Geometry Question re: Hexagons

Malintent

Veteran Member
For a woodworking project... I want to cut a smooth circle from the outer perimeter of a constructed hexagon. I need to calculate how wide a board I need to construct the hexagon with, such that there is sufficient material to cut away to form a circle, while maintaining integrity of the construction..

It is a geometry question, stated simply as, "what is the maximum distance of a circle's circumference from an inscribed hexagon"? No luck finding an answer to that, so here is a more detailed means of expressing what I need to find out:

inscribe a hexagon within a circle, such that each of the 6 vertices of the hexagon are touching the circle. The radius of circle is equal to the length of each segment of the hexagon, as well as its radius.

At every center point of each segment of the hexagon, the circle's circumference is at its maximum distance from the hexagon's segment.

What is that distance?

That distance would be the exact width of the board, if used to construct the hex, where cutting the circle would just exactly separate the attached segments (not good). So I would take that distance and add the amount of material I need for structural integrity. I suppose the length of each board (segment of the hex) would need to be my desired final radius, plus 1/2 the distance I am looking for.

right?

Sarpedon

Veteran Member
I don't know how to do this mathematically, so I fired up the ol' AutoCad and drew it. Frankly, I suggest you start with the piece of wood you want and work backwards graphically rather than mathematically.

Anyway: I ended up with the following: With a Circle 120 units in radius, the midpoint of the hexagon was 104 units from the center and 16 units from the edge of the circle. The amount was not exact, but this difference will be insignificant compared to your tool thickness at this scale.

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beero1000

Veteran Member
If I'm understanding your setup correctly, you want the difference between the circumradius and inradius of a regular hexagon. If the hexagon has side length S, the distance is $$(1 - \frac{\sqrt{3}}{2})S$$

TeX seems to be out right now, so I'm stuck typesetting like some sort of savage

It's (2 - √3)S/2

Treedbear

Veteran Member
I don't know how to do this mathematically, so I fired up the ol' AutoCad and drew it. Frankly, I suggest you start with the piece of wood you want and work backwards graphically rather than mathematically.

Anyway: I ended up with the following: With a Circle 120 units in radius, the midpoint of the hexagon was 104 units from the center and 16 units from the edge of the circle. The amount was not exact, but this difference will be insignificant compared to your tool thickness at this scale.

~0.5% smaller than exact

Junior Member
It is a geometry question, stated simply as, "what is the maximum distance of a circle's circumference from an inscribed hexagon"?

I tackled this with pencil, paper, and calculator, assuming a 1 unit length for the radius of the circle and the sides of the hexagon. Calculating the altitude of an equilateral triangle with sides of 1 gives a little over 0.866; this would be the the distance from the center of the circle/hexagon to the midpoint of the side of the hexagon. Subtracting 0.866 from 1 gives 0.134, the distance from the midpoint of the side of the hexagon out to the circumference of the circle. The proportions are similar to what Sarpedon came up with.

Malintent

Veteran Member
If I'm understanding your setup correctly, you want the difference between the circumradius and inradius of a regular hexagon. If the hexagon has side length S, the distance is $$(1 - \frac{\sqrt{3}}{2})S$$

TeX seems to be out right now, so I'm stuck typesetting like some sort of savage

It's (2 - √3)S/2

"circumradius and inradius" are the terms I didn't know that resulted in not finding my answer on my own. Thank you for the education!

so, now that I was able to construct a concise question for Google, I easily found this regarding the relationship between the two values:

"since such a ratio is just the ratio between the side and the height of an equilateral triangle, (R/r) = (2/sqrt(3))"

which is apparently a constant ratio... approximately (way more precise than needed) R - r (the value I need) = R * 0.134

.. just like ya'll said. Thanks again!

so, with a 1.5 foot segment length, I will need 2.41 inches of "spare room" on each board for the arc of the circle. Since I want 1 inch of material inside the finished piece (for structural integrity), I will need to use at least a 3.1 inch wide board. so, 4 inch lumber boards will be perfect (they are 3.5").

I imagined I would have been able to do this with 3 inch boards (2.5), but I was wrong. THAT'S why I like to calculate before cutting materials

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Sarpedon

Veteran Member
Tool thickness. Those saws at the sawmill are massive.

Malintent

Veteran Member
Tool thickness. Those saws at the sawmill are massive.

Well, I won't be milling my own lumber... my blades are the standard 1/8" kerf. But that isn't a measure of accuracy, just waste.

Malintent

Veteran Member
4 inch lumber boards will be perfect (they are 3.5")

Obviously.

Is it? It was never obvious to me that a "2 by 4" actually measures 1.75x3.5. Or were you being sarcastic and didn't realize (like most, I would imagine) that lumber is prepared that way commercially?

Malintent

Veteran Member
anyway... thank you all again for the "magic ratio" of 1 : 0.134... such an easily workable answer.

Jimmy Higgins

Contributor
Why don't you just get a hexagonal saw bit?

beero1000

Veteran Member
Tool thickness. Those saws at the sawmill are massive.

It's not tool thickness. Dimensional lumber is cut to those dimensions at the mill, then dried and planed. The drying fibers contract and the planing takes off more, so you lose some size.

bilby

Fair dinkum thinkum
Tool thickness. Those saws at the sawmill are massive.

