Flat Fretboard

[NamelessOne]

well, the geometry of it states that if you have a cylindrical surface, such as a standard non-compound radius fretboard, strings that are not parallel to each other can never all be parallel to the board. you will end up with the board being closer to the string in the middle of the string's length, especially with the strings closer to the edges of the fretboard. the middle (D and G, or A for people like me) strings will generally be just fine, and able to get a nice low action. If there is no taper between the strings (unlike most guitars), you can get nice low action until you start bending.

Enter a compound radius. An ideal compound radius follows the shape of a cone, and the strings should ideally be parallel to the cone. This means that with more string taper (different bringe and nut spacings), the cone should be more promenent. this also means that there's only one coefficeint of cone-ness (I don't know the word) that works with each string spacing, for maximum effect. However, a imperfect compound radius is usually better that a cylindrical radius. When you bend a string, the point of attachment to the cone moves off a point where it's parallel, and the string may fret out, or buzz.

When it's a flat board, it's simply a plane. Nothing fancy here, just every line that is the same distance from the surface of the plane at two points is by definition parallel to it. Even if you can bend so far that the two Es touch, you'll never fret out, and it's easier to get low action without buzzes.

BTW, Perry made me mad too when I was new here. But, I simply shut up, and now value his advice and experience. Hydrogeoman, stick around for a while, and you won't care anymore. Often, the people with the best advice and questions on a forum manage to get everyone else mad at the same time. I've seen people with REALLY interesting opinions and questions get banned on another forum because nobody seemed to want to hear some of his opinions. But, nobody's going to do anything of the sort to perry, because the rest of us want to hear what he has to say.

[rhoads56]

But, nobody's going to do anything of the sort to perry

haha, yeah right!! Wait until the mods from hell read this.

[sexybeast]

pmarlin,

believe it or not, I like my action liwer so when I bend strings, all the others come with my fingers, otherwise when I release the string, my fingers allow the string to slide off the top and make noise. Just personal preference, I guess. Higher actio makes for better tone, I believe, so I'm REEEEALY picky about string height.

[Ben]

well, the geometry of it states that if you have a cylindrical surface, such as a standard non-compound radius fretboard, strings that are not parallel to each other can never all be parallel to the board. you will end up with the board being closer to the string in the middle of the string's length, especially with the strings closer to the edges of the fretboard. the middle (D and G, or A for people like me) strings will generally be just fine, and able to get a nice low action. If there is no taper between the strings (unlike most guitars), you can get nice low action until you start bending.

That is a very interesting point that seems to have been completely over looked...

Perry, the way I understand this, this would suggest that you may actually be wrong.

Even if you never bent a string whilst playing, a flat board would allow for a lower action than a board with any (non-compound) radius

[/pedantry] (ok so im a hypocrite, sorry )

[rhoads56]

Perry, the way I understand this, this would suggest that you may actually be wrong.

Even if you never bent a string whilst playing, a flat board would allow for a lower action than a board with any (non-compound) radius

[/pedantry] (ok so im a hypocrite, sorry )

This was one of the things we measured when i spent all that money on scientific research. The measurement, for a 12" radius, with each E string (the worst), and a 24.75" scale, was 0.0024 millimeter = 0.000094488 inch. Now, go off, find that set of digital calipers, and see if you can measure that size.

<< smiley to show im having fun with you...

So, theoretically, you are 110% correct....

<< smilie cause i know you're thinking... thats great!

But you havent taken into account we will be fret levelling the neck, cause we do that around here, as we are all wanting the best guitar, right?

<<< damn, you just about had me didnt ya...

Now, how do you fret level a standard radius neck? Oh, yes, in a straight line, the same plane as the strings. DAMN, that just stuffed up your "Perry's wrong" theory...

You gotta get up pretty early dude...

... sorry, damn forum software wont allow the emoticons any more...

Ben ten points for thinking. But, like i said to sexy earlier, you have to consider the entire job at hand, not just one part of it (re: his drawings, my comments about fingering)

[marksound]

I really should stay out of this, but what the hell.

Do you guys with all these ideas and theories think that someone like Perry hasn't tried some or all of these ideas and theories before he came to his conclusions about what works and what doesn't? I think Perry and his peers probably have, and that's why there are so many givens in electric guitar construction. Personalities aside, Perry knows his stuff and the proof is in the product.

