Recently I came up with interesting discovery how to get equally tempered tones by using a stack of just intervals 3/2 and 5/4. Since they are a part of most harmonic cord, major triad, should be caught very accurate.

The kernel of my theory is using the stack of 7 perfect fifths (3/2) and 1 major third (5/4) to get a single equally tempered tone.

(3/2)^7 * 5/4 * 1/16 = 1.3348388671875

The ratio must be divided by 16 (or 2^4) because the tone I'm looking for went 4 octaves up.

In this example the result is perfect fourth. Mathematical precision is till 5th digit after the decimal point. The error is 0.00128 cents. Repeating same stack 11 times gives a final error of 11 * 0.00128 = 0.01408 cents.

I wrote an article about this discovery but many musicians claim this is not possible in terms of acoustics because lot more events occur, for example inharmonicity and octave stretching. Due to imperfection of the instruments this remains just theoretical.


Although this may never be achieved in practice, I believe the math behind this theory is exceptional. Many roots of 2 could be expressed by small fractions of the first 3 prime numbers (2, 3 and 5). Everything from 2^(1/12) to 2^(11/12) could be calculate very accurate by stacking above expression 11 times. Reversing the ratios is legit too.

Is it possible to tune a piano using this method of stacking 7 P5s and 1 M3? I'm expecting better accuracy up to 246 times but what would be the actual results?

So far the best procedure I can offer for tuning is this one:


Is it efficient enough? Is it going to accumulate errors or it's error free?

Another rough procedure that pushes the tuning process to its maximum is:


Since the stack contains twice less steps but larger ratios, is there any possibility to get it done without accumulating errors?

  • 3
    Is there a question here? In any event, even though it's to within the fifth decimal of a true equal-tempered fourth, it's a little bit different. If you use this interval in a 12-tone chromatic scale, the resulting intervals won't be equal. And the fact that you got to this interval by stacking just intervals doesn't change the fact that it's somewhat different from a just fourth of 4/3. And suppose you start on F and end up on B-flat (well, A sharp). If you use those seven stacked white notes for your scale, you'll still have Pythagorean tuning there with its 81:64 thirds, not 5:4.
    – phoog
    Aug 26, 2021 at 23:35
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    I’m voting to close this question because it's not a question.
    – phoog
    Aug 26, 2021 at 23:45
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    I've voted to reopen this question because it now contains a question.
    – phoog
    Aug 27, 2021 at 0:33
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    Note that, mathematically speaking, this cannot be error free as 2^(1/12) is an irrational: it can be estimated but not calculated with infinite precision using fractions (hence the name irrational).
    – Tom
    Aug 27, 2021 at 6:04
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    @Tom "irrational" doesn't mean "impossible"; for example, you can create a line segment with a length of exactly √2 by creating a unit square and using its diagonal to define your line segment. The fact that you can't represent it as the sum of a finite series of rationals doesn't mean that it can't exist.
    – phoog
    Aug 27, 2021 at 12:32

1 Answer 1


The problem isn't the math. The problem is the efficiency of the tuning algorithm (or, lack of efficiency, as the case may be).

Since the algorithm is monotonically increasing (up to the point of octave correction), let's begin by tuning the lowest pitch, A0, which we'll accept as given.

Since our goal initially is to tune all 12 chromatic pitches as accurately as possible, we can forego the octave corrections for now. This won't affect the math, since multiplication is commutative. We can do all the fifths and thirds first, and deal with the octaves later.

Thus, to tune the each of the fourths most accurately (without regard to octave) requires 8 operations: 7 fifths and 1 third.

Thus, tuning each of the 12 chromatic pitches to the greatest accuracy requires a minimum of 8 * 12 = 96 operations. (And, since tuning each fourth ascends four octaves, we would need a piano with 12 * 4 = 36 octaves.) Limiting our piano to 7 total octaves, and having tuned the equivalent of one of them, we now need an additional 12 * 6 = 72 octave tunings.

Hence, a total of 168 operations involving 32 working octaves is required.

Further, since shortening our keyboard to an actual 7 working octaves only requires changing the order of operations (doing the octave corrections as needed), we still need a minimum of 168 total operations.

In other words, on a seven octave keyboard (7 * 12 = 84 keys), every key must be tuned twice, on average, to achieve the ideal result.

  • In my best case scenario you need only 2 octaves to work on. Take a look supplementary articles, i.e. Fast Inverse Stack nearequaltemperament.com/inverse-stack . Explains the whole process efficiently by working on 2 octaves only, C3-C4, the ones we can hear very accurate.
    – Vinkelman
    Aug 27, 2021 at 4:27
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    @Vinkelman If that's the question you want answered, please put it in your post.
    – Aaron
    Aug 27, 2021 at 4:28
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    @Vinkelman Looking at the "fast inverse stack", I see it takes 99 steps to tune a single octave. That leaves 6 octaves left to be tuned, which requires another 6*12 = 72 steps, for a grand total of 171 steps. Less efficient than the procedure outlined above.
    – Aaron
    Aug 27, 2021 at 4:41
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    @Vinkelman I should reiterate that I see no problem with the math (i.e., the accuracy of the tuning), only the efficiency compared to standard practice.
    – Aaron
    Aug 27, 2021 at 4:43

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