Any geometry experts out there? Conical geometry

[wizard333]

Unless the string spacing at the bridge on a guitar is the same as the nut, guitar strings necessarily follow a cone between the nut and the bridge, not a cylinder. Contrary to manufacturer's claims, almost all guitar necks are "compound radius" if you actually measure them (I've seen a few hundred on a Plek analysis, I recall maybe 2 that were actually close to a straight radius). So most fretboards are actually some kind of approximation of a cone at least

That said, I'm curious whether anyone actually tapers that cone properly from the nut to the last fret. I've searched on the web and found basic conical geometry tutorials, most of which I remember from 7th grade, but nothing that answers the question.

A guitar fretboard should be a section of a truncated cone. The known variables we have are L, the length of the side of the cone, and we can pick R at any two points, say 1st and last fret, 1st and 12th, whatever. We could also measure the width of the neck at any point, which gives us a segment at any given point. Given those known quantities, can we then compute what R should be at any other point on that cone? Does it vary in a linear fashion with L? I.e., if we know L between nut and 12th fret is 12.75", and say R at the 12th is 14" and R at the nut is 12", can we compute the proper radius at fret 7 by taking the ratio of L' (nut to 7th fret) over L? I have no idea, I can't find the correct equations to lead me to that or any other conclusion. Anyone know for sure? Show your math please!

[Paleo Pete]

Don't know, but if you don't mind trudging through a few pages of sites to try and find it,

Google Scholar

Have a dictionary handy, you just might need it if you don't have an impressive vocabulary...

[eschertron]

You are right, a cone is nothing more than straight lines and circular cross-sections, all entirely regular. The relationship of length and radius is expressed by a right triangle, where the length along the fretboard is the hypotenuse, and the radius is the base of the triangle (or of any similar triangle with the same angle at the apex). Nothing that a ratio and and good approximation of pi won't handle.

Having said that, I'm pretty sure that guitar fingerboards are either a simple approximation of this conical surface, or a shape that deliberately diverges from it. Aren't the 'compound radius' fingerboards designed so that the cone has a much larger radius near the upped frets in order to avoid fretting out on string bends? I don't know, but I suspect as much. And with adjustable bridges, all bets are off when comparing the radius of the strings to the radius of the fingerboard.

It's interesting to note that guitars and their progenitors differed from the violins in that the plucked instruments had flat fingerboards while the bowed instruments had curved fingerboards. Why the trend toward radiused fingerboards? Is it an ergonomic thing?

Edit: I submit that we don't know L, the length of the cone. We know the length of the segment, but must come up with the apex of the cone at some arbitrary length away from the nut. Your ratio of radii then can be related to the unknown L over L plus the length of the fretboard. The results you come up with are determined by the accuracy of the measurements you take of the curvature of the neck.

[wizard333]

Aren't the 'compound radius' fingerboards designed so that the cone has a much larger radius near the upped frets in order to avoid fretting out on string bends? I don't know, but I suspect as much.

Well yes, that is the definition of a cone, the radius gets larger as you move from the apex. Also, with adjustable bridges, you can certainly F it up and ruin the cone, but I already know that most techs have no idea how to do that right. I do it correctly but I take great care in that adjustment and measure everything. No metric and your process is necessarily out of control.

Which is the point of this excercise and question really. We should be able to measure the fretboard at any two points and know what the radius should be at any other point. A Plek for instance could be programmed to do that exactly, it gives options for 1st, 12th, and last fret radius, but unfortunately even the guy who installs all those and trains the users regularly makes the 12th fret rounder than the 1st, or the last fret the same or rounder than the 12th, or things like that trying to preserve a few ten thousandths of fret height. My concern is not with fret height, the best way to maintain fret height is to not level them. My concern is with playability.

The question isn't all the ways it can be messed up, it's what is the calculation to do it corectly.

Also, we do indeed know L. That is the lengh of the side truncate cone, which is why I specified truncated. Truncated cones have known geometrical properties, of which L is one, just like a full cone measured to the apex.

The reason that is important is that your strings form a truncated cone. If your fretboard doesn't form a regular truncated cone, but some odd approximation or (as often happens) some weird geometric shape where the radius increases and decreases multiple times along the length of the neck (often see this problem around the 12th fret), the strings will still want to be a cone and will happily buzz and fret out on bends etc.

Don't know, but if you don't mind trudging through a few pages of sites to try and find it,

Google Scholar

Have a dictionary handy, you just might need it if you don't have an impressive vocabulary...

Been there done that, didn't find an answer just basic conical geometry that didn't address this, which is why i was hoping for a geometry expert that knew more than what generally gets posted online. I'm nearly sure there is a way to answer the question with the known givens, but I don't know what that answer is.

