Figuring out correct radii to fit fretboard taper - created 03-18-2010
Atienza, Louie - 03/18/2010.19:49:48
The previous discussion on fretboard sanding blocks ended up becoming a heated debate over the relationship of taper to fretboard radius. It got me wondering how one could figure out what fretboard end radius would produce a (theoretically) true conical section, given the length of the board, width of the nut and fretboard end, and starting radius.
I'll define the terms as such: (all units should be consistent)
n= width of nut end
e= width of fretboard end
l= length of fretboard
rn= radius at nut
Using these four variables we shall find:
diff= difference of fretboard end width to nut width
f= multiplying factor
dn= distance from nut to theoretical apex point
de= distance from fretboard end to theoretical apex point
And finally
re= radius at fretboard end
First thing we need to do is find the difference between fretboard end and nut. Simply,
diff = e - n
We then divide the width of the fretboard end, e, by the the difference, diff, to get the the
multiplying factor:
f = e / diff
Multiplying f by the fretboard length, l, gives us de, the distance fom the fretboard end to the
apex:
de = l * f
The distance from the nut to the apex, dn, is simply the distance from the fretboard end,
minus the fretboard length:
dn = de - l
We now know everything we need to figure out the radius of the fretboard end. Basically ina
cone the radius grows proportional to the distance from the apex. Therefore,
dn / rn = de / re
Solving for re (the radius of the fretboard end) we getL
re = rs * (de / ds)
So let's say you have a fretboard that's 1-3/4" at the nut, 2-1/4" at the fretboard end, 17-1/2"
long, and you want a starting radius of 10"
Then,
diff = 2.25" - 1.75" = .5"
f = .5"/2.25" = 4.5 (units cancel out)
de = 17.5" * 4.5 = 78.75"
dn = 78.75" - 17.5" = 61.25"
Finally,
re = 10" * (78.75"/61.25") = 12.857"
To find the radius if the fretboard ended at the bridge, just add the scale length to the dn; we'll
call it db
db = dn + s
So in the above example, if the scale length is 25.5"
db = 61.25" + 25.5" = 86.75"
rb = rs * (db / ds)
So
rb = 10 *(86.75"/61.25") = 14.163"
As you can see, most compound radius boards we make are actually flatter than the "theoretically coreect" conical section.
It would be a simple thing to plug these equations into a spreadsheet, and you'd have yourself a fretboard radius calculator!
Yup!
Just as an observation, there is a very (very!) small cosine error in your result, because cos(theta) != 2 x cos(theta/2), though it's very (very!) close for small angles. Your apex is not on the fretboard centerline, and this is where the error manifests.
Another approach is to consider the fretboard profile as an arc of three parameters: the radius at some point on the fretboard, the width at that point, and the height of the fretboard center above the edges. The equations for this are not too hard to work out or find somewhere, for finding the third value from the other two.
This lets you start by finding what I call the "arc height" anywhere along the fretboard, for example, choosing the 7th fret as a nice, playable reference position, and then using the widths at the nut and end to figure out what the radii should be.
It also has the advantage of working for cylindrical (parallel-side) and even inverted-taper fretboards, if that is ever interesting, where your de will go to infinity, and thus things start to get a little difficult to deal with.
And, as another side-effect, you're working directly with the height of the center versus the height at the edges, which might be helpful in physically constructing the thing, depending on what kind of tooling you use.
Donl, thanks for pointing that out. I decided not to figure from the center because that would have made the equations a little more complicated than practical! It would be a problem, however, if one were to make a fretboard with an unusually narrow nut and unusually wide bridge.
Actually the formula's correct if the nut and frets follow the cone's surface (they would be slightly curved) instead of being cut perpendicular to the centerline. Though nobody builds necks that way! One could easily approximate the percentage of error by considering the nut or fretboard end as a chord of a circle the same radius of that end, find the sagitta, and divide that by the length of your reference point to the apex.
Your method of using the center height is a good one if you have a board where you want a certain thickness at the edge and center. You would just consider the widths n and e as chords of a circle, and the difference from the center and edge thicknesses as the sagitta, and it's not too hard to figure out what radii you need to keep both edge and center consistent.
Well.. but now I'm not sure if the error is actually present, after all! Here's a sketch of what I was thinking about: the difference in length between the red and green centerlines. This problem has shown up often in issues I'm working on for the vfret program, but I'm not sure it applies to your model. I think I'm too tired to think clearly. :)
And now I have put on my "todo list" to make sure my equations still work right with a flat fretboard. It's always something!
I'm working on a compound radius jig that includes the surface/axis angle in the radius blocks that it rides on, to avoid the errors similar to those that apply to perpendicular frets -- mostly to try to make sure the motion is smooth, rather than worrying about the dimensions of the finished product. Think of... a matched pair of inside/outside radius blocks, but mounted at an angle equal to the cone's surface. The blocks are effectively sections of a disk perpendicular to the radius, rather than the surface.
Fun problems. :)
Louie, excellent topic !
My first thought was that it would be even more useful to define the parameters as nut width, nut radius, scale length, and string spacing at the bridge, since it is the string spacing which defines the fretboard taper. That way the cone would extend from nut to saddle, the radius at the saddle would be calculated and the radius at the 20th (or whatever) fret could be derived very easily.
Try Gregory's radius calculator spreadsheet when it gets into the library - he's already done all the hard work. Just plug in scale length, string spread at nut and bridge and radius at nut and it will give you the radius at any fret.
Murray, that would be easy to do, as you could just substitute the string spread at the nut and bridge for the nut and theoretical fretboard width at the bridge. The reason I did it this way is because most people I know have an established nut and 20th fret width (or even 12th or 14th fret width) and are jigged up that way. So ii's very easy to substitue the length from the nut to the 12th or 14th fret, and the formulae still work.
Donl, the radius jig I maade and use is along that idea, though I did not cant the radiused blocks along the perpendicular to the radius. but the error is still very small, and as long as you move either the router or workpiece in a lengthwise manner, the string paths will remain straight.
Brian, I realize that one could use a spreadsheet. I prefer to sit down with pen and paper and do my calculations longhand, only because it keeps my mind active. But the whole point of this is to illustrate that everything is proportionate, regardless of how the nut and fretboard end are shaped. Also, I'm one who deoesn't take anything on blind faith. I want to know and see for myself the mechanics of things.
To note, there are exactly two cones that satisfy any given fretboard taper, though the second on is a very narrow cone with radii half the nut and fretboard end width.
Donl I just thinking about your jig idea in post 5, and I think the toughest part is keeping the router perpendicular relative to the fretboard surface while also following the radius wanted at a particular point along the perpendicualr of the centerline; in effect it's a cone with the center smudged toward the bridge, so to speak. The trick is keeping the router pointed that way as the radius changes...
To note, there are exactly two cones that satisfy any given fretboard taper, though the second on is a very narrow cone with radii half the nut and fretboard end width.
Louie, you've lost me. Just as I thought I had seen the light regarding fretboard taper and compound radii, you throw a curved ball.
How can there be two cones which satisfy a given fretboard taper ? If you specify the nut radius there is only one cone, but as there are an infinite number of nut radii then there are an infinite number of cones for any fretboard taper.
I hope I am not heading for another stroll into the woods ...
Sorry Murray! The other cone is useless for guitar work. Think of a cone whose outer walls are the fretboard edges; a long narrow cone. But obviously, not applicable to guitar (unless you like a 7/8"-1-1/8" radius!)
There are an infinite number of surfaces that will give you straight string paths, though not all of them would have the string paths converge at the cone's theoretical apex.
Louie... I'm not sure I follow you, and it's probably because I didn't describe things very well.
In my design, I'm rocking the fretboard rather than the cutter. I have a base plate and top plate, with several concave/convex pairs of radiused cauls along the length for good support. Each pair has its upper and lower surfaces cut at the cone's surface angle, so when mounted, the "working surfaces" of the cauls lie in planes perpendicular to the axes.
I will lock it at some specific angle, and make a lengthwise pass under a fixed-height cutter. The the facet being cut is always parallel to the baseplate on both axes, and at a fixed height above the baseplate. So the cutter must be square to the table the jig is sitting on, at a fixed height for all facets as the fretboard is rocked underneath it.
The assembly will be small enough to run through my thickness sander, which I've been able to get good accuracy from. I can start "high" making passes to knock the edges off, and then progressively work toward a final series of passes to get the right thickness.
A router could also be used, held at a fixed height above the thing. The router could slide along the length, maybe on side rails, or the jig could slide underneath it. Or... I've thought about putting tall side plates or end plates on the baseplate, turning the whole thing over, and running it along the router table. (Having both side and end plates would allow for perfect dust collection, too!)
Aaah OK I see now. Looks like a lot of fun to build and align! But I see the virtues of the design.
The design I'm working on will be an atachment to my revised neck shaping jig. The router rails remain fixed, but now on ball bearing slides. It will be fully adjustable for any start and end radius, as well as thickness. It's just a matter of fine-tuning the mechanism, not yet ready for prime-time!
Murray, if you're wrong, we'll have to stroll in the woods together. I think there are an infinite number of cones that fit a given fretboard taper. There is only one apex for a given taper, but the narrowness or "squatness" of the cone will vary with the radius at (eg) the nut or bridge.
The difference will be in how many degrees around the cone the fretboard section is.
Additional comments...
Think of each caul pair as the outer edge of rotating disk perpendicular to the axis of the cone.
If the cauls are mounted vertically, not in a plane perpendicular to the cone's axis, they can't be treated as rotating disks -- in other words, the sides of the cauls do not remain in the same plane as the cauls rock back and forth. (I am using plates (not shown) on both sides of each caul pair to keep them aligned. This is, in fact, a crucial part of the design!)
Also, I'm cutting cylindrical surfaces on each caul, even though, ideally, they would have a slightly different radius on their leading versus trailing edges. I haven't thought of a simple design for the caul-making jig to do that, so for now, I'll either ignore it, or sand a few thousandths off the high edge of each caul surface if the math shows the error is noticeable.
For those in the "infinite number of cones" camp, you're not considering how to fit the cone to the taper of the fretboard edges. If the goal is a fixed height along the centerline, and a fixed height at the edge (which I think
s
our goal), then there is only one cone that fits. For this cone, the surface of the fretboard is indeed the surface of the virtual cone.
For any other cone, the strings will not each lie on straight lines converging at the apex. The cone that the strings imply will be different than the cone that the fretboard implements.
Donl It shouldn't matter if your cauls have canted edges, think of the stacked rings puzzle... the only think I can think of that may ruin your day, is as you move the top plate across, one end will move more than the other end. So you'll have to take that into account and make the cauls wider to accomodate that.
Also, to add above, while the strings may not point at the apex, the string paths will be on a straight line. This is true for all proportionately derived surfaces. Even is the fretboard is concave at the nut and convex at the 20th fret, you could make it with straight string paths... or an S-shaped board, etc..
If you could, however, cut the angle of the cone into each caul then their surfaces will be parallel to your bottom plate, which might make it easier to keep things solid...
Brian, that's why I quantified that the formula gives the one "correct" fretboard end radius, given the nut and end width, length, and starting radius, where the fretboard sides meet at the cone's apex. There are an infinite amount of surfaces that will fit a given fretboard taper, but the fretboard edges may not necessarily meet at the cone's apex given the radii used. Depending on the style of player using the neck, you may be making the fretboard flatter than it needs to be toward the fretboard end. To me it depends on the player's style. If they have their thimb in the back of the neck higher up, they may prefer the flatter radius. If they play more chordal stuff high up (or not play higher up at all) then making the board flatter there is not necessary, and you make it a little bit harder to flatpick, in general. If the player does a lot of bends, then it's beneficial to make the board flatter than the "true" cone at the end because each string produces it's one cone.
But yes, since you could quantify any radii for the nut then there are an infinite amount of cones. The problem is, most of them are not useful, since you're limited eventually by your fretboard thickness and fretboard edge thickness. And yes, between 10" ans 12" radius there are an infinte amount of radii, though it would be impractical to exploit that, as we like using round numbers! You can't have a radii smaller than half the width of the nut. On the other end, the radius gets so large that the fretboard is virtually flat. So while theroretically it's unlimited, in practice it's limited by our tooling, material dimensions, range of comfort, practicality, etc...
I must add that on a "theoretically correct" fretboard radius, as you bend the strings still go uphill, which is on ereason why the fretboard end radius is flatter than it should. The only way you could have a conical board and not have the strings go uphill when you bend is if the nut width is 0!
Donl, are you saying that for a given fretboard taper and scale length there is only one radius at the nut that will have the strings lying on straight lines converging at the apex?
Louie, I agree with you that in reality, having the board flatter toward the bridge will compensate for the fact that the string bends with the fulcrum at the nut, and not at the apex of the cone. The starting point for the compound radius (which you might compensate later) is an infinite number of cones with the apex at a fixed point for a fixed fretboard taper.
