearing a comma of its own

A piano tuner’s first octave, the bearing, has two tests inside it. Each fourth and fifth is meant to beat at a rate a printed table gives, and each major third is meant to beat a twelfth root of two faster than the third a semitone below. On an ideal string the two tests are the same test. On a real one they disagree, and two essays have measured the disagreement from the outside.

A bearing counted correctly is stretched by its route counted every interval to the table and found that the order of the notes decided the bearing’s stretch, 1.62 cents an octave by one plan and 0.51 by another, with the thirds zigzagging by up to seventeen per cent. A bearing set by its thirds carries five eighths of its octave reversed the authority, solved every note at once so that the thirds were exact, and left the fourths and fifths off their table by ten per cent on average and sixteen at worst. It ended by proposing the obvious softening. A tuner’s ear does not need the progression exact; it needs each step within what can be told. Relax the thirds to a band, pull the fourths and fifths as near their table as the band allows, and ask how wide the band has to be, at what stretch, and whether some width lets everything pass at once.

The computation is easy to run and its answer does not have the shape the question expected. A band of 0.8 per cent does everything. Anything wider buys nothing at all, because the fourths and fifths are not free to go anywhere else. They are one object.

Twelve checks that are one circle #

Write out the fourths and fifths available between F3 and E4. There are seven fourths, whose lower notes run from F3 to B3, and five fifths, whose lower notes run from F3 to A3. Starting from A3 and taking whichever interval stays in the octave gives A3, E4, B3, F♯3, C♯4, A♭3, E♭4, B♭3, F3, C4, G3, D4, and the twelfth step, a fourth from D4, lands back on A3. Twelve intervals, twelve notes, one closed loop, with no fourth or fifth left over.

That is the fact the argument turns on, and it is not new. Twelve fifths and seven octaves is the same circle in its oldest form. Twelve pure fifths overshoot seven octaves by the Pythagorean comma, about 23.5 cents; equal temperament narrows each fifth by a twelfth of it; and the printed beat rates are that twelfth converted to beats per second, interval by interval. They are arithmetic on the assumption that the coinciding partials sit at whole multiples of their fundamentals.

They do not. A piano string’s stiffness sends its partials sharp, by an amount that rises as the square of the partial number, and a stiff string slows the fourths and quickens the fifths has priced that for each interval. So an interval counted to the printed rate on a real string is a slightly different size from the interval the table had in mind.

Walk the circle counting each of the twelve to its printed rate. On the modelled grand, whose strings carry a stiffness set by register, the walk comes back to A3 2.22 cents away from where it left. A comma is what a circle of intervals fails to close by, and this is one: small, with a physical cause rather than an arithmetical one, and scaled directly by the strings. Temperament was invented to distribute the first comma. The piano has quietly supplied a second, and the printed table, which already distributes the first, has made no provision for it.

The consequence follows from there being twelve intervals and eleven free notes. The fourths and fifths cannot all beat at the printed rate, on any bearing, whatever the tuner does. The sum of what they temper is fixed. The only freedom is how the shortfall is divided.

What the band actually does #

Now run the question as it was asked. Solve the eleven free notes together. Let each of the seven steps in the thirds’ progression miss its printed ratio by up to the band at no cost and beyond it at a heavy one, and pull the twelve fourths and fifths as near their printed rates as that leaves room for, in least squares.

At a band of 0.8 per cent the problem is over. The worst fourth or fifth falls from 16.1 per cent off its table to 9.5, the thirds use 0.79 per cent of the room they are given, and at a band of one per cent, of five, or of thirty, the answer is the same to the second decimal. The stretch barely notices: 1.15 cents an octave with the thirds exact, 1.14 with them free.

So the lead’s question has an answer it did not anticipate. How wide a band brings the fourths and fifths back to their table? None does, since they cannot go back. How much stretch does the bearing carry? The same as before. Is there a band at which every check inside the bearing passes at once? Yes, on the grand, and it is less than one per cent, because at that point all twelve intervals sit at their common minimum, each 9.4 per cent slow.

