On Sat, 28 Oct 2006 22:39:23 -0500, ()
wrote:

In article ,
Prometheus wrote:
On 28 Oct 2006 15:06:12 -0700, "
wrote:

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

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.

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?

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.