Date: Tue, 29 Aug 2006 19:52:09 -0700 (MST)
From: Spencer W Hunter <shunter@U.Arizona.EDU>
Subject: Re: How to Construct a Hexagonal Parabolic Dome

Thank you for your reply, which is the first feedback I've received on
it [ "How to Construct an Hexagonal Parabolic Dome" ] since I wrote it
for James Fischer in July of 1999!  I am rather fond of the parabola
generator image, which someone forwarded to me after I described my
high school invention, but I can see where it might not be that
helpful.  Perhaps a small text image will help you grasp the general
idea, which follows.

If I were to construct a 2v paraboloid instead of the 6v one described
in the instructions, I would need these templates:

1) a'aa

2) a'b'b'
   abb'

...and the plan-view diagram would look like this (the periods are the
points of the triangles, and the letters between the periods represent
the triangle side lengths):

      .
     b  a
   .  b' .
 a   a' b' a
.  a  .  b  .

This equilateral triangle is one-sixth of the paraboloid, with the
center being the point on the lower left.  You will notice that it is
symmetrical about an axis that extends from the point on the lower
left to the point in the middle of the upper right side, so that
triangle abb' on the lower right is the mirror image of triangle ab'b
on the top.

One could extend this triangle indefinitely to the right for greater
frequency divisions, but because of the symmetry, you only need the
30-60-90 degree lower half to determine the unique templates.

Instead of actually drawing the parabola and measuring distances as
described in the instructions, you could use distance formulae to
determine the lengths of a', a, b', and b, using the equation
.5(x^2)=y from x=0 [x=-.25 for the obvious a'] to x=1:

a'= SQRT((.25 - -.25)^2 + (.03125 - .03125)^2) = .5
a = SQRT((.5 - 0)^2 + (.125 - 0)^2)            = .515 approx.
b'= SQRT((.75 - .25)^2 + (.28125 - .03125)^2)  = .559 approx.
b = SQRT((1 - .5)^2 + (.5 - .125)^2)           = .625

Since all this is probably more confusing than the original document,
feel free to ask any further questions you may have!  :-)

Spencer Hunter, Library Specialist
gopher://www.u.arizona.edu:80/hGET%20/%7Eshunter

> Hi,

> I've searched for plans to make a parabolic reflector, and your site
> is the only one I could find at this time. I am familiar with 6v
> geodesic geometry and I believe that I have a tenuous grasp of
> determining the line segments, but the arrangement and number of
> triangles has me confused.

> The photos and diagram of "apparatus for drawing parabolas" are
> nice, but not necessary.  A diagram of how to create the line
> segments and triangle arrangement would be worth a thousand words!