On the other hand, I'm sure Leo Fender was told by a lot of professionals that his ideas wouldn't work. Failure is a huge part of innovation, and some of Leo's ideas failed miserably. Others are now industry standards.

What I'm trying to say is, use the tried and true as a baseline and go from there. Don't reinvent the wheel.

I'm done now.

[sexybeast]

Alright, enough. Let's all admit that I know everything there is to know and be done with it.

Sorry.

The fret levelling thing never came to mind, because I've never done it. Hmmm.

[drezdin]

I don't see what the fuss is all about. Perry knows his stuff. If he says it doesn't effect action then I believe him.

I'm building a guitar with a flat radius now, but the only reason is I like the way it feels. I spent a long time playing classical and the flat board just feels right.

So if the radius doesn't have an affect on action then maybe we should focus on the fret leveling (as Perry mentioned) and proper setup to get the low action?

Edited October 13, 2006 by drezdin

[Ben]

This was one of the things we measured when i spent all that money on scientific research. The measurement, for a 12" radius, with each E string (the worst), and a 24.75" scale, was 0.0024 millimeter = 0.000094488 inch. Now, go off, find that set of digital calipers, and see if you can measure that size.

That was why I wrote the pedantry thing, I was sure the effect would be miniscule

what do you use to level your frets then?

Do you use a radiused block, or a flat surface?

The way im interpreting it from what you wrote above, is that you use something flat to level in the same direction as the string, creating a compound radius in effect...?

I thought that when you leveled the frets, you'd just level them to the same radius as the board.

I know this thread has gone past the point of usefulness (is that even a word?) to you sexybeast, but its still kinda interesting. (or at least I think so)

Edited October 13, 2006 by Ben

[sexybeast]

Nope, I'm still in all the way.

[fryovanni]

Now I am more curious about the taper.

I modeled a straight forward 12"r and the difference due to tapering from center to outside(based on low or high E travel). I found the fret would be .019" lower if you measured from a common horizontal plane equal to the top of the 1st fret. If you look at it from a absolutely horizontal plane(strings are all set parallel to that plane) the frets would limit the lowest action. Nothing would stop you from elevating the position the string at the string nut and lowering the bridge saddle to return the tapered string to parallel to the top of the frets. This does not improve the bending issue(actually probably makes it worse). However as you pull a string tward the center of the radius no issues would be encountered actually clearance would increase. Going over the top of course creates a loss of clearance as the string crosses frets on the opposite side of center. Other factor will come into play of course because a neck usually has a bit of relief. Which can help this situation until you reach your higher frets. A compound radius or working a slight taper into your higher frets can assist in the higher fret range. None of this is taking into account string vibration and min. clearance required (which may make over shadow just about all of this unless you deal with the most extream bending- and if I understand Perry this is the case).

I hope that I am not overlooking factors. I find I sometimes do not see all factors that are in play when I look at things like this.

Peace,Rich

Oh I almost forgot we need smilies-

Edited October 13, 2006 by fryovanni

[Guest AlexVDL]

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[NamelessOne]

fryovanni-

you mention the "top" of the fretboard. Might I remind you that while the fretboard has a centerline, it is a cylindrical surface, and they are basically uninterrupted- no top. the "ideal" position I mentioned in my previous post is ther closest thing to a "top" you can find. In a cylindrical raduis board, the if you bend a string towards the near edge of the board, you actually will end up raising the center of the string away from the frets.

Btw, I'm only a person with a decent degree of reasoning, and a gift for visualising geometry, but not a person with the knowledge of how to calculate that kind of thing numerically. I have my grade 11 math already (not bad for a 14-year old), but no real access to any way of simulating this other than pencil and paper. I'll try sometime, though.

I'm sure that this makes very little difference, but this discussion should mention both the thoery, and how much it affects practice. I mean, if there's some guy who can bend across the board on a tight-radius, highly fanned-string bass, it would make a difference to him. Some people push the limits, and they need instruments that do also.

Remember kids, in theory, there's no difference between theory and practice. In practice, there is.

already formulating thoughts on how to solve this...

edit:

alright, my calculations agree mainly with perry... I have yet to calculate how bending affects this, but for now...

I got the same answer as perry, but, I didn't spend years and hundreds of thousands of dollars doing it... I got the calculator from a sibling, the ruler was given to my parents, and I'm borrowing the bass... that accounts to... FREE!

therefore, I am infinity percent more cost-effective than perry.

