Okay I will first start off by saying I'm a big math and physics guy, I love them almost as much as music. I was was just thinking one day about the math of find where the frets go on the neck. The first thing that came to mind was just a ratio with frequency and length of the neck. So I tried just that 55hz/110hz=32in/x, I did this first cause I know that it should be 1/2 the length. I looked at it then I realize that it would double the length so it wasn't right. Then I looked over at my bass on its stand and the first thing that came to my mind after looking the the way the frets are placed was a graph of an inverse function http://aycu24.webshots.com/image/43343/2005178457740970894_th.jpg (http://allyoucanupload.webshots.com/v/2005178457740970894) (told you im a geek when it comes to this stuff also that is not my bass I have Dean Razor). So I inversed the length of the equation 55hz/110hz=1/32in/x or just 55hz/110hz=x/32in(x=16in). So far I think its coming out right is the first fret on a 32in scaled bass about 30.20422172645in away from the saddle(55hz/58.27hz=x/32in)? and do people use a similar equation? or do people use the proportions Pythagoras found for the interval of frequencies?

...my brain is melting.

DistanceFromBridge = ScaleLength/(2^(FretNumber/NotesInOctave))

or for a standard 34 inch bass...

DistanceFromBridge = 34/(2^(FretNumber/12))
Fret number zero is the nut itself.

Awsome man, I am a physics lover too!

I think most luthiers use a template (someone will be along to correct me, I'm sure).

People do not use a Pythagorean tuning. We use a scale with 12 equal steps to the octave. As for the math, an octave = twice the pitch, two octaves = four times, and so on. In other words where f is your starting frequency,

one octave = f*2^1
two octaves = f*2^2.

That's the same as
one octave = f^(12/12)
two octaves = f^(24/12),
right?

You just have to change the numerator of that fraction to the number of semitones above the root you want to find the note's sounding pitch.

Of course, to double the frequency of a vibrating string, you halve its length rather than doubling it. Where p is the distance from the bridge to a given interval and l is the length of the neck,

one octave: p = l/2 = l/2^(12/12)
two octaves: p = l/4 = l/2^(24/12)

Usually you want to measure from the nut rather than the bridge, so you subtract the distance above from the total neck length, l. p = l - l/2^(x/12), where x is the fret number, counting from the nut. Of course, this generalizes to any number of equal divisions of the octave. You want 30 out of 72 steps on a 34-inch neck? p = 34 - 34/2^(30/72) = 8.528 inches. A fourth: exactly what we wanted.

If you wanted to calculate the fret locations for a non-equal tuning, you would replace 2^(x/y) with the frequency ratio for each pitch in the scale. E.g., the note a fourth (4:3) up from the open string on a 34-inch neck would be 34 - 34/(4/3) = 8.5 inches.

Just made this Fret Calculator Excel Spreadsheet...

You just change the scale length and the number of notes in an octave (for like Indian Music or Middle Eastern Music etc) to whatever you want and the spreadsheet adjust all the numbers for the frets. This will handle ONLY equal temperament scales (Our American and most modern Western music scales are equal temperament.)

I like this discussion. I want to throw in another complication. The tuning on a guitar is 'approximate', best compromise, hence the evolution of the Earvana compensated nut and the Buzz Feiten tuning system which is supposed to compensate for string length/tone. I don't understand what the problem is or how these compensated nuts help - it shouldn't be related to string diameter, it must be length/tension. All I know is that on a 6-string the B and the G played open seldom sound bang-on, or if they do, the moment a chord is fretted, they sound a fraction off again.

I will not get started about BFTS. I will not get started about BFTS. I will not get started about BFTS. I will not get started about BFTS. I will not get started about BFTS. I will not get started about BFTS. I will not get started about BFTS. I will not get started about BFTS. I will not get started about BFTS. I will not get started about BFTS.


Bassguy, if you're a math/physics guy too, you can get all the relevant equations from the formulas in this spreadsheet.
http://www.talkbass.com/forum/showpost.php?p=1728593&postcount=16

I don't understand what the problem is or how these compensated nuts help - it shouldn't be related to string diameter, it must be length/tension. All I know is that on a 6-string the B and the G played open seldom sound bang-on, or if they do, the moment a chord is fretted, they sound a fraction off again.
As I understand it, the BFTS compensates for the [what should be well known] deficiencies of a scale composed of 12 equal steps. Here's a video where he talks about it:
http://gedgreen.co.uk/fileadmin/vid3.html

Some of his wordings are odd. He seems to consider the 12-tone fourth, fifth, third, &c. to be what you might call the "mathematically proper" tuning, and he talks about the ear being more tolerant of deviations from the perfect intervals, such as the fifth and fourth, than it is of deviations from the third and sixth.

