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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.

Prometheus wrote:

That is incorrect. By using the radius, you lay out six equilateral triangles around the centre, each side being equal to the radius, with internal angles being 60 degrees. The difference between pi and 3 is the difference in length between the straight line and the arc.

Bingo!

Give that man a cigar.

If "several" is not too many, I would just lay them out with pencil lines and align the cut lines with the edge of a sled.

One of the handiest things I have made is a small sled, about 18" front to back, with a single runner. For the width, I just ran runner first in the left miter slot, then the right, so that either side corresponds with the cut of the saw blade. I have several slots routed in it to position hold downs, or sometimes just screw down a toggle clamp where it is most needed.

It is easy to draw a hexagon with just a compass, if you don't know the method google or ask.

#### Site Timeline

- posted on October 29, 2006, 9:20 pm

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.

- posted on October 29, 2006, 11:48 am

That is incorrect. By using the radius, you lay out six equilateral triangles around the centre, each side being equal to the radius, with internal angles being 60 degrees. The difference between pi and 3 is the difference in length between the straight line and the arc.

--

BigEgg

Hack to size. Hammer to fit. Weld to join. Grind to shape. Paint to cover.

BigEgg

Hack to size. Hammer to fit. Weld to join. Grind to shape. Paint to cover.

Click to see the full signature.

- posted on October 29, 2006, 1:02 pm

Bingo!

Give that man a cigar.

--

Regards,

Doug Miller (alphageek at milmac dot com)

Regards,

Doug Miller (alphageek at milmac dot com)

Click to see the full signature.

- posted on October 29, 2006, 12:31 pm

[snip description of arc-and-compass construction of hexagon]

Nonsense. This produces a perfect hexagon, regardless of size.

Get out a piece of paper, a compass, a straightedge, and a pencil. Mark a point in the middle of the paper. We'll call that point A. Set the compass to any arbitrary radius, and draw a circle with its center at A. Now make a mark at any arbitrary point on the circumference of the circle. Call it B. Draw a straight line connecting A and B. Call its length R.

Set your compass on B and draw an arc intersecting the circle at one point. Label it C. Draw a straight line between B and C. Its length is also R.

Likewise, connect A and C.

Now the interesting part: since A is the center of the circle, and C is a point on its circumference, the distance between A and C is***also*** R, and
therefore ABC is an equilateral triangle, and the angle BAC is 60 degrees.

Exactly one-sixth of a circle.

Put the point of the compass at C, and construct a new point D. Repeat four more times. You've constructed a total of six equilateral triangles, each with one vertex at the center of the circle, and its other two vertices on the circumference.

Perfect hexagon.

[snip description of workable but clumsy method]

That, of course, is why most folks just use a compass.

Nonsense. This produces a perfect hexagon, regardless of size.

Get out a piece of paper, a compass, a straightedge, and a pencil. Mark a point in the middle of the paper. We'll call that point A. Set the compass to any arbitrary radius, and draw a circle with its center at A. Now make a mark at any arbitrary point on the circumference of the circle. Call it B. Draw a straight line connecting A and B. Call its length R.

Set your compass on B and draw an arc intersecting the circle at one point. Label it C. Draw a straight line between B and C. Its length is also R.

Likewise, connect A and C.

Now the interesting part: since A is the center of the circle, and C is a point on its circumference, the distance between A and C is

Exactly one-sixth of a circle.

Put the point of the compass at C, and construct a new point D. Repeat four more times. You've constructed a total of six equilateral triangles, each with one vertex at the center of the circle, and its other two vertices on the circumference.

Perfect hexagon.

[snip description of workable but clumsy method]

That, of course, is why most folks just use a compass.

--

Regards,

Doug Miller (alphageek at milmac dot com)

Regards,

Doug Miller (alphageek at milmac dot com)

Click to see the full signature.

- posted on October 30, 2006, 6:38 am

Prometheus wrote:

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

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

- posted on October 31, 2006, 10:00 am

On 29 Oct 2006 22:38:47 -0800, snipped-for-privacy@yahoo.com wrote:

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

Should be there fairly quickly.

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

Should be there fairly quickly.

- posted on October 31, 2006, 11:22 am

On 29 Oct 2006 22:38:47 -0800, snipped-for-privacy@yahoo.com wrote:

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.

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.

- posted on October 29, 2006, 2:54 am

If "several" is not too many, I would just lay them out with pencil lines and align the cut lines with the edge of a sled.

One of the handiest things I have made is a small sled, about 18" front to back, with a single runner. For the width, I just ran runner first in the left miter slot, then the right, so that either side corresponds with the cut of the saw blade. I have several slots routed in it to position hold downs, or sometimes just screw down a toggle clamp where it is most needed.

It is easy to draw a hexagon with just a compass, if you don't know the method google or ask.

--

No dumb questions, just dumb answers.

Larry Wasserman - Baltimore, Maryland - snipped-for-privacy@charm.net

No dumb questions, just dumb answers.

Larry Wasserman - Baltimore, Maryland - snipped-for-privacy@charm.net

Click to see the full signature.

- posted on October 29, 2006, 3:57 am

Neal wrote:

The easiest way to draw an hexagon is...

The board you are cutting it out of find its middle down opposites sides a&b and draw a line across,do the same with sides c&d. Now draw a mark, 5" either side of each line at the edge of the board,now if you draw straight lines with a ruler or straight edge from each 5" point to point this will give you an hexagon with six equal 10" sides

The easiest way to draw an hexagon is...

The board you are cutting it out of find its middle down opposites sides a&b and draw a line across,do the same with sides c&d. Now draw a mark, 5" either side of each line at the edge of the board,now if you draw straight lines with a ruler or straight edge from each 5" point to point this will give you an hexagon with six equal 10" sides

--

Sir Benjamin Middlethwaite

Sir Benjamin Middlethwaite

- posted on October 29, 2006, 4:01 am

The3rd Earl Of Derby wrote:

oops! sorry its getting late for me. :-(

That'll give you eight sides.

oops! sorry its getting late for me. :-(

That'll give you eight sides.

--

Sir Benjamin Middlethwaite

Sir Benjamin Middlethwaite

- posted on October 29, 2006, 7:06 am

On Sun, 29 Oct 2006 04:01:32 +0000, The3rd Earl Of Derby wrote:

(clip happens)

So he can just use the two extra sides to start the next hexagon, right?

}-)

(ducking!!!)

Bill

(clip happens)

So he can just use the two extra sides to start the next hexagon, right?

}-)

(ducking!!!)

Bill

- posted on October 30, 2006, 12:04 am

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

Then I'd use a tablesaw pattern jig to make as many as needed. IMHO this is a far simpler solution than a sled. http://www.woodcentral.com/cgi-bin/readarticle.pl?dir=jigs&file=articles_146.shtml http://www.popularwoodworking.com/features/fea.asp?id 34

Art

Then I'd use a tablesaw pattern jig to make as many as needed. IMHO this is a far simpler solution than a sled. http://www.woodcentral.com/cgi-bin/readarticle.pl?dir=jigs&file=articles_146.shtml http://www.popularwoodworking.com/features/fea.asp?id 34

Art

- posted on October 30, 2006, 5:56 am

On Sun, 29 Oct 2006 16:04:55 -0800, "Wood Butcher"

That's pretty slick. I'm going to have to try that one out.

That's pretty slick. I'm going to have to try that one out.

- posted on October 30, 2006, 4:30 pm

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:

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:

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