Jump to content

Compound Radius


Saber

Recommended Posts

If you have a fretboard with a compound radius of 10" at the nut and 16" at the heal (such as the Warmoth neck for example), since the radius is increasing as you're going toward the bridge, won't the radius be be even greater than 16" at the bridge? Won't you need a bridge with a radius of around 19-20"?

Link to comment
Share on other sites

If your talking about a Stratocastor in general, think about most of the floating bridges on them. The saddles have 2 adjustable screws which enable you to raise and lower each string individually so you can do almost any radius with them.

It gets harder when you start using necks like that using a Floyd Rose system, then you may have to shim each saddle according to the needs of the guitar set up.

I'm not positve about the majority of Tunematic bridges, but I would think most can be adjusted depending on the saddles they have installed.

The way I see it, the majority of available bridges out there are ready to handle different radius's.

Link to comment
Share on other sites

Yes. The shape of the strings will be conical. I don't think the radius is as high as 19" but I'm not sure. I set the bridge radius so that the strings fit the upper frets optimally. I don't actually set the bridge to a particular radius though - so I never actually measured what it is.

Link to comment
Share on other sites

Just for laughs I jumped into AutoCad (easier than math in this case) and found size for radius of bridge as asked.

Assume, 25 1/2" scale, 10" at nut, and 16" at 18.75" from nut this results in about 18.16" radius at bridge. Of course since the bridge for each string is at a different distance from nut this isn't really a true number but gives you an idea, again just for laughs.

Cheers

John

Link to comment
Share on other sites

Assume, 25 1/2" scale, 10" at nut, and 16"  at 18.75" from nut this results in about 18.16" radius at bridge.

What if the radius was 10" at the nut and totally flat at the 22nd fret. Then the bridge would need an "inverse radius" (just for laughs?) I wonder how that would play.

Link to comment
Share on other sites

errr, that one is difficult.

Problem is flat mean infinite radius and this doesn't work in a graphical solution. I'm at work and don't have time to figure this out for real but I did a check with very large radius nuts (4000 and 8000 inch). I got two drastically different answers for 22nd fret but both were in the thousands of inches. I think its safe to say you still almost flat at the 22nd fret.

Let me think about this one some more. Problem is going to be that the shape of the fretboard has a drastic impact on answer I think but still I think we are going to see that we are basically still flat. I'll think about it and get back to ya.

I can tell you that the closests guess we had was 15" (grin). I didn't guess.

Cheers

John

Link to comment
Share on other sites

After having more than 2 minutes to think about it (at work) this wasn't difficult at all. Here goes

25 1/2" scale, 10" radius nut, flat bridge. Using nut sting spread 1 3/8" and bridge strin spread of 2 3/32" you need .... drum roll ... right at 26.3" radius at 22nd fret to get flat at bridge. Sorry for wrong first answer. What I get for not thinking a problem through enough. :D

Cheers

John

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...