Actually, I don't think that's true at all. In the 12-tone scale, the fourths and fifths are extremely close to their purely tuned counterparts. The thirds and sixths, on the other hand, are around 15 cents (15/100 of a semitone) from where they ought to be! A world that tunes religiously to the 12-tone scale suggests that people are very tolerant of mistuned thirds and sixths, and prefer to have their fifths and fourths in tune.

However, some people are (rightfully) bothered by those mistuned intervals. The BFTS worsens the fifths and fourths slightly in order to improve the thirds and sixths slightly. I believe the effect is not quite constant across the neck, so that the exact degree of correction varies depending on where you play. I'm just going by the video with regard to the BFTS; I have no hands-on experience with it.

Looking at the patents for BFTS, my interpretation is that it has nothing to do with resolving 12TET temperament deficiencies. Anything that attempts to do so (I believe) necessarily creates an instrument that sounds better in one key and worse in others. I believe BFTS is a combination of compensated nut, stretch tuning, and slightly off intonation.

Ehhhhhhh... what does BFTS means? :hiding:

Ehhhhhhh... what does BFTS means? :hiding:

Buzz Feiten Tuning System

Buzz Feiten Tuning System.






And fretless ftw.

Looking at the patents for BFTS, my interpretation is that it has nothing to do with resolving 12TET temperament deficiencies. Anything that attempts to do so (I believe) necessarily creates an instrument that sounds better in one key and worse in others. I believe BFTS is a combination of compensated nut, stretch tuning, and slightly off intonation.
I made my assumption based on the way Feiten described it in the video. With all his talk about altering the perfect intervals in order to get flatter thirds, I don't see what else it could be. He's very explicit. I think its detrimental effects on certain keys are mitigated by the fact that it mostly (just?) acts on the G and B strings.

I could very well be wrong, but I heard that the BFTS does an adjustment because when you fret a note you lengthen the string. Because the lengthening of the string is slightly different for different frets (see my pic below,) the intonation is not perfect. If you wanted to have a bass with perfect intonation you would have to design it so that instead of pressing the string down you bring the fret up when you wanted to play a note. The BFTS is supposed to adjust for this.

However, the lower you have your action at the first fret (based on the depth of nut slots and if the nut is correctly cut) the less problem you have with this. The other thing that puzzles me is that depending on the depth of the nut slots (which affects the distance between the first fret and the string when it's not fretted... the gap) it seems to me that you would need a slightly different BFTS for each bass based on the size that first fret gap. That is unless once you install it you are supposed to set it up to a standardized gap (but even with this you would still need to do this for every fret which means that you have to have the correct bridge saddles height and this one is the kicker... the correct curvature in the neck. This is akin to trying to make one size and shape pair of pants fit everyone in the world since truss rods don't adjust every neck witht the same curvature.) So if this is what the BFTS is for, it would get no closer to perfect tuning for the standard bass than not having it. That is unless someone designed a seperate BFTS for every bass.

OK so here is my pic that shows why the string is lengthened different amounts based on which fret you press the string to. I exagerated the height of the nut in order to show the problem more easily.

There's a free downloadable app called wfret (google...) that has a couple different ways of calculating it...

It gives the distance from nut, distance between frets, and prints a template.

Thanks lemur821 for that equation and Thank you WarriorJoe7 for the excel. My way does work but its to find the distance from the bridge. Also I do understand the excel sheet pilotjones. I also wanted to ask how did you come about this equation, Its interesting also looks helpful for every string instrument.

Looking at the patents for BFTS, my interpretation is that it has nothing to do with resolving 12TET temperament deficiencies. Anything that attempts to do so (I believe) necessarily creates an instrument that sounds better in one key and worse in others. I believe BFTS is a combination of compensated nut, stretch tuning, and slightly off intonation.

Don't get us started on stretch tuning ;)

Im sure I would have to read this thread numerous times to understand exactly what's going on, but this has got to be the most interesting thread on TB ive read so far. subscribed. Is it possible, however, that for a bass, its not as important (please dont throw me any stones) since bassists dont do (or most of the time) many chords? I mean surely bassists jump from string to string, note to note, but chances of hearing a big difference are'nt as great as with a g****r (stones again, please) since two strings are rarely played at the same time? I need to take out my school books and stay up till 4am....

