Level and true a compound radius fingerboard

[Bruce Johnson]

Yeah, that's what we're doing, but in steps. The process we're talking about above, cutting the hourglass shape, is done with the truss rod tightened to make the fingerboard flat. That gets the basic surface geometry correct. Then, you loosen the truss rod to bring in a bit of relief. Now you have the natural curve to match the vibrating string. That's how you get the lowest action without buzzing.

On a neck without a truss rod, particularly one with a round radius like an upright bass fingerboard, it's much trickier to get that perfect curve. It's much harder to make a perfectly smooth shallow concave curve, with no lumps or ripples. Another really good reason to have a truss rod. Use it to temporarily make the neck flat. Then it's much easier to cut the string paths flat, using something flat. Once the surface is flat, adjust the curve with the truss rod.

I speak from experience that precisely truing up an EUB fingerboard with no truss rod is a lot of fussy work.

 

[Paulsar]

I have been using this formula on different guitars and basses for a couple years now. I used to buy and sell guitars and basses and sometimes I would pull the frets out and re-radius the boards and compound them, then re-fret and set them up again.

I don't know why anyone hasn't come up with this before. Well, I shouldn't assume noone has. Someone probably has because it makes so much sense mathematically and geometrically , but I haven't seen it. The formula will work for any new guitar or bass, fretted or fretless, as long as you have 100% control over how the thing is built.

All you need to know or plan on to start with are:

1. The radius at the nut
2. The average center to center string spacing at the nut
3. The average center to center string spacing at the bridge

Our example bass for which the calculations below are done will have a 12" nut radius, 10mm string spacing at the nut and 19mm spacing at the bridge, and 24 frets. I've also rounded up numbers at the nut and down at the bridge for simplicity sake.

A compound radius fretboard is a section of a cone. The radius of a circle (cross section of the cone) is half the diameter, and the circumference of the circle is found using pi (3.1416). So we use that to figure out the following:

Here we go

12" nut radius = 305mm X 2 = 610 diameter X pi (3.1416) = 1919 or 192 string spaces around the top circumference of our section of the cone.

Now that we have that number 192 to work with, we transfer it to the bridge where the strings average 19mm (3/4") center to center.

192 X 19mm = 3648mm circumference at the bridge section of the cone, divided by pi (3.1416) = 1161mm diameter divided by 2 = 580mm or a 22.83" radius at the bridge.

To get the radius at the 24th fret of the fretboard, we use the octaves. The 24th fret is the third octave and is 3/4 of the distance from the nut to the bridge. So we add 3/4 of the difference between the nut radius and bridge radius, to the nut radius, like this:

10.83" X .75 = 8.1225" + 12" = 20.1225

(if your bass only has 21 or 22 frets, use percentage by measuring from the nut)

So a nearly ideal compound radius fretboard for a 34" scale bass with a 12" radius and average 10mm string spacing at the nut, and 19mm at the bridge, would be 12" and 20.1225" or just round it to either 20" or 20 1/8 at the 24th fret.

Yup, I'm not a math expert, but what I explained above would eliminate the need for "hourglassing" or creating a waist on compound radius fretboards.

 

[Harry Cahyadi]

Paulsar,

I have got 2 questions (sorry, I haven't been dealing with math formula for a while),

192 X 19mm = 3648mm circumference at the bridge section of the cone, divided by pi (3.1416) = 1161mm diameter divided by 2 = 580mm or a 22.83" radius at the bridge.

Why the 192 is being used for the bridge calculation?

To get the radius at the 24th fret of the fretboard, we use the octaves. The 24th fret is the third octave and is 3/4 of the distance from the nut to the bridge. So we add 3/4 of the difference between the nut radius and bridge radius, to the nut radius, like this:

10.83" X .75 = 8.1225" + 12" = 20.1225

(if your bass only has 21 or 22 frets, use percentage by measuring from the nut)

What is 10.83"?


Thank you.

 

[Paulsar]

Paulsar,

I have got 2 questions (sorry, I haven't been dealing with math formula for a while),

Why the 192 is being used for the bridge calculation?

What is 10.83"?

Thank you.

192 is the number of spaces between the strings around the top of the cone at the nut end. Since a bass or guitar has the same number of strings at the bridge and nut, just spread out, thats where that number comes into play.

10.83 is the difference between the nut radius and bridge radius and is used to get the radius at the octaves by splitting it up or using a percentage of it and adding it to the nut radius to find the radius at the neck heel end of the fret board.

The formula I used above is just an example. You can simply plug any radius in and different numbers come out. Like if you wanted to use a 14" radius at the nut on a guitar with 7mm string spacing and 10.5mm at the bridge, you use the same formula and principal, but your numbers will end up a lot different.

 

[Harry Cahyadi]

192 is the number of spaces between the strings around the top of the cone at the nut end. Since a bass or guitar has the same number of strings at the bridge and nut, just spread out, thats where that number comes into play.

10.83 is the difference between the nut radius and bridge radius and is used to get the radius at the octaves by splitting it up or using a percentage of it and adding it to the nut radius to find the radius at the neck heel end of the fret board.

The formula I used above is just an example. You can simply plug any radius in and different numbers come out. Like if you wanted to use a 14" radius at the nut on a guitar with 7mm string spacing and 10.5mm at the bridge, you use the same formula and principal, but your numbers will end up a lot different.

Exciting stuff.

Thanks for sharing.

 

[Paulsar]

Exciting stuff.

Thanks for sharing.

I hope it makes sense. I'm one of those people who can picture this stuff in my mind, do the calculations, but I have a heck of a time explaining it to others.

 

[Harry Cahyadi]

I hope it makes sense. I'm one of those people who can picture this stuff in my mind, do the calculations, but I have a heck of a time explaining it to others.

It makes sense. Now I just need put it into practice

 

[SteveCS]

I understood you explanation from your previous posts. I was just asking the question to confirm my whether I understood it correctly.

So in short, the answer to my question is "Yes", which means that when the FB is perfectly levelled, it will not be perfectly round in shape. Instead, it will have "x number of flat sides", where x equals to the number of strings or (number of strings -1) if the bass has odd number of strings.

Am I right?

To help visualise the concept, check out the 'Romberg' pattern for Cello and DB fingerboards...