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Posted by dow on March 22, 2010, 1:54 am
 
I posted an enquiry in the "sci.math" newsgroup, and got the following
reply:

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Robert Israel    View profile
  More options Mar 21, 9:39 pm

Newsgroups: sci.math
Date: Sun, 21 Mar 2010 20:39:59 -0500
Local: Sun, Mar 21 2010 9:39 pm
Subject: Re: paraboloid centre of gravity

The focus of the paraboloid z = r^2 (in cylindrical coordinates) is
at
z = f = 1/4.  The surface area of this paraboloid from z=0 to z=c^2
(and thus r = 0 to r=c) is
A = int_0^c 2 pi r sqrt(1 + 4 r^2) dr = (pi/6) ((1+4c^2)^(3/2) - 1)
The corresponding z moment is
M_z = int_0^c 2 pi r^3 sqrt(1 + 4 r^2) dr
    = (pi/60) ((6 c^2 - 1)(1+4c^2)^(3/2) + 1)
To make the centre of mass at the focus we need M_z = A/4.  This
is equivalent to
  (1 + 4 c^2)^(3/2) = 7/(7 - 12 c^2)
Square both sides, discard the solution c=0, and you get
  2304 c^8 - 960 c^6 - 800 c^4 + 120 c^2 + 105 = 0
This polynomial has two positive solutions, approximately
  .6796708312 and .8163613643
but the second is not a solution of the original equation (it would
have
 (1 + 4 c^2)^(3/2) = -7/(7 - 12 c^2)).  The first solution has
c/f = 2.718683325 approximately.  Yes, I think it is a coincidence
that
this is close to e.
--
Robert Israel              isr...@math.MyUniversitysInitials.ca
Department of Mathematics        http://www.math.ubc.ca/~israel
University of British Columbia            Vancouver, BC, Canada



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This guy obviously knows a lot more math than I do. He apparently
agrees with my numerical conclusion about the dimensions of the
reflector, but the closeness of the rim-radius : focal-length ratio to
"e" is apparently just accidental coincidence.

Oh well....

                     dow

Posted by dow on March 22, 2010, 2:06 am
 
Incidentally, using Robert Israel's value for the rim-radius:focal-
length ratio, the depth of the bowl is 1.847809755 times the focal
length. To five significant digits, that's exactly the same as my
little program calculated. That's gratifying!

                       dow

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Posted by Morris Dovey on March 22, 2010, 3:16 am
 On 3/21/2010 9:06 PM, dow wrote:

Yuppers - not quite as intellectually exciting as seeing e appear
unexpectedly - but a very useful ratio nonetheless.

Thank you!

--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto/

Posted by dow on March 22, 2010, 8:33 pm
 
Yes. Initially, I wasn't very confident that my value for the ratio
was correct. But now that a professional mathematician has calculated
the exact same value, using a method that was very different from
mine, I think we can be sure that it's right.

                             dow

Posted by dow on March 23, 2010, 2:04 am
 Here's another thought... Suppose a fixed arm, with the cooking pot at
its end, enters the open end of the dish along the axis of the
paraboloid and holds the pot at the focus. It turns out that the dish
would have to rotate about the polar axis by more than 72 degrees
before the edge of the dish strikes the arm. At 15 degrees per hour,
that would take nearly 5 hours. If the arm is set up slong the line
that is the axis of the paraboloid when it is pointing at the sun at
noon, the cooker can be used from about 7:15 a.m. to 4:45 p.m. without
the dish hitting the arm. In the tropics, that's pretty well the whole
part of the day when the sun is high enough for solar cooking to be
practical.

So there's no need to cut holes or slots in the dish for the arm to
pass through. The machine can be extremely simple - just one dish
balanced about its focus, rotating at 15 degrees per hour about a
polar axis that passes through the focus. A stationary arm, aligned as
described above, holds the cooking pot. That's all!

                         dow

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