Geometry help needed by guitar builder

[pjdeklerk]

 

[pjdeklerk]

Wondering why am I getting no response on this? Is it a stupid question? Is it unreadable?

 

[Otis]

Your question is not stupid at all, but the diagram is a bit blurry, when enlarged. This forum hasn't been set up to accommodate images with large dimensions or low resolution. It's better to use the imgbb hosting site. For future reference, please see this notice, for more information.

I can read the fuzzy text, but I've had to guess that each end of the rectangle's base (i.e., side W closest to the circle's center) lies on the circle. Assuming that's correct, then we call length W a 'chord', and we call height H an 'arc height'. (Here's a reference, if you'd like to revisit fond memories of college, heh.)

H = R - 1/2∙√[4∙R² - W²]

Cheers :cool:

PS: This forum is run by volunteers who have no set schedule.

 

[pjdeklerk]

Your question is not stupid at all, but the diagram is a bit blurry, when enlarged. This forum hasn't been set up to accommodate images with large dimensions or low resolution. It's better to use the imgbb hosting site. For future reference, please see this notice, for more information.

I can read the fuzzy text, but I've had to guess that each end of the rectangle's base (i.e., side W closest to the circle's center) lies on the circle. Assuming that's correct, then we call length W a 'chord', and we call height H an 'arc height'. (Here's a reference, if you'd like to revisit fond memories of college, heh.)

H = R - 1/2∙√[4∙R² - W²]

Cheers :cool:

PS: This forum is run by volunteers who have no set schedule.

Thanks for the reply. I figured readers would either zoom in on the picture or click the link to the pdf version. Actually none of the corners of the rectangle are on the circle. the circle touches the center of the top of the rectangle. The circle intersects somewhere on the sides of rectangle. where on the sides is what I need to be able to calculate.

 

[Otis]

Sorry, the zoomed diagram is blurry; and puzzling over the diagram, I didn't read the entire text. (Dumb)

Didn't notice the pdf extension, either; the attached-image link usually appears there. Please pardon me.

Let's call the distance you seek little 'h'. (Big 'H' is the rectangle's height.)

h = R - √[R² - (W/2)²]

This is valid, as long as W is less than 2∙√[2∙R∙H - H²]


If you'd like to check my work (I suggest you do, before making sawdust :smile, here's what I did.

I turned the diagram upside down and introduced an xy-coordinate system, such that the tangent line is the x-axis and the y-axis bisects the rectangle.

Now the rectangle's (x,y) coordinates at the upper-right corner are (W/2, H)

The coordinates at the lower-right corner are (W/2, 0)

The circle intersects the rectangle somewhere vertically between these points (as long as the rectangle is not too wide).

The y-coordinate of that intersection point is h (i.e., the height of the intersection point above the x-axis).

The equation for a circle of radius R whose center has been shifted vertically up R units from (0,0) is:

x² + (y - R)² = R²

Solving for y, we get two equations (one plots the circle's upper-half and the other plots the lower-half).

For the circle's lower-half (tangent to the x-axis), the equation is:

y = R - √[R² - x²]

We're interested in y, when x = W/2, because that's the height of the intersection point (h).

h = R - √[R² - (W/2)²]

 

[pjdeklerk]

Sorry, the zoomed diagram is blurry; and puzzling over the diagram, I didn't read the entire text. (Dumb)

Didn't notice the pdf extension, either; the attached-image link usually appears there. Please pardon me.

Let's call the distance you seek little 'h'. (Big 'H' is the rectangle's height.)

h = R - √[R² - (W/2)²]

This is valid, as long as W is less than 2∙√[2∙R∙H - H²]


If you'd like to check my work (I suggest you do, before making sawdust :smile, here's what I did.

I turned the diagram upside down and introduced an xy-coordinate system, such that the tangent line is the x-axis and the y-axis bisects the rectangle.

Now the rectangle's (x,y) coordinates at the upper-right corner are (W/2, H)

The coordinates at the lower-right corner are (W/2, 0)

The circle intersects the rectangle somewhere vertically between these points (as long as the rectangle is not too wide).

The y-coordinate of that intersection point is h (i.e., the height of the intersection point above the x-axis).

The equation for a circle of radius R whose center has been shifted vertically up R units from (0,0) is:

x² + (y - R)² = R²

Solving for y, we get two equations (one plots the circle's upper-half and the other plots the lower-half).

For the circle's lower-half (tangent to the x-axis), the equation is:

y = R - √[R² - x²]

We're interested in y, when x = W/2, because that's the height of the intersection point (h).

h = R - √[R² - (W/2)²]

Otis,
Thank you so much!. I tried it with the appropriate constants and the answer totally makes sense. It explains some results I was seeing and now I know just how to adjust the radius. This is very satisfying to me. I no longer even care that I couldn't figure it out for myself LOL!.

If you're interested in the real world problem you solved go and look at a (non-classical) guitar. On most guitars, if you look at the side of the fretboard you'll see that each fret has a little part on its underside that embeds into the surface of the fretboard so as to attach it (this is called the tang). You'll also notice that there is curvature across surface of the fretboard (and the frets) from side to side. The degree of curvature is the radius of the fretboard. The problem is how deep to cut the slots that the frets will sit in so that there is enough depth to accommodate the tang. Because of production constraints when using a table saw the fret slots are all cut to the same depth before the fretboard is "radiused". After you "radius" the fretboard, the fret slots get shallower the further away you go from the center of the fretboard. If you don't cut the fret slots deep enough the fret won't be able to sit flush with the fretboard on the outer edges. Perhaps this is hard to envision but it's a huge help to be able to calculate this. Thanks again. Peter

 

[Otis]

… when using a table saw the fret slots are all cut to the same depth before the fretboard is "radiused" …

Interesting, but I didn't realize a table saw could be so precise; that is, I would have thought the fret slots are measured to a fraction of a millimeter. I've cut only plywood sheets and 2×4s on a table saw, where plus-or-minus 3/8th-inch was good enough, heh. So, after you cut the slots, the wood is still shaped like a rectangular solid. What kind of tool do you use, to round the upper surface to a specific radius?

 

[pjdeklerk]

Interesting, but I didn't realize a table saw could be so precise; that is, I would have thought the fret slots are measured to a fraction of a millimeter. I've cut only plywood sheets and 2×4s on a table saw, where plus-or-minus 3/8th-inch was good enough, heh. So, after you cut the slots, the wood is still shaped like a rectangular solid. What kind of tool do you use, to round the upper surface to a specific radius?

So glad you asked since it's about my favorite thing to talk about.

I just acquired a table saw blade designed for cutting fret slots which has teeth .023 inches wide. You then have to build a crosscut sled for the table saw that allows you to make precise 90 degree cuts. You also need a kind of template that allows you to make your cuts at exactly the right spacing (so the instrument will be in tune with itself). Once you have all that it is very easy to cut precise fret slots in an un-radiused fretboard. Of course you only cut part way through the board. The formula you provided allows me to know exactly how high the blade needs to be.

The usual way to radius the fretboard is to use a sanding block which has a sanding surface carved out to the desired radius (it is concave of course while the fretboard will be convex.) You can buy these but they are overpriced like most specialty tools. My next project is build a contraption that is basically a pendulum that holds a router. This allows you to cut the sanding blocks to the desired radius. It's all great fun!.