Posted by *Frogwatch* on July 24, 2009, 4:16 am

*> Frogwatch wrote:*

*> > In general, when light is reflected, reflectivity is a function of*

*> > wavelength, angle of incidence and material properties.*

*> Thank you! One of the things I was after was identification of the*

*> variables, and you've filled that in rather nicely. If "material*

*> properties" includes more than smoothness/roughness, then I still have*

*> digging to do, but this all is a good start.*

*> > For a*

*> > material that partially transmits light, the proper relationships are*

*> > given by the Fresnel equations (google em).*

*> Ok - will do. Thank you again because I wouldn't have known to look*

*> without your help.*

*> > This probably does not*

*> > apply if your reflector is good or for most metal surfaces. In these*

*> > cases, you might find tables of reflectivity in the Handbook of Chem*

*> > and Physics or can look up reflectivity of the material surface.*

*> > These tables generally assume nearly perpendicular rays to the*

*> > surface.*

*> I've been looking at metal and ceramic surfaces, but will still do the*

*> reading on the Fresnel equations.*

*> The info I've found so far is for perpendicular rays, and I'm going to*

*> need to find some way to derive at least a close approximation of*

*> reflectivity for other angles.*

*> > As the angle of incidence goes up, the reflectivity goes up*

*> > just as predicted by the Fresnel equations. Generally, the wavelength*

*> > does NOT change on reflection.*

*> Ok - I'm taking your word for the first, and I've observed the second.*

*> Now I'm beginning to crave knowledge of /why/ - and trying to not let*

*> myself become distracted... :)*

*> > Light that is not reflected is either absorbed, or transmitted into*

*> > the material OR is scattered from the surface although scatter is a*

*> > special case of reflection.*

*> Makes sense to me.*

*> > Light that is absorbed has its energy go*

*> > into the motion of the electrons of the material.*

*> > Consider a beam of light being reflected back and forth between two*

*> > parallel plates. Assume the reflectivity is 95% which is good for*

*> > most materials. After 20 reflections, the fraction of energy left in*

*> > the beam is then .95 raised to the 20th power or just under 36%*

*> Right - my little software model says 35.8485922% - and this is*

*> essentially the game I play with my flat panel with "stacks" of formed*

*> reflecting black aluminum ribbons. The surfaces aren't parallel (because*

*> they're curved), but the geometry guarantees a minimum number of*

*> reflections over a useful range of sun azimuths.*

*> > Most light absorbers are rough surfaces so as to scatter light into*

*> > large angles where it has another chance to be absorbed by the rough*

*> > surface. Consider a metal surface, a highly polished surface reflects*

*> > well but a rough one scatters the light and absorbs some too. A very*

*> > smooth surface of carbon is highly reflective whereas a rough carbon*

*> > surface is highly absorbing.*

*> Hmm - I think I need to learn more about how to quantify and model this*

*> behavior (and about how to economically produce surfaces to match*

*> calculated roughness/smoothness values).*

*> I suspect there may be a optimum trade-off between reflectivity (and*

*> number of reflections) and absorptivity (is that a word?) of the surface...*

*> > The scattered amount of a beam of light is given by: i=Iexp(-4*pi*

*> > (sigma*sin(theta)/lambda)^2) where sigma is the roughness of the*

*> > surface in the same units as your wavelength lambda. Theta is the*

*> > angle WITH RESPECT TO THE SURFACE, not the angle of incidence.*

*> I understand the relevance (and don't have any difficulty accepting the*

*> correctness) but I think I have a bit of studying to do before I'll be*

*> ready to apply this usefully with any degree of confidence.*

*> Again, thanks for your time and effort to help me understand more about*

*> this stuff. It's very helpful!*

*> --*

*> Morris Dovey*

*> DeSoto Solar*

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

One problem you will find with almost all examples of using the

fresnel equations is that they neglect absorption. The reason for

this is because it involves an imaginary part of the waves causing the

equations coefficients to be 2x2 matrices making the whole thing

messy. You may be able to find a derivation somewhere with the

absorption. In general, I would simply use the equation for

absorption I gave that is called the Debye-waller factor

Posted by *Morris Dovey* on July 24, 2009, 3:54 pm

Frogwatch wrote:

*> One problem you will find with almost all examples of using the*

*> fresnel equations is that they neglect absorption. The reason for*

*> this is because it involves an imaginary part of the waves causing the*

*> equations coefficients to be 2x2 matrices making the whole thing*

*> messy. You may be able to find a derivation somewhere with the*

*> absorption. In general, I would simply use the equation for*

*> absorption I gave that is called the Debye-waller factor*

That's exactly what I found (or rather, didn't find). I worked with what

I could find until I thought my head was going to explode, then took a

break to try to digest what I had, and to attempt to catalog what I

still needed...

