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 Post subject: Compound Radius Boards
PostPosted: Wed Nov 07, 2012 11:21 pm 
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Walnut
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Apparently, Gibson has been offering compound radius board equipted LP's for some time.

Question's are:

1. Why did they choose a 10"x16" compound radius combo, rather then the standard 12"x16"? What's the benefit of a 10" radius vs the standard 12"radi?

2. If compounds are so great, why do they not offer such on every model?


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PostPosted: Thu Nov 08, 2012 12:08 am 
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I can answer question 2: tradition!

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PostPosted: Thu Nov 08, 2012 2:26 am 
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http://gretschpages.com/forum/general-t ... 626/page1/

Study that. It's easy to use and gives precise values for an arbitrary radius along your fretboard (or bridge) when you have one prescribed radius, a scale length, and your string taper designed. In other words...if you already know what radius you want at the nut it'll give you the bridge radius, or the 12th or 22nd fret....whatever you want.

1. If you don't use the proper radii in every position, as determined by your scale length and string taper...your results are going to be somewhat arbitrary...and the likelihood of being able to run lower action without buzzing is reduced. Fact is....being wrong about the radii could result in a worse situation than a reasonably well leveled constant radius fretboard.

2. Once you know the proper radii for your scale length and taper...if you are not capable of accurately producing it...the likelihood of being able to...see above.

3. Compound radius fretboards offer the lowest possible string action when using a string taper...but it's not the simplest thing to pull off. Constant radius fretboards are far easier to make from a high production standpoint.

Btw...my 60's reissue Goldtop LP has a constant radius.

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Last edited by Stuart Gort on Thu Nov 08, 2012 2:14 pm, edited 1 time in total.

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PostPosted: Thu Nov 08, 2012 5:59 am 
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I'm sure that the saddle radius at the bridge factored into the decision as well. Assuming they want to use the same bridge as other models, that sets the radius at that end of the strings. If you want to have radius X at fret 12, those two parameters define the slope of the cone and will determine the radius at the nut.

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PostPosted: Thu Nov 08, 2012 2:46 pm 
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So, if someone's building a Gibson-like guitar(scale length, nut width, string spread + bridge type) should they trust Gibson's design specs and go with a 10" x 16" compound board, rather then a 12"x 16"?


Last edited by QueZee on Thu Nov 08, 2012 2:58 pm, edited 2 times in total.

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PostPosted: Thu Nov 08, 2012 2:53 pm 
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QueZee wrote:
So, if someone's building a Gibson-like guitar(scale length, nut width, string spread + bridge type) should they trust Gibson's design specs and go with a 10" x 16" compound board, rather then a 12"x 16"?


Design the taper, scale length, and nut radius.....then do the math. Trusting a set of numbers you found somewhere is EXACTLY the wrong way to go about creating a compound radius fretboard.

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PostPosted: Thu Nov 08, 2012 2:57 pm 
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We used a CAD app to calculate the 12"x16"compound radius values based on the fretboards nut/heel width, length, string spread and nut radius. It just calculates it automatically when you input those values


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PostPosted: Thu Nov 08, 2012 4:54 pm 
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QueZee wrote:
We used a CAD app to calculate the 12"x16"compound radius values based on the fretboards nut/heel width, length, string spread and nut radius. It just calculates it automatically when you input those values


Ah...honestly....that's the way I do it too...but I've checked and the formula is spot on to the cad values.

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PostPosted: Thu Nov 08, 2012 6:01 pm 
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Quote:
Trusting a set of numbers you found somewhere is EXACTLY the wrong way to go about creating a compound radius fretboard.


Isn't that what I said above?
Tradition is the same as, "we've always done it that way".

Back in the 80's I was a certified Fender repair guy. We had tons of vintage series Strats coming in with bumps in the fingerboard around the 9th to 14th frets. Fender wouldn't pay for more than $50, so they sent me an new neck, and I sent the old one back. The new one had the SAME dang BUMP.

