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The frets on a guitar, like the positions on a trombone slide, are not evenly spaced. To make the pitch of a note go down by one semitone you must lengthen the vibrating part of the string or the effective length of the trombone by a certain amount; the length must be multiplied by a number which I call “S”. Multiplying the length by “S” twelve times must double the string length, lowering the pitch by one octave. Therefore “S” to the 12th power equals 2, so “S” is the twelfth root of 2, or approximately 1.059463.

Half a semitone is a quarter-tone. If we wanted to add new frets to a guitar fingerboard so that quarter-tone intervals could be played, we might try placing a new fret halfway between a consecutive pair of old frets. If one fret is at distance “L” from the bridge and the next fret is at distance S × L from the bridge then the quarter-tone fret would be placed at the halfway point between them, which would be at the * average* of those two distances from the bridge:

Now imagine subdividing one semitone into 100 smaller intervals, so that in the space between one fret and the next we create 100 smaller spaces. If we arrange the new “mini-frets” to have equal spaces between them then their separation will be 0.059463 ÷ 100 = 0.00059463 times the original string length. But we really shouldn’t space them equally, for the same reason that the original frets are not evenly spaced. To make them divide the semitone into equal musical intervals the string lengths must have equal * ratios*.

As we go from any one mini-fret to the next the vibrating string length must always be multiplied by the same number, which I shall call “C”. That causes the pitch to go down by a tiny interval called a “cent”. Multiplying the string length by “C” one hundred times in a row must cause that length to increase by a factor of “S”, causing the pitch to go down by one semitone. Therefore, C^{100} = S, so C = the * hundredth root* of S, which comes out to roughly 1.000577789. If the mini-frets are correctly spaced then the space between one mini-fret and the next one is then 0.000577789 times the string length, instead of the 0.00059463 that we found with evenly-spaced mini-frets. There is a slight difference between those two numbers, but it’s not very much. In other words, the mini-frets should be

There is an excellent **Wikipedia article** which includes more mathematical details about "cents" and also provides some audio samples.

Back to **How to Calculate Pitch Discrepancies:
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