# Simple SPS ??

Reading about the drawbacks of conventional Solar Power Satellites and the comments in response to Chris eventually triggered an idea, or perhaps a memory of something hinted at in something I read once. I’m somewhat less certain of complete originality of my ideas than I used to be.

The standard concept SPS has a few drawbacks that Chris brought out quite well. I hadn’t really given it that much thought and found the problems to be more interesting than the idea itself. There are several conversions in getting from sunlight to the terrestrial power  grid. Each conversion has some efficiency loss which increases the required SPS size. The four or five conversions times efficiency of each jacks up the SPS size to several times the value that one would think of without doing the trades. The kilowatt per meter making a square kilometer a gigawatt facility becomes several square kilometers of SPS to net a gigawatt on the grid.

Another problem is the heat that must be disposed of to keep the solar cells and transmitters cool enough to work properly. The mass on orbit doubles again to net your gigawatt on the grid. A couple of unsettling problems if you happen to be an SPS fan.

In comments it was suggested that it would be better to simply orbit a mirror to reflect sunlight to the desired location. That doesn’t work because the sun is not a point source of light. Sunlight is converging at about 1% to a mirror in GEO and will diverge at the same 1% when reflected to Earth. The reflected sunlight would cover a disk of well over 300 kilometers on the ground so that a one km mirror would light the ground at 1/90,000 of solar power.

So my thought is to use one mirror to focus solar radiation on a hot spot that would then be a point source of radiation for a second mirror to send to Earth. The hot spot could be thought of as similar in intent to the tungsten filament in a light bulb in a flash light. This cartoon is not to scale and is meant to show the intent only.

This possibly could reduce beam spread to something reasonable at the expense of the beam being smeared across many frequencies. The visible light SPS could serve a few functions sometimes suggested by reflected sunlight advocates. If one km of sunlight could be focused such that 50% of the light was in a 10 km diameter, 1/200 of sunlight would be considerably brighter than a full moon. City lights for a large city without any conversion at all, and both storm and strike proof. Battlefield illumination as desired out of reach of interdiction. Operations lighting in the arctic for commercial and military uses. Night search and rescue. And so on for illumination as the beam would be too weak for power collection.

If a full sunlight focus is possible, then zenith solar cells would be considerably more productive than current usage.

The main advantage of a scheme like this, if feasible, is that it would be relatively light, cheap, and simple with a real likelihood of being implemented with mostly ET sources such as asteroids or the moon.

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#### johnhare

I do construction for a living and aerospace as an occasional hobby. I am an inventor and a bit of an entrepreneur. I've been self employed since the 1980s and working in concrete since the 1970s. When I grow up, I want to work with rockets and spacecraft. I did a stupid rocket trick a few decades back and decided not to try another hot fire without adult supervision. Haven't located much of that as we are all big kids when working with our passions.

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### 35 Responses to Simple SPS ??

1. Paul451 says:

It’s hard to see that your “point source” would have a smaller arc-section than the sun, half a degree, and thus be more focusable.

Even in your image, you’ve got the hot-spot turned sideways to the second array. So most of the heat/light will be radiated away from the secondary array, wasted.

Ie, the narrower the arc-section of the hot spot, the more radiation you are radiating away from the secondary array. The wider the arc-section, the fuzzier the focus of the secondary array.

2. Paul451 says:

Or to put it another way: Replace your “hot spot” with a white surface at a 45-deg angle to the primary and secondary arrays. That’s essentially what you’ve created, in a round-about way.

3. Peterh says:

Working with imaging optics you won’t get the incandescent element brighter than the surface of the sun. So unless you’re willing to throw away a lot of light you won’t get a narrower secondary beam.

non-imaging optics allows you to cheat a little, but not nearly as much as you’d want.

4. johnhare says:

The cartoon is to illustrate the idea rather than act as a blueprint. There are a number of bells and whistles that would be added in a real design. Such as a two way mirror surrounding the hot spot except for the emitting face such that radiation at undesired angles is reflected back to the hot spot while still letting the concentrated sunlight in. And multiple hot spots with reflectors all aimed at the same place on Earth. Optically flat emitting surface of the hot spot. And so on for things not useful for a first look at an idea as they clutter up the concept and make it harder to evaluate the base thought.

