Forget the 50 mpg carb; this guy has a 80 mpg Mustang

[moot]

I understood that not only the top area was collecting.

[Holden McGroin]

It is the geomerty that is important.  If I take a collector and put it at 45 Deg angle of incidence to the suns rays, it will collect only 70.7% (sine of 45 Deg) of the rays per sq ft of the collector than if the collector is perpinducular to the rays.  If the panel is 60 Deg to the suns rays, then it collects only half.

The most efficient way to have a solar collector arranged is to impinge in the suns rays at perpindicular (hence the tracking collector technology) and at perpindicular the sun only delivers 1 KW/sqM.   

[Rolex]

Well, looks like those years we spent studying calculus-based Physics and Thermodynamics were wasted, Holden. 

[moot]

No Holden, I mean the only way to break the insolation limit for any given area is for any given area to put the insolation to work on more than just the topmost layer.

|   |      |  |    |
 v  v      v  v   v
--------|---------
 v  v    | v      v
--------|---------
     v   |          v
--------|---------
          |
____________
/   /     /    /    /

How that can happen is beyond me, but that's the only way I can think to make sense of the math.

[Holden McGroin]

No Holden, I mean the only way to break the insolation limit for any given area is for any given area to put the insolation to work on more than just the topmost layer.

But... the insolation limit of 1 kw/sq meter is the limit.  You cannot harvest more from the sun than it sends us.

Putting multi layer collectors can work to increase the collecting efficiency (there is a company out there who make a window film collector so the windows you look thru can generate some electricity)

Trying to get more than 1 kw/m2 with your multi layer idea is like harvesting the wheat field and then doubling the harvest by running the combine over the cut feild again.  There ain't no more wheat in a harvested field.  You have to wait for it to grow again.

You can more efficiently harvest sunlight, up to the efficiency limit of 100%.  Once you harvest all there is, (100% efficiency) you have no more to gather and you have gathered 1 kw/m2.  That's all there is, there ain't no more.

[moot]

Yes I know Holden, I took the first 3 years of engineering like everyone - but that's all the data there is. That's what I have to base it on..  The next question from that basis is "Is there any way to exploit photons without entirely stopping them from continuing on to the next panel (to be exploited)?".  I'm not arguing anything but what the math premise is supposed to imply.  That's what needs to be answered before any speculation about how it actualy happens. 

E generated by total panel area = X watts/cm2
Insolation = Y W/cm2
Bones' setup : X/Y equal >1

This implies either that energy is made by the panels without photons being drained of all their energy (E drained from photons per solar panel < Y ), or that the insolation figure you found is wrong, or that Bones' numbers are somehow wrong. 

Of course I may be completely mistaken thanks to too much beer.

[Holden McGroin]

The next question from that basis is "Is there any way to exploit photons without entirely stopping them from continuing on to the next panel (to be exploited)?". 

Einstein recieved his Nobel for work in the photoelectric effect which is the physics behind PV panel technology. In the photoelectric process, if an electron absorbs the energy of one photon and has more energy than a specific amount, it is ejected from the material. If it recieves less than a specific amount (The work function) the electron remains in the matrix material.

The photon is a quantum amount of light. How do you siphon off only some of the energy of a quantum amount of energy?  If you could do so, it would not be a quantum amount.

So the answer to your question is no.

That's for PV.  If your process is lenses or mirrors, then the photons are directed towards some collecting system and by design do not travel thru unmolested to areas below the lenses or mirrors.

Regardless of the process, the sun delvers only a specific amount of light per sq area to the surface of the earth (area measured perpindicular to the suns rays) and that is appx 1 KW/M2

[CAP1]

Holden I see what you're getting at but I think you're missing the point. What you state is true IF you are using conventional collection techniques.
What Bones is alluding to is that although the physical size of his apparatus is only 10sq.ft. the effective collector area in his design is 30sq.ft.

exactly. this way, he could....using your example........triple the amount of energy in a given space.

i think

[Holden McGroin]

exactly. this way, he could....using your example........triple the amount of energy in a given space.

i think

sigh...

[Holden McGroin]

OK lets see if I can explain it more clearly.

The sun only sends us 1000W/M^2, the M^2 measured perpendicular to the suns rays.

Now if one can imagine a bunch of straws with a 1M^2 cross section gathered up in a bunch and pointed at the sun, each of those straws delivers us 1 KW of energy.

In any one of these tubes that we gather all the radiation and do not allow to hit the earth, 1 KW can be gathered.

No matter how we gather the energy in any one of these straws, we cannot go over 1000 watts, because after that the straw is empty.

Now after we gather the light from the straws, there is a shadow 'downwind' of that straw.

When the shadow is cast on the earth, it will have AT LEAST one square meter of area that will be in shadow.  At high noon on a certain day in the tropics, that shadow will be 1 square meter.  At any time other than high noon local time on specific days in the tropics, the shadow will measure more than 1 M^2, as the light hits the spherical earth at an angle different from perpendicular.

The only way to gather more than 1 KW is to harvest more shafts of light... more straws.

In order to gather 3 KW (as claimed) we must gather the light of at least 3 straws.  As a matter of practice, the world's best PV cells would need to gather the light of 7.5 straws, because 40% is presently the world record for gathering efficiency.

So 7.5 M^2 of solar collector will cast a shadow on the earth of more than 7.5 M^2 except in very specific geographical and temporal locations.

But lets give him the benifit of the doubt and say he has 100% effeciency on the gathering of light energy.

I can build a 3 M^2 collector and mount it to a superstructure that mounts to a pole sunk into the ground with a footprint on the earth only a foot square, so now is my effective collection 3 KW per sq foot? 

No, it is still 1 KW /  M^2 

And 1 KW / M2 is all that I could dream for.  I can never get better than that, unless I go to orbit.

[DiabloTX]

Wow, I just walked into a real live episode of the Big Bang Theory.

[soda72]

Speaking of solar power does anyone think this idea has any possiblities?  I had always wonder if this would work...

IBM Uses Magnifying Glass to Boost Solar Output

[SD67]

5 times the output ? 230 watts of energy on just one centimetre? That can't be possible since that would work out to more than 1KW/M²

[mietla]

exactly. this way, he could....using your example........triple the amount of energy in a given space.

i think

why only tripple? I'd go for at least a million


There is a 100 fish in a small pond. How many fish can you catch assuming that your rod has a million hooks?

[Maverick]

Cap,

Go back to the example I gave you. Holden is stating that there is a fixed and finite amount of sunshine hitting any given square meter of Earth period.

Like the water example I gave you. A spigot flows 1 gallon per hour. No matter what size or design of bucket you use under it you cannot collect more than one gallon of water per hour.