Also wanted to clarify that there are 2 equations you can use depending on whther you want the distance between the fret and the nut or between the fret and the bridge.

I already did incorporate this into the spreadsheet I made but here are the formulas I used to make the spreadsheet...

DistanceFromBridge = ScaleLength/(2^(FretNumber/NotesInOctave))
DistanceFromNut = ScaleLength-ScaleLength/(2^(FretNumber/NotesInOctave))

or for a standard 34 inch bass...

DistanceFromBridge = 34/(2^(FretNumber/12))
DistanceFromNut = 34-34/(2^(FretNumber/12))
Fret number zero is the nut itself.

another one you can use is the old rule of 18.

The distance between the frets of a stringed instrument is determined by a mathematical rule called “the rule of 18”, determined by Vicenzo Gallelei, in the 16th century. The rule states that:

L¬i = vsl/c
where Li is the distance from the nut to the fret number represented by the letter i in millimetres
c is the divisor, which is set to 18
vsl is vibrating string length also in millimetres

This rule is used to determine the distance from the nut to the first fret. Subsequent fret positions can be calculated using the following rule:

L¬i = L1 + Li-1 – (c-1)/c.

that also works.

Yes, the spreadsheet will work for any fretted string instrument in the world that is equal temperament (I am not aware if there any fretted ones that are not equal temperament.)

I came up with the equation by knowing that 440Hz is A, and also 220Hz, and 110Hz. Here was my line of reasoning.

So if 110Hz is A then 110x2^0=110=A, 110x2^1=220=A, 110x2^2=440=A, 110x2^3=880=A and so forth. I figured since there are 12 notes in an octave they are probably evenly spaced by the exponent. So B would then be 2/12 higher (2 steps) than each exponent listed above for A. 110x2^(14/12)=247Hz=B. Then I did this for a few other frequencies. I looked it up and saw that they were correct so I knew I had the right formula.

Now I knew from the fretboard that (on a 34 inch scale bass) the octave was at 17 inches from the BRIDGE and the 2nd octave was 8.5 inches from the BRIDGE. So it seemed that the frets worked the same way (only in reverse... see next paragraph) as the frequencies (same spacing) it was just a matter of finding the formula that would account for the scale length.

What I meant by in reverse was that when you double the frequency the note goes up one octave. When you HALVE the string length (between the BRIDGE and the fret) you also go up an octave. Thus they are reversed (inverse.)

So I knew that I had to incorporate both the scale length and the 2^14/12 to find where the B fret was on the fretboard. So I multiplied and that was not a reasonable answer so I divided and when you divide 34/(2^14/12) you come up with a reasonable answer (something between 0 and 34.) So I measured it and it seemed right. I tried a few more frets and they were right so I knew I had it. The problem was that this was the distance of the frets from the bridge, and the bridge saddles move. So since I wanted the distance from the nut I just subtracted the answer from the scale length. I.E 34-34x(2^14/12) is the distance of the 2nd fret B (on the A string) from the nut.

I then realized that I could plug in any scale length. Later I realized that if there was a scale that had more than 12 chromatic notes that the 12 could be changed to that number. For instance there is some Indian music with 24 notes in an octave so the frets would use this formula but with increments based on 24.

I figured this out probably 5 years ago now and it took me a few hours to find an answer but I did it. I was taking music classes at the time and looking at my basses waveform on a specrum analyzer.

Thanks lemur821 for that equation and Thank you WarriorJoe7 for the excel. My way does work but its to find the distance from the bridge. Also I do understand the excel sheet pilotjones. I also wanted to ask how did you come about this equation, Its interesting also looks helpful for every string instrument.

Yup, rule of 18 is the other one that wfret supports.

I just tested this and it might be for just intonation, but it is not for equal temperament. Thus it wouldn't apply to our modern basses.

It is up to 1/8 off on some frets (17th and 18th fret)

i added it to the spreadsheet to compare

another one you can use is the old rule of 18.

The distance between the frets of a stringed instrument is determined by a mathematical rule called “the rule of 18”, determined by Vicenzo Gallelei, in the 16th century. The rule states that:

L¬i = vsl/c
where Li is the distance from the nut to the fret number represented by the letter i in millimetres
c is the divisor, which is set to 18
vsl is vibrating string length also in millimetres

This rule is used to determine the distance from the nut to the first fret. Subsequent fret positions can be calculated using the following rule:

L¬i = L1 + Li-1 – (c-1)/c.

that also works.

