Many guitarists use harmonics to tune up by ear, or to check their tuning. For example, they might ping the A string over the 7th fret and then ping the bottom E string over the 5th fret. Or ping over the 7th and 5th frets for any other pair of strings (except B and G which are pinged over the 5th and 4th frets respectively). We might expect these pinged tones to sound exactly the same, ie in unison with no "beating". But, as many guitarists know, tuning by harmonics (pinging) can give errors. A simple test is to tune all the open strings carefully to a good quality tuner. Then "check the tuning" by pinging the aforesaid harmonics. On a well-built guitar, they'll be a bit off. Why??

Guitars are built to play scales based on what is called "equal temperament", which means that an octave is "divided" into 12 equal parts. However, books on music rarely explain what "equal" or "parts" __really__ mean - well, not to any great depth anyway. Instead, they wander off into the world of Greek and Medieval tuning and funny scales derived from harmonics and then adjusted a bit by "tempering" certain notes. Way too much information. It's really quite simple: in equally-tempered tuning, the ratio between the frequencies for any given musical interval (i.e. a pair of notes) is always the same, no matter which note in a scale is the lower of the two notes - and this is why the frets on a guitar have a spacing that decreases by an __exact ratio__ as one goes up the fingerboard. If they did not, they could look like this - a special fingerboard for playing non-Western music.

So, what is this magical "exact ratio" and how is it calculated?

As we know, an octave is made up of 12 semitones (the chromatic scale). It is also well-known that, for the interval of one octave, the upper note is exactly twice the frequency of the lower. Therefore, the trick is to find out what ratio multiplied by itself 12 times comes to a value of 2. Since this is not a maths lesson, I'll tell you that this ratio can be calculated by finding the 12th root of 2, i.e. 12√2. Piece of cake ;-)

According to my Casio Scientific, 12√2 equals 1.059463094 but, even though that's a long number, it's not exact. 1.05946309435 is a little more precise but numbers like 12√2 can not be expressed exactly in decimal form. Be that as it may, now we do know the magic ratio precisely enough for our purpose and we can calculate any interval's frequency ratio, assuming that we know the number of semitones contained therein. But first, what has the foregoing to do with harmonics?

When we ping a string, it will vibrate at a tone higher than the fundamental (open) tone. In the opening paragraph, pinging at the 7th fret was used as an example. Pinging on the A string at the 7th fret causes the string to sound at exactly 3 times it's fundamental frequency of 110Hz, i.e. 330Hz. Pinging the bottom E string at the 5th fret causes that string to sound at exactly 4 times it's fundamental frequency (see next para). Musically, these two tones should be in unison but, on a perfectly-built guitar with perfect strings, they'll be a little off - beating about once every 3 seconds. Again, why?

OK, please recall that equal-temperament is the normal tuning for a guitar and recall also that, to be in tune with rest of the band, we tune our A string to exactly 110Hz (two octaves down from the concert standard A=440). So, what __is__ the tonal "distance" from our A open down to E in semitones? Musically, the interval is a perfect fourth, which is 5 semitones or frets. So, on a guitar with it's equal temperament construction and with the A string tuned to a frequency of exactly 110Hz, the frequency of the low E tone would be: 12√2 raised to the power of 5 and then divided into 110Hz (trust me). Which is a fundamental (open string) frequency of about 82.407Hz. Now we can figure those harmonics: with a correctly tuned guitar and with all else being equal, The third harmonic of A is **330Hz** but the 4th harmonic of E (82.407Hz times 4) is **329.628Hz**. There we have it - there is a __difference__ of 0.372Hz. Not much you might say, and you're right - it isn't much - only two cents! In fact, if the E string is stiffer in construction than the A, the real difference could easily be negligible.

However, when we tune higher pairs of strings, the difference is more noticeable. The best example is tuning G to B. Pinging B at the 5th fret and G at the 4th until the beats are longer than say two or three seconds will give a G string that sounds __horribly__ sharp when you play an open E chord. Let's do the calculation . . .

Tuning the B string to a tuner should set it to a frequency of 246.942Hz or thereabouts. Then, when we ping the 5th fret, it should sound at 987.767Hz (2 octaves up, i.e. 246.942Hz times 4). Now pinging the G string at the fourth fret sounds the 5th harmonic (open G frequency times 5) and therefore, if we have tuned it perfectly, the open G is now tuned to 987.767Hz divided by 5 = **197.553Hz**. However, this frequency is __not__ what your electronic tuner gives out, which is actually **195.998Hz**. So, by pinging, we've made the G too sharp. In terms of musical cents, the difference is almost 14 cents - very noticeable to most people.

Now that we know tuning by pinging gives a slightly sharp lower string, what if we started with the top string set to a tuner and then tuned by pinging each pair downward i.e. E/B, B/G, G/D, etc? The errors would add up through these 5 tuning steps. How sharp would the bottom string be? I figure it to be sharp by 21.5 cents. If you then pinged the bottom E at the 5th fret and played the top E open, it wouldn't sound at all good!

Conclusion: Because our scale is based on octaves, it would appear that the only safe harmonics to use while tuning up a guitar are the octaves, (over the 12th or 5th frets). Higher harmonic octaves are less useful because, on real strings, the higher harmonics play sharper than they should due to inharmonicity. Also, I conclude that tuning from an inner string toward the outer strings would give less error that tuning from the top string down to the bottom - for example, see a method used by Kevin Ryan here.

*Note: number of cents different = 1200 * log (f2/f1) / log 2*

Best regards,

Ted