I disagree that practically the radius at the nut should be between 10" and 12" (with an infinite number of possibilities between)although that may be most popular. I think anything between about 7.5" to flat would be appropriate depending on playing style (someone with a classical background, for example, may prefer a 20"(nut) to 31" (20th fret) compound radius.
Brian, I was just trying to explain that between 10" and 12" there are an infinite amout of radii that can be used, though it would be impreactical to use, say 10.327"
You're statement about fretboard taper and scale length is incorrect. For any given radius at the nut, at least half the width of the nut, there is a corresponding radius at the fretboard end that will give you string lines meeting at the apex of the theoretical "cone."
Now the opposite question: given a fretboard, with known length and width at nut an end, and two arbritrary radii, is it a cone section? Yes, if you use the fixed axis method or similar method of fretboard shaping. But do you have a straight string paths for all strings? Not necessarily. To prove it, think of my example above. If you increase the radius of the fretboard end, then the radius "grows" faster than theoretically correct, for the fretboard taper.
Think of a spinning cone, which is in effect the way compound radius boards are usually made. Now think of a router in a carriage going in a straight line, parallel to the cone centerline. If you set the radii to the "correct" values for the taper, then when you run the center of the bit on the fretboard edge, it will follow it exactly. If you increase the radius at the end, it will move "faster" than at the nut; or, it will leave the edge of the fretboard as you go from the nut to the end. Some people may not notice this; one reason is that a lot of times the board is radius before tapered.
Or another way: Think of any cone. There is only one way to cut a section where all lines point toward the apex. If you skew it in any way, your string lines will never match the lines to the apex.
It IS of course possible to make any compound radius with straight string paths. But they are not necessarily conical sections. That is the reason I've been careful to say infinte "surfaces", not cones.
Sorry Louie, which statement about fretboard taper and scale length is incorrect?
Sorry it was the first comment on post 18... Murray's correct on that...
Lost me there Louie...My first line on post #18 was a question to Donl.
FWIW, I completely agree with your statement "For any given radius at the nut, at least half the width of the nut, there is a corresponding radius at the fretboard end that will give you string lines meeting at the apex of the theoretical "cone." If, and only if, there is a set taper. That apex will necessarily be the intersection of the extension of the edges of the fretboard past the nut.
You write:"given a fretboard, with known length and width at nut an end, and two arbritrary radii, is it a cone section? Yes, if you use the fixed axis method or similar method of fretboard shaping. But do you have a straight string paths for all strings? Not necessarily. "
In the above example, are you assuming a set taper? If so, then you can't choose two arbitrary radii and assume it's a cone. Once you have the 2 dimensional triangle set by the fretboard taper, the apex is set. If you then define the third dimension with a radius at any point on the length of the cone, you define a line from the apex, through that radius - from there the radius at every other point is determined. i.e. you only get to choose one radius to define the cone - if you choose two different radii, the apex changes, so it's not a cone.
Louie -- ah, not so, my friend! Imagine in your mind (so I don't have to make a drawing!) first the cone's axis; let's make it vertical for now. Now... make rotating disks that meet up with the surface of the cone. Remove the cone, and glue a board straight up and down across the edges of several of the disks. You can swing the board left and right, and it will move just fine along the virtual cone surface. That's the geometry I've created, and it is also why the cauls are tilted.
It is almost equivalent to make the disks "v-shaped", i.e, to make them inverted cones with radii perpendicular to the cone's surface, which would be the geometry if the cauls were vertical in my jig. However, there are practical problems with that geometry -- the cauls would need to be curved to really do it right.
And yes, the motion will be larger at the larger radiuses, but what will happen is that edge of the fingerboard will directly along the centerline of the jig at a certain angle.
Brian -- Yes, mostly, more or less. You can change some things, but not others, and keep the same cone. I wish I was better at CAD so I could quickly draw this, but...
Suppose the cone is vertical. If you slice it with a plane that runs through the axis, you'll get what I would call a "vertical" line along the surface (because I don't know the proper term for such a line).
If you make two such axis slices, you'll get a wedge-shaped subset of the cone, and two of those vertical lines marking the edges of it.
Now slice the cone horizontally at two different heights, representing the ends of the fretboard. Slice off an outer layer of the wedge and turn it into ebony, and turn the rest of the cone into air.
You can slice out a wider piece of the same cone, wrapped farther around the thing, as long as the edges are still made by "axis slices." Spread the strings out proportionally.
You can make the end cuts both four inches lower, and get a fingerboard of the same length that is wider. If you want it narrower again, bring the two axis slices closer together. The angle between them will decrease -- the fretboard taper angle will be reduced.
If you were to "spread out" the upper end of the fingerboard to widen it, leaving the lower corners alone, you'd be forming an edge line from the lower corner to the upper corner that no longer lies on a vertical line -- that edge (or outer string) would have to wrap around the cone, which is the geometry we're trying to avoid.
(I sincerely hope my word pictures are a net gain in communication!)
Donl I understand that hehehe. My thinking behind your jig is, say you built it the way you have drawn, except that you cut the bearing edges of the cauls to the theoretical angle from the cone center to the surface. Now thin the cauls till they're almost zero. The jig would obviously still work. Now widen each edge perpendicular to your angle, making the square-edged cauls pictured. It should move just as easily as if the bearing surfaces of the caul were shaped to match the cone surface.
Brian it is a section of a cone! Take any arbitrary large hollow cone. Take your fretboard and trace its shape onto the cone, and cut it keeping the cutting tool perpendicular to the tangent of the center of the piece. You now have a fretboard blank that from the front satisfies the taper of the fretboard you want, with two arbitrary radii, yet the string paths don't point to the apex, since those "string paths" are no longer flat. This is exactly what happens when a radius jig is set to other than the "correct" taper.
Bryan, I am so empathizing with you, as what you are saying is exactly the way I was thinking before I underwent my epiphany at the hands of Mario.
I too had previously thought that there are an infinite number of cones which will fit a given taper and given nut radius, and indeed there are in fact an infinite number, but what I wasn't grasping is that only one of these cones allows the strings to lie perfectly against the cones. I believe now that this "perfect" cone must be the best possible solution to attaining maximum playability, although as others have said, different cones could be made to work, but I do not believe they could work as efficiently.
If I could expand on Louie's post above, it might also be helpful to point out that on any arbitrary large hollow cone, you can also locate the fretboard template at a point on the cone so that your desired nut radius coincides with the actual radius of the cone.
You will then have a conic section with the desired nut radius, the desired nut width, the desired fretboard taper, but as Louie says, the individual string paths will not lie correctly.
I wish I knew what the correct mathematical terminology is for such a slice of a cone ...
Et tu, Murray?
Hang on....we don't disagree. I agree that there is only one cone that fits a given taper AND nut radius. I have said that the taper defines the apex of an infinite number of cones, but as soon as you define the radius at any point, you define the radius at all other points and thus there is only one cone that fits that.
Did any of you understand that I meant anything different? If so, I defy you to show me where :-)
In case you still don't agree with what I've said, you've forced me to make diagrams and photograph them (my wife just moved her office into my study, and the scanner is not connected yet).
Lets start with diagrams of cross sections of a few cones, with the fretboard (always same nut width, scale length and width at end of fretboard) superimposed on the surface of the cone.
If it's too hard to understand in 2D, I'll post some 3D examples in a few minutes.
Now in 3D....
Imagine a hollow cone with an infinite radius at the "nut". i.e. a flat piece of paper. Ink drawn on the paper is as close as we can get to lying exactly on the surface of the cone in a straight line. Thusly (note on the diagram, the fretboard taper extended to define the apex or origin of the cone, and the distance from the "nut" to the apex)
Now, if I roll this piece of paper into a cone other than flat, for any given radius at the nut, there is only one possible radius at the fretboard end with the ink still laying flat on the paper. (Note that the apex of the cone has not changed, nor have the dimensions of the fretboard.)
Now, if I make a wider cone, keep the apex the same, and the fretboard dimensions the same, the ink still stays on the surface of the cone!!! All that changes is the proportion of the cone (the number of degrees of the 360 degrees of the cone's circle) that is involved in fretboard surface.
cool , Brian, and jeez, I hope we're not going to fall out over this, but since you asked, I have to disagree with your earlier statement ..."The starting point for the compound radius (which you might compensate later) is an infinite number of cones with the apex at a fixed point for a fixed fretboard taper." ...
Now, I could be wrong here, (it wouldn't be the first time, and I am keeping a weather eye open for Mario coming up on my blind side with his Winchester '73 blazing ...) but as I understand it now, there is actually only one cone which satisfies any given fretboard taper, not an infinite number.
The crucial point is this : on this given cone, you can slide your fretboard template along it until your selected nut radius coincides with the radius of the cone at that point, ( thus giving you an infinite number of conic sections, depending upon which nut radius you are specifying.)
You then trace out your fretboard taper at that point, and that gives you your perfect conic section.
That is the flaw in your pictures above, Brian, when you roll the second cone you are actually increasing the nut radius; the section you trace out will still be a perfect conic section, just with a different nut radius from the previous cone. (and a different 20th fret radius as well, for that matter !)
N'est-ce pas?
Murray, looking at the photos I posted, there are 3 possible cones with the same given fretboard taper (it's the same piece of paper with the same diagram of the fretboard), and I guess you could imagine that there are an infinite number of cones with the same fretboard taper, same apex, same distance from the apex to the nut. (otherwise I'll have to take a whole lot more photographs)
I would love it if Mario would chime in on this - hopefully he'll be able to explain it better than I can and either set you or me straight. Or maybe Greg Robinson could chime in - his spreadsheet helped me understand this concept - maybe the fact that he and I are both from Australia helped. Is geometry different in the Northern Hemisphere:-)
BTW, no need to fall out over any of this, except maybe that it's Brian, not Bryan.
As you will now have spotted, Brian, I did in fact edit my previous post to correct my earlier misspelling of your name (for which, many apologies !)
I am empathizing with you even more, btw, as you are saying exactly what I believed to be the case three days ago, before I saw the light ...
Murray, it looks like you edited your last message while I was posting mine.... Thanks for correcting my name spelling.
I don't think there's any flaw in my pictures or my statements. You said:"as I understand it now, there is actually only one cone which satisfies any given fretboard taper, not an infinite number." I say as long as you have not defined the radius at any point, there are an infinite number of cones. And yes, as you say, I have changed the nut radius from cone to cone, but that's my point. There is an infinite number of cones which are defined by the fretboard taper and an infinite number of radii at the nut. The fretboard taper defines the apex, and the chosen radius at the nut (or the radius at any other single distance from the apex for that matter) defines the angle at the apex.
What that means is that once you know your fretboard taper and scale length, you can choose whatever reasonable radius you want at the nut (I would say anything from 7.5" to infinity - flat- is reasonable) and from that calculate the ONE fretboard end radius that will give a section of a perfect cone.
If you had said:"as I understand it now, there is actually only one cone which satisfies any given fretboard taper AND NUT RADIUS, not an infinite number." I would agree with you.
It's 2 am here now...but I'm enjoying this so much, I'll wait a few more minutes for a reply before I go to bed.
Brian, NO,NO,NO !!!
Seriously, there is only ONE cone which satisfies any given fretboard taper, but along this cone (which goes from zero at the *vertex* (anybody mind if we call it the vertex rather than the apex ?)) to infinity, there are an infinite number of nut radii to choose from.
You move your template along the cone until the nut line coincides with the required nut radius, and that defines the perfect conic section.
Honestly, Brian, in a couple of days, or maybe even less, you are going to smite your forehead and say "I geddit now !"
That's what happened to me .
I gotta go now, but sleep well !
Murray, I had the same smite of the forehead in June 2007 when Greg originally offered his spreadsheet, and I wanted to know why I couldn't have a 10" nut radius with a 20" bridge radius. Eventually he set me straight.
Murray, all cones can have a vertex with a zero radius and go to a radius of infinity at a point infinity from the vertex. There isn't just one cone which fits these parameters. The thing that differentiates all these cones, is the angle at the vertex. That angle is defined by the radius at a given distance from the vertex.
Looking at the pictures (1,000 words or about 300,000 pixels each), why do you think I can't have the same fretboard taper and scale length with different cones for different nut radii and keep the ink in straight lines?
How would you define the ONE cone which satisfies a given fretboard taper?
My wife's already asleep, so I can stay up a little longer :-)
Brian, the taper changes in your example above, because as you widen the cone, you widen the the chord (or straight line from the the corners of the nut or fretboard end) thus changing the taper. But if you cut a piece of cardboard with a triangular slot matching the taper of your fretboard, and lay it over any cone whsoe slant length equals the length of the slot, both edges of the slot would touch the cone.
To prove that there are an infinte amount of cones, let's take the triangle produced from the apex and the two corners of the fretboard end. As long as any cone has an end radius at least the width of the fretboard end then those three points of the triangle will always lay on the cone. Thus there are an infinite amount of cones that would satisfy any fretboard taper. As the radius approaches infinity the board approaches flat.