Also I do understand the excel sheet pilotjones. I also wanted to ask how did you come about this equation, Its interesting also looks helpful for every string instrument.Well, it depends on which equation you're talking about.

The tension-scale length-pitch-linear mass equation is a standard one, in any physics book, or from the D'Addario site if you wish.

The fret placement sheet is using the 12th root of two equation, as is also standard, and which Warrior Joe seems to have independently derived (bravo!). I took JP's fret sheet, and modified it so it has the distance from nut and distance from bridge and incremental distance and 36-fret capacity and will do non-12-tone (but still equal temperament only) scales. BTW, the Rule of 18 is an approximation of the 12th root of 2, from before people knew how to do that kind of math. You will see the number 17-point-something around, which is a corrected value.

The calculations for tension rise due to string stretch during plucking, with and without extra string beyond the nut and bridge, were derived by me, and the derivations checked by others (at least two other engineers and a physicist).

I've noticed that as string diameter increases I have to move bridge elements farther from the nut in order to intonate my basses, and that nut-to-bridge distance is appreciably greater for my B string than it is for my G string.

I don't understand how fret spacing calculations work when they don't take string gauge into account. To someone like me who doesn't understand the math and physics used to perform these calculations, fanned frets seem to be strongly suggested as a companion to the nut-to-bridge distance.

Can someone explain the way string gauge and fret spacing effect one another in language a [middle-aged] child can understand?

Actually this is an excellent question that I wish I knew the answer to also. I know that it has to do with the thickness of the string. Also I know that the G string gets the most accurate intonation up and down the neck and that you rarely if ever get an E or B string that will intonate the whole way perfectly... I have had one bass (of 100s I set up) that intonated the E string almost perfectly. And amazingly the intonation was pretty damn good on all of the strings. (My Read short Scale.) Most of the time I just make sure it is pretty close up to the 15th fret for the E string. I know that this also has to do with string diameter.

So because of the string diameter, the scale length has to be made slightly longer to account for this, and then all the frets aren't quite right either. I guess the only way around this that I can think of is to replace each fret with 4 mini frets that are in the exact right place. That would do it (but then you would always have to use the exact guages of strings that this was set up for forever in order to take advantage of it.)

I am not sure if fanned frets would fix this but thinking about it it might. I think it might work by slightly exagerating the fan on the E side. For instance if you took a regular fretted bass and fanned the frets ever SO slightly it might fix it but I am not sure. When I mean slightly, I mean however far you moved the bridge saddle on the E string, you would fan the E side of the fret about half of that distance on the 12th fret and less as you get closer to the nut, but I am guessing.

I've noticed that as string diameter increases I have to move bridge elements farther from the nut in order to intonate my basses, and that nut-to-bridge distance is appreciably greater for my B string than it is for my G string.

I don't understand how fret spacing calculations work when they don't take string gauge into account. To someone like me who doesn't understand the math and physics used to perform these calculations, fanned frets seem to be strongly suggested as a companion to the nut-to-bridge distance.

Can someone explain the way string gauge and fret spacing effect one another in language a [middle-aged] child can understand?

I've noticed that as string diameter increases I have to move bridge elements farther from the nut in order to intonate my basses, and that nut-to-bridge distance is appreciably greater for my B string than it is for my G string.

I don't understand how fret spacing calculations work when they don't take string gauge into account. To someone like me who doesn't understand the math and physics used to perform these calculations, fanned frets seem to be strongly suggested as a companion to the nut-to-bridge distance.

Can someone explain the way string gauge and fret spacing effect one another in language a [middle-aged] child can understand?
Fatter strings do indeed need more compensation than thinner ones. I've got about a 1/4 inch difference between my highest and lowest string.

The reason is that when you push a string down to the fret, it has to stretch a little, and that changes its pitch from the ideal. The exact amount of change depends on the actual individual string and probably also on how broken in it is. Stiffness also plays a role, although probably less on bass strings than guitar strings.

It's possible to compensate for this stretching by moving all your frets back by some small percentage of their position. Luckily, moving the saddle back will achieve the same effect. If you move the saddle 1/8 of an inch, that's a pretty small change relative to its distance from the nut (about 0.4% on a 34 inch neck), but as you move up the neck it becomes more significant (at the 12th fret of a 34 inch neck it would be more like a 0.7% change). And it turns out that this increasing percentage of correction on higher frets is exactly what is needed. You are correct that fanned frets could be used as a method of compensation, but you would actually have to fan them backwards (slightly longer scale on the high strings) if you wanted your saddles to lie along a line perpendicular to the neck. On a guitar, due to the change from unwound to wound strings, you would need zigzag frets.