I don’t see where it would be necessary to get the hot spot hotter than the surface of the sun. Just creating a useful point source should be sufficient. If the general concept is simple, cheap, and light enough, efficiency of sunlight use becomes secondary to efficiency of investment use.

5. Pete Zaitcev says:

Another option is a cloud of Pete Worden’s free-fliers, each orienting itself independently. They may orbit in a belt 2000 km up. The receiving mirror still needs to be gigantic, but perhaps not ouside of realm of reasonable. The orbiting mass is enormous, too, but putting up bundles of little sats is probably cheaper than a few battle station designs. Heat dissipation is less of an issue.

6. Andrew_W says:

I don’t think it’ll work John, getting your point source down to a point as seen from the reflector means it needs to be very small relative to the distance, so you end up needing a massive reflector area relative to the amount of energy that can be produced.

A flat surface point source 0.1m across at 2273K emits ~15kw, from GEO to get the footprint on Earth down to 3km the reflector needs to be 1km from the point source, so to catch most of that 15kw you’ll need a parabolic reflector, what? something like 3km across? And it’ll need to be a very accurately shaped reflector.

I don’t see you getting the point source much hotter than 2273K without it subliming, and even at that temperature most of its emissions are going to be in the IR.

7. Andrew_W says:

I’ll revise the temperature up to 2773K, the temperature of the filament in an incandescent light bulb, energy output goes to 33kw/0.01m^2. But it’s still as inefficient as an incandescent light bulb.

8. gbaikie says:

In comments it was suggested that it would be better to simply orbit a mirror to reflect sunlight to the desired location. That doesn’t work because the sun is not a point source of light.
Wiki:
“A point source is a single identifiable localised source of something. A point source has negligible extent, distinguishing it from other source geometries. Sources are called point sources because in mathematical modeling, these sources can usually be approximated as a mathematical point to simplify analysis.

The actual source need not be physically small, if its size is negligible relative to other length scales in the problem. For example, in astronomy, stars are routinely treated as point sources, even though they are in actuality much larger than the Earth.”

The sun is a star.

“Sunlight is converging at about 1% to a mirror in GEO and will diverge at the same 1% when reflected to Earth. The reflected sunlight would cover a disk of well over 300 kilometers on the ground so that a one km mirror would light the ground at 1/90,000 of solar power.”
I don’t think this is correct if you were using a flat mirror, and if using a slightly concave mirror, the reflected light on surface of the earth could be smaller than the size of the mirror [and have more intensity than sunlight.
And not sure why one would have mirror at GEO. A purpose of at GEO is to always be above one spot on earth. So with solar panels one can always face the sun, because you can rotate the panels, and be over one spot on Earth. With a mirror you need to be at 45 degree or greater relative to Earth surface and the sun. And so with mirror you going to spend time between the sun and the surface which you are suppose to reflect to. Or quite simply it’s not the same advantage.
If you always want to point the mirror at somewhere/anywhere on the surface, one should have a sun synchronous orbit. If you want to use a lot mirrors for one location [and mirror can be light weight and cheap] one can uses a lot of mirror satellites, so use some kind highly elliptical orbit- such as in some polar orbit, if target in or near polar regions [includng Europe and Canada being as being “near” polar regions- whereas say, Mexico isn’t near]. And with mirror satellites one can harden the any electronic used and fly thru Van Alan Belts.

9. Andrew_W says:

“I don’t think this is correct if you were using a flat mirror, and if using a slightly concave mirror, the reflected light on surface of the earth could be smaller than the size of the mirror [and have more intensity than sunlight.”

Look at it this way, two beams of light (one from each side of the Sun) strike the same point on a mirror, devise a way in which they are reflected as one beam.

10. gbaikie says:

–“I don’t think this is correct if you were using a flat mirror, and if using a slightly concave mirror, the reflected light on surface of the earth could be smaller than the size of the mirror [and have more intensity than sunlight.”

