I need to cut several hexagons about 10" across. Being a novice woodworker, I
have not been able to cut them accurately and have been searching for a jig and
have not been able to find one. Can anyone direct me to instructions how to cut
accurate hexagons or where I might be able to find jig plans?
How accurately do you need them? It shouldn't be to difficult to do
with a miter gauge. If you need more accuracy, and you're going to do
a bunch of them, a dedicated sled could be made.
Either a dedicated sled or a Incra 1000SC miter gauge. The dedicated sled
would be a lot cheaper and probably quicker to come by. My geometry not
being fresh, I don't know the exact numbers to use on your framing square,
but someone on this list can give you the numbers to use. Then all you
have to do is line up the proper marks on the framing square with the blade
side of your sled, mark, attach a board along the line and (as they say in
"Short Circuit) "bimbo."
AS for making the sled , all you have to do is get a piece of 1/2" MDF or
birch plywood (don't use CDX - sheathing), cut it about 18 x 24. Then get
a piece of hardwood, 3/4 x 3/4 x 18, and attach it to what will be the
front top of the sled (make absolutely sure it is square to the edge.
After that, measure over a little more than the distance from your blade to
the miter slot, (using your framing square, or any other square) draw a
line, and attach a hardwood runner (3/4 x /3/8 x24) along that line with
screws and glue - again making absolutely sure it is square to the leading
edge of the board. After the glue dries, run the sled through the saw
blade and you will have an edge that is exactly fitted for your saw.
(This one is for a 90 degree cut, but the same principle applies for a 60
degree - except you do not cut the leading edge of the board at 60 degrees
and you place your backer board with the blade edge toward the middle of
the length of the sled. About 8 inches back with one going the other way
(opposite angle) back at about 14 inches)
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.
How are you laying it out? Are you using a compass to mark the six
sides? The diameter of the compass place anywhere on the circle,
rotated around the circle will give you all the six points to be used
for the straight lines. Then use the miter gauge of the table saw to
follow the lines. 30 degrees
On 28 Oct 2006 15:06:12 -0700, " email@example.com"
Yep- that's what they show you in Geometry class (except it's the
radius, not the diameter) when all you have is a straightedge and a
compass. It's not entirely accurate, though. If you do this, the
last side will be longer than the first five because PI is not exactly
three. The larger the hexagon, the greater the error.
I'd lay it out on plywood with a protractor, and then make a rectangle
around it (at 60 degrees to the edge) the size of the blank. It'll
look like a rectangle with a 60* wedge cut out of one side, if you lay
it out the way I'm feebly attempting to describe.
(note that when you cut your blanks and lay out the corresponding
rectangle, the width must be the distance between two parallel sides
of the hexagon and the length must be the distance between two
opposite points of the hexagon or longer, or this will not work.)
Cut that out, then set your fence on the table saw so that the plywood
just fits between the edge and the blade. There's your jig.
Fit the blank into the jig, and cut off the first corner. Remove the
blank, flip it over and set it back into the jig. Cut again. Remove
the blank and rotate it 180 degrees and seat the point against the
back of the cutout. Cut again. Flip it over, and make your final
Should be a perfect hexagon, assuming you laid it out carefully. Same
technique works for an octagon as well.
Damn, that took longer to describe that it would have to just make the
sucker and do the job! Hope it helps. Picture would have been worth
more- if you can't piece this together, let me know, and I'll draw it
up and post to ABPW.
On Sat, 28 Oct 2006 22:39:23 -0500, firstname.lastname@example.org ()
That's what they called it, anyhow. To get the circumference of a
circle, you use the equation Pi(3.1415) x Diameter. You're tracing
around the circumference of the circle using a length that is one-half
of the diameter. So if you rework the equation a little, it ends up
that you are assuming that Pi = 3, which it does not. (See below for
details, if you care) When you're doing this on a little circle, the
difference does not matter much- you end up with what appears to be a
"perfect" hexagon. When you use a larger circle to make a hexagon
using this method, the error is magnified, and it becomes necessary to
either make the last side longer to meet the starting point, or add a
short seventh side to complete the polygon.
