Opinions wanted on compound radius for an F octive

[joehendrick]

Hi All- I'm working on my first F octive. Just finishing up the binding so neck and fret board are next. I am doing a 22" scale. I had planned to just do a flat fret board but now am thinking radius. Any advice on the range of a compound radius for this scale?

Thanks

Joe

[Jim Hilburn]

A radius on any larger instrument is just the way to go IMO and most instrument builders concur. Benefit of compound is that you don't have a big radius at the bridge which is better for picking. On electric guitars compound makes for better string bending but not a concern on an octave, at least the way I play it.

[Jim Hilburn]

Didn't really answer the question, did I.
There's a formula that Don Mc Rostie came up with on how to figure out the different radii along the length of a cone. But I'd suggest something on the order of 8" at the nut, 12" at the 12th and 20" on the saddle.

[Jim Hilburn]

https://books.google.com/books?id=zF...radius&f=false

[murrmac]

But I'd suggest something on the order of 8" at the nut, 12" at the 12th and 20" on the saddle.

Wouldn't the 12th fret radius have to be halfway between the nut radius and the saddle radius to conform to a conical conformation? So if the board had an 8" radius at the nut, and the saddle had a 20" radius, then the 12th fret radius would be 14" ?

[j. condino]

On a stratocaster, I would not have anythng but a compound radius, because I bend those strings like crazy. I make octave mandolins with approx. 22.5" scale and never felt a need for the compound radius, so I use a uniform radius across the entire fingerboard. If you want to play BB King licks on your octave mandolin, it may be a concern.

J.
www.condino.com

[HoGo]

Wouldn't the 12th fret radius have to be halfway between the nut radius and the saddle radius to conform to a conical conformation? So if the board had an 8" radius at the nut, and the saddle had a 20" radius, then the 12th fret radius would be 14" ?

NO, it is not that simple. You just cannot arbitrarily choose both radiuses (nut and saddle) and expect the neck surface be part of cone (you can choose one radius and the other can be calculated to be part of cone). The taper of fretboard (or better said of strings) is part of teh equation. If you start with two radiuses and connect them with straight lines in the direction of strings you'll end up with parts of ellipses or parabolic curves at all other positions. But in practice the difference between those elliptical curves and circles is very small, though can be problem on some long scale instruments.

[murrmac]

. You just cannot arbitrarily choose both radiuses (nut and saddle) and expect the neck surface be part of cone (you can choose one radius and the other can be calculated to be part of cone).

Actually, to be nitpickingly pedantic, you can always choose two arbitrary radiuses and have the fretboard surface form part of a cone.... as long as you don't care about the string spacing at the saddle.... an unlikely scenario, I know.

The problem lies in the fact that once you have defined your nut radius and your saddle radius, then the taper (and hence the string spacing at the saddle) is predetermined, and that taper will almost certainly end up being wider than you want it to be.

In the case of a 8" nut radius and a 20" saddle radius, if the width of the nut is 1 1/8" , then in order for the strings to lie on a perfectly conical surface, the string spacing at the saddle would have to be 2.8".

As you say, in practical terms it matter not a whit whether the radius is compound or truly conical.

[Wes Brandt]

NO, it is not that simple. You just cannot arbitrarily choose both radiuses (nut and saddle) and expect the neck surface be part of cone...

If I visualize this, I cannot see why not… if you have the two radius, two diameters and think of what's between them as tapered round shaft, otherwise known as a cone… is not the fingerboard just a "fillet" from the side of that cone?

For viols, which have 6 and sometimes 7 strings, the taper is actually in reverse. The nut end starts out with the proper radius for fingering and at the bridge end you add more radius in order to more closely match the bridge curve, which you choose with only the ease of bowing in mind.

So my bass viols end up with 8o mm radius at the nut and about 65 at the bridge end, even through the fingerboard goes from something like 60mm wide at the nut to 90mm wide at the bridge… I actually do draw the two radius on the ends of the fingerboards and plane away, it really takes some careful work to pull off using a long plane and following the string paths. Then you have to veneer the dang thing.

It's really put me in tune with fingerboard radius making though… and I don't get it… isn't this just very simple geometry? involving only straight lines and circles?

[j. condino]

'Gotta agree with Wes on this one. Make a few 34" long double bass fingerboards and you'll never stress about a mandolin again. It is more about a steady hand and the ability to visualize the finished product than complicated mathematics.....

j.
www.condino.com
www.kaybassrepair.com

[murrmac]

If I visualize this, I cannot see why not… if you have the two radius, two diameters and think of what's between them as tapered round shaft, otherwise known as a cone… is not the fingerboard just a "fillet" from the side of that cone?

For years, I used to think exactly the same way , until one day the penny dropped.