Look at it this way, two beams of light (one from each side of the Sun) strike the same point on a mirror, devise a way in which they are reflected as one beam.–

At focus point of concave mirror, the image is inverted. Or camera lens invert the image. Or a pin hole inverts the image.
But to do what talking about is done with cameras and telescopes, so not a matter of me devising a way.
https://www.lhup.edu/~dsimanek/scenario/lenses.htm
“Three kinds of telescopes. (A) Keplerian (astronomical), (B) Galilean, (C) Newtonian.”
The “(B) Galilean” illustration indicate how magnify and make the light travel parallel by using two lens at correct position relative to the focus point of first lens.
But “C) Newtonian” would be using a mirror rather than a len. Or large telescopes use large mirrors to capture light or they are one sided concave mirrors which direct the image to other lens. But what I mean is flat and less concave mirror than used with large telescopes. Or the large telescopes in use have focus point of tens of feet,
and for mirror in space one want a focus point near infinity [which is impractical for observational telescopes].

11. We do not need the light to come as a single beam. We just need the destinations to be close together.

We may not be able to make very large curved lenses but we can make large Fresnel lenses.

12. Andrew_W says:

Maybe the example I gave was misleading simple, it’s the whole of the solar disc that’s shining on the whole of your mirror.. every square mm of the reflector zapped with sunlight from multiple directions.
Perhaps someone else can explain it better.

13. We can live with half the light turning up on the Earth. It can be a rainbow effect rather than a pure white light source. This is not a telescope so the image of the sun does not need to be round.

14. gbaikie says:

–Maybe the example I gave was misleading simple, it’s the whole of the solar disc that’s shining on the whole of your mirror.. every square mm of the reflector zapped with sunlight from multiple directions.
Perhaps someone else can explain it better.–

Well if you want the full sun in the lens/mirror, it’s got to size of the Moon at lunar distance [or bigger]. GEO is about 1/10th of lunar distance from Earth. So it’s inverse square law- one needs something 1/100th the area of the disk of Moon. If it’s 3800 km it’s 1/100th the distance to Moon and so need something 1/10,000th the disk area of the Moon.
So if you do this [GEO or 3800 distance] it will look like another sun, from a small portion of the earth’s surface. So with flat mirror from 3400 it will be a smaller portion of Earth as compared to large flat mirror at GEO, but from a location on Earth they will look the same [and be hard to detect the difference between the reflection and the real sun].
If same size mirror is instead conclave mirror, you will not see the entire sun- it would magnify the the image- so it looks bigger and the sunlight is more intense as compared to “real” sun- and a smaller region will have this more intense sunlight.
And the opposite of this is true with convex mirror- one could use smaller diameter mirror and include the entire sun in the image- and it covers large region though sunlight will be weaker as compare to the real sun.

Now say have big enough concave mirror to include the entire sun and you are outside the region of the amplified sunlight, but close enough to the region to see the sun. And look at this reflection of the sun with binoculars- one could see things like sunspots on the Sun. But if do same thing inside the region, it will blind you, as it’s like looking at the sun with binoculars [not a good idea]. So being outside the region defeat the illusion that one is looking at something which looks like the sun.
Now if use a contex mirror or a reflective sphere. The sun will always look smaller and and be weaker- it will not be a very good illusion of the sun, but one will see it from a larger region of Earth.

15. johnhare says:

I think the sun divergence argument has become theoretically disjointed. For a simple understanding of the focus problem, get a mirror and reflect the direct sunlight on an object 100 feet away. The measured reflection will be about 1 foot larger in diameter than the mirror you use. If you can get a measured reflection at 1,000 feet, you will find that the reflected sunlight is about 10 feet larger in diameter than your mirror. At a mile 50 feet larger, and at 22,300 miles over 200 miles larger. This is something that anyone can verify experimentally tomorrow.

The near distance we are used to working with makes the far distance effects counter intuitive. If the effect wasn’t real, we wouldn’t be having this discussion in the first place.

16. Chris Stelter says:

An array of mirrors at 6000km altitude wouldn’t be too bad, assuming you had some PV farms that are on the order of 100s of GW (otherwise you’re wasting nearly all the reflected light).

