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?
Hi Neal,
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 go> I need to cut several hexagons about 10" across. Being a novice woodworker, I
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)
Deb
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
-Lee
Neal wrote:
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 cut.
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.
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.
It's been a while since I took geometry, but when I did, this method produced a "perfect" hexagon, withing the limits of accuracy of the layout tools of course. What does Pi != 3 have to do with it?
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
oops! sorry its getting late for me. :-(
That'll give you eight sides.
Yes, it is exact. The error often seen when doing it with a compass is due to placement of the compass point. Incremental error. It stacks up around the circle.
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 more accurate.
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.
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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 Example: 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) Then: 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, .954.
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.
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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 hexagon.
Set your compass to that length, and then draw an arc up from each corner.
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 arcs cross.
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.
You have based your calculation on arc length. You are looking for chord length, not arc length. Trig it out, you will see it.
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Agreed.
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)
where: 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)
(clip happens)
So he can just use the two extra sides to start the next hexagon, right?
}-)
(ducking!!!)
Bill
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.
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 lengthNow for the missing link:
s = [5 x 60 x 3.1415] / 180 s = 5.2358
Finally.
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 out.
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 segemented turnings.
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 nicely.
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.
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.
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