Yes, you can visualize a cone, the length of your fingerboard, with two assigned radiuses at either end, one corresponding to the nut radius, and one corresponding to the saddle radius, and you could draw a trapezoid on the cone, in the shape of the fingerboard, with the nut width at one end and whatever width you wanted for the end of the fingerboard at the other.

The problem with doing that is that the string paths will not lie in a straight line on the surface of the cone... leastways not if you space out the strings at the nut and the saddle equally, as is conventional.

In the example of your viol fingerboards, the surface described by the measurements is most definitely not the surface of a cone ... it is impossible to have the wider end with a narrower radius , with all the string paths tangent to the surface, and still have a conic surface. What you are doing is producing a sophisticated compound radius surface ... but a section of a cone it most assuredly is not.

Does any of this matter in practical terms ? Not in the slightest. It's just a quibble about terminology.

t's really put me in tune with fingerboard radius making though… and I don't get it… isn't this just very simple geometry? involving only straight lines and circles?

Exactly. Draw the radius at either end, plane in the string paths and sweeten the areas in between ... job done...perfect compound radius ... (but still not a surface of a cone )

[Wes Brandt]

The problem with doing that is that the string paths will not lie in a straight line on the surface of the cone... leastways not if you space out the strings at the nut and the saddle equally, as is conventional.

In the example of your viol fingerboards, the surface described by the measurements is most definitely not the surface of a cone ... it is impossible to have the wider end with a narrower radius , with all the string paths tangent to the surface, and still have a conic surface. What you are doing is producing a sophisticated compound radius surface ... but a section of a cone it most assuredly is not.

Draw the radius at either end, plane in the string paths and sweeten the areas in between ... job done...perfect compound radius ... (but still not a surface of a cone )

Of course the stings do not lie on a straight line and do spread out… just like the diameter of the cone "spreads out"… they may not be exactly proportional ...but they could be if you wanted them to be.

It is possible to have a wider end with a smaller radius, I do it all the time… if I had a modeling graphic so I can visually slice up a cone, I could show you. With viol fingerboards I'm dealing with a cone that tapers relatively gradually, thats the key I think. If the taper was more acute for the given length, you would be right, there would be a point where the bridge end could not be wider than the nut end and maintain the length you want. But at the measurements I mention, it works.

No sweetening of the areas is required but I have left out part about putting in the relief, more on the bass side than the treble, which I do by bending the fingerboard slug (while squeezed/clamped firmly, lengthwise in a vice) into a back bow, using a wedge to push up the bass side before I do the final planing …flat, straight, following the string paths and which, once released, creates more relief on the bass side but still some on the treble, which is what you want but all of which is irrelevant for this discussion.

I happen to be making a bass fingerboard today, so I will put up some shots so you can see this and with exact numbers.

[murrmac]

Of course the strings do not lie on a straight line and do spread out… just like the diameter of the cone "spreads out"…

When I said that the strings wouldn't lie straight, I meant that they wouldn't be contiguous with the surface of a cone ... I wasn't implying that the string paths should be parallel ... apologies for the misunderstanding.

Looking forward to seeing the pics ( and the measurements ) of your bass fingerboard.

[Wes Brandt]

When I said that the strings wouldn't lie straight, I meant that they wouldn't be contiguous with the surface of a cone ... I wasn't implying that the string paths should be parallel ... apologies for the misunderstanding.

Looking forward to seeing the pics ( and the measurements ) of your bass fingerboard.

But think about it ...if the line of the strings are truly running straight up and down the side of the cone, where ever you put a straight edge on the cones surface, as long as it is running straight with the center of the cone, you will have a straight line… if you twist the straight edge in relation to the center of the cone so it is crossing the centerline of the cone, then you are passing across what is no longer a straight line.

You can then slice or "fillet" what ever you compound radius you want off that cone, in either direction, within it's dimensional limitations.

[Marty Jacobson]

This is really not that complicated to actually draw up in CAD. Talking about it is pretty complicated, but in practice you just draw it and it works. I use a 8" - 14" compound radius on my mandolins, and I'd use something pretty similar on an octave mandolin, unless the customer asked for something different.

I have a parametric CAD file I use to generate any fretboard at any scale length with any version of compound radius and width variations. I can send it to anyone who wants it.

And yes, it is possible, as Wes stated above, to put a wider or smaller radius at either end. The width of the board at each end is irrelevant. This is a boolean operation between a lofted section of a cone and a "fretboard-shaped" solid.

[murrmac]

Nobody is disputing that it is possible to generate a compound radius of any desired specification, with any two selected radiuses at either end, and any two selected fingerboard widths ... builders do it all the time... not a problem.

The point I am try to make is that these surfaces, symmetrical and totally functional though they may be, are not sections of the surface of a cone, so referring to them as "conical" is incorrect .

I have a feeling that Don McRostie has a lot to answer for regarding about this conflation of "compound " and "conical" ... the Stewmac article on compound radius is misleading, implying as it does that the surface of a compound radiused fretboard is equivalent to a section of a conic surface.