Of course, I’d like to see the environmental impact statement on such a project:
“Oh, we’d like to basically eliminate night time for hundreds of square miles…”

BTW, the angular diameter problem is less on Mars, since the Sun is further away. Also, because areosynchronous orbit altitude is just 17,000km, you’re also better off in that regard. Still need a ridiculously huge solar farm.

17. N/A says:

The basic description implies a pseudo bent pipe approach though, right? Could a near equivalent of a bent pipe approach be a solar pumped laser though? How passively could one cool such a laser though?

18. gbaikie says:

–For a simple understanding of the focus problem, get a mirror and reflect the direct sunlight on an object 100 feet away. The measured reflection will be about 1 foot larger in diameter than the mirror you use. If you can get a measured reflection at 1,000 feet, you will find that the reflected sunlight is about 10 feet larger in diameter than your mirror. At a mile 50 feet larger, and at 22,300 miles over 200 miles larger. This is something that anyone can verify experimentally tomorrow. —

How big is the mirror which at 100′ has reflected light 1 foot larger?

Here is something to consider:
“The station uses 173,500 heliostats (each with two mirrors) to concentrate sunlight on the 459-foot towers. The towers were built this high to increase efficiency and reduce the already large footprint of the site; still, the furthest heliostats out are more than half a mile away from their tower. Four types of heliostats were used depending on the distance. ”
http://www.powermag.com/ivanpah-solar-electric-generating-system-earns-powers-highest-honor/

So these mirrors are as much as 1/2 mile away [5280 / 2 = 2620 feet].
I don’t know what these four types are. But if a 1 meter square mirror [flat/heliostats mirror] were to make 2 meter square reflection, one is quartering the power of the sunlight.
I would guess the different type of mirrors further away might be somewhat concave and/as using sunlight which is 1/4 or less the intensity would be nearly useless to heat up the molten salt- not that I am saying that such a silly project would not do things which are utterly useless.

19. gbaikie says:

“For a simple understanding of the focus problem, get a mirror and reflect the direct sunlight on an object 100 feet away. The measured reflection will be about 1 foot larger in diameter than the mirror you use. If you can get a measured reflection at 1,000 feet, you will find that the reflected sunlight is about 10 feet larger in diameter than your mirror. At a mile 50 feet larger, and at 22,300 miles over 200 miles larger. ”

Logically it seems to me, you is starting with mirror which is 1 square foot.
And it seems were the mirror were to be 10 foot square [100 square feet], you saying
that with 10 foot square at 100 feet would be 20 foot square [400 square feet] at 100 feet distance. And at 1000 feet it would 100 foot square [10,000 square feet].
If true with the heliostats [mentioned above] are bigger than 10 foot square and the tower which is 459-foot high does not appear wider than 100 feet and no heliostat can be closer than 450 feet distance from mirror to top of the tower. And one could say the majority of the heliostats are further than 1000 feet from the tower. And the closer ones are reflecting sunlight at steeper angle- which results in light being more spread out.
So at the best of times with 1000 meter per square meter of solar flux reaching the heliostats from the sun, only about 10 watts per square meter reaches the tower from the heliostats.

I would also add that sunlight at the surface of Earth is more diffused as compared to sunlight in the space environment. Another factor is that mirrors aren’t perfect, but a larger mirror would lessen the imperfection of a given mirror in terms of light reflected. So imagine the heliostats for the tower are pretty good mirrors and one could use less good mirrors in space and have the sunlight not spread out as much as these heliostats for this tower.

And also I think the mirror could be slightly curved- something like 1 inch or less over 100 feet [which is close to flat]. Though there other things one could do, like have 1 km mirror comprised of 100 separate mirrors which 100 by 100 meter square, and so each could point at different spot- or they act like tower [and also I guess, heat salt, but also if you did this one would also kill some birds- though it’s likely that much less birds should be killed]

20. johnhare says:

Very roughly, the 10′ mirror would reflect 20′ diameter at 1,000 feet. So while all of the sunlight reflected is hitting the target, the intensity per square foot is roughly 25% of the original on average. Heavier in the center, lighter on the edges.

