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I found it inside an italian music book:

enter image description here

It [artistically] represents the music scale. I think it is a series of trigonometry sin equation. With some difference in the amplitude and in the frequency.

What are the exact parameters of the curves?

Alexandre
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nkint
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  • As you can see from the answers, there is really no good way to correspond all the sine waves on this page with the names of notes shown here. Perhaps the page before this, or after, has some kind of explanation of what this figure is supposed to show. We have several people here (not I!) who are fluent in Italian, so perhaps one of them could translate. I think it is quite possible, however, that the figure is simply utter nonsense. – David K Nov 19 '15 at 23:56
  • @DavidK Well, "utter nonsense" is perhaps uncharitable :) But the figure is... more romantic comedy than documentary, let's say. Notice that it doesn't make too big a point of connecting the solfeggio names (do, re, ...) with particular curves: they're not callouts. There's a very good reason for that! – BrianO Nov 20 '15 at 00:03
  • @BrianO: You may be onto something there: They all get each other in the end! – hmakholm left over Monica Nov 20 '15 at 00:07
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    @BrianO Utter nonsense can be beautiful. It just doesn't pay to try to make sense out of it. – David K Nov 20 '15 at 00:07

2 Answers2

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The figure appears to shows 12 superposed sine curves with wavelenghts $1$, $\frac12$, $\frac13$, down to $\frac1{12}$, and amplitude proportional to the wavelenghts, such that the slopes at the nodes are equal. In more elementary mathematical terms, it is a plot of the 12 functions $$ f_n(x) = \frac1n \sin(nx), \qquad\qquad 1\le n\le 12 $$ on the interval $x\in [0,2\pi]$ (and then everything turned 90­° such that the $x$-axis is vertical).

This is not directly related to the diatonic scale alluded to on the left.

If the vertical (on the page) axis represents time, the notes corresponding to the sine waves are the first 11 harmonics of the one represented by the largest wave -- many of these do correspond to "pure" intervals in music, but don't directly map to the seven steps of a diatonic scale.

Assuming that the base tone is a C ("do"), the 12 tones depicted are $$ C, C, G, C, E, G, A^{(\sharp)}, C, D, E, F^{(\sharp)}, G $$ spanning $3\frac12$ octaves, where the bracketed sharp signs indicate in-between notes that are not usually used in Western music.

  • ^ Not usually used in equally tempered western music. An acapella group with just intonation would definitely use the I7 chord, for example. – pancini Nov 20 '15 at 00:36
  • @ElliotG: Does an just-intoned I7 really have that low a seventh? A ratio of 7:4 is around 967 cents, much lower than the seventh in V7, which is 16:9 at 996 cents. (Edit: Hmm, Wikipedia does repeat that claim about barbershop groups, sourced to unnamed "some theorists"). – hmakholm left over Monica Nov 20 '15 at 01:07
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These are the 12 harmonics of a given frequency, as Henning Makholm points out in his answer. All the sine waves shown are integer multiples of some base frequency, one cycle of which is shown. Many of these correspond to pitches of the just intonation scale, which is not the scale to which modern Western instruments e.g. keyboards are tuned. In that tuning, the "tempered scale", pitches are adjusted ("tempered") so that all keys are equally, subtly out of tune :) and so that modulations become possible and none sound sour.

The diagram shows one cycle of "do", whatever frequency that might be, and all the other harmonics plotted against it (plus the octave above "do"). No base frequency is given, so the diagram is relative to some/any chosen reference. If $\theta_{do}$ is that frequency in Hz (cycles per second), then this shows one cycle of $\sin \theta_{do}$, plotted against $\sin$ of various multiples of $\theta_{do}$. The amplitudes seem a little arbitrary — happened to make for a legible diagram. (They're not entirely arbitrary, of course: sine waves with more cycles per the period shown have smaller amplitude.)

