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January 5th, 2011, 10:06 PM   #1
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Simplify Formula for Greatest Number of Guitar Frets

Out of curiosity I wondered what the greatest number of frets that would be practical on a guitar. There are 3 variables involved to do this:

N=the Number of frets on the guitar
L=Length for the guitar strings
G=the Gap span length between the last fret (closest to the bridge) and the previous fret closest to it

For my taste I'd say that the smallest practical gap (G) is 0.25" and the longest string Length (L) that I could manage to play chords on would be about 40"; so I want to solve for N in my formula.

Now I already know by playing around with the formulas that the number for N is going to be 39 with some decimal points, but it bothers me that I can't simplify my formula to solve for N.

Here is how I solve my problem with a formula:

The distance from the Bridge to any fret number is: L / 2^(N/12)
Note: Fret 1 is at the top of the guitar neck nearest to what is called the Nut; and Fret 39 in this example would be practically next to where you are strumming the guitar strings.

So if I assume that N will be the last possible fret on the guitar (e.g. 39) then I can calculate the Gap distance (G) by subtracting the distances of Fret (N) from Fret (N-1). E.g.

G=L/2^((N-1)/12) - L/2^(N/12)

So by playing around with this formula I was able to find my answer, but I'd rather have a formula to calculate it instead. When I played with the numbers I was able to calculate that a 40" string length and 39 frets will have a gap span of 0.250011514" between fret 38 and 39 so that would be my practical limit for the largest number of frets that I could use.

Any help with solving for N with my formula is appreciated.

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January 5th, 2011, 11:28 PM   #2
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Multiplying by (2^(N/12))/G gives 2^(N/12) = (L/G)(2^(1/12) - 1),
so N = 12ln((L/G)(2^(1/12) - 1))/ln(2) = 12ln(L/G)/ln(2) - 48.86234 approximately.

Hence N = 39.000797... if L/G = 160.
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