When I toured the Fender factory during a NAMM show, we happened to pass by the tracer tools that cut the vintage necks. Our guide mentioned that they were using the same tooling from the old days. I ran my fingers over the steel tool and there were the stupid bumps in the tool! No wonder they had a problem. I sang out immediately and explained the problem. No response..... they'd always done it that way. They would NOT depart from the sacred specs and tools. NO.

When I got back home, I told all my clients who brought in their Strats we could play "change the crappy necks game" with Fender, or I could pull the frets and plane the fingerboard... for a goodly sum, of course.

Most of them went for the refret, and were very happy.

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PostPosted: Thu Nov 08, 2012 6:10 pm 
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Update:
Reworked the numbers using the supplied formula and still getting 12"x 16", when the bridges radius is NOT included. In other words, the 12"x16"radi figure is just the fretboards compound (12" at the nut, and 16" at the bottom of the fretboard, when "D" = arbitrary distance from the nut to the fretboards bottom, or 18")

However, if we work the numbers to INCLUDE the bridges radius, to determin what the actual radius of the compound fretboard should be, we get a whole different set of numbers.
They are: 12"radius at the nut, 14"radius at the saddle, 13+"radius at the fretboards bottom. This occurs regardless of string spread.
And, if we input Gibson's 10"radius nut, you get a 12+"radius at the fretboards bottom when using a 14"radius saddle. No where near their 10"x16" compound figures. The only way a 10x16 compound will work properly is if you paired it with a 18"+radius bridge.

Any idea what the deal is? Gibson must be using a specially radiused bridge to accomidate their 10x16 compound, right?


Last edited by QueZee on Sat Nov 10, 2012 4:14 pm, edited 7 times in total.

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PostPosted: Sat Nov 10, 2012 8:55 am 
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Figures lie and liars figure.... laughing6-hehe

But I have seen more than a few guitars where the bridge saddle radius was not a match to what was done on the fretboard. Usually leading to some real set up issues with string heights. Either the e strings or the d & g strings wind up ridiculously high to avoid having them fret out. These have not necessarily been cheap guitars either. Rather than try and calculate these things I would measure an actual specimen. Put your radius guages in your pocket and head off to GC and measure. Only then will you know for sure what Gibson is doing and how well it works.

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PostPosted: Sat Nov 10, 2012 4:20 pm 
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QueZee wrote:
Update:
Reworked the numbers using the supplied formula and still getting 12"x 16", when the bridges radius is NOT included. In other words, the 12"x16"radi figure is just the fretboards compound (12" at the nut, and 16" at the bottom of the fretboard, when "D" = arbitrary distance from the nut to the fretboards bottom, or 18")

However, if we work the numbers to INCLUDE the bridges radius, to determin what the actual radius of the compound fretboard should be, we get a whole different set of numbers.
They are: 12"radius at the nut, 14"radius at the saddle, 13+"radius at the fretboards bottom. This occurs regardless of string spread.
And, if we input Gibson's 10"radius nut, you get a 12+"radius at the fretboards bottom when using a 14"radius saddle. No where near their 10"x16" compound figures. The only way a 10x16 compound will work properly is if you paired it with a 17"+radius bridge.

Any idea what the deal is? Gibson must be using a specially radiused bridge to accomidate their 10x16 compound, right?


I can't say what is going wrong with your numbers without sitting down with you but these cad sketches will deliver the precise values supplied by the Gretch formula. I've double checked this many times.

I could only speculate about things. Like...that your measurements shouldn't include the width of the nut, but rather, the width of the strings at the nut. I don't know what's happening really.

Note on the sketches that the circle is the nut radius and the furthermost vertical line on the right represents the arbitrary radius of the 22nd fret. Extend it out to the bridge or any other fret station to obtain the correct radius at that point.