[Wes Brandt]

The point I am try to make is that these surfaces, symmetrical and totally functional though they may be, are not sections of the surface of a cone, so referring to them as "conical" is incorrect .

You say that they are not sections of a cone but you do not explain why…. what happens to change them to "not sections of a cone"? Why, if the string paths follow a line straight with the cones center are they NOT laying on a straight surface/line …that shouldn't be to hard to conceptualize and explain if it is true.

Edit… or another approach, explain what exactly is different about these fingerboards to make them not sections of cones.

[Wes Brandt]

I doubt these photos help explain anything but I'll put them up anyway.

I had used the radius for a tenor viol in my previous posts. For a bass I use 90mm at the nut end and 75mm at the bridge end. The nut end is 60 wide and bridge end is 90 wide.

Hopefully all the layout markings are clear… all I do is plane from end to end trying to follow the imaginary center of the cone or the imaginary string lines ...either will do.

I frequently mark the fingerboard across with pencil lines so I can see what the true path of the plane is. Also I check the radius along the way with the template sizes in between and the ends for true roundness as I go. The final truing is with a 11 inch sanding block.

I realize I don't have the length with me… which is of course important… I'll put it up when I get home from this coffee shop I'm at just now.

[Marty Jacobson]

They're not sections of a cone because the circles are not concentric (i.e. what would be the axis of revolution is at an angle, so it's not a true revolved solid). It's a cone-like solid, not a true cone.

[Wes Brandt]

They're not sections of a cone because the circles are not concentric (i.e. what would be the axis of revolution is at an angle, so it's not a true revolved solid). It's a cone-like solid, not a true cone.

Why would you use the distorted cone on the right to slice your fingerboard off of, as opposed to the undistorted cone on the left?

EDIT… the language you use is for professionals and I would need a deep course in geometry… I'm of the belief that most things can be explained in much more common terms, in simpler concepts and analogies.

[Rob Sharer]

On a stratocaster, I would not have anythng but a compound radius, because I bend those strings like crazy. I make octave mandolins with approx. 22.5" scale and never felt a need for the compound radius, so I use a uniform radius across the entire fingerboard. If you want to play BB King licks on your octave mandolin, it may be a concern.

J.
www.condino.com

Point/counterpoint:

I'm basically with James on this. Leo Fender put a tight radius on his guitars to suit chording, perhaps not imagining the string-bending excesses that were to come. Flat radii suit bending better, and a compound radius is a great compromise, comfortable in the chord zone and flat where the bent strings need room to swing. If you're not bending strings, you really don't need a compound radius.

However...

A customer of mine, having spent too much time reading the interwebs, simply insisted that I take his brand-new Eastman dreadnought and give it a stainless refret and a compound radius treatment, cost no object. I suspended my skepticism long enough to do the job, and while I didn't go wild bending strings on a dreadnought, I did find the result to be strangely enjoyable to play, quite comfortable for reasons I'm not sure I can explain.

No science, just an anecdote.

Cheers,
Rob

[murrmac]

You say that they are not sections of a cone but you do not explain why…. what happens to change them to "not sections of a cone"? Why, if the string paths follow a line straight with the cones center are they NOT laying on a straight surface/line …that shouldn't be to hard to conceptualize and explain if it is true.

Edit… or another approach, explain what exactly is different about these fingerboards to make them not sections of cones.

OK, this is how I first realized that what I had always thought were conic surfaces, were not, in fact conic surfaces.

Forgive me if I talk about guitar fretboards here as i think it is easier to explain.

Let's assume that a builder wants to build a guitar with a standard length fretboard 18" long, and that he wants the radius to go from say, 10" at the nut to , say, 16" at the end of the board. The scale length is 25.5"

So, we can visualize a cone, 25.5" long (measured along the side of the cone ...not along the central axis) ... the radius at the narrow end is 10", and the radius 18" along is 16".

Let us further assume that the distance between the centers of the E strings at the nut is 1.5" (which is pretty close to the actual distance on a standard 1.75" wide nut)

So, on this cone, using a beveled straight edge, we draw a line 25.5" in length long along the center of the cone ... this line is of course in the same plane as the central axis of the cone.

Next, on the narrow end of the cone, we mark a point 1.5" along from the line we have just drawn. From this point, we draw another straight line on the cone's surface, again in the same plane as the central axis of the cone. This line obviously tapers away from the first line the further it goes ... the question is ... how much?

Now obviously nobody is going to actually make a cone to find all this out, so we can calculate it instead.

Now, elementary Euclidean geometry tells us that on a cone, the ratio of [the distance between the lines at any point] to the [circumference of the circle at that point] will be constant throughout...again using elementary Euclidean geometry it is easy to calculate that the ratio of the 1.5" arc at the narrow end of the cone to the circumference at the narrow end (which is of course PI*20") comes out to be 41.88.