Edited October 13, 2006 by NamelessOne

[fryovanni]

fryovanni-

you mention the "top" of the fretboard. Might I remind you that while the fretboard has a centerline, it is a cylindrical surface, and they are basically uninterrupted- no top. the "ideal" position I mentioned in my previous post is ther closest thing to a "top" you can find. In a cylindrical raduis board, the if you bend a string towards the near edge of the board, you actually will end up raising the center of the string away from the frets.

I think you are missing what I said.

"I modeled a straight forward 12"r and the difference due to tapering from center to outside(based on low or high E travel). I found the fret would be .019" lower if you measured from a common horizontal plane equal to the top of the 1st fret."

I followed the line a string travels from nut slot to bridge saddle. I compaired the height differencial between the 1st fret and 24th fret based on a plane set to the 1st fret(as though travel down the radius was in the same relative path, not along a tapered path). Mind you I am using auto cad to model this and simply referencing it to find distances(no magic calculations, or gifted visualization).

As far as the strings distance from frets as you bend tward the edge of the fretboard. The string will get closer to the frets(the bridge saddle height is fixed). If you bend the string twards the peak of the radius you will raised the string(until you go over the peak). This is all based on viewing the fretboard as points on a cylindrical object (no accounting for relief or any other factors).

Check back with me if you still think I am off here(I am certainly more than capable of making a mistake). Like I said I only made a model in cad and was using the model to reference (this was not done in my head).

Peace,Rich

[NamelessOne]

okay... first paragraph... you found the distance between the edge of the fretboard and a tangent to the conter of it. repeated this for frets 1 and 24? then gave the difference between the distances, or no?

the quoted text, however, refers to the second paragraph there. the raduis has no peak, but the board has a center, which I beleive is what you are referring to. what I think is the weak link in your reasoning is that you assume the bridge has no radius. If that were true, I'd agree completely with you. Or, if the bridge had a longer radius that the fingerboard. I view the fretboard and bridge as points on a cylindrical object also. it's just, you must remember that the curved face of a cylinder is the same everywhere. If you only have a segment of it, such as a fretboard, that still applies within the segment. For example, look down the fretboard of your guitar (would be easier with just a neck, or even just in your head) and tilt it a couple degrees to one side. There's now a new highest point on it, beside where the old one was. now, the same thing happens if you bend past this peak as any other possible peak on the board. the "peak", or point where a string can only be bent in a way that brings it closer to the fretboard, is defined by the path of the string, and differs between strings.

I'm sorry if that's hard to follow, my brain works a LOT differently than most people's, and I explain things differently than most people would.

Also, after I posted my other calculations (made by reducing the radius and string to a circle and chord, thanks to the similar triangles or something like that theorem, then using the radius-bisects-chord-at-90-degrees theorem and the pythagoream theorem), I did a worst-case scenario. I looked at my bass (string spacing about doubles between ends) and checked how far I could bend. I got my low E up to the dot inlays on the center of the board, and I'm not a good bender. So, it moved just over an inch between the point I fretted it at and the bridge. at a 7.5 inch radius, the tightest I've ever heard of used (on a bass), the string is about .02 inch closer at the middle than it would be on a perfect board. This also supports the second paragraph I wrote here, as that's a lot more than what I got for just where it was, and I hadn't bent it over the center of the board. I imagine, due to the properties of circles and sine and such, as you bend further, this effect increases much faster.

also, for anyone who wants to calculate this, scale length doesn't matter, due to the whole "similar triangles" law (the same thing that makes frets work with fanned strings, and stops f*nned frets from working with fanned strings). So, simply calculate is as the fretboard being an arc of a circle, and the strings being chords on the arc.

[fryovanni]

Ok, I am going to go through your post slowly and as carefully as I can.

"okay... first paragraph... you found the distance between the edge of the fretboard and a tangent to the conter of it. repeated this for frets 1 and 24? then gave the difference between the distances, or no?"

No, That is not what I did. I made a 3-D solid in auto cad (a cylinder length was the distance between the nut and 24th fret. the cylinder had a radius of 12"). I created an arch at the bridge location and marked points appropriate to give me my string spacing. I marked points at the nut with appropriate string spacing. I then connected those points with lines. I located the 1st and 24th fret locations and marked these points at their intersections. I used the distance tool to identify the distances along the Z axis. Like I said I did not calculate anything this was a 3-D model built in auto cad. My measurements are all made by sampling a 1:1 model. (No magic)

"the quoted text, however, refers to the second paragraph there. the raduis has no peak, but the board has a center, which I beleive is what you are referring to."

