Acoustics & Music Theory
why it sounds like that
Series

Excitation point — the series

11 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. A string struck at one 7th of its length. The amplitude of each partial of an ideal string excited at 0.1429 of its length. The mode shape is a sine, so a partial with a node at the excitation point cannot be set moving at all: partials 7, 14 are silent here. The envelope over the rest is one over n, a struck string's.

    Where the hammer lands

    Strike a string at exactly one over n and the nth partial is silent, because the hammer has landed on that mode's node and cannot move it. A piano's hammers strike between a seventh and a ninth of the way along, which puts the seventh partial — the most dissonant member of the series — at or near a null. That is a design decision made in wood, and it takes one line of trigonometry.

    part 1 · instruments
  2. The same note, hit at a middling dynamic. The spectrum of a struck string with the hammer's own contact time applied as a low-pass. Contact lasts 1.60 ms at this force, against 2.26 ms at the softest and 0.95 ms at the loudest drawn — felt is a nonlinear spring, so a harder blow is a shorter contact and a brighter note. The spectral centroid moves from partial 1.5 to partial 2.2, which is a change of timbre and not of loudness.

    A hammer is not an impulse

    Contact lasts a couple of milliseconds, which low-passes the note — any partial whose half-period is shorter than the contact is barely excited. Piano felt is a spring that stiffens as it compresses, so a harder blow makes the contact shorter, the corner higher and the note brighter. A loud note is not a scaled-up quiet one, and no linear model gives that.

    part 2 · instruments
  3. How near the node a hammer has to land. The level of partial 7, relative to the loudest partial of the same note, against how far the hammer is from that partial's node — for a contact 2.6 per cent of the speaking length wide, which is 16 mm on a 620 mm string. At the node exactly the partial is absent whatever the width, because a mode shape is odd about its own node and integrating an odd function symmetrically gives zero. Twenty decibels of suppression needs the hammer within 2.82 mm, and thirty needs 0.89 mm. The width itself barely matters in the musical range: its sinc reaches its first zero only at partial 77.

    The hammer is not a point either

    The comb of missing partials is computed throughout for a contact of no width, and a piano hammer is sixteen millimetres of felt on a string of six hundred. Integrating the mode shape across the felt turns out not to fill the null in: a mode is odd about its own node, so a symmetric contact centred on it gives an exact zero however wide it is. What destroys the null is being in the wrong place, and the tolerance is 2.8 millimetres for twenty decibels of suppression and 0.89 for thirty. The design decision found at the outset is a manufacturing tolerance.

    part 3 · instruments
  4. Two clocks on one contact, and which of them runs out first. Two times that could end a hammer's contact with a string, across the compass. The flat line is the felt's own recoil, which this site models as a fixed 1.6 milliseconds at a stated force and which does not know what note is being played. The falling line is the string's: a mass m driving a string of impedance Z loses its momentum in m/2Z, which is 11.9 milliseconds at the bottom of the keyboard and 0.76 at the top. They cross at G3. Below the crossing the felt lets go first and above it the string gets there first, and over six octaves the two are within a factor of two of each other — so the contact time is a coupled quantity and not a property of the felt.

    The hammer that is heavier than its string

    Three earlier essays varied where the hammer lands, how long it stays and how wide it is, and the third named the fourth variable: the exciter's own mass. It inverts across one instrument — the hammer is lighter than the wire it hits at the bottom of a piano and thirteen times heavier at the top — and it turns the contact time into a quantity the string decides rather than the felt.

    part 4 · instruments
  5. C4: the pulse computed and the pulse assumed. Above, the force the hammer delivers to the string at C4, integrated forward against the felt's nonlinear force and the string's returning corner, drawn against the half-sine of 1.60 milliseconds that every earlier figure assumed. The computed contact lasts 2.21 milliseconds and the corner comes home 4.6 times inside it. Below, the excitation each pulse gives to each partial. They agree at the bottom and part company higher up — worst at partial 6, by 34 decibels — because the assumed pulse has nulls the computed one does not.

    The pulse that was assumed

    Every figure until now low-passes the string's excitation with the spectrum of a half-sine, which is what a hammer would deliver against a rigid wall. An earlier essay said so and declined to do better. Doing better takes forty lines and refuses the prediction that came with it: the corner's round trips govern the spectrum as expected, and the contact time is governed by something else entirely — the mass ratio discovered one essay earlier.

    part 5 · timbre
  6. The partial above which the returning corner has lost its phase. Up a piano, the lowest partial whose extra travel after one round trip to the near end exceeds a quarter of a cycle — above it, the component comes home somewhere other than where a corner needs it. The curve follows the inharmonicity coefficient, which is smallest in the tenor and rises both ways, so the corner survives best at A2 — to the 24th partial — and worst at A7, where only 5 partials are still in step. The earlier coupled model assumes a clean corner at every partial it uses.

    The corner does not come back a corner

    Stepping a hammer forward against the string's own returning wave showed that the coupling matters most in the bass. It assumed the wave that came back was the shape that left. On a stiff string it is not: partials travel at different speeds, and by the top octave of a piano the sixth partial is the first to arrive out of place. But a periodic wave can only see phase modulo a cycle, and counted that way twenty-one of forty-eight partials are still in step — which is why the returning wave is degraded rather than destroyed.

    part 6 · timbre
  7. Which of the four terms is doing the removing, note by note. Each of three terms has a corner — a partial number above which it is the one taking the energy away — and the lowest corner at each pitch is the term that binds. The comb is flat at 8, because where the hammer lands is a fraction of the length and does not change with the note. The contact time's corner falls from 34 at A0 to 0.5 at A7. They cross at about E3, 165 hertz: below it the strike point decides the spectrum and above it the contact time does. The dispersion's corner is above both at every pitch on the instrument — 5.8 at the top against a contact corner of 0.5 — so that earlier term is real and is nowhere the binding one. That is a statement no one of them could make on its own.

