Hexagon cutting tablesaw jig

Actually, to four decimal places, it's 3.1416, but that's only a minor quibble.

No, you're not. You're constructing *chords* using a length that is one-half of the diameter.

No, you aren't. Think about it some more. Draw if if necessary.

There is no error, regardless of the size of the hexagon, and no mythical seventh side is ever needed.

[snip]

Helloooooooo!!! If you've just used the compass to draw that circle, it's

*already* set at exactly the distance between the center and the circumference. No "adjustment" is necessary, and any change in the setting of the compass is guaranteed to introduce error.

This should be interesting... a "proof" of a false proposition...

Hold it right there, cowboy. The sides of a hexagon are _straight_lines_, not circular arcs.

The remainder of the "proof" is ignored, since it rests on a demonstrably false proposition.

[snip]
[snip *another* clumsy construction]

Holy smokes! Talk about making things more complicated than they need to be! Yes, there is a geometric method for drafting a perfect hexagon. Several, in fact. The simplest one is the method using points around the circumference of a circle. Here's another:

Mark a point, A. Set the compass to any arbitrary radius. Draw an arc. Mark any point, B, on the arc. Connect A and B. Draw another arc, centered at B, that intersects the first arc. Mark the intersection C. ABC is an equilateral triangle.

Draw a third arc, centered at C. Draw fourth arc, centered at A, that intersects the third arc. Mark the intersection D. ACD is also an equilateral triangle.

Draw a fifth arc, centered at D, and a sixth, centered at A, that intersects the fifth arc. Mark the intersection E. ADE is also an equilateral triangle... and BCDE is half a hexagon.

Continue in a similar manner to construct points F and G, which define the equilateral triangles AEF, AFG, *and* AGB -- *and* the hexagon BCDEFG.

Now put your compass point at A and draw a circle.

Where are points B, C, D, E, F, and G?

Seems that you could use a refresher on the fundamentals yourself.

Reply to
Doug Miller
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What the hell are you talking about? Proposition 15 has *nothing* to do with this discussion. Further, Euclid provides a clear and simple proof of it anyway. If you have difficulty accepting it, well, tha's not *Euclid's* fault...

Operator error.

Operator error.

No, it doesn't...

.. and no, it's not. _There_is_no_error_ except in your understanding.

That'd be my guess...

I'm looking forward to reading your soon-to-be-released geometry textbook, in which you point out more of Euclid's errors.

Reply to
Doug Miller

Bingo!

Give that man a cigar.

Reply to
Doug Miller

Forget arc length. Neither pi or arc length have any bearing on the problem. No, using trigonometry is not assuming a hex and then constructing a circle about it, it is assuming a circle and inscribing a hex. The basis of the formulas is the radius of the existing circle. Care for some proof by AutoCAD? I can provide that too (I'm really surprised that you don't have a CAD program of some type).

Reply to
CW

Your proof unfortunately is predicated on the incorrect assumption you've made above; You are _NOT_ "tracing around the circumference..." You are using the _points of intersection_ of a chord length equal to the radius, with the circumference, to locate the position of the sides of the hexagon. Afer all, it is the _straight_ chord that makes up the side of the hexagon, not the _curved_ arc of the circle. Look at it this way: Forget the circle. If you start with the length of a hexagon side and start constructing equilateral triangles that share the appropriate sides, you will end up with the same hexagon as the circle method. The serendipitous nature of plane geometry provides us with an elegant shortcut.

(Never thought I'd use 'serendipitous' in a woodworking post, though I suppose we've strayed slightly from being on topic by now)

Reply to
lwasserm

I'll try making this type of jig today. So far, just using the miter gauge has not been very accurate due to my low quality gauge. Sleds seem to be the solution to many woodworking problems.

After reading about the the issue of the inaccuracy of drawing a hexagon using the radius was discussed, I checked the dimensions in my CAD program and found that using the radius produces a perfect hexagon.

Thanks everyone for your help and I'll let you know how it turns out.

Neal

Reply to
Neal

I do have AutoCAD, I just like to fiddle around without it whenever possible. It's great when I need it and have a computer around, but I don't always have a computer around when I'm working in the shop.

I see where the error was now (obviously on my end), I just had to take the long, long way around it to get it straight in my head. But at least I got it squared away, which is more important (to me) than having made an ass of myself about it.

Reply to
Prometheus

Yep. I messed it up in my head. I had Geometry in school a long time ago, but was too busy screwing off to really pay attention and ask questions. Since then, I've been working at it off and on with a copy of Euclid's Elements for general knowledge purposes, and that was the one proposition I couldn't figure out. Kind of a dumb block to have, but it was persistant. Should make some of the more complex propositions a little easier now. I just wish I knew why I can't get the damn thing to draw out properly- everything else comes out fine.

Reply to
Prometheus

Use Lee's ( snipped-for-privacy@gmail.com) method (except use the radius instead of the diameter) and make your first one.

Reply to
Wood Butcher

Reply to
Prometheus

sounds like a good tutorial. I for one would like to see a pics-and-words .pdf about it...

Reply to
bridgerfafc

Hi,

How many do you need? Will there be a hole in the center that can be used as an index? Below is my first guess.

I would start by ripping the material into 10 inch wide strips, this produces sides 1 and 4 of the hexagon.

If there is a center hole, one could make a sled with a pin whose center is 5 inches from the near edge of the blade. Add a fence that is

5" from the center of the pin and at 30 degrees ( 60 degrees from the blade.) With side 1 against the fence cut side 2. You could then either rotate the piece to cut side 3 or do a flip so the side 4 is against the fence. There might be some small operational advantage in always using side 1 or 4 against the angled fence to avoid a cumulative error from inaccuracy in setting the angle of the fence.

Without a central hole things are a bit harder. Cut the 10" wide strips into pieces to a bit longer than the point-to-point width of the hexagon ( about 10.77 inches I think.) Set the miter (or sled fence) to

30 degrees and with the length stop on the miter to a bit over the point-to-point width. Cut side 2. Without changing the length stop flip the piece over top-to-bottom and cut side 3. Check that sides 2 and 3 are the same length. (Repeat this procedure for all of your hexagons before changing the stop. And make a few test pieces for tuning the last part) Here is where things take some care. Use a length stop that reaches the point formed by sides 2 and 3 after the piece is flipped end-to-end. Set the length stop very carefully so that the so that the distance from point formed by sides 2 and 3 is the point-to-point width of your hexagon ( 10.77") Cut side 5, flip, and cut side 6. Use test pieces first and test everything. I would start and bit long and slowly shorten the length because I could keep using the same test piece. I suppose one could try to get the length stop set before cutting sides 2 and 3, but I think the above procedure is allows on to fix things if something is off, rather than cutting the pieces undersized.

Hope this makes some sense.

Roger Haar

Neal wrote:

Reply to
Roger Haar

Don't have a camera, but I'll draw it up and post on alt.binaries.pictures.woodworking.

Should be there fairly quickly.

Reply to
Prometheus

It's posted to ABPW as a jpg, under the heading "Hexagon Jig" Didn't seem worth making a .pdf, as it's only one page.

If you don't have access to the binary groups for some reason (Google, free newsserver, etc.) I can e-mail it to you if you like.

Reply to
Prometheus

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