Brian, as faras your example, I believe yes, you can of course have a board with a 10" starting radius and a 20" radius at the bridge and still have straight string paths. It just won't necessarily be a cone. Make two pieces one of 10" radius and the other at 20" radius, and make them the appropriate widths. Mark 12 or so equidistant points on the surface of each one. Place them the distance of your scale. Now throw some mud, cement, plaster-of-paris, whatever between them. Get a straightedge that will reach across both points and use it as a screed, moving proportinately across using the points you marked as a guide. You've now made a complex surface, where all string paths (and all derivative paths in between) are an exact straight line, though it is not necessarily a conical section. This is basically what we do when we use a long sanding block and true the stringpaths on a radiused fretboard; we actually "correct" the errors we machine onto the board.
The Charles Fox method does produce a surface where all string paths are straight, because it in effect does exactly what I described above, albeit with sandpaper. I'm working on a router jig that takes this into account, but in a way that's not totally convoluted.
I'll try to sketch up a mathematical proof and post it later on.
I have to add that when the fretboard end radius equals the distance from the fretboard end to the apex, then the fretboar is exactly flat, which gives us a range of radii that we could work with.
I guess you are sound asleep by now, Brian, but I trust you will catch up on this later
How to define the ONE cone ? Extremely simple, (and I wish somebody had explained it to me in such basic terms on the other thread )
The ONE cone can be defined very simply as the cone whose angle at the vertex is the same as the angle produced by extending the two edges of the fret board until they meet at a single point.
Now, this cone can be extended ad infinitum, from the surface of the Earth out past Jupiter if you wish, simply select your nut radius, anything from 7" to maybe 3 light years, move the template until the selected nut radius coincides with the radius of the cone, mark out your template on the surface of the cone, and there you have your perfect cone, ( or rather "conic section" ), " perfect" being defined as the surface on which the strings will lie in their correct orientation, and be contiguous with the surface of the cone at all points.
If you do this near Jupiter btw, do watch out for asteroids, these suckers can really hurt.
Just re-reading Louie's post above, there is one thing needs commenting on.
Louie, you say ..."Brian, the taper changes in your example above, because as you widen the cone, you widen the the chord (or straight line from the the corners of the nut or fretboard end) thus changing the taper... "
And this is in fact true, given the way Brian has chosen to depict it.
There is also another way to look at it, and that is to regard the fretboard template as rigid, and to get the outline you simply lay the rigid template on the surface of the cone, and sight down vertically. That way, the taper doesn't vary, HOWEVER, it still doesn't produce a perfect conic surface on which the strings will lie in perfect alignment.
Exactly Murray! Which is precisely what I'm trying to get at! Not all conic sections produce stratight string paths, an d not all surfaces with straight string pull are conic.
Some of the methods we currently use, like the "cone center as pivot" method, does indeed produce conical sections just like yuou describe above, where the strings are not in perfect alignment, and requre further manipulation by hand.
I think the attempts that Donl and I) have is to eliminate that "error"...
But if the radius changes in proportion to the distance on the fretboard, then there will alwasys be straight string paths.
When I get back later, I'll sketch this out, and it should clarify my thoughts on this.
I'm only understanding half of each post as it get complex, but I think you are agreeing with each other that __once__ you have a radius at any point is defined and the taper is defined only one string path is possible. It makes sense that there would be an infinite number of possibilities otherwise, but as was stated, for a given taper, defining a radius anywhere fixes the rest of the radii as far as string path goes.
I believe you're correct Bill, I think we have a grasp of what's going on, though explaining a 3-dimensional thing in 2-dimensional words is not easy.
To convey what I mean and the "mechanics" of the formulas I presented, I made a couple models out of paper.
This represents the fretboard taper. The angle is arbitrary; it could be whatever you need to satisfy the taper you're using. I've marked the variables for reference. In practice, you would already know the taper, fretboard length, and distances from apex to nut and fretboard end:
Here is my cone model, unfolded (uncurled?). As you can see the radius is the maximum radius you can use, which is exactly the distance from the apex to the fretboard end, and this produces a flat surface.
The cone, constructed...
Overlaying the taper window over the cone, one can see that both edges of the window sit on the cone. I have shaded the area to illustrate this.
This holds true even when the cone is as narrow as possible, meaning the cone base equals the fretboard end
Opening up the two cones shows the amount of surface area difference. One conclusion I draw from this is, in general, if you make the fretboard flatter toward the end (or 20th fret or whatever) you should increase the width there as well (widen teh taper), otherwise the string spacing may feel tighter than necessary.
Even a non-circular cone would work, as long as the slant height is the same. Here's an ovaloid cone:
As well as the negative of the cone, or a concave surface
So Murray, I misspoke and I owe you an apology!
So I shall reword my theorem or conjecture to include my correction:
For any fretboard taper, there are an infinite number of starting radii that will produce string lines that converge at the apex. However, given any starting radius (or radius anywhere along the fretboard) in addition to the taper will yield only one correct fretboard end (and bridge) radius that satisfies the above condition.
Even a radial surface with a sawtooth cross section or any shape), can produce straight string paths:
Louie, I think we are on the same page here
btw I am totally with you on the negative cone/ concave radius thing, rock on, no problem there ...
I think there might be a fundamental problem in basic conception here, inasmuch as I and some other posters are envisaging a perfect cone, much as Euclid would have done if he had been a luthier, and others (you included) are visualising alternative scenarios which nonetheless yield straight string paths, and of course there are an infinite (or maybe even greater) number of these alternative scenarios (scenarii ???) which bear no relation whatsoever to what we are designating a "perfect cone"
The problem, (if in fact it is a problem), arises when you bend the strings in the upper register. The true conic section, in my humble estimation, appears to me to provide the best potential for ease of string bending compared to any other scenario.
However, I wouldn't be dogmatic on that point, and if there is any experimental, practical evidence to the contrary, (which would involve world -class players playing on fretboards with varying compound radii) then I for one would be fascinated to hear about it.
Louie, you are quite right about the chord of the section changing in my cones - mine were just to illustrate the concept, and yours are more true to reality. The concept however, remains the same.
Your statement:"For any fretboard taper, there are an infinite number of starting radii that will produce string lines that converge at the apex. However, given any starting radius (or radius anywhere along the fretboard) in addition to the taper will yield only one correct fretboard end (and bridge) radius that satisfies the above condition." is exactly what I've been saying (assuming that by "starting radius" you mean radius anywhere other than at the vertex, where the radius is always zero).
This being the crux of this discussion, we should just leave it there, but I must disagree with a couple of your other statements.
Firstly, in post #45, you state the maximum radius is the distance from the apex to the fretboard end. I would say that the maximum radius is infinity, i.e. draw a full circle on a flat piece of paper and mark the centre as the circle as the apex. When you place your window with its point at the apex, you have straight string paths parallel to the FB surface, converging at the vertex.
I also disagree that you can have 10" nut radius and 20" bridge radius for a given fretboard and have proportionally spaced strings lie flat on the surface. It's the same as not being able to have strings lie flat on a cylinder.
Before you define your fretboard taper, just using the scale length and 2 radii, you do define a cone. Its apex will be different to the one that is defined by the fretboard taper.
If you use your window and place it anywhere on any of your cones in a way that the apex of your window doesn't coincide with the apex of your cone, you will see that you can't get proportional string spacing with the strings lying flat in a straight line on the surface.
Having said that, I'm not going to argue it any further as I've lost way to much sleep over this already, but will read with interest and an open mind any info which might change my mind.
have there ever been two more fascinating threads on MIMF ?
I would imagine that "nut compensation" would be the only other topic which would stimulate so much controversy ...
"Louie, I think we are on the same page here
Murray the best way to think about this is, think of an arbitrary fretboard taper. Take the two corners of the fretboard end and the apex. You have an isocoles triangle. Now take any cone whose slant length is the same as the legth from the fretboard end to its converging point or apex, and whose radius is between half the width of the fretboard end to the length of the apex to the fretboard end. No matter what value you pick in that range, you will always be able to inscribe that triangle in the cone. So yes, no matter what starting radii you use, you can always find the terminal radii that describes a perfect cone, with all lines converging to the apex.
I believe that Jim and Ken Warmoth took bending into account when they decided to use a 10"-16" compound radius. If your nut width is 0" then you should be able to bend anywhere and the string will always follow the cone. But since the nut keeps the strings spaced out, they do not pivot at the apex, they each create their own "cones" so to speak.
A way to visualize this is, let's say your max bending potential at the 12th fret, where it's the easiest, is 2 string-space-widths (Jimi could probably bend the whole board!) If you took that line to the 20th fret that's a 3.3 string-space-width spread in either direcction. Now if you tried to level each string along this "triangle" you you end up making the fretboard lower in the center. It would be difficult to make a fretboard this way precisely without an elaborate jig or CNC. The larger-than-correct fretboard end radius solves that to an extent. String height and relief take care of the rest. and it's really not a lot, since as you move up the fretboard the strings get higher and the break angle from the fret to the bridge increases.
By the way, ever wonder why a straight 12" (cylindrical) radiused board works well? Think of the 12" radius at the nut and fretboard end. Now if we wanted a true conical surface, using the same parameters above but changing the radius to 12" at the nut, the teoretically coreect fretboard end radius should be 15.428". (Incidentally my fretboards are intuitively close, as I use a 12"-16" radius.)
At the fretboard edge, the rise from the fretboard edge to fretboard center of a straight 12" tadius is .0588". On a 15.428" radius it's .041" So the difference in height at the center from "true conical" is only .017"! At the 12th fret it's .014"! If you're string height at the 12th fret is .063" it's easy to see that with just .008" of relief, the guitar would still play extremely well, even when bending. With straight 16" tadius (Martin?) it's even less.
"have there ever been two more fascinating threads on MIMF ?"
Murray, to me, it's more my love of math puzzles and geometry... and finding truth amidst the folklore!
"Firstly, in post #45, you state the maximum radius is the distance from the apex to the fretboard end. I would say that the maximum radius is infinity, i.e. draw a full circle on a flat piece of paper and mark the centre as the circle as the apex. When you place your window with its point at the apex, you have straight string paths parallel to the FB surface, converging at the vertex. "
Brian, if you make the end radius the same as the slant height of the cone which is the fretboard end distance from the apex, you have just defined a circular plane! If you define a radius past the fretboard end to apex distance, then the three points of the "taper triangle" could not possibly lie on the fretboard edge lines. Yes, the radius is infinity if you're point of view is right on the plane of the fretboard edge, since it's dead flat. The problem here is, when one cuts the fret slots, they're cut perpendicular to the cone's slant, NOT the true radius line. Take a 10"-16" compound radius board. Stand it on the end, and tilt it back 6 inches - that's the conical surface. Look where the frets point. Up, perpendicular to the surface, not the cone centerline. If you DID cut the slots to the cone centerline, then from the front the frets would have a slight curve in them. At 78.75" in this example, the curve of the frets will be exactly 78.75" radius, and the board will actually be flat!
"I also disagree that you can have 10" nut radius and 20" bridge radius for a given fretboard and have proportionally spaced strings lie flat on the surface. It's the same as not being able to have strings lie flat on a cylinder. "
I shall make a believer out of you! And BTW the surface would not be cylindrical but a complex surface.
"Before you define your fretboard taper, just using the scale length and 2 radii, you do define a cone. Its apex will be different to the one that is defined by the fretboard taper.
If you use your window and place it anywhere on any of your cones in a way that the apex of your window doesn't coincide with the apex of your cone, you will see that you can't get proportional string spacing with the strings lying flat in a straight line on the surface. "
I respectfully disagree. First, you would never define a fretboard as the length and two radii, because we don't build fretboards that way, we pick a taper and thus "fit" the radii onto it. I personally like to have a taper that follows proportionately to the string path, or as close to as possible, since I think it's beneficial to have more room between the string and edge as you go up the neck. It's a PITA when binding, but on my last build I tapered the binding as well so that the string lies exactly over the purfling and the binding taper follows the correct path. Secondly as long as the cones have the same slant height as my triangular window, then it will always coincide with the apex of the cone! It won't fit any other way!
I will add that the formula still works if you just want to consider the string taper, and then make the spacing at the fretboard edge the same. You just subtract the spacing on both sides to derive you new "nut" and "end" widths. But then you're fretboard edges won't converge with the strings.
Louie, you slay me ...
just when I think "hey, Louie's really got it down," then you come out with something that makes me shift in my seat ...
What is with this ..."take any cone whose slant length is the same as the length from the fretboard end to its converging point or apex, and whose radius is between half the width of the fretboard end to the length of the apex to the fretboard end"
Louie, what is a "slant length" ?? Cones are infinite, they extend from their *vertex* out to infinity. And cones per se don't actually have a singular, defining radius, they have an infinite number of radii. Are we talking about the "frustum" of a cone ???
I think that we really need to have a summit meeting to define terms.
And btw this post is not intended to be confrontational, Louie !!
Phew. :)
I would like to personally thank the combatants for the vigor with which they've put forth their arguments. :) Y'all have helped me identify a misconception that I've had about these surfaces, even though none of you ever mentioned it! And I'm not going to repeat my misconception here, either, because that just adds to the confusion.
