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Optical Solar Power Satellite Relay

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At least one; the author seems to have ignored the constant-brightness theorem, which is that the divergence*diameter product of an optical beam is constant. You cannot move more of the Sun's brightness to the Earth's orbit optically.

TTFN
 
I've never heard of the constant brightness theorem. But I've heard of beams with gaussian profiles collimated with diffraction-limited optics. On the latter basis, equal-sized optics ~500 meters in diameter will transmit energy to Earth, 149.6 million km distance, with greater than 90 percent efficiency. The divergence of the beam results in an increase in the spot size at the receiver end. The percent of the spot size that is intercepted is covered the diffraction limited theorem, proportional the product of the areas of the transmitter and receiver, and the wavelength of radiation that is transmitted. So I thought I had this covered. Could I be wrong?

Could you explain how your theorem gets in the way of this mechanism?



 
IRstuff, let me take a guess, in layman's language. The scatter and interference caused by the non-zero diameter of the sun as seen by the first lens would cause the beam to widen, just as though it was being radiated directly by the sun itself.

In other words the best that can be done is to /equal/ the solar light flux from the naked sun.

Cheers

Greg Locock
 
I tried to post a longer reply before, but was having problems submitting, so I shortened the reply.


In order to get the beam to be focused onto the 600m diameter relay located 10^11 km requires a beam divergence of less than 10^-10 radians. Since the input to the collector is almost 2pi radians, that requires a collector optical magnification of about 10^11 times, which means that the exit aperture of the collecting optics needs to be about 600*10^11 m in diameter to get the divergence down. Otherwise, with a smaller exit aperture, all that energy is simply going to rattle around inside the collector and melt it into a slag. But, building a 600*10^11 m aperture is a non trivial exercise.

As a practical example, we deal with laser designators here, and one particular design starts with a 1/2 inch beam that's magnified to get the divergence down to 1/5 of the input beam to meet the spec value. This results in the beam diameter increasing by the same 5x.


TTFN
 
Forgive me if I don't believe your skepticism. The reflection of sunlight from my wristwatch will create a bright spot on the wall. A magnifying glass can focus sunlight sufficiently to ignite paper. Both are examples of sunlight made brighter at a distance using a refracting material. This is based on experience. Do you have at least a reference for your theorem?
 
Any optics book. or even the web.

Your experience is warped by your perceptions. The input to the magnifying glass is nearly collimated. When you focus the beam to a spot, the effective divergence is increased from milliradians to degrees. You can check this by taking the arctangent of the diameter of the glass divided by the distance to the spot.

Now do the reverse. Imagine that the spot is the collecting aperture and that you need to reduce divergence to picoradians, which means that you'll have an exit aperture like the magnifying glass, relative to the spot diameter.

You can believe or not, but the optics is what it is.

TTFN
 
The distance from the Sun to the Earth is 149,600,000 km, not the value you stated.
 
Your distance value is 3-orders of magnitude too large. No wonder you think divergence will prevent it from working! If Earth were 1000-times further from the Sun, I wouldn't expect it to work either!
 
I'm using the value in your paper. And if it's ONLY a 600*10^8-m exit aperture, does that suddenly make it practical?

TTFN
 
I proposed a 600-meter diameter lens. That's sixteen-orders of magnitude smaller in surface area than the aperture you are suggesting would be needed. As stated in my paper, the beaming efficiency is based on the product of the areas of the transmitter and receiver, via the equation:

Ar*At = wavelength^2 * distance^2 * ln(1/(1-n))

where
n = the fraction of the energy received, if the beam has a guassian profile.

Sunlight is composed of a wide range of wavelengths. Integrating across the spectrum yields the results that I described in my paper.

 
Aha! I see the problem. You *were* using the distance value in my paper, which was a typo. I corrected the typo, and updated the paper on my website. You can download the updated paper if you like. Fortunately this typo only affected my paper and not my algorithms. Therefore I believe this concept would work as described in my paper. That is, unless someone can nitpick it some more and find more errors?
 
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