The octave above is twice the frequency, and that's represented in the diagram by the sine wave which completes 2 cycles. The sine wave of the octave is $\sin 2\theta_{do}$.

"so", the 5th, is $\frac 3 2$ times the frequency of "do" in just intonation, so that's represented in the diagram by the sine wave in which completes $3$ cycles. The pitch of this is actually "so" of the next octave.

"mi", the natural $3$rd, has ratio $\frac 5 4$ with "do". It's represented by the sine wave which completes 5 cycles. Note that the frequency of the sine wave shown is thus $2$ octaves above "mi".

The ratios of all notes of the just intonation scale are as follows:

$$\begin{align} &\text{do} \quad&1/1 \\ &\text{♭ re} \quad&16/15 \\ &\text{re} \quad&9/8 \\ &\text{♭ mi} \quad&6/5 \\ &\text{mi} \quad&5/4 \\ &\text{fa} \quad&4/3 \\ &\text{#fa} \quad&45/32 \\ &\text{so} \quad&3/2 \\ &\text{♭ la} \quad&8/5 \\ &\text{la} \quad&5/3 \\ &\text{♭ ti} \quad&9/5 \\ &\text{ti} \quad &15/8 \\ &\text{do'} \quad & 2/1 &\quad \text{(do, an octave up)} \end{align}$$

Some of these can't be octave-adjusted (multiplied or divided by some power of $2$) to fit within a single cycle of "do". This is why tempered tuning was invented! Tempered tuning provides closure; for any interval, some integer multiple of that interval gets you to an octave of "do". For example, three rising Major 3rds brings you to the octave of your starting note, whereas no integer multiple of the just-intonation "mi" is an octave of "do".

BrianO
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  • I don't think the wavelenghts plotted in the diagram match those ratios. For one thing, there are clearly 12 since waves in the diagram, rather than just 8, and the one that varies most quickly has 1/12 of the wavelength of the largest one. Furthermore, since all the ratios in the just intonation sequence (except for the final "do") are strictly between 1 and 2, the intermediate nodes should not return to 0 at the time one cycle of the base node has completed. – hmakholm left over Monica Nov 19 '15 at 23:16
  • @HenningMakholm I didn't see all 12 at first, I had to zoom way in. So, the entire octave. I'll emend accordingly. – BrianO Nov 19 '15 at 23:20
  • No, not an entire octave either -- the period of the smallest curve is clearly much less than half of the largest. In fact the one an octave above the base is the second largest of the curves shown. – hmakholm left over Monica Nov 19 '15 at 23:21
  • @HenningMakholm The second largest is the octave ;) The smallest should be much less than half of the largest, because these frequencies are not octave-adjusted. E.g. the 5th shown is from the octave above, otherwise it would complete 1.5 cycles rather than 3. – BrianO Nov 19 '15 at 23:50
  • x @Brian: Which octave do you take the perfect fourth from such that its period divides the base period? – hmakholm left over Monica Nov 19 '15 at 23:54
  • @HenningMakholm None :) See my revision. – BrianO Nov 19 '15 at 23:58
  • Okay :). Only now I have to take issue with the last sentence -- tempered tuning has nothing in particular to do with wanting every step in the scale to be octave-adjusted harmonics. And equal temperature doesn't even come close to achieving that, anyway; on the contrary all other intervals than octaves have irrational ratios in equal temperature. – hmakholm left over Monica Nov 20 '15 at 00:06
  • Not octave-adjusted harmonics, but closure, in that e.g. four rising Major 3rds get you to the octave, whereas no integer multiple of the just Mi is an octave of Do. I added that to clarify. – BrianO Nov 20 '15 at 00:12
  • Hmm, there are only three major thirds in my octaves, but nevermind :) – hmakholm left over Monica Nov 20 '15 at 00:18
  • @HenningMakholm Um, in mine too -- thanks/sorry, distracted. – BrianO Nov 20 '15 at 00:21