Hybrid Car – More Fun with Less Gas

New mount for stirling motors and satalite dish solar and the liquid piston tracker - Page 2

register ::  Login Password  :: Lost Password?
Posted by dow on March 21, 2010, 3:42 pm

As you know, I have been interested in solar stuff, mainly heliostats,
for decades. My bank account has suffered too! But saving the world is
more important than money!?

Yes. If a symmetrical bowl is used, the cooking pot will usually have
to be at the end of some sort of arm that reaches into the bowl. That
doesn't sound too difficult to do.

The calculation of the depth of the bowl turned out to be truly
horrible. I found myself facing an integral that I had no idea how to
evaluate - and I'm a former math teacher! After asking a few people,
and hunting through tables of integrals, all without success, I
resorted to the brute-force-and-ignorance approach. I wrote a little
computer program that made the machine loop around a few thousand
times, evaluating the integral numerically by adding many terms. This
kind of approach can produce an answer that is as accurate as is
needed, but is never absolutely precise. That's how I came up with the
result that the depth of the bowl should be 1.8478 times the focal

Later, using that result, I calculated the radius of the rim of the
bowl. It came out to approximately 2.718 times the focal length. When
I saw that number, I was startled, because I recognized it as the
value of "e", the base of natural logarithms! So now the question is,
is this just a coincidence, or would a precise evaluation of the
integral lead to the conclusion that the radius of the rim is
precisely e times the focal length? My guess is the latter, but I'm
not sure.

If the radius of the rim is precisely e*F, then the depth of the bowl
must be (e^2)/4 times the focal length. That ratio works out to
1.847264..., i.e. a few parts in ten thousand less than my computer
program's result. It's quite plausible that the program was in error
by that much. For practical purposes such as yours, an error that
small will not be significant, but it is theoretically interesting.

Get rich soon!


Posted by Morris Dovey on March 21, 2010, 4:00 pm
On 3/21/2010 10:42 AM, dow wrote:

Hmm - Is the problem to match the area "under" the latus rectum to the
area between the latus rectum and the rim?

This is just too neat to be just a coincidence!

Wow! "Interesting" is a _huge_ understatement!

Not sure about riches for anyone, but if you can figure out a proof I'm
willing to call it the "DOW's Lemma"... :)

Morris Dovey
DeSoto Solar
DeSoto, Iowa USA

Posted by dow on March 21, 2010, 6:11 pm
The problem is to find the condition that leads the torque that is
exerted in one direction by the weight of the part of the bowl that is
above the focus being equalled by the torque, in the opposite
direction, that is produced by the weight of the part of the bowl that
is below the focus. (Imagine the bowl being somewhat tilted, so the
weights don't pass through the focal point.) So it isn't just the area
of the bowl that has to be integrated, it's the product of the areas
of a whole lot of little elements comprising the bowl and their
distances above or below the focus.

I find it hard to believe that this hasn't been investigated before.
If the "e" thing is correct, there must surely be people who already
know it. I'm going to continue asking around.

I don't know how rich you are, but you are one of he few people I know
of who manages to make a living from solar. That's an achievement!


Posted by Morris Dovey on March 22, 2010, 1:16 am
 On 3/21/2010 1:11 PM, dow wrote:

I was considering the special case where "center of mass" of the bowl
coincides with its focal point. As I picture it, that center of mass
should be independent of bowl orientation - which means that we can
orient the bowl however we find convenient.

It would seem to me that if the bowl is made up of uniform material, a
differential amount of that mass can be represented by a differential
area times the thickness of the bowl material, times the density of the

    dm = dA * thickness * density

and a convenient dA might be a ring-shaped "slice" of the bowl
orthogonal to the line passing through vertex and focus of the parabola,
and that each dA is a function (only) of the displacement of the slice
from the vertex...

Agreed, but I haven't seen it either. Do stick with the problem - it
probably won't make you a pile of money, but it might be fun to be the
first to publish! It's exciting for me just to see e pop up in this
context. :)

In almost all the ways that really matter to me, I'm wealthy beyond
calculation - but if you're talking about just money, the R&D for panel
and engine development has eaten up just about everything not needed for
operating overhead. I have no regrets - there were problems that weren't
being well-addressed, and I believed that I could produce solutions and
deliver benefits that far outweighed the costs. I still see it that way.

Morris Dovey
DeSoto Solar
DeSoto, Iowa USA

Posted by dow on March 22, 2010, 1:30 am
That's pretty much how I went about it, with one difference. If the
material in the ring is not parallel with the axis, for example if the
radius of the ring is increasing with increasing distance above the
focus (which will be the case in this bowl), then the mass of the ring
will be greater than it would be if the radius were constant. So
another factor has to be included, besides the ones you've mentioned.

You're a happy man. That's infinitely more important than money.


This Thread
Bookmark this thread:
  • Subject
  • Author
  • Date
please rate this thread