Space Based Solar Power For the Moon

[Hop_David]

I confess I'm unfamiliar with Ehricke's Solletta proposal, but I worry about mirrors in orbit reflecting sunlight to the ground, simply because the size of the illuminated spot on the ground is at least 0.01 times the distance from the mirror to the ground, 0.01 being the apparent diameter of the sun in radians.  If you manage to place the mirror in a lunar orbit at an altitude of 10 km -- and it would be tough to keep it there for very long, given lunar mascons -- then the spot size is a minimum of 100 m if the mirror is passing directly overhead.  And most of the time, it's not going to be even above the horizon, much less overhead.

In an earlier thread you had written:

  D_i = (L_i/L_s)*D_{s .}

I had asked how that was derived. Well, I've derived it on my own.



Light rays from opposite limbs of a light source hitting a point diverge by an angle, I call it alpha.

Since angle incidence equals angle reflection, the light rays reflected from the mirror also differ by an angle alpha. This makes the two cones similar. Similar cones gives:

D_i = (L_i/L_s)*D_{s .}

Edit: copied and pasted wrong equation.

[JohnFornaro]

Hey.  Good detective work there, Watson.

[Warren Platts]

Since angle incidence equals angle reflection, the light rays reflected from the mirror also differ by an angle alpha. This makes the two cones similar. Similar cones gives:

D_i = (L_i/L_s)*D_{s .}

You could run it through a Fresnell lens after reflecting off the mirror, and than would straighten out the beam some.

[Solman]

I confess I'm unfamiliar with Ehricke's Solletta proposal, but I worry about mirrors in orbit reflecting sunlight to the ground, simply because the size of the illuminated spot on the ground is at least 0.01 times the distance from the mirror to the ground, 0.01 being the apparent diameter of the sun in radians.  If you manage to place the mirror in a lunar orbit at an altitude of 10 km -- and it would be tough to keep it there for very long, given lunar mascons -- then the spot size is a minimum of 100 m if the mirror is passing directly overhead.  And most of the time, it's not going to be even above the horizon, much less overhead.

 Actually I was thinking of something at higher orbital altitude - maybe 100 km and providing 1/100th of a sun or so over a much larger area for general illumination and perhaps emergency PV power for a base that used PV away from the poles. Of course 1% of the daytime power might not be worth bothering with I'll admit.

[Hop_David]

Since angle incidence equals angle reflection, the light rays reflected from the mirror also differ by an angle alpha. This makes the two cones similar. Similar cones gives:

D_i = (L_i/L_s)*D_{s .}

You could run it through a Fresnell lens after reflecting off the mirror, and than would straighten out the beam some.

For an example, I'll use a geosynch mirror beaming to earth. A 336 kilometer mirror 36000 kilometers high would subtend .5 degrees, the same as the sun. Not coincidentally, using the equation proponent provided, 336 km is the minimum spot on earth.

So to provide full sunlight, the mirror must subtend at least the same angle as the sun. (That's assuming the mirror's perfectly reflected)

Is it possible to have a mirror that's just as bright as the sun but subtending a smaller angle?



Above is a mirror that subtends half the angle as the original light source. It's area is 1/4 that of the original light source. To match brightness, it must be 4 times as hot.

I emailed a university professor noting I seem to have recalled a demonstration such a mirror could be used to power a perpetual motion machine. He replied:
"There is a very easy reduction to a perpetual motion machine. It goes like this: The temperature of the sun's surface is about 6000 K, if it were
possible to concentrate sunlight to a mathematical point, the
irradiation would tend to infinity, and so would the temperature of
any object placed at that point. Thus heat, in the form of solar
light, would be flowing from 6000 K into an object at a higher
temperature, which is forbidden by the 2nd law of thermodynamics
(Clausius statement). This would hold even if the temperature isn't
infinite (it's enough if it surpasses 6000 K) or only a fraction of
the light is collected (any sunbeam would do)."

I believe it would take the powder of ground up unicorn horns to get a smaller D_i than Proponent's equation gives.