It's an illustration of an old technique that they teach in schools to
show how Geometry was developed, and it's accessable for most students
and easy to remember- but depending on what you're doing, it's not a
perfect method. Considering that a plastic protractor can be got for
a dollar at most discount stores, it's easier to use that, and often
If you have to use a compass and a straightedge to make a hexagon, the
easiest method I've found was to draw a diameter line across the
circle, set the compass point on where the diameter and circle
intersect, adjust the distance so that the drawing point touches the
center, then swing an arc in each direction, and repeat on the other
side. It's still not "perfect", but it evens out the error and splits
it between two parallel sides. If you want to make a perfect one, the
method is below the proof.
A proof, for those who are unconvinced.
circumference(C) = 3.1415(Pi) x diameter(D)
D = 2 x radius (R)
C/6 = arc length of segments comprising a perfect hexagon
If Pi = 3, then C = Pi x 2R will produce a perfect hexagon
C = 3 x 2R [becomes]
C/3 = 2R [becomes]
C/6 = R
But if Pi = 3.1415.....(ad nauseum, but we'll use 3.1415)
C = 3.1415 x 2R [becomes]
C/3.1415 = 2R [becomes]
C/6.283 = R
So the radius used as a divider for the circumference of the circle
will not work. The first side uses .159 of the circumference, the
second (plus the previous side(s), as with each following step), .318,
the third, .477, the fourth, .636, the fifth, .795, and the sixth,
On a one-inch circumference circle, the error is .046 of an inch- good
enough for illustration purposes, and it looks perfect.
But if your hexagon is to be laid out on in a circle with a diameter
of ten inches, (a circumference of 31.415 inches) the error is
magnified to an unused arc length of 1.445 inches. Easy to see, and
difficult to explain away by noting that Geometry class called it a
method for making a perfect hexagon.
There is a geometric method for drafting a perfect hexagon via a
perfect hexagram, if you're a stickler for accuracy, and only have a
compass and straightedge.
Draw a segement that is the length of the width of the desired
Set your compass to that length, and then draw an arc up from each
Draw a line from the end of each segement to where the arcs meet. You
now have a perfect equilateral triangle that is pointing up.
Drop an altitude line from each point to find the center of the
triangle. (bisect each side of the triangle and draw a line from the
center of each line to the opposite point). Where the three lines
meet is the center of the triangle.
Draw a random length segement off to one side that is more than twice
the distance from the center of the triangle to any side. Set the
compass to the distance from the center of the triangle to any edge
(if you were to draw a circle with this setting, it should touch, but
not cross, each edge of the triangle in the center of each side.)
Set the compass on one end of the random segment, and make a hash mark
on the segment. Move the pivot point to the hash mark, and make
another. We'll call this reference length "A"
Go back to your equilateral triangle. Extend your starting segement a
little with the straightedge, and then draft a perpendicular line (set
the compass on the corner point, make a hash mark on either side, then
open it up a little, set on each hash mark, and make a cresent shape
from each side. Draw a line through the two points where the cresents
cross- that's your perpendicular)
Set your compass to reference length "A", and place the pivot on the
corner you raised the perpendicular from. Make a hash above the first
segement at that length. Repeat this, and the block of steps above on
the other side of the original segment.
Draw a line that connects the two hash marks on the perpendiculars you
just drew. This is the base of the downward pointing triangle.
Set the compass to the length of your base, then draw two arcs
downwards and draw a line from each end of the segement to where the
You now have a perfect hexagram (with a lot of hash marks and layout
lines) To make the perfect hexagon, draw a series of lines connecting
each point to the next around the perimeter.
Or, you could buy a protractor, and call it a day! :)
If anyone really is terribly interested in this method, and my text
description isn't clear, I can draw a picture and post it in the
binary group on request. Doesn't hurt to know the fundimentals.
Too tired to mess around with the trig involved tonight (it's not as
fun as playing with compass and paper), but it would seem immediately
apparent to me that an inaccurate arc length would automatically
result in an inaccurate chord length. After all, the cord is defined
by the points on the arc. Unless the circle is irregular, they should
be in perfect correlation.