    Four terms, and only one of them binds

    Six earlier essays each computed one term of a piano's excitation spectrum and handed it to the next, and nobody had multiplied them. Multiplied, three of the four have a corner — a partial above which they are the term doing the removing — and putting the three corners on one axis says which is in charge at each pitch. Below E3 the strike point decides; above it the contact time does; and the dispersion, which is the newest and most laborious term, is nowhere the binding one.

    part 7 · timbre
  8. Where each exciter's contact corner sits against its comb. The merger's four corners, drawn for three exciters. The comb is the strike point and is a horizontal line, because it is a fraction of the string and has no pitch in it. The contact corner is the exciter's, and it falls up the compass because the contact time is a fixed number of milliseconds against a shrinking period. Where they cross, the binding term changes. piano hammer crosses at E3; harpsichord plectrum crosses nowhere in the compass; dulcimer beater crosses at A5. So the crossing found earlier is not a fact about pianos. It is a fact about compliance: a felt hammer stays on the string for about 1.6 milliseconds and a plectrum for 0.05, and a difference of that size is a corner in a different place.

    The crossing belongs to the felt

    On a piano the strike point decides the top of the spectrum below E3 and the hammer's contact decides it above. The merger's own bookkeeping asked whether that crossing is a fact about pianos or about hammers. A plectrum never crosses at all and a hard beater crosses two octaves higher, and the boundary is a contact time of about an eighth of a millisecond — ten times shorter than felt.

    part 8 · timbre
  9. Three instruments, three wires, and an order of magnitude between them. The inharmonicity coefficient of each instrument's own string design, across its own compass. At middle-of-the-keyboard C a piano's wire is 315 millimetres long and 1.02 thick and gives B = 1.5e-3; a harpsichord's at the same pitch is 355 millimetres and 0.30, and gives 8.1e-5 — a factor of 19. The harpsichord's line is nearly flat across four octaves because its scaling is Pythagorean: the length doubles for every octave down, exactly, until the case runs out. The piano's is a V, because its case runs out two octaves earlier and everything below the break is a compromise. Every column before now used the top line for all three instruments.

    Three exciters and three wires

    Every column drawn so far has run a plectrum and a beater on a piano's string, because a piano's was the only scaling to hand. A harpsichord's wire is an order of magnitude less stiff, for a reason that turns out to be the tensile strength of iron — and with all nine combinations computed, the crossing found before stays exactly where it was, on every wire.

    part 9 · instruments
  10. A plucking point 50 millimetres from the nut, across a harpsichord's compass. The jacks stand in a rail and the rail is one object, so a register plucks at a fixed distance from the nut while the strings shorten by a factor of ten from the bass to the treble. At 50 millimetres that is one 36th of the string at C2 and one 3.5th at C6 — a plucking fraction that changes by a factor of 10.1 without the maker moving anything. The comb corner is one over that fraction and falls with it, from 36 to 3.5; the dispersion corner has the fraction under a cube root and falls only from 106 to 21. They never meet. Setting one over p equal to the cube root of two over three times the inharmonicity and the fraction gives a plucking point at p = √(1.5B), which on this instrument's iron wire is between 3.4 and 9.9 millimetres from the nut — nearer to it than any register a harpsichord has ever carried, the lute stop included. So the maker's one free choice is the binding term at every pitch and every register position, which is exactly what the piano's is not: on a piano the contact time takes the decision away above E3.

    What the second register is for

    Nine earlier essays take the plucking fraction as given — an eighth, a sixth, a seventh — and a harpsichord's jacks stand in a rail, so what is given is a distance and the fraction is a consequence: 50 millimetres is one thirty-sixth of the string in the bass and one three-and-a-halfth in the treble. Engage two registers on one string and the amplitudes add exactly, the near one fills the far one's missing partials, the pair digs holes of its own that neither has, and the combination comes out louder and rounder than either — including the bright one.

    part 10 · instruments
  11. Two registers 50 microseconds apart on one string. The partials of a 355-millimetre string plucked at 32 and 50 millimetres from the nut, drawn twice: once with the two quills releasing together, once with the far one releasing 0.05 milliseconds later, which is 0.026 of this string's period. A delayed release turns partial n through 2·pi·f_n·dt, so the rotation is proportional to the partial number: the fundamental is turned 9 degrees and partial 24 is turned 226. Simultaneous, the pair is missing partials 17 and 20 — holes it digs for itself where the two combs are equal and opposite. Staggered, it is missing none of them: a rotation of anything at all takes two amplitudes out of opposition. The partials neither comb can excite at all — 7, 11, 22 — are filled either way, because where one comb is zero the sum is the other one whatever its phase. The fundamental's gain over the far register alone falls from 4.36 decibels to 4.34.

    The interval between two quills

    Two jacks on one key are voiced separately and do not let go at the same instant. That interval turns each partial of the later pluck through an angle proportional to its number — so it leaves the fundamental alone and inverts the twentieth partial, and the holes the pair digs for itself vanish at a hundredth of a period. What a regulator can tolerate turns out to be one fixed fraction of a period at every pitch, which on a five-octave instrument is a factor of sixteen in milliseconds.

    part 11 · instruments

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