Two routes to one bearing. The band solve knows nothing about circles. The circle’s own answer is a one-line root-find: walk the loop with every interval beating at $(1+s)$ of its printed rate and find the $s$ that closes it. It gives $s = -0.0945$. The notes it lands on agree with the least-squares solve to a hundredth of a cent. Least squares has simply found that with one linear constraint on twelve numbers, the closest point to the origin is the point where they are all equal.

The route is the interval left out #

The circle also dissolves the first of those two essays’ findings, the one that looked most like a mystery.

A counted bearing sets each new note from one it already has, by a fourth or fifth counted to the table, until all twelve notes exist. That is eleven intervals. Since the fourths and fifths form a single ring, a route that uses eleven of them leaves out exactly one, and a route is therefore fully described by naming the interval it omits. There are twelve counted routes through a bearing, not thousands. The order of the notes along the way does not matter; only the gap does.

The earlier essay’s two plans drop straight onto this. Fourths up and fifths down from A3 never tests A3–E4: its stretch is 1.62. Fifths for the sharps and fourths for the flats never tests B♭3–E♭4: its stretch is 0.51. The figures are reproduced to the second decimal by a calculation that has no plan in it. Across all twelve the stretch runs from 0.51 to 2.00, and the shared bearing’s 1.14 is not one of them and is what they spread around, because each route adds the whole comma as a seam at one place on the circle, and the fit through the twelve notes tilts according to where in the octave that seam falls.

The interval left out takes all of it: twelve shares of 9.45 per cent, or 114 per cent of its printed rate, on every route. So it has gone through pure to the other side. Left out of plan A, the fifth A3–E4, printed at 0.74 beats a second narrow, comes out at 0.11 a second wide. This is what the wolf looks like at the scale of a piano’s stiffness, and the wolf at the end of the chain is the same arithmetic: put all of a circle’s error in one link and eleven others come out as intended. The difference is that a keyboard’s old wolf howled and this one is nearly silent. A fifth beating once every nine or ten seconds, on the wrong side of pure, sounds very like a good fifth, which may be why a tuner could lay this bearing and not know.

The zigzag in the thirds has the same origin. Across the seam the two parts of the circle disagree by 2.22 cents, and a third that straddles the seam has one note on each side. Near F♯3 a 2.22-cent error moves the fifth partial of the lower note against the fourth of the upper by about 1.2 beats a second, out of roughly seven, which is the seventeen per cent the earlier essay measured. The next third along does not straddle, and the one after does, so the progression’s steps swing from plus to minus as the notes alternate across the gap.

That removes a contest. The earlier essays treated the fourths-and-fifths test and the thirds’ test as two authorities that a tuner must choose between, and the choice looked expensive. They are nearly the same test, and the apparent conflict came from how the shortfall had been placed. Share the comma and both are satisfied: every fourth and fifth is consistently slow by the same fraction, which a listener comparing beat rates cannot distinguish from the printed pattern, and the thirds’ steps come out correct to better than one per cent without being asked.

What this does to the tuner’s instruction #

The practical form of this is short. A printed table’s rate for each fourth and fifth is about a tenth too fast for a modelled grand’s strings, and it is too fast for all twelve by the same fraction. A tuner counting to the table’s own rates will arrive back at the first note 2.22 cents out, whichever way they go.

The earlier essays’ fifteen per cent, the Weber fraction for comparing two beat rates, says how much of this can be heard, and the shared bearing sits inside it: 9.45 per cent off the printed rate is a difference a tuner comparing to a table would not call an error. So on a grand, the bearing that passes every internal test exists, and it is not exotic.

The same arithmetic also sets a limit, because the shortfall scales with the strings.

Past 1.6 times a grand’s stiffness, no bearing can pass the fourth-and-fifth check on all twelve intervals. That is not a limit on skill or on method, since the shortfall has to go somewhere, and the equal division is already the most favourable. Some interval is at least fifteen per cent off the table on any such instrument, however it is tuned. Instruments with shorter strings have stiffer ones, so this reads as a statement about small pianos. A tuner on one has three choices: accept the deviation as the instrument’s, re-derive the table for that instrument, or set the bearing by ear from the thirds alone. The third is the one the arithmetic favours, because the thirds’ steps stay in tune with a shared bearing, and nothing here contradicts the earlier finding that an upright’s fourths and fifths beat badly under a bearing set by its thirds. It now has a reason: they were always going to.