Pardon the way I put this but it sounds to me like you want to disregard the formula that DEFINES your variables. With a FIXED bridge radius (you seem to have a certain bridge you want to use), and a FIXED nut radius, and a FIXED scale length...you can ONLY adjust the string taper to achieve a proper relationship of the strings to the fretboard. If you have a FIXED bridge radius, a fixed scale, and a fixed taper....the CAD sketches will call out the nut radius but you'll need to know algebra to make a new formula to solve for Rd.

Any any rate...I don't care what Gibson does or why. I'm just telling you how to figure out what is right. The Gretch formula solves solves perfectly...provided it has accurate input. The cad model solves perfectly as well. It sounds like you've identified something that is wrong and want an answer nobody at Gibson is going to give you and no one outside of Gibson is going to know. In the long run Gibson is a business and perhaps there is some financial reason they put a certain radius bridge on a guitar rather than properly engineering it...if I understand this problem correctly. I'm not saying one thing or another. Frankly I'm not even sure what you're trying to calculate anymore. :) But I DO know there is only ONE correct solution as to what compound surface is required with certain fixed variables.


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I read Emerson on the can. A foolish consistency is the hobgoblin of little minds...true...but a consistent reading of Emerson has its uses nevertheless.

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PostPosted: Sat Nov 10, 2012 8:43 pm 
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Any ideas as to how one would go about making their own compound radiused boards? With fairly limited tool access of course. I have a 10" and 12" radius sanding block, and lots of sandpaper.


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PostPosted: Sat Nov 10, 2012 9:27 pm 
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Zlurgh wrote:
I can't say what is going wrong with your numbers without sitting down with you but these cad sketches will deliver the precise values supplied by the Gretch formula. I've double checked this many times.
I could only speculate about things. Like...that your measurements shouldn't include the width of the nut, but rather, the width of the strings at the nut. I don't know what's happening really.

Note on the sketches that the circle is the nut radius and the furthermost vertical line on the right represents the arbitrary radius of the 22nd fret. Extend it out to the bridge or any other fret station to obtain the correct radius at that point.
Pardon the way I put this but it sounds to me like you want to disregard the formula that DEFINES your variables. With a FIXED bridge radius (you seem to have a certain bridge you want to use), and a FIXED nut radius, and a FIXED scale length...you can ONLY adjust the string taper to achieve a proper relationship of the strings to the fretboard. If you have a FIXED bridge radius, a fixed scale, and a fixed taper....the CAD sketches will call out the nut radius but you'll need to know algebra to make a new formula to solve for Rd.

Any any rate...I don't care what Gibson does or why. I'm just telling you how to figure out what is right. The Gretch formula solves solves perfectly...provided it has accurate input. The cad model solves perfectly as well. It sounds like you've identified something that is wrong and want an answer nobody at Gibson is going to give you and no one outside of Gibson is going to know. In the long run Gibson is a business and perhaps there is some financial reason they put a certain radius bridge on a guitar rather than properly engineering it...if I understand this problem correctly. I'm not saying one thing or another. Frankly I'm not even sure what you're trying to calculate anymore. :) But I DO know there is only ONE correct solution as to what compound surface is required with certain fixed variables.



I apologize for the misunderstanding; perhaps we should have further clarified the inquiry.

Basically, what we’ve found is that Gibson’s newest 10x16 compound radius board will not create optimal string action with their preferred tunomatic style bridge, unless the bridge is based upon an 18+” radius. You could call Gibson and ask, I suppose, but not so sure they’d answer. That’s all we’re saying, now that everything else is clarified.

As for the 12x16’ radius we calculated – Yes, the numbers are correct when using a Gibson scale/nut width if you do not input the saddles radius. The formula works fine for that, but that wasn’t what the initial inquiry was about. It was about why Gibson chose the 10x16 route, rather than the standard 12x16” compound. Now, it’s just, “how has Gibson managed to setup proper string action using a compound board with a tunomatic “fixed radius” bridge?” Answer: I’m guessing they’re using custom radius tunomatic saddles.