Louie, I appreciate your comments regarding my jig plans. I
think
I've already considered the points you've raised, in various ways, and without adding more details and confusion, I think I've chosen the best path for my particular implementation. If and when I get it finished, and assuming it works well, I'll present it. It is the next step in making forward progress on my instruments, which is a
very
part-time thing for me, particularly these days.
Hi guys, I haven't been following this discussion, it's long and I have been busy, but I'd like to contribute with the way I see it and do it. I hope I make myself clear and that it's usefull to anybody.
Here it goes:
We need to know where the apex is, according to thales theorem, when two or more parallel lines intersect two angled lines the distances from the vertex are proportional to the segments leghts, so:
ln / lb = wn / wb
substituting: lb = ln+lnb
...
ln / (ln+lnb) = wn / wb
(ln+lnb) / ln = wb / wn
1+ lnb / ln = wb / wn
lnb / ln = wb / wn -1
then:
ln = lnb / (wb / wn -1)
We have located the apex of the cone knowing the width at the nut and bridge (it works the same for any other pair of places)
Now that we know the apex location, if we give the radius at some point the cone is completely known and we can use thales again. If we know the radius at the nut and want to know it at point a, for example:
rn / ra = ln / la
so
ra = rn x la / ln
That's it, that gives us the radius anywhere on that cone, just remember to use the distances from the apex. We could use the formula for the distance from the nut substituting like in the first case.
Murray, no prob! I want this to be a lively discussion. Otherwise it would be just one persons preaching and everyone else nodding blindly in agreement. And I don't want people to just nod blindly; I think we should all at least pause and think, and question what we don't understand, right?
The slant length on a cone is the length from the apex to the bottom of the cone, measured along its edge. The height of the cone is measured from the center.
The reason I don't define an infinite cone is that for our purposes we just need a cone whose slant length is the same as the apex-to-fretboard-end length. Or apex to bridge. The remainder of the cone is irrelevant to the equation!
Of you want to define that as a frustum then fine; but for my intents-and-purposes I'll end the cone at where I need! I think technically however, a frustum is a section of "cone" that may or may not include the apex.
Arturo, great diagram, pretty much sums up everything. Our formulae come to the same conclusions, tough we went about it slightly differently, they're related to each other.
So as rn/ra = ln/la, so does rn/ln = ra/la
great diagram, pretty much sums up everything
Yup. That diagram is what I've been saying all along, no? Why did it have to get complicated?
We were wondering when you were going to show up to the party!
Been out of town...
Murray, here are the definitions of a dome and frustum, as quoted from the fountainof misinformation, Wikipedia..
"A cone is a three-dimensional geometric shape that tapers smoothly from a flat, usually circular base to a point called the apex or vertex. More precisely, it is the solid figure bounded by a plane base and the surface (called the lateral surface) formed by the locus of all straight line segments joining the apex to the perimeter of the base. The term "cone" sometimes refers just to the surface of this solid figure, or just to the lateral surface."
"In geometry, a frustum [1] (plural: frusta or frustums) is the portion of a solid (normally a cone or pyramid) which lies between two parallel planes cutting it."
So Technically I was correct incalling my cone a cone. A cone that goes to infinity I suppose would be called a conical plane...
The fretboard I describe in my (and Arturo's) equations lies within the frustum of the cone we define; and is a section of the frustum.
Interestingly, Arturo's diagram could *not* produce both a fixed height at the fretboard edges and a fixed height down the middle of the fretboard.
In fact, that scenario would be impossible for any (non-flat) cone where the edges and the center line converge at the apex. For that to happen, those three lines would have to be either coplanar (a flat fretboard), or in parallel planes (meaning they don't converge).
That's true Andy! And neither does mine, since it's based on the same principles as Arturo's formulae. There is no conical surface (other than your flat cone) that would give you a constant edge and center height. But there are surfaces that have constant edge and center height, and a radius that grows proportionate to it's distance on the fretboard, and allow straght string paths along the entire fretboard, that are not conical. The conical surface is however easier to produce!
"I also disagree that you can have 10" nut radius and 20" bridge radius for a given fretboard and have proportionally spaced strings lie flat on the surface. It's the same as not being able to have strings lie flat on a cylinder."
Brian it's been 18 years since I sat in math class (I think it was differential equations) and I can't describe in math what pictures and props can. So...
Here's a little rig I threw together in 3 minutes. Basically I have a 5" radius at the nut, and 12" radius at the bridge, with a 24-1/4" string length. I didn't measure the taper, but I have a 3/8" string spread at the nut and 1/2" at the bridge. As one would expect all strings are straight (!) and are proportionately spaced.
A view from the bridge looking down....
Now I'll add 3 strings proportionately spaced in beterrn each string:
If I had string infinitely thin, I could go on "ad infinitum" placing the string infinitely close till I've basically defined a surface, of which one end has a 5" radius and the other a 12" radius, where strings spaced proportionately along the surface would produce a straight path...
The bridge and nut could be whatever radii you desire, or most any curve or shape for that matter - even convex to concave, or canted left to canted right - or 10" to 20" as in your example, you can create a fretboard insuch a manner, ehich is precisely what we do when we "true" a board up...
Just as a visual, i put a piece of wet rag over the strings to better see the "surface"
Of course; now, what's the radius at the end of the "fretboard"?
Mario, it would be directly proportional to the length. So if the end is at 17-1/2", and the scale length (bridge) is 24-1/4", then the ratio of the fretboard end to the scale length is 17.5"/24.25" or .721. The difference between the end and start radius is 12" - 5" or 7". Multiply 7" by the ratio obtained .721 and you get 5.051". Finally add that to the start radius, 5", and your radius at the fretboard end will be 10.051". The radius at the 12th fret is simply the average of the two radii, or 8.5".
Louie, your starting to make me doubt, but there's something in my head still questioning your premise. Imagining your nicely done rig but with the same radius at each end (i.e. conical) you could do the same thing with strings in straight lines with any chosen taper. However, we know that if you put a straight string at an angle across a cylinder, it will not be even with the surface of the cylinder. That makes me feel that there is something missing in the setup.
I'm going to have to rely on those more knowledgeable than I, and I'll keep watching this topic in the hope that the discussion will clarify the issue for me one way or the other.
I know what your math says, but take an actual, physical measurement at the 17.5" mark. Is it a true radius? Or another shape?
See, if I do the same math with a straight radius, it tell me it is correct, also, and we know it isn't. If you have a nut radius of 5" and the saddle radius of 5", the difference is zero. Anything multiplied by zero, is zero, and add zero to 5(the nut radius) and we have 5(what it tell us to use for the end of fretboard radius). But we know that isn't correct. Even if we use a minimal number, like .01(IE: 16" nut radius and 16.01" saddle radius), the math doesn't give us a fretboard we know works correctly.
Your jig works to show we -can- make a fretboard to work with any two given nut and saddle radii, but what it doesn't show is that it's no longer a cone shape, and not a shape we can easily mill and reproduce. I also wonder how far we could bend the strings on that one before they fret out?
Not arguing; I'm learning here, too.
Brian, Brian... I must correct myself. In figuring out a formula for your example , 10" radius at the nut to 20" radius at the bridge, I just proved that this surface is indeed a cone. A cone, that is skewed however. The proof is simple: given any two circles parallel in space, where one is smaller than the other, would define a cone. So let's say
s = 10" (your start radius)
e = 20" (your end radius)
Let's define the difference, d
d = e-s
Now let's plot each circle in three-dimensional coordinates. We'll start with the nut. The x axis will define the plane parallel to the nut, the y axis will go through perpendicular through the fretboard, and the z axis will go along the length of the fretboard. The circle defined by the radius of the nut will be at 0, 0, 0.
I'll define the ratio of any distance z along the fretboard length l as Q.
Q = z / l
The starting circle would be defined as
x^2 * y^2 = s^2
Now if we want the center of the fretboard to remain parallel to the z axis we must displace the circle at the bridge the distance of our difference d. But to factor the proportional rate of change at any given point z we must multiply d by Q. We must do the exactly the same thing to the start radius, as it grows proportionately till it reaches the end radius. This will define any point on the surface of the skewed cone:
x^2 * (y + (d*Q))^2 = (s + (d*Q))^2
On a regualr cone, the end circle is not displaced, therefore
x^2 * y^2 = (s + (d*Q))^2
The only hurdles I see constructing a fretboard in this manner is that it is not easily machined using simple shop fixtures, but could be approximated very easily using the Fox method of fretboard shaping.
Mario, thinking about this now, imagine my jig was actually a jig, where the two radiused ends sat on a flat, and in between is a big belt sander. If one would mark a centerline on the flats, and proportional marks like I have on the two radii, it would simply be a matter of lining up each set of proportional marks tho the centerlines, and you'd produce exactly that surface I represented! And that's how the Fox jig works...
I hear you on the learning thing. I posed this problem because I've thought about this quite a bit, and am learning too as well as throwing stuff out there for others to omment and learn also...
Brian you've brought up another excellent point about the two same radii. If you make the spacing exactly the same at the nut and bridge with identical radii, then most definitely you've defined a cylinder. But if you widen the space proportionately larger at the bridge, you no longer define a cylinder, It becomes some weird shape I can't begin to describe, but if could imagine a narrow strip of paper that you've brought the corners together on one of the short ends, keeping the lines straight...
Mario, on your example, 5, 5, and 5 would only work if your strings are parallel to each other. Like I said, I can't begin to think the math involved when you splay the strings, creating that convoluted surface...
Thinking about this again - dammit I think too fast sometimes - maybe for exact straight string pull on a cone the string spacing should also move in the same proportion as the radius... though I think the Fox way of fretboard shaping takes that guesswork away as well...
Now I can't go to sleep. I just woke up thinking about this. I had to re-evaluate an earlier example, concerning when given the fretboard taper, and thickness at the edge and center, one can figure out the radius needed for both ends, and using the method above can create a surface that satisfies straight string path. It was then asked whether this suface was a cone, citing that it couldn't be since the edge and center are parallel, and thus do not converge.
The answer I believe simply is that the fretboard edge actually falls on the intersection of two cones, of which one or both may be skewed. Regardless, everything should still move proportionately.
Let's say we have the same fretboard taper and length as above, 1-3/4" to 2-1/4", 17-1/2" length. Let's say the fretboard thickness is 1/4" and we want 3/16" constant at the edge. To find the radius, I'll use the intersecting chord theorem. Think of the chord of our circle as the nut, I'll define the ends as a and b. c will be the center of the fretboard at the top, and d will be the diameter of the circle. e will be the intersection of the two chords. Thus
ae * eb = ce * ed
And we know ab, so ae and eb are exactly half, and we know ce is 1/16"
So we need to solve for ed
ed = ae * eb / ce = .875" * .875" / .0625" = 12.25"
cd = ce + ed = .0625" + 12.25" = 12.313"
The radius is simply half, or 6.156"
At the fretboard end
ed = ae * eb / ce = 1.125" * 1.125" / .0625" = 20.25"
cd = ce + ed = .0625" + 20.25" = 20.313"
Halving that, we get 10.156"
Any ruled surface could work, the slanted cone idea seems interesting too. I guess I'll get to the cad program again...
Louie, the surface generated in your pictures above is a slice of what I would term as an "elliptical cone" , ie each radius on the surface (with the exception of the two radii at each end which are arcs of a circle by definition) is the arc of an ellipse, not of a circle.
So far, I think we have established that for any given fretboard taper there is one and only one cone which yields a true conic section for that particular taper, and that cone is the cone with the same angle at the vertex as the angle between the sides of the fretboard.
What I am wondering is why we would actually want to create any other surface to the fretboard other than the true cone ? I mean, has any guitar player ever played a true conical fretboard and said "hey, this sucks, I want a different contour ?"
If I can draw an analogy with the wheel, it is perfectly possible to imagine a wheelbarrow with a slightly elliptical wheel. It would be harder to push than a barrow with a circular wheel however, and the more elliptical the wheel becomes, the harder it would be to push.
The same applies to the fretboard imo. It is perfectly possible to create alternate surfaces other than the true cone, (as per Louie's elliptical surface above) and the string paths will still lie in a straight lie, but as soon as you start to bend the strings, it will become apparent that the true cone is the only way to go.
Murray, that's one point I'm trying to make: when we make a compound radius using whatever pivot method, and then "true" the string paths, unless you used the theoretically correct radii you youln no longer have a conical surface.
The surface above cannot be elliptical, since I used circle segments to establish the shape, and provided a formula for the actual surface that shows the sections are indeed circular. But yes, based or your point of view it could be considered "elliptical" or "ovoloid".
IMO, there are a few factors that would dictate how strings behave when bent. As I explained before, on any pivot-derived compound radius, or conical section, the fret slots are NOT cut along the radius; they are cut actually at a bias. So teh radius actually DECREASES toward the center of teh board. Not much, mind you, but it's there. So when you bend the string still goes "uphill". We actually DO play on an elliptical surface, and not even know it! On my surface, the circle sections are prerpendicular to the fretboard surface, and thus so are the frets, so the error is reduced - you are not bending uphill anymore, you're almost bending flat or even downhill depending on the end radius, which is what we want for bending, right?
