Posted by *dow* on March 21, 2010, 3:42 pm

*> Thanks David!*

*> I think that you are the second person genuinely interested in this.*

*> (Other than me) I was influenced by stuff I read about scheffler*

*> solar kitchens and by my own experience a few years ago when i tried to*

*> make a tracking solar accumulator. I made an off center paraboloid in*

*> late march. Did not think of center of gravity too much! As the season*

*> changed, I moved it to adjust for the changing sun angle and had a devil*

*> of a job keeping the center of gravity close to the axis.*

*> I think that if you have a symmetrical paraboloidal bowl you might end*

*> up having problems supporting the heat collector at certain times of the*

*> year. (I could be wrong on this, of course), perhaps you could cut a*

*> strip out of the bowl where it comes close to the support for the axis?*

*> with the seasonal movement.*

*> Thank you for taking the time to calculate the depth of bowl needed to*

*> have the center of gravity on the axis. That is useful.*

*> I also like the idea that once we get the bowls right they can be*

*> stamped out in large numbers very cheaply.*

*> I cannot work on any of this anymore. I took time off to "concentrate my*

*> thoughts" which worked but which did negative things to my bank balance.*

*> Thank you*

*> Brian*

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

length.

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!

dow

Posted by *Morris Dovey* on March 21, 2010, 4:00 pm

On 3/21/2010 10:42 AM, dow wrote:

*> 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*

*> length.*

Hmm - Is the problem to match the area "under" the latus rectum to the

area between the latus rectum and the rim?

*> 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.*

This is just too neat to be just a coincidence!

*> 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.*

Wow! "Interesting" is a _huge_ understatement!

*> Get rich soon!*

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

http://www.iedu.com/DeSoto/

Posted by *dow* on March 21, 2010, 6:11 pm

*> On 3/21/2010 10:42 AM, dow wrote:*

*> > 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*

*> > length.*

*> Hmm - Is the problem to match the area "under" the latus rectum to the*

*> area between the latus rectum and the rim?*

*> > 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.*

*> This is just too neat to be just a coincidence!*

*> > 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.*

*> Wow! "Interesting" is a _huge_ understatement!*

*> > Get rich soon!*

*> 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 USAhttp://www.iedu.com/DeSoto/ *

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!

dow

Posted by *Morris Dovey* on March 22, 2010, 1:16 am

On 3/21/2010 1:11 PM, dow wrote:

*> 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 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

material:

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...

*> 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.*

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. :)

*> 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!*

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

http://www.iedu.com/DeSoto/

Posted by *dow* on March 22, 2010, 1:30 am

*> 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*

*> material:*

*> 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...*

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.

*> > 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.*

*> 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. :)*

*> > 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!*

*> 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 USAhttp://www.iedu.com/DeSoto/ *

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

dow

> Thanks David!> I think that you are the second person genuinely interested in this.> (Other than me) I was influenced by stuff I read about scheffler> solar kitchens and by my own experience a few years ago when i tried to> make a tracking solar accumulator. I made an off center paraboloid in> late march. Did not think of center of gravity too much! As the season> changed, I moved it to adjust for the changing sun angle and had a devil> of a job keeping the center of gravity close to the axis.> I think that if you have a symmetrical paraboloidal bowl you might end> up having problems supporting the heat collector at certain times of the> year. (I could be wrong on this, of course), perhaps you could cut a> strip out of the bowl where it comes close to the support for the axis?> with the seasonal movement.> Thank you for taking the time to calculate the depth of bowl needed to> have the center of gravity on the axis. That is useful.> I also like the idea that once we get the bowls right they can be> stamped out in large numbers very cheaply.> I cannot work on any of this anymore. I took time off to "concentrate my> thoughts" which worked but which did negative things to my bank balance.> Thank you> Brian