Hmm. Archimedes trisected an angle. You really should keep up.
You are right about "squaring the circle." Can't be done since pi is a
trancendental number. Can't be done with algebra either, unless you simply
express the area as an equation. In other words, it can't be "solved" for a
value.
Possibly. I was drunk the day all that was covered.
HeyBub,
Yes, Archimedes trisected an angle, but his method requires putting marks on
a straight edge, which advocates of pure geometry don't allow.
Kerry
While you guys were trisecting and trancendentaling (sounds painful), I went
ahead and built my bending form. Came out nicely. Thanks again for the
help.
jc
"Advocates of pure geometry"? The challenge issued in ancient times
was to square the circle using Euclidean geometry. Euclidean geometry
is a game with certain rules, one of which is no marks on the
straightedge. If as a practical matter you need to square the circule
then there are many ways to do it, however none of them answer the
original challenge.
This isn't a matter of "advocacy". It's a matter of obeying the rules
of the game. If you don't want to obey the rules of the game that's
fine, but then you're playing som other game, not "Euclidean
Geometry".
Ah, but I bet he used more than just a compass and straightedge.
And yes, I know of the marked straightedge method. It works, but purists
don't like it.
Actually, you can square the circle provided you're not limited to
a straight edge and compass.
Method:
1. Make a wheel the size of the circle to be squared.
2. Mark a point on the wheel.
3. Use wheel to measure distance on line equal to circumference.
Finally, use the standard method of computing the square root of
the length of the line segment you got in step 3.
See? It's easy provided you're not restricted to just a compass and straight
edge.
And what exactly does that have to do with geometry? Your method may
be perfectly valid mathematically but that doesn't make it "geometry".
And that assumes that it in fact squares the circle, which it does
notthe challenge in squaring the circle is to find a square with the
same area and I don't see how knowing the square root of the
circumference helps you in that endeavor. It gives you s=SQRT(2*pi*r)
and what you need is s=r*SQRT(pi).
Yes, he did, but he wasn't using geometry. He was doing something
which might to a layman look like geometry but he was not following
the rules that define geometry.
You might find http://www.jimloy.com/geometry/trisect.htm to be of
interest.
The equation is the general solution. Particular values are of little
interest in mathematics. Squaring the circle was the first great
triumph of analytic geometry.
Joe wrote:
 I need some trig help. I need to create a bending form but I need
 the diameter (or radius, it doesn't matter) based on the following

 The diameter of a circle formed by an arc with a 1/8" apex over a 1
 5/16" base length. Yeah, I know some of the terms are wrong, but
 I'm a long way removed from Mr. Cole's high school trig class.
Sorry to have arrived so late. This gets asked enough that I put up a
web page with diagram and stepbystep algebra at the link below.

Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto/CNC/cove_geom.html
Where A is the apex (or sagitta) and B is the chord (A = .125 and B 1.3125 in the question) then the Radius R is given by:
R =(A^2 + (B/2)^2)/(2A)
derived from:
R(1cosT) = A
RsinT = B/2
where T is the half angle of the arc.
Tom Veatch
Wichita, KS
USA
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