Apparently mirroring your saddle compensation with compensation at the nut is beneficial too, but this is often skipped.

I don't see how anyone can account for string stretch during plucking for several reasons. People pluck strings harder or softer than others and with diffent techniques. Plus when you have different strings they have different elasticities from one string to the next (E vs a vs D vs G) and then compound that with all the different brands and make. Then on top of that (I am guessing but) an open string will stretch on a pluck more than say the 20th fret. Plus if you pluck near the bridge vs near the neck you definitely get less stretch. Plus it also seems that what might be more important is how much the string stretches by just fretting a note, which happens to be different on every fret on every guitar because no 2 necks are exactly alike at this minute level. So unless you are finding some kind of average it seems impossible to do... and even an average is probably pretty negligible improvement on a stringed instrument unless it happens to be exactly right for every fret.

On the other hand where is this program... you did the same exact things in the program that I did in the fretcalc excel sheet. I had distance from bridge and nut, and adjustment for any scale length and any number of notes in a scale... whoops I forgot incremental distance so I will go add that now.

I also had 36 frets but it is set up so that you can add more by copying and pasting from one row to the next. You could take it to 100 but I doubt it is useful much beyond 50. Somewhere past there the frets get too wide for the amount of spacing needed.

Looking for that 17 point something value.




Well, it depends on which equation you're talking about.

The tension-scale length-pitch-linear mass equation is a standard one, in any physics book, or from the D'Addario site if you wish.

The fret placement sheet is using the 12th root of two equation, as is also standard, and which Warrior Joe seems to have independently derived (bravo!). I took JP's fret sheet, and modified it so it has the distance from nut and distance from bridge and incremental distance and 36-fret capacity and will do non-12-tone (but still equal temperament only) scales. BTW, the Rule of 18 is an approximation of the 12th root of 2, from before people knew how to do that kind of math. You will see the number 17-point-something around, which is a corrected value.

The calculations for tension rise due to string stretch during plucking, with and without extra string beyond the nut and bridge, were derived by me, and the derivations checked by others (at least two other engineers and a physicist).

Moving the nut away from the E side of the neck seems like an EXCELLENT idea. I just got an idea. I bet that this will allow the intonation to be better up the neck. if you tilt your nut away on the E string side then you don't have to move your saddles as far away on the other side. I don't have the means to do the experiemtn I want but I have an idea to test this.

I don't see how this "The reason is that when you push a string down to the fret, it has to stretch a little, and that changes its pitch from the ideal" explains why the saddles need to be moved for the E string because all of the strings are strecthing the same amount across any given fret (non-fanned.) I.E. Both the E string and the G string strecth very closeto the same amount at for instance the 10th fret, or the 15th.

Yes and I think you are correct about moving the frets the other direction... I think I had it backwards above but I am not sure. I'd have to try it.

Fatter strings do indeed need more compensation than thinner ones. I've got about a 1/4 inch difference between my highest and lowest string.

The reason is that when you push a string down to the fret, it has to stretch a little, and that changes its pitch from the ideal. The exact amount of change depends on the actual individual string and probably also on how broken in it is. Stiffness also plays a role, although probably less on bass strings than guitar strings.

It's possible to compensate for this stretching by moving all your frets back by some small percentage of their position. Luckily, moving the saddle back will achieve the same effect. If you move the saddle 1/8 of an inch, that's a pretty small change relative to its distance from the nut (about 0.4% on a 34 inch neck), but as you move up the neck it becomes more significant (at the 12th fret of a 34 inch neck it would be more like a 0.7% change). And it turns out that this increasing percentage of correction on higher frets is exactly what is needed. You are correct that fanned frets could be used as a method of compensation, but you would actually have to fan them backwards (slightly longer scale on the high strings) if you wanted your saddles to lie along a line perpendicular to the neck. On a guitar, due to the change from unwound to wound strings, you would need zigzag frets.

Apparently mirroring your saddle compensation with compensation at the nut is beneficial too, but this is often skipped.

Ok the rule of 18 is actually the rule of 17.8171537451057

The excel spreadshhet would not let me figure it out past this number.... I wonder if this number is some kind of root or could be written in another way.

