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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---------------------

*> 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*

*> Reply Reply to author Forward Report spam*

*> Reporting spam*

*> Message reported*

*> Rate this post:*

*> -------------------------------------------------------------------------=*

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Posted by *Morris Dovey* on March 22, 2010, 3:16 am

On 3/21/2010 9:06 PM, dow wrote:

*> 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!*

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

*> On 3/21/2010 9:06 PM, dow wrote:*

*> > 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!*

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

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

> 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> Reply Reply to author Forward Report spam> Reporting spam> Message reported> Rate this post:> -------------------------------------------------------------------------=