[Bruce Johnson]

Here's a thread over on TalkBass covering this exact subject in a lot of detail. The first page doesn't go too deep, but the second and third pages are worth it. This topic, which I call "Hourglassing" a fingerboard, becomes particularly important on small-radius fretless bass fingerboards. The flatter the fingerboard, the less important it is. As we discuss, most experienced Luthiers do the right thing when they level frets and fingerboards, but the terminology gets really confusing.

Even fretless fingerboard levelling tips - TalkBass Forums

Here's a copy of my main post on that thread, but you may need to read the whole thread to understand the context:

I think this discussion is getting confusing because you guys are missing one element. I'll try my hand at explaining the geometry here:

First, a normal, typical bass fingerboard is cylindrical. That is, it has the same radius at the nut as at the heel.

However, as described above, the strings aren't parallel. They are closer together at the nut than they are at the bridge. Even though the radius of the undersides of the strings may be the same at the nut as at the bridge, the underside of the strings actually describe a slightly different shape than a cylinder.

The strings going right down the center of the fingerboard will be true. That is, if you lay a straightedge down the centerline of a normal cylindrical fingerboard, it will lay flat. However, as you move the straightedge off to the side at an angle to the centerline, following the actual paths of the outboard strings, you'll find that the center will be high. The straightedge will rock on a high spot in the center, as if the neck were backbowed. The further off-angle to the centerline you go, the worse this condition will be. Also, the smaller the fingerboard radius (rounder) is, the worse it will be. If you cut the fingerboard to a pure cylinder, you'll end up with high-spot buzzing on the outboard strings, but not the center strings.

So how do you trim the surface of a cylindrical fingerboard to correct for this problem? The process is just like Musiclogic described above; you level-file right along the actual string paths, blending in between them. This process is often called "conical filing", but it isn't really forming a cone shape. That's what gets confusing. It's actually forming a slight "hourglass" shape. The radiuses at either end of the fingerboard are untouched. You are trimming away wood at the middle of the fingerboard (lengthwise), but only on either side of the centerline. So, the radiuses at the nut and the heel may be 12", but the radius at the 7th fret will be slightly less, like 11 3/4". I prefer to call this "hourglass filing" to minimize the confusion.

On a fingerboard that has been properly "hourglass filed" like this, a straightedge placed along all of the string paths will be dead flat. When the straightedge is placed parallel to the centerline, but off to either side of center, there will be a slight gap under the middle, a "relief". But, down the center, there's no gap. This is where owners often get really confused when trying to check the relief and adjust the truss rod. A high quality, hourglass-filed fingerboard can give you confusing relief readings if you don't understand what you are looking for.

If you want to get really technical, the real mathematical description of this hourglass-filed fingerboard is.....wait for it.....an offset hyperbolic paraboloid! I'm sure that brings back some frightening memories from your school days. Picture an hourglass, where you take the narrow waist and push it off to the side enough that one side becomes straight. That's what an "hourglass-filed" fingerboard looks like. The centerline of the fingerboard is right on that straight side. The further off to either side you go, the more hourglass-shaped it becomes. Try not to get a headache.

So, how does this relate to compound radius fingerboards? A "compound radius" fingerboard is just another term for a conical-shaped fingerboard. That is, the radius is larger at the heel than it is at the nut. The surface is a section of a cone. Basically, the more conical you make the fingerboard shape, the less need there will be for the "hourglass filing" correction. There is a point where the surface becomes conical enough that no hourglass filing is needed. However, most commercial compound radius fingerboards aren't that radical, and still need a little bit of correction to make them flat along the string paths.

What confuses things even more is that instruments built with compound radius fingerboards often also use a flatter radius on the bridge than on the nut. So, the underside of the strings are also on a conical shape, but they still aren't parallel, so the fingerboard surface still ends up needing some hourglass-filing correction. Don't hurt yourself picturing that one.

How does this relate to frets? Exactly the same way. A really good fretjob has this hourglass-filing correction as part of it. The typical process involves using a straight file or diamond stick or oilstone, working along the string paths and blending in between. Leveling frets with sandpaper on a radius block is only a way of roughing them in. It takes an extra step to make them really true for the strings.

Let me repeat: All of this hourglass-filing stuff becomes less noticeable and necessary as the fingerboard radius gets flatter (larger radius). A flat fingerboard doesn't have this effect at all. That's one of the reasons why manufacturers and builders like to go with flatter fingerboards; it's less complicated. I personally like to build my basses with very round fingerboards, from 7 1/4" to 4" radius. At a 4" radius, all of these geometry issues become much more obvious. That's why I've spent the time working with them and trying to understand them.

I hope this helps clarify the discussion?