Another way to look at this: Take a paper cone, rest its base on a piece of table. Now draw lines parallel to the table surface onto the cone. Now draw your fretboard. Open the paper up and lay it flat. You'll see that the "elevation" lines you've drawn are curved. The curves represent the radii at a given distance. If you thus cut the fretslots along this curve, you will not have any "elevation" change along the frets, but since the frets are instead cut perpendicular to the centerline they cannot lie on teh same "elevation".
Or, instead of making a fretboard, make a cone of the same radii you'd use for a fretboard your frustum.) Orient it so the surface is level, and make a perpendicular cut to the surface. You now have an elliptical section!
My conclusions?
1 - If you do NOT bend strings, any concical surface derived above would suffice. If you make the end radius flatter than "theoretically correct" you in essence make that end feel "narrower" than it should be, Thus related to this if you DO make the end radius narrower than "theoretically correct" then it might be a good idea to also make the fretboard wider in that area as well.
2 - Even if you used a cylindrical surface, at 12" starting radius the error is only .018" at the 20th fret, and if you thus dressed the string paths true you virtually eliminate that error, in essence creating the surface I describe above. Even if you didn't do that, depending on the relief and action you could STILL bend on this neck cleanly; people have been doing it for decades!
3 - On any conical surface derived in the conventional pivot method, you actually create an elliptical surface in the point of view of the frets. The "true" radius path is always a curved line.
4 - In general, I believe that a compound radius that goes flatter than the "theoretically correct" cone can help as far as bending, but as the end radius grows, so does the "curve error" I described; the board gets more elliptical, and string path trueing is still necessary.
5 - When we sand our string paths true (or use the shaping method I describe above) we "inadvertently" create a complex surface! In fact we approach a fretboard where the perpendicular sections more approach that of a circle rather than ellipse., which in my enlightened vie is a key to clean bending, and possibly better intonation.
Murray after sleepless nights mulling over whether this is worth the little extra effort, I say YES! We all know all our capabilites; at the highest level the playing field levels out a bit; they're a select few bordering on "perfection." If I could imrprove something by 1/20th of 1% it's a lasting improvement, since I only need to build the jig once. And I could offer something that is not normally offered.
I have just realized that my "flash of realization" may have been a trifle premature, and it may be time for another stroll into the woods.
Maybe I am miscalculating somewhere, but taking a nut width of 45mm, a string spacing at the bridge of 60mm, and a scale length of 650mm, (all of which are fairly standard dimensions) there is no way on earth that the cone generated by these parameters can allow for a compound radius of anything like 12" at the nut to 16" at the 20th fret. If you laid the fretboard template on the cone so that the nut coincided with the cone at the 12" radius point, the radius of the cone at the end of the fretboard would only be something like 12 1/8" or even less.
back to the drawing board... (or the woods)
Huh, you guys managed to answer my main question before I could get it typed properly. Ok, so *every* conical fretboard that looks like Arturo's diagram gives you an elliptical radius measurement, just by nature of the fact that we measure radius perpendicular to the fretboard, not perpendicular to the axis of the cone. That makes sense to me.
Louie, I don't think the surface in your jig is a cone at all, but I'm not sure it has a name. To the best of my knowledge, even an elliptical cone doesn't have rules that don't intersect the axis of the cone. As it sounds like you've figured out, even though the two circles at the ends do in fact define a cone, when you spread the strings out at the bigger end, the strings no longer lie on that cone -- they cut through it some. This same thing keeps all the string paths from intersecting in one place. I don't think it's a hyperboloid, either, which was my first guess.
Here's an exaggerated picture I made for kicks this morning, to try and expand your string experiment to see the shape the strings are tracing out:
http://dan.aleph0.com/fb-radius.jpg
http://dan.aleph0.com/from-side.jpg
This picture has 40 strings, with a 12" radius with 1/2" string spacing at the bridge and a 5" radius with the strings spaced 3/8" apart at 24 inches out. I extended the lines out another 4 feet past the nut so you can see that the strings do not all appear to be intersecting at a single point. Looking at the shape straight on from the sides, it doesn't look like this shape has straight sides, either so I'm pretty sure it's not a cone.
This is a neat thread. Who knew "sanding 'till it feels right" could produce such a weird surface?
I see where I was going wrong, and that was in assuming that the taper of the fretboard was the same as the taper of the cone.
It isn't, the taper of the cone is much wider, but at the moment my brain is hurting too much to figure out the math, although I have no doubt inspiration will come in due course.
This is the slanted cone version of the drawing. By the way an elliptical cone and a slanted cone are the same, one definition of an ellipse is the intersection between a cone and a slanted plane which is really what you get on the first drawing when you measure the radius perpendicular to the surface. After this drawing I think it's the same, the tapper fixes the apex and then any radius fixes all others, so it's the same... but better, here we'll measure real circles perpendicular to the surface which is also the way we cut fret slots.
So no cone gives you all you want, but ruled surfaces can have any shape at both ends and still all the "rules" are straight lines. I have in mind the surface we're after but will have to resort to the pro cad software this time.
Just got back from measuring a roof job, and realized that I made an error in my equation! This is what I get for waking in the middle of the night and posting... (mixing up a plus and times sign turned the formula into a parabolic surface!)
The equation for a cone, like Arturo's drawing above should be
x^2 + (y + (d*Q))^2 = (s + (d*Q))^2
On a regualr cone, the end circle is not displaced, therefore
x^2 + y^2 = (s + (d*Q))^2
Arturo, the volumes of the two cones are the same, and I think the surface area as well.
Dan, my thinking now is that this surface is the intersection of two cones, which may be elliptical or round. Think of s cone where you draw lines representing the string paths. Now create planes for each of these toward the "axis" of the cone. Now imagine another cone as a knife, cutting through a surface the same shape as your fretboard taper. You now have a surface, derived from a cone, with straight string paths that do not converge at the original cone's apex!
Murray I see where the confusion lies, and it took me a bit to detach myself from the notion of looking at pure conical shapes. Arturo's drawing explains my concept better than the math equations, though my equations would definitely define a surface with the parameters you lay out.
I'm coining an axiom to summarize what I've learned:
Do not use the surface of the fretboard to determine your string paths, use the string paths to determine the shape of your fretboard! This way, you'll never limit yourself to what you think is possible and what is mathematically (and therefore in my humble opinion physically) possible!
Dan BTW cool pics! There was no way I'm explaining that in words, but in my mond that's what I saw! BTW, if you could make a fretbard that way you'd still be satisfying the requirement of straight string paths!
I've always loved the beauty of complex mathematical surfaces!
sorry Louie, I deleted the post, as I felt I needed more time to think about it (and rightly so !)
Hehehe you made me delete mine! But here's a couple ways to think about it.
Imagine the cone were a stack of infinitely thin circles, attached in the center by an infintely thin ridgid rod. Now push the top of the cone, making a slanted or elliptical cone. Note the shape of the circle discs do not change!
Or think of your frustum, which has a circle on top and bottom. Cut it in half, that cut is a circle...
This is what I was thinking about, displacing the centers of the circles. Is it a ruled surface? (do the strings lay in straight lines?) some mathematics would be needed to demonstrate it. My hypotesis is no.
Yes, I get it.
If you push the cone, you end up with a slanted cone, and for any slanted cone there will be an angle at which you can cut through the cone and the resulting surface will be a circle.
The particular case we are interested in is when the angle which yields a circle is the angle of 90 degrees ...
Arturo, my guess is yes...
(edit) the strings spacing is proportional to the distance from the apex, just as the chords and radii are!
Imagine you divide the largest circle into 360 evenly spaced marks. Draw a line from each to the cone's apex. Now draw an infinite amount of evenly spaced marks... so now on each successive circle has those same proportionately spaced marks. (you have a few on there now.)
Yes Murray, in that instant the frets lie exactly on the edge of a circle, and the board at each location has a cross section of a true circular arc! In any other case the frets will not lie on a true circular arc, it would lie on a elliptical arc!
As far as bending, I would choose an end radius that's greater somewhat than what the "theoretical exact fit" suggests. This would in effect keep the strings from moving "uphill", or better yet get the strings moving "downhill." But as I mentioned earlier this may not be necessary for jazz, chickin' pickin', and other styles where bending is not done. Also once you get to a certain point at the fretboard, the string height would eventually trump any "uphill" bending motion, though to find that "escape" point would require a math I forgot long ago!
Arturo, another good picture!
Would it be possible to post the same picture, maybe in isometric view, so that we don't get the distortions that the perspective view shows?
Here's the iso view
An illuminating thread, indeed.
I have just realized that in a fretboard made from a conventional cone, the frets don't follow a true circular arc, whereas in the one above, they do.
I know you probably pointed this out earlier, Louie, but the penny didn't drop until now.
Murray, pennies are dropping for me as I realize more and more! I'm thinking now, it would not be too difficult to adapt a modification to a pivot-style shaping jig that would actually produce an elliptical cone that would produce fret paths with true circular arcs. In fact I changed my mind as to how to build my fretboard jig, and will build a conventional conical surface jig with this modification so that others interested in trying this could simply modify their own jigs.
Louie, if you can build a pivoting jig which will generate a slanted conic surface you are a better man than I.
I have thought about it long and hard (even before your post above ), and my conclusion is that it is impossible (except to generate a constant radius, ie a cylinder).
A "rolling" fixture, like you described above, is however, eminently feasible, and I am at present working out just such a fixture to use in conjunction with the table saw. (I already have an adjustable jig which generates a conventional cone, but I see no way of adapting it to generate true circular arcs) .
Murray the simplest way I know to do this is to actually produce the elliptical cone that will give you circular slices perpendicular to the one side of the cone.
To do this we would first need to find the length of the center of the cone. That should be easy, since the radius of the end we want perpendicular to the surface. So if our end radius is 'e' and our distance from the end to apex is de, then simply
tan(theta) = e / de
Now the other thing we need to know is the "long radius" of the ellipse formed from the point where our radius e touches the centerline c, going perpendicular from that to the surface of the cone, which I'll call ea
Since we know c, and theta, and ea is perpendicular to c, then
tan(theta) = ea / p, or
ea = p * tan(theta)
I'll call the difference between the two radii, ex.
ex = ea - e
Basically we need to create an elliptical cone with the long radius of ea and short radius of e. The easiest (though not necesarily simplest) way to create this would be to modify a pivot-style conical surface jig. Basically you would set the nut radius first, then either use the above formula to find the end radius, or manually adjust the end radius until both edges of the fretboard line up with the centerline of the router slide when moved. The final "trick" would be to make either an eccentric bearing or cam at the end radius arm that will displace the the pivot arm the distance ex as you move it left or right. To be truly accurate, yo'd need to do the same to the nut radius arm, displacing that point a distance proportional to ex, which I'll call nx. To find that is simple too.
nx / dn = ex / de or
nx = dn * ex / de
If you have the room, you can make the pivot arm a little longer than the length de, and fix one end at the theoretical apex, with a spherical bearing or universal joint. You would then need only one cam or eccentric bearing to displace the pivot arm. Check out my neck-shaping jig in the tools section; I use the cam primciple to create the neck shape on both ends, and it's the same principle I see working with this.
I think the two-cam method would be more practical because it takes up less space, and you could "tweak" the end radius. But I stress the importance of properly setting up the fretboard as I describe above FIRST, otherwise the router will not trace straight paths proportionately along the board.
You may have to tweak the above formulas depending on the geometry of the swing arms; i.e. whether they're perpenicular to the pivot arm, or perpendicular to the fretboard.
yes, I see where you are coming from.
it's really an adaptation of the traditional trammel method of producing an ellipse.
instead of an "eccentric cam"", you could quite easily construct two endplates with intersecting channels at right angles to each other on which each trammel (connected at either end to the router carriage) could move. Once you have worked out the major and minor axes of each ellipse, setting the pins in the trammel is simplicity itself.
yep, I see that it could be done after all.
Great thread guys, thanks for the discussion.
Things I knew, and now have confirmed by this thread (and hopefully helped others with)
-For a straight string lie on section of a cone, the fretboard taper and radius at the nut will determine the radius at any other fret.
-Any cone other than that defined by the taper and one radius will give uneven space between the string and the fretboard.
And, with all your input, I have learned that:
- straight string lie can be obtained for any two radii (at nut and bridge, or nut and last fret for eg), however the resulting shape will not be a cone. In fact, the shapes at the nut and bridge or fb end don't even need to be sections of circles - they can elipses or waves or other undefined shapes. Thanks for helping me get my head around that.
I have come to the conclusion that in reality, we can get really close with the maths, but at the end of the day the slight differences between the some of the shapes are relatively minor compared to the changes with different string gauges and tensions, neck relief and so on that are beyond the realms of worthiness to calculate (at least with my maths ability).
