# A reader asks: How does the circle of fifths work?

A fifth - in musical terms - is really three-halves, and if you go through twelve fifths, you end up seven octaves higher. Wait, I’m starting all wrong.

A musical note has a frequency - the number of times the sound wave it produces vibrates each second. Middle C, for example, has a frequency of 256Hz - each vibration takes $\frac{1}{256}$ seconds. If you double the frequency of a note, you get back to the same note - just an octave higher. If you look at a guitar, you’ll see the 12th fret is halfway between the two ends of the string - if you play the note there, you’re effectively playing half the string (and it vibrates twice as fast).

Of course, doubling isn’t the only possible way you can change the effective length of the string (in fact, you can plonk your finger down anywhere, but let’s restrict ourselves to nice-sounding intervals). If you shorten the string in a ratio of 1:2, playing a third of the way along, you get a “fifth”. Don’t worry about why it’s called a fifth when it’s really a third of the string and seven semitones different. That way lies madness. Instead, notice that you’re now playing two thirds of the string, so the note’s frequency is 1.5 times as large. Going up a fifth from middle C gives you a G-note, with a frequency of 384 Hz.

Funny thing about fifths ((although it also works with fourths, as well as semitones up and down)) - if you keep going up by a fifth twelve times, you get back to the note you started from (in this case, C). Lookit: C-G-D-A-E-B-F#-C#-G#-D#-A#-F-C. You hit every note in the scale on the way. Musicians call this the *circle of fifths*.

Only… it’s not quite mathematically right. Pythagoras would have been furious.

The thing is, if you increase the frequency by a factor of 1.5 twelve times, you get a factor of 129.75. Seven octaves would require a factor of 128 - the circle of fifths, properly done, is off by about 1.4% - almost a quarter of a semitone.

To me, it’s quite surprising that it should get that close (threes and twos generally don’t play nicely together, so the fact that $3^{12} \simeq 2^{19}$ is quite pleasing - and I’m sure the Mathematical Ninja would be able to remember that $\log_2(3) \simeq \frac{19}{12}$ to great effect.

However, to an orchestra, that’s a Serious Problem. Instead of playing notes exactly a third of the way along the string, players need to adjust each note ever so slightly - around a tenth of a percent, in frequency terms - to get a nice resolution at the end.

Disclaimer: I am not a proper musician. I mess around on a guitar, but it’s strictly three chords and the truth. I’m not even a proper mathematician. I’m a sort of an extremely low budget Vi Hart. By which I mean: if you’re a mathematician and/or a musician who’s spotted an error, let me know… nicely!