Euclid's Proposition 15 has always been a pet peeve of mine. I like
my math to work from every angle. You may be correct, and it's
certainly the prevailing wisdom- but no matter how many times I have
tried it, even with locked compasses and large circles, the last arc
never aligns with the start point. If the error isn't there, the
degree of error should always be more or less the same, but larger
circles have a larger regular error- and that shouldn't be the case If
I'm just missing the mark with the compass. A big circle won't make me
consistantly miss the mark by a wider margin. To make it even more
annoying, if you keep following the marks around the circle, the error
increases by exactly the same increment each time you pass the
starting point. If you do it until you fill up the circle, it's as
regular as gear teeth.
Maybe I'm just very consistant with my poor compass placement, but the
method I described comes out much closer to perfect upon measurement
than Euclid's. You'd think the added complexity would be more likely
to mess it up than it would be to increase the accuracy.
That being said, I've long since rushed off the cliff on this from
useful shop technique into geometry nerd senselessness. Doing it by
using the radius is probably close enough for most stuff- but I still
do it my way, or (99% of the time) with a protractor head. I just
don't trust the method they taught us on this one.
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*
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.
Doug Miller (alphageek at milmac dot com)
Actually, to four decimal places, it's 3.1416, but that's only a minor
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.
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
The remainder of the "proof" is ignored, since it rests on a demonstrably
[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
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
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.
Doug Miller (alphageek at milmac dot com)
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)
For every complicated, difficult problem, there is a simple, easy
solution that does not work.
On Sun, 29 Oct 2006 09:30:50 -0600, email@example.com ()
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.
Interior lines from the center of a hexagon to each vertex form 6
Isosceles triangles, each with an included angle of 60 degrees. The
length of each of those interior lines is equal to the radius of the
circle concentric with the hexagon and passing through the vertices.
Length of the base of an Isosceles triangle is 2 X length of side X
sine of 1/2 the included angle.
L = 2 R sin (1/2 A)
L= length of the base
R = length of the sides of the triangle = radius of the circle
A = angle included by the sides
But since A is 60 degrees, 1/2 A = 30 degrees and sin(30) = 0.5000
So, L = 2R X (.5) = R
From which the length of the side of a hexagon is identical to the
radius of the hexagon's circumscribing circle. (and those 6 included
angles are not only isosceles, but are, in fact, equilateral)
I can't argue with the trig- I can immediately see that it is correct.
Here's my only gripe with proving Euclid's method via trig- When you
prove it using trig, you are starting with the hexagon, and
circumscribing a circle around it. It's accurate and mathematically
perfect. And it works just fine. I can draft a regular hexagon, find
the center and circumscribe a circle about it very simply, and it
works great- with no fudging about compass placement error. But I
still can't get Euclid's technique to work, and I haven't seen anybody
do it with full sucess, even when it was being taught.
I am missing something here, though, and just realised it. The arc
length is s = r alpha Pi / 180, right?
So if you do the problem via trig, you come up with a cord length of
five on a circle with a radius of five.
Which is what the compass is drawing, right?
So if I'm looking at the arc length by working backwards from the
formula for the circumference of a circle, I'll come up with:
c = 3.1415 x 10 = 31.415
31.415 / 6 = 5.2358 - the arc length
Now for the missing link:
s = [5 x 60 x 3.1415] / 180
s = 5.2358
You have no idea how long that damn thing has been bugging me. At
least I learned something this evening that will put that problem to
rest. Must have missed that day in class somewhere along the line.
Still not using Euclid for that one, though- I can't get it to come
Anyhow, the method I described for drafting a hexgon from a segement
works too, and you don't need to figure out what size circle you need
in the first place if you know how wide you want it to be. And the
jig works, too- I've used it several times for the bottoms of
Thanks for the correction, CW and Tom. I just have an aversion to
taking things on faith if I can't see all the pieces fitting together
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).
HomeOwnersHub.com is a website for homeowners and building and maintenance pros. It is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.