I think that to compensate for the problems, we just can level the frets in the line of the string pull - that way we know that the fret tops under the string are ideal. In the past I've used an aluminium L with coarse sandpaper to level the frets in under each string. (see photo below) The guitar is strung up to pitch with an extra high nut and the sandpaper part of the sander placed under each string and the frets are levelled string by string. On this guitar I got the action below 1mm at the end of the fretboard (21 frets) with 13-52 strings. I don't think I could have got there without that levelling in line with the strings compared with the cylindrical fb I started with.
Murray, yes! Only reason I suggest the cam or eccentric bearing is that on the conventional trammel method, the pin may not ride smoothly or accurately on the "x" axis, unless you machine it to a high tolerrance. Using a cam allows gravity to "remove" the slop, and wouldn't require as accurate making. (especially in MY shop!)
Brian, when you pick the two radii you in essence define a cone, I'll make a drawing later to explain my "intersecting cones" theory, and it should illustrate it clearly.
I also feel that if one would only need to know he start and end radii, and the jig takes care of the math. I also postulate that making a cone with an end radius close to "correct" will produce the least fret path error possible on a true conical surface. The error increase as the end radius increases, though I also postulate that on a "slant" cone we can manipulate the error in our favor (downhill bending)
But yes, very good achievemen with your technique, you're a human "PLEK" machine! But it illustrates how simply one can make a complex surface!
Brian, here is a crude drawing of my 'intersecting cones" theory.
Imagine two different cones, whatever slant, with circular cross-sections. Now imagine both cones intersectiong at the same starting radius, with the end radius of the larger cone going through both points of the fretboard on the smaller cone. This is what I believe the geometry of my "rig" above describes.
As you can see the resulting surface would still fulfill all the requirements of proportionately growing radii, and straight string paths. And, indeed, derived from a conical surface! But the string paths would not converge of hte apex of the fretboard taper, they'd actually converge at the apex of the "intersecting" cone... Or, the fretboard edges do not converge at the same point as the string paths.
Thus you have a conical derived fretboard surface, whose fretboard edges and string lines converge at different points!
Arturo, I actually drew thos with a sign-making program instead of CAD, since my CAD skills really suck! ButI think with CAD one could easily demonstrate how this works, whther my end radius ea is larger or smaller than the "ideal" cone deawn...
Looking at this now has given birth to another theory... Using CNC, onw could actually produce a board with relief in such a way that would give you as close to the same break angle from the fret to the bridge at any fret! Now that would be cool!
Brian, 13-52's on an electric! Your hands are stronger than mine for sure!
Louie, I think Brian may have been referring to the situation which we discussed earlier, re the "slanted" or "elliptical cone".
Looking at your drawing, it occurs to me that there may be a very quick "down and dirty" way to calculate the fretboard end radius, for the conventional cone you depict.
If you divide the circumference of the "nut radius circle" by the nut width, you will come up with a figure, which I will call the "divisor" (although strictly speaking it should be called the "dividend", but no matter)
taking a 10" radius and a nut width of 1.75", the "divisor" works out at (2*PI*10)/1.75 , which is 35.9039. This is the number of nut widths we could get out of this 10" radius circle.
Now, it is obvious (at least I think it is obvious, I hope I am not wrong here) that the correct radius for the end of the fretboard will be the radius of the circle whose circumference will yield
35.9039 fretboard end widths.
So assuming for the sake of argument that the fretboard end width is 2.25", then the equation reads C/2.25=35.9039, where C is the circumference of the "fretboard end circle", and the equation simplifies to C=35.9039*2.25, which is 80.7838. Dividing this circumference by 2*PI gives the required radius, which turns out to be 12.85". This is rather less of a radius than you came up with, so I have a feeling I must be mistaken in my calculations somewhere, but I cannot see where ...
The problem with your equation, Murray, is that you have to take into account the length of the fretboard, because that along with the nut width and end width is what determines the rate of change! (See post #1)
So let's define all the parameters we know:
nut radius = 10"
nut width = 1.75"
end width = 2.25"
fretboard length along its edge (it actually doesn't matter it could be the centerline, the proportions are the same) = 17.5"
The rate of change of the taper is the difference of the nut width divided by the fretboard length. So
(2.25"-1.75")/17.5 = .02857
The distance from the apex to the nut point dn is thus the nut width, divided by the ratio rate of change
1.75/.02857 = 61.25036 (compare that with the result I got in post one, an error of 3/10000 of an inch!)
Now you can use that for the end point to apex de as well
2.25/.02857 = 78.75393 (or an error of 4/1000"!)
The error comes from rounding the ratio rate of change number.
So to find the end radius, you only need to know:
rn / dn = re / de, or
rn * de = dn * re. Solving for re:
re = (rn * de) / dn
And this works regardless of whther the cone is slanted or not! (to an extent of what's humanly possible at least!)
So now let's say you want 10" radius at the nut, and 16" at the fretboard end. Given a cylindrical cone, the rate of change is
6" / 17.5 " = .34275
Another way to put this is, from the apex to 17.5", the radius at that point is 6".
To find the apex to end distance de we would use a ratio again:
.34275 = 16" / de or
de = 16" / .34725
de = 46.07631"
ds = de - 17.5 = 28.57631"
Double checking:
10" / 28.57631" (should equal) 16" / 46.07631"
.34994 and .34725 (or 1/400", again the error in rounding)
I should note, that trigonometry is really in essence a mathematics of proportions, and as such, one could eliminate the angles and just deal with the proportions.
So then what about angles? Let's take the above example. We know that at 16" radius, our cone is 16" in diameter, so
sin(theta) = 16" / 46.07631" = .34725 which is exactly the rate of change of the taper of the cone! So:
sin^-1(.34725) = 20.319 degrees
So if you set your conical jig with the pivot arm 20.319 degrees from the center of the fretboard surface, you should be able to create a conical surface of 10"-16" radius within the confines of your fretboard dimensions.
Louie, the intersecting cones theory wont work, the surface you describe is simply a part of the second cone. I think that a cone simply can't make what we want, we're asking for a surface made from lines proportionally drawn from one arbitrary circle section of arbitrary radius to another, that's only a cone if all the lines converge in a point and that sets the relationship of tapper and radius we already know. What we need is a ruled surface with whatever extreme profiles we want and a jig to make it, or a cnc machine. In the end, does it matter? that's what the cad program (or a lot of mathematics) can tell us.
This is an interesting pic of a ruled surface roof in Gaudi's Sagrada Familia school: http://www.gaudiclub.com/ingles/I_VIDA/fotobras/escoles/escoles11.jpg
By the way, the drawings where made with sketchup, it's free and great for this kind of quick 3d drafting.
Arturo you can define any cone as two circles in space where one circle is smaller than the other, because if you drew a centirline through both circles, you could rotate (transform) either circle until it's angle matched the other!
Here's a way to look at it. Make a cone-shaped cake. Now make 6 equal slices. This will define the string paths. Now take another cone (a birthday hat) cut it in half, and horizontally slice the suface ACROSS the cuts you made on the cake, following the birthday hat's radius. I guarantee you that on the resulting surface those cuts still form a straight line!
I don't know the math to prove my theory, but if you try my cake experiment, even in cyberworld using SketchUp, you'll see what I mean... it's the SHAPE of the second one relative to the first cone, i.e. the orientation of the circle planes relative to the surface, that gives me what I want!
Or put it simply. Look at your last drawing. We know the centerline and lines you've defined the fretboard edge are dead straight! If you then move the ends of the lines an infinitely small amount keeping the apex center, the lines will remain straight. And you can do this ad infinitum till you get to the other side of the cone, thus tracing a radially ruled surface!
Or think of it fron a bird's eye view, looking from the top of the cone. The lines would be straight, and if you make an infinite amount of lines you have just defined a circular or elliptical plane!
May I suggest you do the math according to fretboard width OR string width, but not mix them? A 1-3/4" wide neck is a center to center string spacing just under 1.5", for example. So, use 1.5" wide at the nut, and 2.250" or 2.312" at the saddle. The fretboard width and taper is inconsequential to the radii, but the radii are dependent on the string width taper.
The lines on the hat are either not straight or not proportional.
My last drawing is not a cone, the circle centers lay in a curved line, so it's some kind of wobbly cone.
In a cone, the only straight lines are those that pass through the apex, the rest are ellipses, parabolas or hyperboles. In your intersecting cones the surface is a section of the second cone, and the "strings" don't pass through the apex of this cone, so they're not straight lines.
We simply need a way to slide a guide on any two profiles in a proportional way and pass a router lengthwise through this guide, that will give us the ruled surface we want, but if we want a cone the tapper and one radius define the rest. I hope I have time to sit at the computer and model the thing on cad soon.
Your last drawing then is incorrect.. I already described a way of achieving the "ruled" surface above. Draw the cone first, then derive the sections and lines from the cone. Or draw any two circles in space as I mentioned above where one is smaller.
Mario, I don't intend to mix fret board or string spread width. But one could just as easily input the string spreads and achieve a proportional result! I taper the string "inset" as well on my builds, but for those who have constant inset, you would just subtract both sides to get the string spreads, derive your radius, and them add them back afterwards. No matter, you could still use the formulas.
"In your intersecting cones the surface is a section of the second cone, and the "strings" don't pass through the apex of this cone, so they're not straight lines.:
Wrong... the strings pass through teh apex of its own cone. and the fretboard edges to its own... once you establish the conical surface you can make the fretboard edges any shape you want!
Point is, Louie, the math you already did doesn't jibe with the real world because your math doesn't follow the real string taper, which is the true taper you need to get everything correct. Use the string spacing's taper as your constant.
Just FWIW, the 'worst case' deviations from proper on a normal FB (1.5" string spread at nut, 2.125" at the bridge) are about 0.021" at the end of the fretboard on a 10" radius and 0.014" on a 16" radius. Not huge in the grand scheme of things, but every little bit matters.
So far as having the break angle the same over all frets goes...that's what happens when you take away all relief and lay the strings across the frets, so it only occurs in the degenerate case (and that makes the instrument somewhat unplayable :) )
Arturro, just to re-emphasize, the centers do in fact lie on a straight line.
The situation is entirely analogous to the 2D sketch below. The deformation of the cone from a true cone to a slanted cone in no way alters the fact that the centers will still be in a straight line.
All the horizontal lines are bisected by the line from the vertex (or the "apex" as Louie likes to call it
sorry, forgot the attachment ...
Mario, its very simple to do it your way and still come up with the right answer. So then let's say your fretboard is 1.75" at the nut and 2.25" at the 20th fret. We'll use a 17.5" fretboard length. You want 1/8" string offset from the fretboard edge. So then your effective "nut" and "end" widths are 1.5" and 2", right? We'll use the 10" nut radius again.
We figure out the difference, .5" and divide our end width 2 by .5 and get 4. Multiply that by 17.5" and you get your apex to end length. or 70". You now know the apex to nut distance is 52.5", so:
52.5" / 10" = 70" / re
re = 70 * 10 / 52.5 = 13.333", your fretboard end radius.
Only problem with this, is if you cut your fretboard to size, then shape it in this fashion, the fretboard edges won't be flat. Of course, if you bind the edge AFTER shaping the board, you'd obviously follow the plane of the outside strings.
Arturo, all strings lie on a straight line. Everything is moving proportional, so by definition it HAS to! I didn't see the curve before, thus by definition your drawing is not a cone - at least, not a proportionately growing cone, but an exponentially or logarythmically growing cone, wither of which we couldn't or wouldnt want to produce.
Bob good info. I had demonstrated earlier that on a 1.75"-2.25" fretboard with a length of 17.5", if you made a straight 12" cylindrical radius, your marginof error at the 20th fret from ideal is only .018... Sure we dont want to remove over a third of the fret, but it's easily removed by "trueing" the string paths...
Murray, good drawings! Now, lets use the drawing at the left. Let's overlay this on a graph, giving the cone base the x axis and the cone centerline the y axis. The base will be the end radius. Let's say it's 13.333". Let's say the fretboard end to apex is 70". The angle from centerline to slant line is:
theta = sin^-1(13.333"/70") = approximately 10.98 deg.
The height of the cone is therefore
h = 70" * cos(10.98 deg.) = 68.719"
The ratio of y to x is
68.719"/13.333" = 5.154
We thus have an equation for the side of the cone.
y = -5.154x + 13.333 (x >= 0 and x <= 13.333) which we know is a straight line.
All line originating at any point on the base and ending at the apex is a straight line, and can be defined by an equation similar to the one I just demonstrated.
Louie, in your diagram in post #107, by slanting the cones in relation to each other, (i.e. the axes are not co-incident), the plane for a circular radius for one cone at s is cutting through the other cone at an angle, so the radius there must be an ellipse. Therefore, there cannot be a true intersection for the area of the fretboard. Put it this way, if you make a really wide fretboard - say 2(s) wide, the narrow cone is clearly away from the wide one at the fretboard end. As the fretboard gets narrower or closer to the nut, the difference is smaller, perhaps to the extent of being insignificant.
BTW, 13 to 52 jazz flats with a <1mm action are not so hard to play.
