This has been puzzling me for a while and probably needs someone with a maths A-level/ degree to solve.

How do you work out the spaces between frets. I know that to go up one octave you half the length of the string, which doubles the frequency. But how much do you shorten the string by to get a semitone or tone (inproportion to the string of course) it's faily obviously not just 1/12 per semitone and i was thinking you could work out a general formula for it but my maths just isn't good enough. I got as far as log12...

There are quite a few geniuses on UG and i hope one of you guys can help
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im not exactaly sure, but i think it all breaks down to a simple quadratic formula. I'm not sure wat the formula is, but im goin to go try to find out. this is very interesting actually, lol.
there's a factor you reduce the fret distance by the whole way along the fretboard

fret spacing is determined something like this.

you take the guitar's scale and divide it by 'bout 17.817154 for the distance from the nut to first fret. Ya take that distance and subtract it from the total scale and divide what's left again by 17.817154 for the distance from the centre of the 1st fret to centre of the second.
It's something to do with the 12th root of 2...you need a scientific calculator.

if scale length = S, then first fret distance = .

And so on. Work it out. I can't think now.
A 6% change in length gives a semitone change in pitch. I think that's what you are asking.
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Originally posted by Stratwizard
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God I never knew this either. I dont get how frets higher up the board have to be closer together, it just seems a bit illlogical. Its something to do with high end sound physics I guess.

Quote by Robbie n strat
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There's a constant ratio (we'll call it x) between the length from the bridge to each fret and the distance between the bridge and the next fret. The 12th fret is half way along the fretboard. Therefore, x^12=0.5
So the ratio is the twelth root of 0.5, which is 0.94387 approximately. That's the length from the first fret to the bridge, divided by the length from the nut to the bridge.

And I've got one and a half maths a-levels
well, i found that its not a quadratic formula because the "x's" of the formula would repeat, but i did find that the average length between frets from 0 to 12 is 30/13 cm. but if it means anything, the lengths that repeat to rule out a quadratic formula are
nut to fret 1= 3.5 cm,
fret 1 to 2= 3.5cm
2 to 3= 3cm,
3 to 4= 3cm
4 to 5= 2.8cm
5 to 6= 2.8cm
ect.
all of this is on a 22 fret, 24 inch scale guitar
Wow.

Isn't it wierd how we all have different equations?

There was this really advanced fret calculator somewhere on the net which let you calculate fret intervals according to 3 different formulas.

http://fretfind.ekips.org/

I dunno what the other ones are for...12th root of 2 seems to work for me...
No, it's exponential.
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Quote by notoriousnumber
God I never knew this either. I dont get how frets higher up the board have to be closer together, it just seems a bit illlogical. Its something to do with high end sound physics I guess.

not really, the length of the string from the bridge to the nut in longer than to the 20th fret and 6% of something big is longer than 6% of something small, so the higher frets are going to be closer together.

The change of pitch in cents is where the exponential comes in "Interval in cents = 3986 X log(frequency ratio)". 100cents is a semitone in equal temperment and 1200 cents in an octave.
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Originally posted by Stratwizard
John Petrucci overrated? Is God overrated?
Last edited by kingal at Jul 12, 2006,
So are you making a guitar or something?
The math is interesting really, too bad I'm terrible @ math.
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the higher frets are going to be closer together because its closer together to the neck.

Lets make a line
``````
A-----C-----B
``````

if we know half the distance of AB is an octave, we know half the distance of CB is an octave.
The frets have to be smaller in order to fit an octave's worth of frets in smaller distance.

The ratio's stay the same, but the distance they get to use differs because the higher the fret goes, the less space you have to work with. You can use the ratio to find out any note, and nowadays I dont think they really think about that much anymore because someone else did for them and the neck works perfectly as it is designed. Yeah. lol
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Well, it depends of the fundamental mode of vibration.
I think, correct me if I'm wrong, that normally a first overtone(sinusodial curve) is obtained when we pluck....
The frequency for 1st overtone is velocity divide by wavelength....
Therefore if a graph of frequency against wavelength is plotted, then a curve would be obtained.... That's why it the length vary to change a semitone...
For harmonics, 2nd overtone is obtained. The frequency is equal to 3(velocity)/2(wavelength)...
That's the only logical explanation I came to....
correct me if needed :P
Quote by smb
There's a constant ratio (we'll call it x) between the length from the bridge to each fret and the distance between the bridge and the next fret. The 12th fret is half way along the fretboard. Therefore, x^12=0.5
So the ratio is the twelth root of 0.5, which is 0.94387 approximately. That's the length from the first fret to the bridge, divided by the length from the nut to the bridge.

And I've got one and a half maths a-levels

SMB is certainly on the right track but i'm struggling to understand in real terms what his answer actually means - is it just me thats being incredibly dim witted

If i understand you right a particular value of x when put to the power 12 should equal 0.5 and you have to find x from the length of string remaining...?

Is there a way of working out more then 1 fret at a time...and how did you come up with that formula i'm starting a-level maths next year but am keen to get ahead so it's easier and i can spend my first year partying and having fun and being a general know it all
^ You can use simple calculus for it

The guitar's neck actually isn't perfect but it is based on the equal temper system. The easiest way to calculate standard fretspacing is with the 17.x scale system posted real early in the thread. Then it just becomes a simple repetitive function.
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Well say your scale length is 65cm. To find the distance from the bridge to the first fret you work out 65*x. (I'm using * for multiplied if that's not clear)

If you want to work out the length from the second fret to the bridge it's 65*x*x, that is 65 times x squared. For the third fret it is 65*x*x*x, or 65 times x cubed. And so on.

That a better explaination? Good plan for your a-levels by the way. That's pretty much what I did.
Quote by kingal
not really, the length of the string from the bridge to the nut in longer than to the 20th fret and 6% of something big is longer than 6% of something small, so the higher frets are going to be closer together.

Thanks, I get it now.

Quote by Robbie n strat
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^^ in that case i would get a pro to do it, cause i would more then likely f*ck it up.
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The same applies to fanned frets, but you calculate for one scale length on the top of the fretboard and another on the bottom and the put the frets at the angle that fits both.
ya know, i was thinking the exact same thing the other day when i wanted to build my own guitar

but then i realized, i can buy a premade custom neck : P lmao
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