It's not tool thickness. Dimensional lumber is cut to those dimensions at the mill, then dried and planed. The drying fibers contract and the planing takes off more, so you lose some size.

Excuses, excuses.

There is absolutely nothing that prevents the production of a board that is 4in wide; and even if there was, there would be nothing preventing the lumber yard from calling boards that are 3.5in wide '3.5 inch boards'.

The reason why they don't is tradition - it's a hangover from the guild system, wherein the 'mysteries' of the trade are deliberately obscure and confusing, to make it difficult for people who didn't go through the guild apprenticeship system to compete against guild carpenters.

All pre-industrial trades, from carpenters to lawyers, have deliberately imposed barriers to understanding, erected to keep their monopoly.

Of course, there remains some value in the argument that a person who doesn't have enough experience with working timber to know how wide a four inch board actually is, should be discouraged from attempting any serious carpentry work.

But now that the information is so widely available, it seems a bit pointless to keep up this silliness.

Bronzeage

Super Moderator
Staff member
It's not tool thickness. Dimensional lumber is cut to those dimensions at the mill, then dried and planed. The drying fibers contract and the planing takes off more, so you lose some size.

Excuses, excuses.

There is absolutely nothing that prevents the production of a board that is 4in wide; and even if there was, there would be nothing preventing the lumber yard from calling boards that are 3.5in wide '3.5 inch boards'.

The reason why they don't is tradition - it's a hangover from the guild system, wherein the 'mysteries' of the trade are deliberately obscure and confusing, to make it difficult for people who didn't go through the guild apprenticeship system to compete against guild carpenters.

All pre-industrial trades, from carpenters to lawyers, have deliberately imposed barriers to understanding, erected to keep their monopoly.

Of course, there remains some value in the argument that a person who doesn't have enough experience with working timber to know how wide a four inch board actually is, should be discouraged from attempting any serious carpentry work.

But now that the information is so widely available, it seems a bit pointless to keep up this silliness.

Anyone who has ever worked with "rough" lumber will happily pay for a 4 inch board and take a 3.5 board.

In any case, whatever lumber one buys in the US needs to be measured carefully. They may say a sheet of plywood is 96 inches by 48 inches, but it actually cut to the nearest centimeter past 96 and 48.

beero1000

Veteran Member
It's not tool thickness. Dimensional lumber is cut to those dimensions at the mill, then dried and planed. The drying fibers contract and the planing takes off more, so you lose some size.

Excuses, excuses.

There is absolutely nothing that prevents the production of a board that is 4in wide; and even if there was, there would be nothing preventing the lumber yard from calling boards that are 3.5in wide '3.5 inch boards'.

The reason why they don't is tradition - it's a hangover from the guild system, wherein the 'mysteries' of the trade are deliberately obscure and confusing, to make it difficult for people who didn't go through the guild apprenticeship system to compete against guild carpenters.

All pre-industrial trades, from carpenters to lawyers, have deliberately imposed barriers to understanding, erected to keep their monopoly.

Of course, there remains some value in the argument that a person who doesn't have enough experience with working timber to know how wide a four inch board actually is, should be discouraged from attempting any serious carpentry work.

But now that the information is so widely available, it seems a bit pointless to keep up this silliness.

Nothing except marketing - bigger is better, and if they can sell you a 3.5" board and make you think you're buying a 4" board then they don't mind at all. Even if they get sued for it.

Kosh

Junior Member
Excuses, excuses.

There is absolutely nothing that prevents the production of a board that is 4in wide; and even if there was, there would be nothing preventing the lumber yard from calling boards that are 3.5in wide '3.5 inch boards'.

The reason why they don't is tradition - it's a hangover from the guild system, wherein the 'mysteries' of the trade are deliberately obscure and confusing, to make it difficult for people who didn't go through the guild apprenticeship system to compete against guild carpenters.

All pre-industrial trades, from carpenters to lawyers, have deliberately imposed barriers to understanding, erected to keep their monopoly.

Of course, there remains some value in the argument that a person who doesn't have enough experience with working timber to know how wide a four inch board actually is, should be discouraged from attempting any serious carpentry work.

But now that the information is so widely available, it seems a bit pointless to keep up this silliness.

Anyone who has ever worked with "rough" lumber will happily pay for a 4 inch board and take a 3.5 board.

In any case, whatever lumber one buys in the US needs to be measured carefully. They may say a sheet of plywood is 96 inches by 48 inches, but it actually cut to the nearest centimeter past 96 and 48.

I've recently built a cnc machine, and it's pretty annoying to be dealing with sheet goods that are supposedly 19 mm thick but I'm finding them from 17 mm on up......I have to measure each piece and adjust the program accordingly.

Cheerful Charlie

Contributor
To understand the relationships of hexagons to circles, google for Unit Circle. Then you will want a nice trig capable calculator.

Treedbear

Veteran Member
... I need to calculate how wide a board I need to construct the hexagon ...

Are you talking about a standard 5 sided hexagon or one that actually has 6 sides?

Malintent

Veteran Member
... I need to calculate how wide a board I need to construct the hexagon ...

Are you talking about a standard 5 sided hexagon or one that actually has 6 sides?

take a 3-legged dog and feed it 1 square meal. Then take it's poop and use it to fertilize a 5-sided pentagonal hexagon. It will eventually grow another leg, which you can either use to make an actual hexagon, or just give it to the dog to use.

I thought everyone knew this...