...and realized that I already had sufficient understanding to formulate

a practical solution for the problem at hand. It's strange how it

sometimes works out that way when one has the right mentors. Kahlil

Gibran wrote this about teachers:

"If he is indeed wise he does not bid you enter the

house of wisdom, but rather leads you to the threshold

of your own mind."

And that's exactly the way it's played from this end. (Full text at

http://www.iedu.com/DeSoto/Misc/Verse.html#teach for anyone who's

interested.)

The step across the threshold was a very small one, and the solution was

verified with an almost trivially simple experiment conducted and

photographically recorded on my living-room floor.

The solution doesn't lend itself to DIY implementation, so I took what I

had to a patent attorney. Last week he completed his search and

concluded that my solution and the information I provided meets the

"novel" and "non-obvious" requirements, and that my photos demonstrated

that it was reasonably reducible to practice.

I've asked him to proceed with the filing and asked him to exercise the

"no disclosure for one year" option so I can pursue additional patent

protection outside the US.

I'm near bursting at the seams to tell all (I even have a web page with

description, drawings, and photos all ready to install on my web site),

but my friendly PA cautions me that I should deal with the patent and

commercialization issues first.

More on all this later (in a year or less)...

...for now, many thanks to all who've been so patiently helpful!

--

Morris Dovey

DeSoto Solar

DeSoto, Iowa USA

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

Posted by *Frogwatch* on July 24, 2009, 5:07 pm

*> Frogwatch wrote:*

*> > One problem you will find with almost all examples of using the*

*> > fresnel equations is that they neglect absorption. The reason for*

*> > this is because it involves an imaginary part of the waves causing the*

*> > equations coefficients to be 2x2 matrices making the whole thing*

*> > messy. You may be able to find a derivation somewhere with the*

*> > absorption. In general, I would simply use the equation for*

*> > absorption I gave that is called the Debye-waller factor*

*> That's exactly what I found (or rather, didn't find). I worked with what*

*> I could find until I thought my head was going to explode, then took a*

*> break to try to digest what I had, and to attempt to catalog what I*

*> still needed...*

*> ...and realized that I already had sufficient understanding to formulate*

*> a practical solution for the problem at hand. It's strange how it*

*> sometimes works out that way when one has the right mentors. Kahlil*

*> Gibran wrote this about teachers:*

*> "If he is indeed wise he does not bid you enter the*

*> house of wisdom, but rather leads you to the threshold*

*> of your own mind."*

*> And that's exactly the way it's played from this end. (Full text athttp://www.iedu.com/DeSoto/Misc/Verse.html#teachfor anyone who's*

*> interested.)*

*> The step across the threshold was a very small one, and the solution was*

*> verified with an almost trivially simple experiment conducted and*

*> photographically recorded on my living-room floor.*

*> The solution doesn't lend itself to DIY implementation, so I took what I*

*> had to a patent attorney. Last week he completed his search and*

*> concluded that my solution and the information I provided meets the*

*> "novel" and "non-obvious" requirements, and that my photos demonstrated*

*> that it was reasonably reducible to practice.*

*> I've asked him to proceed with the filing and asked him to exercise the*

*> "no disclosure for one year" option so I can pursue additional patent*

*> protection outside the US.*

*> I'm near bursting at the seams to tell all (I even have a web page with*

*> description, drawings, and photos all ready to install on my web site),*

*> but my friendly PA cautions me that I should deal with the patent and*

*> commercialization issues first.*

*> More on all this later (in a year or less)...*

*> ...for now, many thanks to all who've been so patiently helpful!*

*> --*

*> Morris Dovey*

*> DeSoto Solar*

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

You may consider filing a "provisional or pre-patent application"

which is generally two pages and then you have a year to decide to

file a ful application. It is generally much less costly than a full

application and gives you a year to shop the idea around and decide if

you want to spend the money (roughly $0,000 using an attorney) for a

full application.

Being a small businessman with limited resources, I generally only

patent in the USA unless whoever I license to requires foreign

patents. My attitude is that the USA and maybe Canada represents most

of the high tech market anyway. You might consider patenting in only

a single major Euro country thus preventing somebody else from having

a real Euro market unless they work with you.

Good luck and I love to see people doing this.

David OHara (aka Frogwatch)

Posted by *daestrom* on July 24, 2009, 10:13 pm

Morris Dovey wrote:

*> Frogwatch wrote:*

*> *

*>> One problem you will find with almost all examples of using the*

*>> fresnel equations is that they neglect absorption. The reason for*

*>> this is because it involves an imaginary part of the waves causing the*

*>> equations coefficients to be 2x2 matrices making the whole thing*

*>> messy. You may be able to find a derivation somewhere with the*

*>> absorption. In general, I would simply use the equation for*

*>> absorption I gave that is called the Debye-waller factor*

*> *

*> That's exactly what I found (or rather, didn't find). I worked with what *

*> I could find until I thought my head was going to explode, then took a *

*> break to try to digest what I had, and to attempt to catalog what I *

*> still needed...*

*> *

*> ...and realized that I already had sufficient understanding to formulate *

*> a practical solution for the problem at hand. It's strange how it *

*> sometimes works out that way when one has the right mentors. Kahlil *

*> Gibran wrote this about teachers:*

*> *

*> "If he is indeed wise he does not bid you enter the*

*> house of wisdom, but rather leads you to the threshold*

*> of your own mind."*

*> *

*> And that's exactly the way it's played from this end. (Full text at *

*> http://www.iedu.com/DeSoto/Misc/Verse.html#teach for anyone who's *

*> interested.)*

*> *

*> The step across the threshold was a very small one, and the solution was *

*> verified with an almost trivially simple experiment conducted and *

*> photographically recorded on my living-room floor.*

*> *

*> The solution doesn't lend itself to DIY implementation, so I took what I *

*> had to a patent attorney. Last week he completed his search and *

*> concluded that my solution and the information I provided meets the *

*> "novel" and "non-obvious" requirements, and that my photos demonstrated *

*> that it was reasonably reducible to practice.*

*> *

*> I've asked him to proceed with the filing and asked him to exercise the *

*> "no disclosure for one year" option so I can pursue additional patent *

*> protection outside the US.*

*> *

*> I'm near bursting at the seams to tell all (I even have a web page with *

*> description, drawings, and photos all ready to install on my web site), *

*> but my friendly PA cautions me that I should deal with the patent and *

*> commercialization issues first.*

*> *

*> More on all this later (in a year or less)...*

*> *

*> ...for now, many thanks to all who've been so patiently helpful!*

*> *

Well, best of luck. We'll be looking for you in the news :-)

daestrom

> Frogwatch wrote:> > In general, when light is reflected, reflectivity is a function of> > wavelength, angle of incidence and material properties.> Thank you! One of the things I was after was identification of the> variables, and you've filled that in rather nicely. If "material> properties" includes more than smoothness/roughness, then I still have> digging to do, but this all is a good start.> > For a> > material that partially transmits light, the proper relationships are> > given by the Fresnel equations (google em).> Ok - will do. Thank you again because I wouldn't have known to look> without your help.> > This probably does not> > apply if your reflector is good or for most metal surfaces. In these> > cases, you might find tables of reflectivity in the Handbook of Chem> > and Physics or can look up reflectivity of the material surface.> > These tables generally assume nearly perpendicular rays to the> > surface.> I've been looking at metal and ceramic surfaces, but will still do the> reading on the Fresnel equations.> The info I've found so far is for perpendicular rays, and I'm going to> need to find some way to derive at least a close approximation of> reflectivity for other angles.> > As the angle of incidence goes up, the reflectivity goes up> > just as predicted by the Fresnel equations. Generally, the wavelength> > does NOT change on reflection.> Ok - I'm taking your word for the first, and I've observed the second.> Now I'm beginning to crave knowledge of /why/ - and trying to not let> myself become distracted... :)> > Light that is not reflected is either absorbed, or transmitted into> > the material OR is scattered from the surface although scatter is a> > special case of reflection.> Makes sense to me.> > Light that is absorbed has its energy go> > into the motion of the electrons of the material.> > Consider a beam of light being reflected back and forth between two> > parallel plates. Assume the reflectivity is 95% which is good for> > most materials. After 20 reflections, the fraction of energy left in> > the beam is then .95 raised to the 20th power or just under 36%> Right - my little software model says 35.8485922% - and this is> essentially the game I play with my flat panel with "stacks" of formed> reflecting black aluminum ribbons. The surfaces aren't parallel (because> they're curved), but the geometry guarantees a minimum number of> reflections over a useful range of sun azimuths.> > Most light absorbers are rough surfaces so as to scatter light into> > large angles where it has another chance to be absorbed by the rough> > surface. Consider a metal surface, a highly polished surface reflects> > well but a rough one scatters the light and absorbs some too. A very> > smooth surface of carbon is highly reflective whereas a rough carbon> > surface is highly absorbing.> Hmm - I think I need to learn more about how to quantify and model this> behavior (and about how to economically produce surfaces to match> calculated roughness/smoothness values).> I suspect there may be a optimum trade-off between reflectivity (and> number of reflections) and absorptivity (is that a word?) of the surface...> > The scattered amount of a beam of light is given by: i=Iexp(-4*pi> > (sigma*sin(theta)/lambda)^2) where sigma is the roughness of the> > surface in the same units as your wavelength lambda. Theta is the> > angle WITH RESPECT TO THE SURFACE, not the angle of incidence.> I understand the relevance (and don't have any difficulty accepting the> correctness) but I think I have a bit of studying to do before I'll be> ready to apply this usefully with any degree of confidence.> Again, thanks for your time and effort to help me understand more about> this stuff. It's very helpful!> --> Morris Dovey> DeSoto Solar> DeSoto, Iowa USAhttp://www.iedu.com/DeSoto/