# Why do frets get closer as the notes get higher?

Why do spaces between guitar frets get smaller as the notes get higher?

• The over-arching rule for fret size is the Rule of 18, which states that a fret is placed 1/18th of the distance from the nut/previous fret to the bridge. Actually, 1/17.817. And that's not quite right in every case, as the G on the low E will tend to be sharp because of the string bending implicit in fretting and the width of the string. You can't really tune a guitar, but you can get closer... – Dave Jacoby Dec 9 '13 at 17:20

This is more clear when you think of it in terms of the length of the string. Ignoring for a second that strings need to be a minimum length in order to vibrate, we can produce a full octave on the first half of any length of string. Open for the "base" note, midpoint for the octave, one third of the way up for the fifth, etc.

So on a guitar you have the first half of the string to make up an octave. Then on the second half, you have another half of that (a quarter) to make up the next octave. Then another half of a half of a half (eighth) to make the next octave. Each time, you only have half the space for the octave. Thus, each fret has to be half the size of the fret an octave lower.

This mirrors my comment above: While maintaining frequency ratios, the absolute frequency differences changes by a factor of 2, and that is represented in the physical frets as well.

• The answer is totally off. Is it not the exponential increase that makes the higher note frets get closer. See my answer. – Bogdan Alexandru May 1 '19 at 12:34
• @BogdanAlexandru I fail to see how our answers are in disagreement. You've simply provided the general case. – user28 May 1 '19 at 22:34

Because frequency with regard to pitch is exponential, not linear.

Exponential (frequency doubles for each octave -- each higher octave fits perfectly into the lower one with the least possible interference -- 2:1 ratio):

```A3: 220hz
A4: 440hz
A5: 880hz```

Linear (same value added to each successive frequency, making the top note a 3:2 ratio with the previous -- much more interference than an octave):

``` A3 : 220hz
A4 : 440hz
'A5': 660hz -- **nope**```

In reality, this 660hz would be a just intonation fifth above A4, or E5.

• Actually frequency is Exponential with respect to pitch, not Geometric. Exponential is `P(n) = P(n-1)*K` (which is also `P(n) = P0 * K^n`), so "frequency doubles for each octave" would be `Freq(octave) = Freq(octave-1)*2`. Geometric is between Linear and Exponential and is expressed as some fixed power of an index: `P(n) = n^K`. – RBarryYoung Dec 3 '13 at 4:08
• Even if frequency was linear, the frets would still get closer for higher notes. In fact, the distance between the frets would go down much faster in that case. – Bogdan Alexandru May 1 '19 at 12:36
• This answer talks about frequency. The reason why the frets get closer is because frequency is inversely proportional to vibrating string length, which is known as Mersenne's First Law of Acoustics or Pythagoras's Law. – awe lotta Jun 20 at 3:36

The frets are getting arithmetically closer together (their absolute distance decreases). But they are not getting geometrically closer, in terms of their distance from the bridge.

If you take the distance from the bridge to any fret (call that X) and also the distance from the bridge to the next higher lower (call that Y), then the ratio X/Y is the same whether or not X and Y are frets 2 or 1, or frets 20 and 19.

This ratio produces the frequency ratio that corresponds to the (equal temperament) semitone. It is 21/12, or about 1.0595. (The two denotes that frequency doubles per octave, and the 1/12 indicates one half-step out of a potential twelve in an octave.)

Fret 19 is 1.0595 times farther away from the bridge than fret 20.

Fret 1 is 1.0595 times farther away from the bridge than fret 2.

This 1.0595 ratio is also the frequency ratio. If you know the frequency of some given note, like A = 440 Hz, you can figure out what is the frequency of A#, one semitone higher. Just multiply 440 x 1.0595 = 466.18. The A# above the 440A has a frequency of about 466.2.

There are 12 semitones in an octave. If we multiply a number by 1.0595 and do that 11 more times, we will get about twice the original number. You can try this on your calculator. Type 1 X 1.0595. Then hit the = key 12 times. You should get a number very close to 2.

For anyone interested the common ratio between the frets would be the twelfth root of 2.

• Only for a well-tempered, nonoffset fretting :-). I forget who, but someone makes guitars with frets that look a bit like lightning bolts, allegedly to enhance the resonances between strings. – Carl Witthoft Dec 2 '13 at 21:15
• If the fretting is non-offset then it can't be well-tempered. Those frets you refer to are actually called "true temperament" frets. – Andrew James Dec 2 '13 at 22:13
• And in particular the word you are looking for is equal-tempered, for traditional fretting. – NReilingh Dec 3 '13 at 4:19
• @CarlWitthoft, you are probably referring to the "True temperament" system (truetemperament.com/site/index.php). They make fretboards with both (more) equal tempered and well tempered variants. No pure/just intonation from what I know, that would probably be to impractical. – Johan Dec 4 '13 at 15:26

Interestingly, all the answers are completely missing the point. Everyone brings up the exponential increase of frequency as the reason ("you double the frequency for each octave"), but that is a red herring.

Even if the frequencies of consecutive pitches increased linearly, the higher note frets would still be closer together.

The actual reason why this happens is because the length of the vibrating string is inversely proportional to the frequency of the sound it produces. This very simple physical explanation is the answer to the question.

Given two "consecutive" frequencies f1 and f2, the distance between the two frets is proportional to (1/f1-1/f2) or (f2-f1)/(f1*f2).

So even if (f2-f1) were constant (i.e., frequencies increase linearly) or outright increasing, the denominator (f1*f2) still increases very fast as f1 and f2 go higher, which means that no matter what formula you pick for the frequencies, the distance between frets will get smaller.

• Obviously you could pick a formula that results in equal-spaced frets; that formula for frequencies would match the frequencies found on a guitar built with equal-spaced frets. I wouldn't say that "you double the frequency for each octave" is a red herring; certainly it is a result of the relationship between frequency and string length, but it nicely illustrates why the frets must be closer when octaves correspond to frequency doubling. – ex nihilo May 1 '19 at 12:52
• While this is true, the same formula holds whether the pitches increase linearly or exponentially. Since fretted instruments have their frets placed to produce our 12TET pitches, both explanations are correct - each octave doubles the frequency, which halves the remaining string length. Luthiers think in terms of the exponential relationship because that tells us where the next fret needs to go - and in music that relationship is exponential. – Tom Serb May 1 '19 at 13:04