So ... we next divide the circumference at the 18" mark (which is of course PI*32") by 41.88, to find out what the width of the end of the fretboard will be. It turns out that this width will be exactly 2.4"

So what is the string spacing at the saddle going to be ? We don't have a given radius for the saddle, so we need to calculate it . Again it is easy to figure out that the radius is 19.33", so the circumference is PI*38.66". Dividing this by 41.88 gives a string spacing at the saddle of 2.9" .

Now, these are the dimensions which are the inevitable result of the surface being a section of a cone and as you will instantly see, these sizes would be unrealistic in a real life guitar ...and if you were to alter the surface so that you had a sensible saddle string spacing of say 2 1/4" , then that reshaped surface would no longer be a true conic surface.

[murrmac]

They're not sections of a cone because the circles are not concentric (i.e. what would be the axis of revolution is at an angle, so it's not a true revolved solid). It's a cone-like solid, not a true cone.

Marty's got it right ...

[Graham McDonald]

There is a simple, if not quite so scientific/mathematical way to make a compound radius board. For an octave start with a 6mm/.25" thickfretboard, tapered to shape. Mark a line along each edge 1mm/.04" down from the top. With a plane and a sanding board (a piece of 3/4" melamine faced mdf or ply half the size of s sheet of sandpaper) faced with 80 grit sandpaper, plane or sand down the fretboard in an arc to the line marked on the edges, leaving the middle of the board at the 6mm thickness. I have no idea what radii you might end up with, but it works fine for a guitar, OM or bouzouki. Same approach for a mandolin, but only take the edges down about .5mm/.02". I got the method from Jim Williams, author of A Guitar Maker's Manual, who got it from Charles Fox when he was running a guitar making school in Vermont in the 1970s.

Cheers

[Wes Brandt]

OK, this is how I first realized that what I had always thought were conic surfaces, were not, in fact conic surfaces.

Forgive me if I talk about guitar fretboards here as i think it is easier to explain.

Let's assume that a builder wants to build a guitar with a standard length fretboard 18" long, and that he wants the radius to go from say, 10" at the nut to , say, 16" at the end of the board. The scale length is 25.5"

So, we can visualize a cone, 25.5" long (measured along the side of the cone ...not along the central axis) ... the radius at the narrow end is 10", and the radius 18" along is 16".

Let us further assume that the distance between the centers of the E strings at the nut is 1.5" (which is pretty close to the actual distance on a standard 1.75" wide nut)

So, on this cone, using a beveled straight edge, we draw a line 25.5" in length long along the center of the cone ... this line is of course in the same plane as the central axis of the cone.

Next, on the narrow end of the cone, we mark a point 1.5" along from the line we have just drawn. From this point, we draw another straight line on the cone's surface, again in the same plane as the central axis of the cone. This line obviously tapers away from the first line the further it goes ... the question is ... how much?

Now obviously nobody is going to actually make a cone to find all this out, so we can calculate it instead.

Now, elementary Euclidean geometry tells us that on a cone, the ratio of [the distance between the lines at any point] to the [circumference of the circle at that point] will be constant throughout...again using elementary Euclidean geometry it is easy to calculate that the ratio of the 1.5" arc at the narrow end of the cone to the circumference at the narrow end (which is of course PI*20") comes out to be 41.88.

So ... we next divide the circumference at the 18" mark (which is of course PI*32") by 41.88, to find out what the width of the end of the fretboard will be. It turns out that this width will be exactly 2.4"

So what is the string spacing at the saddle going to be ? We don't have a given radius for the saddle, so we need to calculate it . Again it is easy to figure out that the radius is 19.33", so the circumference is PI*38.66". Dividing this by 41.88 gives a string spacing at the saddle of 2.9" .

Now, these are the dimensions which are the inevitable result of the surface being a section of a cone and as you will instantly see, these sizes would be unrealistic in a real life guitar ...and if you were to alter the surface so that you had a sensible saddle string spacing of say 2 1/4" , then that reshaped surface would no longer be a true conic surface.

If I follow you correctly.. then the isn't the surface of your fingerboard example actually still conic, but it's the string layout itself that doesn't match increasing spread of the cone? However, the strings are not off that much though, the worst two would be the two outside strings …but is it enough to actually cause a problem, especially after you introduce relief? Can you calculate how much deviation from a straight line a string path would have in the middle? .001? .002?

And then, though I first set individual string heights on guitars and mandolins by measurement, I test by playing and adjust accordingly… could it be that we naturally fudge the string heights to compensate for the strings not truly following the "natural" lines of the cone?

Out of curiosity, how would you go about adjusting your fingerboard to make it work anyway? Remove material from where?

I'm starting to feel pretty geeky here, but I still don't see it. Sorry.