So if you use that number it is the same as the 12th root of 2... whereas if you use 18 it will be off up to 1/8 of an inch. I am in the middle of revising my spreadsheet.

Here is the new spreadsheet. It does the exact same thing as JPs program from what I am guessing. You can change the scale length, numbe of notes in an octave and number of frets and every thing will auto adjust.

I also redid the rule of 18s but used 17.8171537451057 from above

I don't see how this "The reason is that when you push a string down to the fret, it has to stretch a little, and that changes its pitch from the ideal" explains why the saddles need to be moved for the E string because all of the strings are strecthing the same amount across any given fret (non-fanned.) I.E. Both the E string and the G string strecth very closeto the same amount at for instance the 10th fret, or the 15th.
Because strings of different mass and different construction respond differently to stretching and bending, and there may be additional differences in action. You can see this if you look at a properly intonated guitar's saddles. Going from the high end, the saddles get progressively further from the nut, but then the pattern changes when the wound strings with their different construction and thinner cores start.

P.S. Is there any way you could avoid top-posting?

I'll reply to what I can accurately address. And what I can understand, since, with no disrespect intended, I can't quite make out the full idea in one or two posts.

Jazzdog asks, basically, "why do you have to do more intonation compensation on the fatter strings?" I'll start by saying that I don't think it's due to the fact that they are fat, per se.

First, why do we do intonation compensation (the shifting back of saddles)? Because, as you fret the string, you stretch it towards the fretboard. This makes it go sharp. So, as one of the previous posts said in different words, by increasing the scale length (moving the saddle back), you have effectively made the frets all be too close to the nut (for the new string length), so they will sound the note flat. This compensation does not exactly fix the problem. It provides pretty good results, with variation depending how far up or down the neck you are. It is good enough for most people and circumstances, but it leaves the door open for other ideas that are, or try to be, better.

The amount of compensation required is governed (I know) by the action height, the core elasticity (which is based on core material and cross section size), and the initial string tension, and (I think perhaps) by the core mass/wrap mass ratio. There may be other factors, too. The first three of these all directly affect what will be the resultant tension rise (and therefore pitch rise) when you push the string towards the fretboard, so they they directly affect how much compensation will be necessary.

A tight string goes up less in pitch when you stretch it than a loose one does. So part of the reason the E string always requires more saddle compensation is that it is normally a looser string, so it goes sharper when you fret it.

A string with higher action has to be stretched farther to reach the fretboard when you fret it. The E string is normally set up with a higher action than the G string, so it goes sharper when fretted, so there is a second reason why it requires more compensation.

Funny how by doing a longer scale length for your B or E string by fanning, you not only get more even tension across the set, which feels better to many people. You get a higher tension on the low strings; so, on one hand, it should rise less during fretting, and need less compensation; plus, at a higher tension and possibly lighter gage, it may be (not sure on this) able to be set up with a lower action, which would also require less compensation. And remember, since compensation does not perfectly counter to out-of-tune-ness of the notes up and down the neck, the less out-of-tune produced, the better.


Joe, as far as compensating for the tension rise during plucking, I wasn't suggesting that anyone try to do that. It's in the spreadsheet in response to other questions that were raised a year or two ago. Compensation is for the tension and pitch raise from fretting, not from plucking.

A tight string goes up less in pitch when you stretch it than a loose one does. So part of the reason the E string always requires more saddle compensation is that it is normally a looser string, so it goes sharper when you fret it.
I'm using extremely light gauges for my high strings and their saddles still need to be a good bit closer to the nut than the low strings. Of course, my action is fairly low, but I question the real-world magnitude of this particular effect.

I think it was asked, why 17.817...? The exact number is: 1 / [1 - 1/(2^(1/12))]

I'll try that in Ascii
1
-----------
1
1 - -------
1/12
2

well ha for all these compensating problems is more reason why I prefer fret-less :smug:
PioletJones I have been thinking about your equations and I need to brush up on my stuff get my old physics book and relearn it.

I think it was asked, why 17.817...? The exact number is: 1 / [1 - 1/(2^(1/12))]

I'll try that in Ascii
1
-----------
1
1 - -------
1/12
2

got it... so I did it on a calc and got 17.817153745105767553490105576143

LOL

got it... so I did it on a calc and got 17.817153745105767553490105576143

LOLSo now you can position your frets with nanoscale accuracy... if you can get Cornell or Harvard to do it for you. :)