Another of my posts, from the same thread:

Yeah, upright bass fingerboards really exaggerate all of this geometry. I do a lot with hot rodded Baby Bass necks. Dramatic conical fingerboards, combined with a wide taper in the string spacing and relief cuts that have to rotate around the cone! Shaping out an upright bass fingerboard is a lot harder than most people understand.

You've got to keep the terminology straight though:
Doing an "hourglass" trim does not, by itself, change a cylindrical fingerboard into a conical fingerboard. You can do an hourglass trim on a cylindrical fingerboard, or you can do an hourglass trim on a conical fingerboard.

If you're changing a cylindrical fingerboard into a conical (or "compound radius") fingerboard, that would correctly be called a "conical trim". You are changing it into a cone, so that the radiuses are now different at the two ends.

Upright bass fingerboards start out as a conical shape, around 2 1/2" radius at the nut and 4" radius at the heel. To true it up for low action, it needs to be "hourglass filed", which makes the middle a little skinnier, but doesn't change the radius at either end. That would correctly be called an "hourglassed conical" fingerboard.

Now, most electric basses have cylindrical fingerboards, with the same radius on both ends. To trick it out for low action, you can "hourglass file" it. But that doesn't make it into a conical (compound radius) fingerboard. It just makes it into an "hourglassed cylindrical" fingerboard.

Of course if you want to, you can do both: First "conical file" it to convert it from a cylinder into a cone (that is, making it into a compound radius fingerboard), then "hourglass file" it to correct for the taper in string width.

Is that more understandable?

I know this all sounds complicated, but it comes down to the same basic thing: In terms of playability, the only thing that matters is the 1/4" wide strip of wood directly underneath each string. The rest of the fingerboard is just there for decoration. That's what Big B is doing with his technique with the jointer: Get the string paths straight, and then blend off the excess between them.

A few years back, I worked on some really radical instruments that had reverse compound fingerboards! They were 7 1/4" radius at the nut and 2 3/4" radius at the heel, if you can imagine that. Six strings, short scale length. The fingerboard geometry was crazy. They required a significant amount of hourglass filing, close to 1/8" off in the middle of the fingerboard. As I mentioned above, the more radical the radius, the more hourglassing is required. For those necks, I ended up making a special router fixture that cut six round-bottom flutes directly along the string paths. Once those paths were created, I would round off the spines between the flutes. It was surprising how hourglassed the final shape of the fingerboard was.

[wizard333]

It's the idea, but there is no math there, so its purely conceptual. I'm looking for the math.

Also, they are talking about basses, where you don't bend strings much like on a guitar. Notice they say "you just level under the string then blend in between". If you do that on a guitar, you're probably going to get buzzing and fretting out on bends. A guitar is more complicated because the way the string crosses the board as you bend becomes critical. I have seen plenty of examples of guitars that played reasonably clean as long as you didn't bend, but fretted out badly on bends.

[Bruce Johnson]

What do you mean, it's purely conceptual? That's how real, actual fingerboards and frets are trued up to the best form geometrically possible. That's how we do it on guitars, basses, whatever. You start with the cylindrical or conical form of the fingerboard, and lay out the lines of the actual string paths, with their taper in spacing. If you cut the string paths straight, and then blend evenly between them, you end up with the "hourglassed" offset hyperbolic paraboloid. That's basic geometry that you can do in any CAD program.

For a guitar where the strings are going to be bending, you use the same basic geometry, but sweep the string paths through the expected angles of bending. However, you have to compromise, because there isn't any single fingerboard surface shape that's going to give you a flat string path along all the possible combinations of bend angle and note fretted. Unless the fingerboard is actually flat, of course. So, you work the hourglassed shape to get flat string paths along the furthest-expected bend paths. That prevents the fretting out during the bend. But the compromise is that, when the string isn't bent, it has a small amount relief under it, so the action has to be just a bit higher. As long as there's some radius to the fingerboard, you can't get a flat string path at every angle in relation to the centerline.

And that's why compound radius (that is, conical) fingerboards are popular for guitars that get bent. Transitioning to a flatter radius at the heel reduces the compromise. In particular, it helps because most bending is done at the high end of the fingerboard, towards the centerline. But there still isn't any perfect solution. Except for a completely flat fingerboard. There isn't any conical or parabolic shape that's going to work for all the string paths that you can get with bending. The farther away from flat the shape is, the bigger the compromise is, requiring higher relief and action.

[frus]

What do you mean, it's purely conceptual?

I believe he's looking for the exact formula or something

[wizard333]

Yes, like I said, where's the math?

Also, I'm not convinced the hour glass shape is better than a conical where bending is concerned. When you push a string across a fretboard that isn't flat, the string height above the next fret decreases the farther the string goes. The flatter the radius, especially as you go up the board, the better. The rounder radius part of the hour glass shape will cause a bend that is clean on another part of the board to tend to fret out there. That has, in fact, been my observation in practice where that shape was used.