[Proponent]

For an example, I'll use a geosynch mirror beaming to earth. A 336 kilometer mirror 36000 kilometers high would subtend .5 degrees, the same as the sun. Not coincidentally, using the equation proponent provided, 336 km is the minimum spot on earth.

I think you mean 360 km rather than 336 km.

I emailed a university professor noting I seem to have recalled a demonstration such a mirror could be used to power a perpetual motion machine. He replied:

"There is a very easy reduction to a perpetual motion machine. It goes like this: The temperature of the sun's surface is about 6000 K, if it were possible to concentrate sunlight to a mathematical point, the irradiation would tend to infinity, and so would the temperature of any object placed at that point. Thus heat, in the form of solar light, would be flowing from 6000 K into an object at a higher temperature, which is forbidden by the 2nd law of thermodynamics (Clausius statement). This would hold even if the temperature isn't infinite (it's enough if it surpasses 6000 K) or only a fraction of the light is collected (any sunbeam would do)."

I know little about non-imaging optics, but according to the relevant Wikapedia entry it is actually possible to produce a spot that is brighter than the surface of the source.  What remains impossible, and prevents a violation of the 2nd law, is to illuminate an object over its entire surface with an intensity greater than that of the source.

Unfortunately, as far as I can tell, non-imaging optical technology seems to work only over short distances.

[Hop_David]

For an example, I'll use a geosynch mirror beaming to earth. A 336 kilometer mirror 36000 kilometers high would subtend .5 degrees, the same as the sun. Not coincidentally, using the equation proponent provided, 336 km is the minimum spot on earth.

I think you mean 360 km rather than 336 km.

I call diameter sun 1.4e6 km and distance to sun 1.5e8. Perhaps these roundings have led to our different results. I attached the spreadsheet I whomped up for this.

I emailed a university professor noting I seem to have recalled a demonstration such a mirror could be used to power a perpetual motion machine. He replied:

"There is a very easy reduction to a perpetual motion machine. It goes like this: The temperature of the sun's surface is about 6000 K, if it were possible to concentrate sunlight to a mathematical point, the irradiation would tend to infinity, and so would the temperature of any object placed at that point. Thus heat, in the form of solar light, would be flowing from 6000 K into an object at a higher temperature, which is forbidden by the 2nd law of thermodynamics (Clausius statement). This would hold even if the temperature isn't infinite (it's enough if it surpasses 6000 K) or only a fraction of the light is collected (any sunbeam would do)."

I know little about non-imaging optics, but according to the relevant Wikapedia entry it is actually possible to produce a spot that is brighter than the surface of the source.  What remains impossible, and prevents a violation of the 2nd law, is to illuminate an object over its entire surface with an intensity greater than that of the source.

I will have to read that article carefully. I suspect it will cause me to change my opinions/models.

Unfortunately, as far as I can tell, non-imaging optical technology seems to work only over short distances.

[Proponent]

Yes indeed, the angular size you use is more accurate and somewhat less than 0.01 radians.

I agree with your spreadsheet.  By the way, for the very small angles involved here, you can approximate tan x as x, for x in radians.

I will have to read that article carefully. I suspect it will cause me to change my opinions/models.

As I read it, it's not going to change things much.  It sounds like non-imaging techniques can make for somewhat cheaper or more-efficient solar collectors and radiators, but as far as I can tell they're not going to produce a big boost in achievable efficiencies or temperatures.

[Hop_David]

By the way, for the very small angles involved here, you can approximate tan x as x, for x in radians.

(looking at triangles....) Yes I can see that. Length leg adjacent gets closer and closer to length hypotenuse as the right triangle looks more and more isosceles. And sin(x)/x --> 1 as x--> 0.

Now I can see why you like to call the sun's angle .01 radians.

[JohnFornaro]

I just checked out a bit of Hop's page, and noticed the "sinus" drawing of the guy's nose.

A memnonic that I remenber from Junior High School regarding these angles is the famous Indian Chief: Soh-Cah-Toa.

Sine = Opposite over Hypotenuse
Cosine = Adjacent over Hypotenuse
Tangent = Opposite over Adjacent