Center line, quadrant, thickest point of the fretboard... You get my point.

"what I think is the weak link in your reasoning is that you assume the bridge has no radius. If that were true, I'd agree completely with you"

No, I do not assume the bridge has no radius.

"Or, if the bridge had a longer radius that the fingerboard."

A bridge with independantly adjustable saddles can be adjusted to a longer,shorter, equal or actually any thing you want(within the limits of it's adjuments).

" I view the fretboard and bridge as points on a cylindrical object also."

You can look at it that way,but it certainly is not going to be set up that way. You would have strings setting on frets.

"it's just, you must remember that the curved face of a cylinder is the same everywhere. If you only have a segment of it, such as a fretboard, that still applies within the segment."

I have no misconceptions about the curved surface of a cylinder.

"For example, look down the fretboard of your guitar (would be easier with just a neck, or even just in your head) and tilt it a couple degrees to one side. There's now a new highest point on it, beside where the old one was."

Sure, So you are pointing out that no matter what point along that 12" radius I am 12" away from a given point.

"now, the same thing happens if you bend past this peak as any other possible peak on the board. the "peak", or point where a string can only be bent in a way that brings it closer to the fretboard, is defined by the path of the string, and differs between strings."

You miss the fact that a bridge saddle will lock the string to a point. In pulling a string you do not rotate it about the radius. You are pulling it across at an angle, but the string remains locked at bridge saddle. So you do have a "high point" along the travel of the string as it relates to the saddle or fret(if the string bottoms out).

I am tired now.... I may come back and update this. Your brain may work differently than most, but thats cool. I am betting I am not comunicating very well here.

Edit: Quick experiment foryou. Take a cylinder (jar,can whatever you have handy). Take a straight edge and place it flat down a center line. This simulates a zero clearance (no action from string nut to bridge) nice and flat.Now turn the straight edge to one side or the other to simulate the angle created by the difference in string spread from nut to bridge. Notice the straight edge will not set flat anymore. This may show you what I am saying.

Later,Rich

You know after re-reading this

-"alright, my calculations agree mainly with perry... I have yet to calculate how bending affects this, but for now...

I got the same answer as perry, but, I didn't spend years and hundreds of thousands of dollars doing it... I got the calculator from a sibling, the ruler was given to my parents, and I'm borrowing the bass... that accounts to... FREE!

therefore, I am infinity percent more cost-effective than perry.".

I think you have this figured out. Use the information well. I will get back to my binding.

Edited October 14, 2006 by fryovanni

[NamelessOne]

As far as I can tell...

We agree. We've agreed the whole time. Don't you hate it when that happens?

Except, I claim that the "peak" of the fretboard, the point where the ruler lies completely touching the jar, as you put it, varies from string to string, and has nothing to do with the thickest point on the fretboard. It's the individual string's ideal path, not the center of the fretboard, that defines where it ceases becoming farther from the fretboard when you bend. This ideal path may be closer to the center, farther from the the center, or even not on the board at all.

the flat bridge stuff and other cases you (rather diligently) argued against were ideas of situations where the thickest point on the fretboard actually defines where a string starts getting closer to the fretboard on a bend. My examples given were trying to illustrate my point, made above.

The first bit you quoted, about the model, sort of agreeing with me. You're just using CAD-speak when I'm using geometry-speak. I lost a bit in translation, but it appears you calculated (using a rather powerful calculator) the distance between a plane (?) (finding Z-coordinate, since Z=0 is a plane) and the fretboard at various points, or is it a string and the fretboard?

Have fun with your binding, the friend I'm working on my guitar with hates it...

THE CONCLUSION OF THE THREAD:

flat fretboards: never cause complex mathematical discussions. they should be made standard on all guitars.

[rhoads56]

bla bla bla

Hold your horses buddy... you posted another very short answer without any explenation saying there is absolutely no connection. But there actually IS a connection, as you can read from my post!!!!!

Only when you bend strings the radius matters if you want a low action. I have never met anyone who don't bend strings on a electric guitar (maybe some jazz players), so yes my post is relevant to his question.

Alex, no-one ever denied what you had to say. We were just wondering why you felt you had to pretend to be the only one saying it. Re-read the entire thread.

[Guest AlexVDL]

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