Brian, I see where the confusion lies. I'm not suggesting that any two cones that intersect will produce a straight line at the intersection. Maybe my drawing is a bit inaccurate, but I'm trying to illustrate that both bases lie on teh same plane. What I'm trying to get at here is this. Take the first cone, with the fretboard I inscribed onto it. Now create a second cone, one that I can shrink, stretch or squeeze as I wish, but in doing so keep its proportionality, just as if you've selected it in a CAD progam and done so. Now give the second cone a larger base radius and place the base on the first base such that the second base touches both points of the fretboard end, just as the first cone. Now squeeze down the cone until the point where both cones have the same start radius on the same plane. You'll find that on the second cone it's offset back from the first cone. Now push the top of the second cone until both circles align. The fretboard now has the shape of a slanted or eliptical cone. Since that surface also moves proportionately, all proportionately spaced lines must be straight as well.
The problem in conception is that the surface of the elliptical or slant cone surface behaves differently than a straight cone, It's hard for me to explain, but I see it as clear as day, and I assume others could as well. Think of the rig I showed earlier. It represents a slant cone. The fretboard I depict in the above drawing is akin to the surface in the rig. So yes it's a special case, but a special case that believe can be manipulated to our advantage! What I look to do is create a compound radius board in such a way that eliminates the error of fret shape I've discussed above. Again, this is most prevalent when the end radius is significantly greater than the "theoretically correct" conical radius, but it's there on any circular cone-derived fretboard.
The only case I can think of where two regular circular cones can do this is if they're exactly the same.
Mathematically you cannot draw a straight line on the surface of a cone that doesn't pass through the apex, the surface on your drawing is a section of the second cone, and its borders don't converge on the apex, so they're not straight lines. The intersection of the two cones are not straight lines. In my second drawing of my post #90 the cone is elliptical and the fret slots circular, yet you cannot freely chose radiuses and tapper, they're linked by the fact that it's a cone.
It's interesting that nobody seems interested in the ruler surface, which we could easily produce with all our requirements.
Oi, I haven't been around here for a while. I briefly skimmed most of the posts, and there's a lot of fun and interesting math going on. I'm sure it's already been summed up at some point or another, but in practical terms if you want to use a true conical section that results in a straight line under each string, it doesn't need to get that complicated. All you need to do is ensure that the radius remains directly proportional to the string spacing.
(Radius A / String Spacing A) = (Radius B / String Spacing B)
Or (Radius A / String Spacing A) * String Spacing B = Radius B
So to toss in some arbitrary numbers, if you want to start with a 12" Radius at the nut and have 1.5" string spacing, and end up with a 2" string spacing at the end of the board, then
(12/1.5)*2=16
Shazam. 12" radius at the nut, 16" radius at the end of the board, true conical section, each string over a straight line. Though very academically interesting, there's really no need to make it more complicated than this in practical terms.
Of course there's really no rule that says you must follow a true conical section in order to get a straight line under each string. You can start and end with any radius you please, and in proper leveling of the board and frets end up with a straight line under each string. You just wouldn't end up with a true conical-shaped board, and would have no more than two points along the length of the strings where the board or saddle beneath them would form a theoretically true radius.
Interesting stuff, and I am in no way trying to discourage academic arguments (I love this stuff myself), but in practical terms it doesn't need to be very complicated to make it perfect.
Arturo the point I've been making all along is that given two radii, as long as the surface and radii grow proportionately, you have in essence defined a ruled surface. Such is the case with the rig I used earlier to demonstrate the fact.
I understand what you're saying, yet I'm trying to explain what I see in my head. In fact the whole reason I started this thread was to show the potential errors in shaping fretboards the conventional way, which produce straight lines at the fretboard edge, remember?!
David that's incorrect, because you need the distance between the nut and the fretboard end to determine the taper! You must take the taper into account to get the "true" end radius. Shazam!
My argument is that when you pick an end radius that doesn't coincide for the taper, you do not end up with straight string paths. And even when you DO find the correct rasius, the frets do NOT lie on a circular plane, they lie on an elliptical plane. I thus came to the conclusion that a slanted cone surface is the only way to have both straight string paths AND circular fret paths. I postulate that you can choose ANY end radius (equal or greater than the "correct" end radius) provided you tilt the cone in such a way that the frets lie on a circular path.
Can we go back a step Louie? Why do you say that the frets do not lie on a circular plane? The frets are cut like slices through the cone, perpendicular to the axis of the cone. The radius of the fret tops is larger than the radius of the fingerboard by exactly the height of the frets, but otherwise, just follows the radius.
If you have a circular radius at the nut, a circular other radius at the fretboard end, and a cone that joins them (whether or not it's the ideal cone for the taper), and the frets are cut parallel to the ends of the fretboard, then the frets must also be on circular arc.
Brian, I'll refer to Murray's diagram above. In each case, we'll consider the right side of the cone as the fretboard surface. On the cone at the left, you can see that the radii are perpendicular to the centerline of the cone, or at an angle to the fretboard surface. Since we do not cut fret slots in that manner - we cut them perpendicular to the surface of the "cone" - the fretslots thus do not sit on a circular path; they instead sit on an elliptical path.
Now look at the cross section on the right. As you can see, since the radii are perpendicular to the surface of the "cone" or fretboard, your fretslots will lie on circular paths. But from the view of the cone centerline, the elliptical sections are at an angle to the cone surface on that side.
You are correct in what you say, provided you shape the fretboard in the manner we described earlier. If you just use a pivot-style radius jig, even if you find tht "true" end radius, you'll still encounter the "error" I mention. But I postulate (and shown above previously) that you can figure out the "tilt" you need for any given end radius that will produce an elliptical cone, where the circular cross sections are prerpendicular to the fretboard surface.
*edit* Reading your post again.... Yes, if you slice a slanted cone as Murray drew above, you will achieve circular cross sections, but if you look at the left cone, we do not cut fretslots going along the radius line, we essentially cut them at a bias..
Louie, I'm not entirely clear on your meaning in "you need the distance between the nut and the fretboard end to determine the taper! You must take the taper into account to get the "true" end radius." Are you talking about the height of the string above the fingerboard at the nut? If this is what you're referring to, I can see the reasoning but think you would be better served by looking at the fret height (which in practical terms would be the same as the nut slot height above the fingerboard, just using a more easily measurable reference). In this case, yes, it would be more accurate (only realizable in the abstract) to refer to the radius of the surface of the frets than of the board. This would change my formula listed above to:
(((Board Radius A + Fret Height) / String Spacing A) * String Spacing B) - Fret Height = Board Radius B
or if we want to start with a 12" radius and 1.5 string spacing at the nut, and have frets with a .045" crown, for the actual shaped board radius at the end of the board (with string spacing measuring 2" at that point) you would have
((12+.045)/1.5)-.045= 16.015
So .015" difference in the end board radius when you consider the height of the actual playing surface above the board, which though interesting in the abstract is not really noticeable in a final product.
If by saying you must take the taper in to account you mean in terms of angles measured in degrees, you really don't. Go by the simple rule of keeping the keeping the radius of the playing surface directly proportional to the string spacing, and you get each string above a straight line of a conical section. The actual degree of taper can be factored out of the equation.
If you move far beyond the practical in to the pure abstract (which I think this discussion did long ago), you may be able to argue some slight modifications to this, but these factors would be entirely dependent on the completely unreliable factor of at what angle the player depresses a string toward the fingerboard. Are they pushing toward the center of the cone, or perpendicular to the average plane of the neck? This would be really, really pushing the limits of theoretical, academic-only interests, and have no realizable effect on the final product.
Plus, I don't think I've seen anyone talk about introducing relief yet. Adjust the neck from straight in to relief, and each line under each string does not bend the same, as the neck is adjusting perpendicular to the plane of the board/neck joint, or the center of the board (roughly and in theory anyways), and not shifting exactly the same way at it's edges.
There are a lot of different scenarios that can be argued as to the theoretical shape of the frets at any point, whether they be elliptical or radiused, how slots are cut, etc. In theory, if you cut the board on an axle/pendulum radius sander then cut the slots perpendicular to the bottom of the board, then no, even though the board will be a perfect conical section, no fret will form in perfect theoretical radius (though you'd be hard pressed to actually measure the difference between an ellipse and radius here unless we were making it in to a lens). Follow the rule of constant radius to spacing ratio though, and you will get a straight line under each string, regardless of whether you call any point an ellipse or radius.
Louie, you wrote:
Since we do not cut fret slots in that manner - we cut them perpendicular to the surface of the "cone" - the fretslots thus do not sit on a circular path; they instead sit on an elliptical path.
I cut the slots perpendicular to underside of the fretboard. I then develop my cone. The result looks like Arturo's diagram in post #90. If the situation was like the cross section on the right of Murray's diagram, the end of the fretboard would not be perpendicular to the top of the fretboard. i.e. the nut would slope.
The diagram on the left looks wrong, because the taper is so exaggerated, if we imagine the bottom surface of our fingerboard parallel to the axis, our mind looks at that and thinks that the fingerboard can't be so much thinner toward the nut than the end. In reality, the cones we use are so long and narrow that the difference in thickness at each end is so small we don't notice it. Don't forget the apex of the cone on a fretboard is about 30 inches behind the nut - that's one long cone!
It's simple to know that the radii at the nut and end of fretboard are circular cross-sections, not elliptical, because we measure them and produce them with circular radius gauges. And if our fret slots are cut perpendicular to those two circular arcs defining a cone, then all our frets must be cut as circular arcs. Again, looking at Arturo's diagram in #90 demonstrates this.
And just as a reminder, you don't need any true conical section to get a straight line beneath each string. Make a 12" radius at the nut, and 12" radius at the 12th fret, and if the leveling process is done carefully you can easily achieve a straight line under each string. You just end up with the nut and the 12th fret being the only true radii, with the frets in between those shaped in to a steeper non-radiused curve (would they be elliptical, parabolic?? dunno..) and the surface of the frets beyond the 12th flattening out. It's very, very minor, and simple enough to do in the leveling of a board or frets.
Of course again, it's all abstract. In practice I prefer to abandon these theoretical shapes all together, and in some cases intentionally shape the board differently under the bass strings than the trebles, as well as differently in the main section of the neck than the extension or upper frets. Bass strings can benefit from more relief than trebles for many setup scenarios, as well as fall away or an exaggerated flattening of the "radius" on the upper frets. Theory is fun, and understanding can certainly help you in problem solving and achieving more consistent results, but in the end there are a lot of other things to compensate for that take priority in practical terms.
David, I realize that one can make any proportional surface that produces straight string paths. I've alrady gone over this many posts ago, and actually rigged up a display to show here. The fact remains that if you make a conical fretboard using a pivot-arm method (which is normally the case) then no matter how you machine the board, the strings will always go uphill when you bend them because the surface of the board in relation to the fretslots is actually an ellipse. I then propesed a mechanical approach that takes this into account. But this was over 30 posts ago, and the debate now is how to describe the fretboard you refer to, which is in fact a conical section.
Brian, as I mentioned any pivot-arm means of producing a compound radius will always have an "ellipse" error, regardless of whwther you use the theoretical "correct" end radius. Murray's left diagram exaggerates the error, yes, but it does show that there IS a nerror there. If you look at Murray's diagram, you would simply cut the nut and fretboard end along a parallel, and the cut is straight not sloped. At least to the extent we can make it. On a pivot-arm generated compound radius board, the nut slots, and end are actually sloped to the true radius lines. Murray's diagram is simply a cross section of Arturop's diagram, which is IMPOSSIBLE to make using a pivot-arm jig because Arturo's diagram in post 90 is actually a SLANTED or ELLIPTICAL cone, of which a circular cross section is derived. So unless you can create an ELLIPTICAL cone, you cannot cut fretslots that coincide with the circumferences of the circular cross-sections.
I concede that the error is small. I've gone over this in post 57. Even the error between a 12" conical and 12" cylindrical is quite small. But I've already said this is a refinement that I personally look to achieve, and hopefully share.
I'm not saying this is the end-all-be-all way to do it, and billions of dollars have been made by the people who make AND play the instruments built with fretboards made the conventional way. But if you have some down time or R&D time, the modification to a pivot-arm jig is not that difficult to do - the concept is actually simple. So I feel it could be a practical thing for some, but not all. Personally, I don't have any dust collection system in my shop other than the Shop-Vac, so any fretboard shaping that involves a wide belt sander is out. I've done it by hand with my scraper plane, but to me I'd rather do what's most accurately done by mechanical means in that way, instead of tuning by hand and checking with a straight edge, so I can spend more time on the aspects that I normally find difficult.
another way to visualise it is this : the frets follow the shortest path between two points on the cone which are equidistant from the vertex of the cone (or "apex" if you prefer), right ?
a moment's thought will bring about the realisation that the shortest distance between these two points (on a conventional cone) cannot be the arc of a circle. It's impossible.
on a cylinder, the shortest distance would be the arc of a circle, and on the slanted cone, the shortest distance is also the arc of a circle, as long as the two points are also equidistant from the "vertical" line on the surface (see earlier diagram ).