As you know, a 10x16 radius, or any other compound board, will work perfectly with any bridge that allows individual saddle height adjustment. Tunomatics do not, unless you make longer saddles to accommodate the compounds flat radius.

Rd = [Rn x (X + D)] X
X = 54.17 = focal point of Gibson fretboard (string spread matters not, because the strings run parellel with the fretboards edge)
D = 18 = approximate length of Gibson fretboard
Rn = (10” radius and 12” radius) at the nut

When solving for the correct radius at the fretboards bottom, you get:

Rd(10”) = 13.32” radius at the fretboards bottom
Rd(12”) = 15.98” radius at the fretboards bottom

When solving for the correct bridge radius, you get:

When D = 24.5625 = distance from nut to bridge (Gibson standard)
Rd(10”) = 14.53” bridge radius = close to the 12” or 14” tunomatic bridge radius
Rd(12”) = 17.44” bridge radius = even further from the tunomatic standard


Capice?


Last edited by QueZee on Sat Nov 10, 2012 10:48 pm, edited 5 times in total.

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PostPosted: Sat Nov 10, 2012 9:30 pm 
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msween wrote:
Any ideas as to how one would go about making their own compound radiused boards? With fairly limited tool access of course. I have a 10" and 12" radius sanding block, and lots of sandpaper.


You could find quite a few YouTube videos detailing various methods. The best one, in my opinion, shows a guy using a benchtop belt sander outfitted with custom bracing that allows him to make virtually any radius combo, quickly.


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PostPosted: Sat Nov 10, 2012 9:35 pm 
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msween wrote:
Any ideas as to how one would go about making their own compound radiused boards? With fairly limited tool access of course. I have a 10" and 12" radius sanding block, and lots of sandpaper.


I made one by marking out the radius at the nut end and the bridge end and then used a 24" long sanding block to sand to both of them. Took some elbow grease but it ended up fine. If I were to do a lot of them I'd look to set something up with a sander or routers or cnc, but for one this way worked just fine. I did a 10 X 14 compound with an abr TOM bridge and haven't had any problems with set up, of course I'm not as engineering oriented as a lot of people around here and really just got lucky.


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PostPosted: Sun Nov 11, 2012 12:34 am 
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QueZee wrote:
Rd = [Rn x (X + D)] X
X = 54.17 = focal point of Gibson fretboard (string spread matters not, because the strings run parellel with the fretboards edge)


Actually...using the fretboard edge instead of the actual string taper will throw the focal point out further and your calculations will be subtly incorrect. The way to visualize the error is to imagine the two 10-16 fretboard arcs extending to full half circles and then drawing straight lines to connect the ends of one arc to the correlating end of the other arc. Then draw two parallel lines just inside of those two lines.

If you are right....the endpoints of the two new parallel lines (your 'strings') will have the same vertical distance to each arc....but they won't.

Of course the error produced by measuring the fretboard edge will be pretty dang minute...but if you're using cad there isn't any reason not to be as accurate as possible.

QueZee wrote:

D = 18 = approximate length of Gibson fretboard
Rn = (10” radius and 12” radius) at the nut

When solving for the correct radius at the fretboards bottom, you get:

Rd(10”) = 13.32” radius at the fretboards bottom
Rd(12”) = 15.98” radius at the fretboards bottom

When solving for the correct bridge radius, you get:

When D = 24.5625 = distance from nut to bridge (Gibson standard)
Rd(10”) = 14.53” bridge radius = close to the 12” or 14” tunomatic bridge radius
Rd(12”) = 17.44” bridge radius = even further from the tunomatic standard


Capice?


Ya...those numbers look right (roughly). I designed my complex fretboards exactly after my Les Paul fretboard with the exception of using a 10.5" radius at the nut. I believe it was 13.7 or so at fret 22.

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