The sign of the error, and two readings it rules out #

It is worth being precise about what the 9.45 per cent is, because it is easy to read it as the old statement that stiffness makes a piano’s beats disagree with the table, and it is a different statement.

The old statement is about a single interval placed at its tempered size: its real beat is not its printed beat, by an amount with opposite signs for fourths and fifths. That is why a bearing cannot simply be corrected interval by interval, with a table of individual fudge factors. At the shared bearing, however, every fourth and every fifth is off by the same signed share, and the two signs of the single-interval correction no longer appear, because the notes have moved to absorb them. In the printed rate’s own sign, all twelve are 9.45 per cent closer to pure than the table says.

It also rules out a reading of the thirds and the fourths check different octaves in which the two families of check are independent witnesses to be weighed against each other. Inside one octave they are nearly one witness, since the thirds within the bearing are set by the same circle. The disagreement that essay measures is at the octave’s edge, between partials the bearing does not contain, and nothing here touches it: the octave that fails both checks least still has its own problem to solve outside this ring. What changes is that the bearing brings no quarrel of its own to that problem.

What the circle assumes #

The model. Every string’s stiffness comes from one smooth function of pitch on one modelled grand, each partial from the standard stiff-string formula, and a beat from the difference between the two coinciding partials and nothing else. Stiffness is multiplied by a single factor for the rest of the axis, so “four times the grand’s” stands for no particular piano.

Real strings are not smooth. A bearing on a real piano may cross a change of string gauge, and neighbouring strings can differ in stiffness by more than the function allows. That would make the shortfall unequal in its effect from interval to interval, without changing the total. The circle’s argument depends on the sum and not on the smoothness, so the 2.22 cents is the part that survives; the even 9.45 per cent is a feature of a smooth string scale.

Fifteen per cent is borrowed. It is the Weber fraction for comparing two beat rates, inherited from earlier essays, and it is doing the work of the limit at 1.6 times the grand’s stiffness. At ten per cent the grand would barely pass and the limit would fall to about 1.06; at twenty it would rise to about 2.1. The argument that some interval has to take the shortfall is unaffected by any of those. Only the sentence about passing depends on the fraction.

A tuner does not count to two decimal places. Nobody hears 0.67 beats a second against 0.74 by counting. The rates in the middle of a bearing are slow, and tuners compare them with one another more than they time them against a printed number. So the table’s error may never be heard as an error at all, and a good tuner’s bearing may already be close to the shared one because comparison is what they do. Whether it is has not been measured here; the calculation says what a bearing would be if it were, which is the thing a recording of several professional bearings could test.

And the shared bearing is not claimed to be the right tuning. It is the one that keeps all twelve internal tests as near their table as possible. A tuner may want more stretch for the octaves above, and nothing here argues against adding it; the progression of thirds votes for the bearing has already shown the bearing’s stretch deciding what the octaves want.

Whose piano, and whose table #

The rate table belongs to equal temperament as it is taught to piano tuners working by ear in the Western trade, and it is that trade’s method rather than a universal one. An electronic tuner following an inharmonicity measurement is not counting to it at all and may place the notes differently. A historical temperament has its own unequal rates and the same circle: its fourths and fifths close it too, and a stiff string would leave the same kind of shortfall in it, on a harpsichord or fortepiano smaller by however much less stiff their strings are. The claim about 9.45 per cent is about one equal-tempered table on one modelled modern grand.

Still open: a table for the instrument in front of the tuner #

The shortfall is one number per piano and it can be measured in the time it takes to play a circle, since a tuner counting twelve intervals to the table and coming home 2.22 cents out has measured it. The next computation is a tuner who uses that: walk the circle once, read the closure error, divide by twelve, and correct every printed rate by that fraction before setting the bearing. How much of the shared bearing does one pass recover, how much does the tuner’s own counting error on the way round contaminate the reading, and does a second pass improve on the first or merely add noise? The answer would say whether the printed table can be kept, with one correction per instrument, or whether a tuner is better off abandoning it for the comparison they already make.