Ok, my conclusion of all this is that we can make flat fretboards, cylindrical ones in which the strings tapper could give some action problems, conical ones, either normal or slanted/elliptic in which the tapper defines the radius variation and the slanted one give's you really circular fret slots (but I guess the difference is so small it wont matter), or a ruled surface with any arbitrary end shapes.
In practice my choices would be flat, cylindrical and ruled, the two conical methods seem too much of a trouble to calculate and jig up, compared to the ruled which can also give us conical, I was thinking of building a jig for making conical fretboards and now have changed my mind towards the ruled surface jig.
I think many of us have learned from this discussion and we all pretty much agree. This is what theoretical discussions are good for, to establish common knowledge and draw practical uses. What's the next theoretical topic?...
Arturo, I have to say I agree. It's all a matter of point-of-view. Two people standing in two separate locations looking at the same thing can perceive that thing differently. Heck, being human, they could stand at teh same exact place and have different perceptions!
I also am with you on the idea of using a ruled way to derive the surfaces, while moving the board proportionaltely on an arc based on the string splay.
I thank all who participated in this thread, and it's refreshing to have a vibrant discussion that at no point got ugly or personal!
I leave you all with a srange twist on Greek mythology:
After Xeno lost his race, Zeus condemns him to building a guiar for him. Zeus tells Xeno he can go on with his life upon competing the guitar.
The only stipulation is that Xeno must make a fretboard that extends all the way to the bridge, and install every fret that occurs between the nut and bridge.
If it takes Xeno an average of 1 second to install and dress each fret, and he can make the fret as small as needed to fit, how old will Xeno be when he completes the fretboard if he started on his 35th birthday?
I truly understand the desire for maniacally precise machining, but do keep in mind how terribly miniscule the differences are that we are talking about here. Let's look at about the most extreme example you're likely to find. Take a 7.25" constant radius neck, 24 frets, with a dramatic taper of 1.375" string spacing at the nut to 2.125" spacing at the bridge. If my quick math is correct (I've been at the shop since 4 AM, so it may be a bit foggy), if you adjusted the neck to dead straight in the center from the nut to the 24th fret, you would end up with about five and a half thousandths backbow in the center of the outer strings. Not an insignificant amount, but keep in mind this is all variables at their extremes (but still part of the reason I don't care for constant 7.25" radius boards).
Come back to the more reasonable, say a 16" constant radius board, with 1.5" spacing at the nut and 2.125" at the bridge, and a 20 fret board adjusted dead straight in the center, and now you're down below one and a half thousandth backbow at the center of the outer strings. Of course that's with the theoretically straight neck from nut to 20th fret. Consider the same board leveled with a bit of fall away at the extension, and adjusted straight in the center from the nut to the 12th fret, and you're looking at about three-quarter of a thousandth backbow under the outer strings between the nut and the 12th.
By this point, even with a constant 16" radius board, we are at least bordering on, if not well in to the abstract of machining a board. Your methods of glueing the board to the neck, glueing the neck to the body, fretting, and any dressing of the board or the frets you may do will play a more dominant role than this. Move in to just about any reasonable compound ratio, and the differences will be even less significant.
My point about the constant spacing/radius ratio is that even if only important in theory, it will reliably create a perfectly straight playing surface beneath each string. Yes, if your fret slots are cut perpendicular to the bottom of the board each fret will form a theoretical ellipse rather than radius, but regardless of how their end shapes are defined they will still form a perfectly straight line under each string. You just won't be able to call them a perfect theoretical radius, but the amount by which they would differ would be better discussed in millionths of an inch rather than thousandths, which of course is entirely abstract in this application.
I applaud your efforts for seeking the perfect method, and greatly enjoy the mathematical riddles. In terms of any practical benefit however, there is really just none to be realized in improvements over the axle-style radiusing tool set up to the constant radius/spacing ratio. This is true even in the most abstract sense in terms of establishing a straight line beneath each string, and the only benefit you have to gain is in making each fret shaped to a theoretically perfect radius.
I tried to find your description of the mechanical means of achieving this goal, but wasn't sure which post it was. One that has been discussed before is to shape the board with a router setup, run on rails along the length of the board and guided on radius forms on the ends (or by moving the board across it's width on these radius forms beneath the router). If as you moved the rails across the board you made sure to move each end in line with the string spacing (possible to do by belts or gears if you set it up strategically), this would deliver your goal of a straight line under each string and true radius at each fret. Maybe that's what you already proposed, and I just couldn't find it.
In any case, the tolerances of such tooling would render most of the precision moot, as even if you set it up on a sturdy base with a fresh cutter on a Hardinge mill, creating end lines to half a thousandth tolerance on machined wood would be a pretty lofty goal.
All great ideas, and worthy of considering whenever you're designing any tool. In the end though, there are just a lot of other variables that overshadow such considerations in my experience.
Thanks David.. The other consideration however, is what the string does when you bend on a typically made compound radius board. When you bend, no matter what you do the string will always travel uphill on the cone. Granted when the end radius is very close to "correct" the error is indeed minimal. It's when you make the end radius much wider than "correct", the error gets larger. Using a slant or elliptical cone, you can manipulate the cone to keep the string "level" or even move downhill" relative to the cone when bending, and thus theoretically you could lower the action a "tiny bit" more and use a "tiny bit" less relief. Yes you can true the board or the frets to the string paths, but then you got to do it. I personally don't want to, and I don't think it's much trouble to "get it good" the first time, whatever way it's done. In fact, once you set it up, it would take just as long either way.
Again, to me, it's personal satisfaction that I created a fretboard surface as close to ideal as I possibly can. Plus it does not take much effort, nor do I need CNC accuracy, to build or modify a jig to "gain" the accuracy, so why not?
I'll be the first to admit that I am a newbie to acoustic guitar building, though I have built quite a few electrics. I'm still green when it comes to binding, rosettes, bracing, etc. but I feel I can improve my skills, and have tried to "up" the skill level with each successive build I've done so far. So if I can spend less time trueing up a board and more time on the other aspects I'm down for it, even if I have to take the time to do it, because it'sonly once.
The beautiful irony of this is that I'm almost always on the side of going too far with attempts at precision, often with others insisting the degree of attempted precision is in vain and beyond the capabilities of the materials. My argument is typically, "it doesn't take that much more effort to try and hold these tolerances though, so why settle for good enough". It's laughable that I find myself arguing the other side here.
So I truly do understand where you're going with this, and am pretty much in the same boat. I just spent several hours this weekend lapping my fret leveling blocks on a granite surface plate that's been trued to .000025" (25 millionths of an inch), so I'd say we're not that far apart in our motivations here. "Hey, I have the resources - why not?"
When I see someone else doing it though, it's funny how quick the pragmatic side of me seems to kick in.
Do you guys need to find a hobby, like building guitars or something?
Just kiddin'.
Nelson, sometimes math is my hobby (placing hand-sign 'L' on my forehead...)
My apologies Louie, I meant the diagram in post #62 which is circular cross-sections. Either way, for either a circular cone or an elliptical cone, the shape of the arc is the same at each fret as is is at the ends. (BTW that also means that if each end are equilateral triangles of different heights, then all frets will follow equilateral triangles, but if you cut them not perpendicular to the bottom of the fretboard, like Murray's right diagram, then you'd get isosceles triangles.)
Picture this -the nut end of the fretboard is circular in cross-section. If this was a zero fret instead of a nut, wouldn't this also be circular in cross section? It's the same cut- just cut all the way through for a nut compared to a zero fret. And if all the frets are parallel to this circular arc on a cone, then they too are circular arcs.
I can see how a pendulum induced compound radius can bring in an elliptical error, but that doesn't mean that a cone that doesn't have an elliptical error (i.e true circle radius at each end) has elliptical errors. (nor does it mean that the ellipse isn't a "better" shape for the fretboard surface).
Brian, think of it this way... Looking at the diagram in #62, the only way you would have frets that run in a true circular parth is if you made the fretboard half the thickness of the cone. Since the fretboard back is more-or-less parallel to the surface of the cone, then any cut made perpendicular to the back of the fretboard would be cut at a bias to the cone's centerline. In fact I postulate that we've been playing on elliptical surfaces all this time!
The reverse is also true. One can slant the cone, so that at the short side of the cone, the circular cross sections are perpendicular to the cone's surface (see post #125, right). But if you took a cross section of this cone perpenicular to its centerline, you'll find that it is elliptical, thus one could define this cone as an elliptical cone with its base cut at a bias to its centeline. By the way it is interesting to note that the slant angle relative to the centerline is different for both cones, from the view of this section.
Arturo, I'm ready to eat crow, but I must first recover from foot-in-mouth disease. Though the third diagram in #107 is the surface I want to depict, you were right that it is not a conical surface, since the radius of the larger end I inscribed, along with the starting radius, suggests a rate of change different than the taper of the fretboard. But it is still a ruled surface....
It's a conical surface Louie, The surface is determined by the S circle and the ea circle, so it's a slice of the second cone, but the edges of this slice don't pass through the apex, so they're not straight lines. Once again, the intersection of two cones is not a straight line in general. the strings wouldn't pass through the apex of the first cone either, as you have modified the end line, neither the string paths would be straight, as they're on a conical surface (slice of the second cone) and don't pass through it's apex. just lay a ruler on a cone and see what happens is the line doesn't go through the apex, the conical fretboard would have a 'hump' in the middle for the string. Just the fact that you sliced the cone at a different tapper doesn't change it's surface. What you need is to find the surface the strings will follow and draw that surface (a little bellow), and in this case it won't be the slice of your cone.
Or maybe you're talking about a surface that gradually changes from that of the first cone to that of the second, that would be it. One way to visualize it would be to extend a very flexible material through both end shapes.
Hey Louie, I re-read the whole thread (admittedly got lost with some of the maths), drew some diagrams to demonstrate my point and managed to prove yours. Thanks.
Arturo, that's what I'm talking about, and that's exactly what happens when the second radius is larger than "theoretically correct..." This is the "ruled surface" that my little "string art" experiment showed (remember the days of string art?) I forsee a problem, however, in machining a ruled surface such as this conventionally. Since as you go from the middle toward the edge, the surface gets "helical" and you'd need the router to "average" the different surface angles relative to the straight line path. Or build a carriage that allows the router bit to follow a helical path
Brian, thank you for partcipating (and challenging MY mind!) If there's any of the math you'd like me to explain drop me a line, the math is really easier than it looks; though admittedly I got stuck a couple times, that's why I double-and triple-checked everything, and derived the same results obtaining the same resuls using proportions, and angles...
Couldn't sleep again last night with thoughts of ellipses and cones dancing in my head. After finally "getting it" that the frets take an elliptical shape on section of a circular cone's surface, I couldn't help thinking about fretboards I have made where I know the surface is made of circular arcs.
Here's what I'm thinking - it depends on the method of fabrication of the fretboard.
If we take a slotted fretboard and use a swing arm jig to make get the compound radius, we get a section of a round cone. Because the frets are cut at an angle other than 90deg to the axis of this round cone, they follow an ellipse at the surface.
However, if we take a slotted fretboard and rough in a circular radius at the nut and an "ideal" circular radius at the end and join those two radii, all the frets will be sections of circular arcs. BUT, the frets are still cut at an angle other than the axis of the cone. Therefore, the cone we produce from this method of construction fulfils all the criteria for an "ideal" cone (string paths would meet at the origin of the cone as would the extensions of the fretboard sides, straight string lie etc.) and would be an elliptical cone - a particular ellipse which gives round sections for the relevant angle of the fret "slices".
Have I got it?
I think that's it Brian. This is the "ruled" surface that Arturo and I have been dicussing. But it may or may not be a conical surface, depending on the nut and end radius used. If you use the "theoretically correct" radius derived from the equation in post #1, then yes, it would be a slant cone. BUT, if you use a radius greater than the "theoretically coreect" radius, then it is no longer conical, but is still a proportional surface that yields straight string paths and circular fret paths.
So yes, in my opinion the method you describe would produce what I consider an "ideal" cone, as long as you use the theoretically "correct" end radius. And since when the strings bend, they stay on that circular path, one can reduce or eliminate the need to make the end radius "flatter" than theoretically "correct."
The error in thought I had in my diagram was that I assumed that the surface I produced was conical. But since my end radius is larger than the "ideal" radius, the rate of change of the radius along the fretboard length is faster than the rate of change of the width, therefore it cannot be a cone. But since they still do move proportionately, it is still a ruled surface....
Thanks to all the participants in this thread. I'm interested how much difference there really is between the circular frets on elliptical cone compared to elliptical frets on circular cone - but not interested enough to do the maths :-)
Brian, I am in the middle of building my jig. I will be preparing a spreadsheet as well so there will be no math needed to set things up...
It's my personal goal to post the 200th message. Keep it up boys.
Hehehe Bill only 43 posts away...
Hey guys, I saw that the latest issue from GAL has an article on cylindrical fretboards... Anysone see this? Wondering what the article contains...