Originally Posted by

**Victor Redseal**
Pythagoras is credited with discovering the musical comma. He discovered the musical notes were ratios related to length and frequency. If one took a lyre string tightened at both ends, plucked open we could arbitrarily call it 1. If we halve the length of the string, it will play exactly one octave higher (i.e. it vibrates exactly twice the rate of 1). So the octave can be expressed as a ratio of 2:1 or 1:2. However, if we shorten 1 by a third of its length, it produces a note that plays a fifth higher. The ratio of a fifth is then 2:3 or 3:2. These ratios are called intervals and Pythagoras calculated them thus:

1. C – Fundamental (1)

2. C# – Minor 2nd (15/16)

3. D – Major 2nd (8/9)

4. D# – Minor 3rd (5/6)

5. E – Major 3rd (4/5)

6. F – Perfect 4th (3/4)

7. F# – Tritone (5/7)

8. G – Perfect 5th (2/3)

9. G# – Minor 6th (5/8)

10. A – Major 6th (3/5)

11. A# – Minor 7th (5/9)

12. B – Major 7th (8/15)

13. C’ – Octave (1/2)

Just remember that the ratios can be inverted depending on whether you lengthen or shorten the string.

Now, since there are six whole steps in a scale (e.g. in the space of an octave) and a whole step is 9/8, then if we raise that ratio by the power of 6, it should come out to exactly 2. Does it? No. It works out to 2.073 which is only slightly off but the human ear can perceive it and it sounds wrong to us. What is means is that there is a tiny but noticeable drift between enharmonic equivalents such as A# and Bb or E# and F. This is the Pythagorean or ditonic comma.

Suppose we measure the 5ths in an octave. There’s only one 5th in an octave. Two 5ths will pass out of the octave. So, if we start measuring 5ths, we have to find a way to keep the tones within the octave. Once the 5th is out of the octave, its value must be halved to keep it within the octave. Starting at C, for example, the first 5th interval ends at G and we know that the ratio is 3/2 (or 2/3). The next 5th takes us to D and so we would square 3/2 to obtain 9/4 but that passes out of the octave (is greater than 2 and an octave must be exactly 2/1). So we multiply 9/4 by 1/2 to obtain 9/8, which is less than 2 and so is within the octave. Next, we jump up to A which is mathematically obtained by multiplying 9/8 by 3/2 or 27/16 which is within the octave. If we keep going through 5ths until we pass through all 12 semitones (after A, we go to E, B, F#, C#, G#, D#, A#, F and C) we end up with a final value of 262144/531441.

The true octave, however, would yield 262144/524288. Again, the actual length of the string would be somewhat shorter than the true octave string length and so would be sharp. Our differential is the ratio of 531441/524288 which is the ditonic comma (not “diatonic”). That is the Pythagorean comma.

Another comma is called *syntonic*:

If we form a circle of the C major octave where the full octave is 360 degrees exactly, then C=0 and 360, D=320 (360 x 8/9), E=288 (360 x 4/5), F=270 (360 x 3/4), G=240 (360 x 2/3), A=216, B=192, and C’=180. From D to F is a minor 3rd (3 half-steps) with a ratio of 320/270 or 6.4/5.4 even though the true ratio should 6/5 or 324/270. So the actual difference is 324/320 or 81/80. That ratio is called the syntonic comma. From C to G is a perfect 5th of 360/240 or 3/2. However, if we were to measure a perfect 5th in the next octave from D to A or 320/216 ratio, we notice that it is an 80/54 ratio. A true perfect 5th would be 81/54 or 3/2 and so, again, we end up with a discrepancy of 81/80 (oddly, the reciprocal of this number is 0.987654321). The next octave after that would yield the major 6th (F-D) and the perfect 4th (A-D) also off by the syntonic comma. The comma is very noticeable and must be dispersed in some manner.

__Method I __

One way to disperse the comma is through adjusting the major 3rds in the octave. In a 12-tone octave, there are three major 3rds (4 half-steps x 3 = 12 half-steps). A true major 3rd has a 5:4 ratio, that is, if you shorten a string by 4/5 but retain the same tension, the string will play a major 3rd interval higher. If we start at middle C, our three major 3rds would be C-E, E-G#, Ab-C (remember that Ab is the enharmonic equivalent of G#). Since the octave interval must always be an exact 2:1 ratio, the Ab-C major 3rd will be a bit flat. Why?

Since a major 3rd is 4/5, then we would cube that ratio to obtain 64/125 for the full octave. But a full octave is 64/128 (1:2 ratio). Since 64/125 represents a longer string length than 1/2, the major 3rds will be noticeably flatter than in a true octave since a string’s length is proportional to the pitch. The discrepancy is the ratio of 125/128 or 0.9765625, which is called a diesis. Each major 3rd interval in the octave must be sharpened slightly by a third of the diesis or about 0.3255208333.

__Method II __

We may also measure the minor 3rds in an octave of which there will be four. Using C major as an example and starting at middle C, our minor 3rds will be C-Eb, Eb-F#, F#-A and A-C’. Since a true minor 3rd would have 5/6 ratio, then the total value of an octave in minor 3rds is 5/6 raised to the power of 4 or 625/1296. The actual value of a full octave is 648/1296 or 1/2. Since the string length of an octave of minor 3rds is somewhat shorter than a true octave resulting in a higher pitch, the minor 3rds will be a bit sharp and must be uniformly flattened. So, the ratio of 648/625 or 1.0368 tells us the total difference in tone and so each minor 3rd interval must be flattened by a quarter of 1.0368 which is 0.2592. While other ratios are called a comma or *diesis*, this 648:625 ratio has never been named for some reason.

__Method III __

In this method, we solve the ditonic comma which also solves the syntonic comma. Pythagoras was said to have solved his comma but we are not told how. We do it today by tempering. The value of the ratio of the ditonic comma 531441/524288 is approximately 1.01364327. So we would flatten our 5ths by 1.01364327/12 or about 0.08447. This is the method for tempering the 12-tone scale.