21. gbaikie says:

–Very roughly, the 10′ mirror would reflect 20′ diameter at 1,000 feet. So while all of the sunlight reflected is hitting the target, the intensity per square foot is roughly 25% of the original on average. Heavier in the center, lighter on the edges.–
What I tried:
9 inch mirror round diameter flat mirror at 190 feet, had reflected area of about 21″ in
diameter with the sun lower on horizon- about 5 pm.
Also at about 5 pm a smaller 3 1/2 by 3 rectangle mirror, creates circular reflection at about 60 feet with lit area of about 6 inches diameter, and same mirror at 190 feet was quite faint but about size of 15″ in diameter.
It seems that generally the smaller mirrors increase the size of reflection more dramatically over shorter distances.
It also seems that if reflecting over enough distance, the square glass of those heliostats is a waste of glass.

22. Paul451 says:

Andrew Swallow,
The formula is W = D * tan(A)

Where W is the minimum possible focal width, D is the distance between the mirror and the focus, A is the arc-angle of the light source. (At 1 AU, A = 0.5 degrees. Tan(A) = ~0.0087. At Mars, Tan(A)=~0.0057, so focal width is 2/3rds.)

That’s it. Basic optics. There’s no trick that lets you shrink W any more that that. Note that the size of the mirror is irrelevant to the focal width, it only affects the intensity. Likewise, it makes no difference whether you have one large mirror, or a thousand small ones, the focal width will be the same (assuming they are all perfectly aligned.)

John,
“and at 22,300 miles over 200 miles larger. This is something that anyone can verify experimentally tomorrow.”

Well, maybe not that last one.

23. “Likewise, it makes no difference whether you have one large mirror, or a thousand small ones, the focal width will be the same (assuming they are all perfectly aligned.)”

It is the alignment that I am messing around with. I want the say 100 mirrors to point to the same point. This should produce a high intensity. On Earth we harvest the high intensity area(s).

24. johnhare says:

Each mirror will spread its beam by ten meters per kilometer. If you can figure out how to focus reflected sunlight to a smaller target than that, you have something valuable there. You could start the project by trying to get a focus of less than a foot a a hundred feet of distance.

25. Peterh says:

An illustration that may help: https://upload.wikimedia.org/wikipedia/commons/thumb/7/71/Lens3.svg/542px-Lens3.svg.png

As observed from the center of the optical element, the object and image are the same angular size. Lens or mirror the same rule applies.

26. John hare says:

Pete

I’m not sure who you are responding to. Your link shows a point source being faithfully reproduced by a lens. Since the sun is not a point source, I think you are confirming that the reflection would have the same half degree divergence as the source which is the sun in this case.

27. Peterh says:

note how the angular size from base to point of Object arrow is the same as the angular size from base to point of the image.

28. “Each mirror will spread its beam by ten meters per kilometer. If you can figure out how to focus reflected sunlight to a smaller target than that, you have something valuable there. You could start the project by trying to get a focus of less than a foot a a hundred feet of distance.”

Do not bother. Just align the mirrors to hit the same one foot target. Then the energy received by the target is proportional to the number of mirrors.

29. George Turner says:

I’ll go about the calculation a different way, just scaling up a simple Iridium flare.

An Iridium satellite with a reflector area of 1.6 square meters can shine at visual magnitude -8.1 at the center of the flare. Going by either the formula for relative brightness versus magnitude and comparing it to noon sun, or using the formula for illuminance I versus visual magnitude m of I = 2.1E-6 * 10^(-0.4*m) lux, an Iridium flare provides an illumination of 0.0036 lux, which would be 0.00228 lux per square meter of reflector. (The Iridiums orbital altitude is 782 km, btw).

A class 1 stadium for most sports is 500 lux, whereas class 3 for many sports is about 75 lux. 75 lux would require a reflector area of 32,800 square meters (181 meters on a side), while 500 lux would require 219,000 square meters (468 meters on a side).

30. Andrew_W says:

I’ll just point out that while stadium lighting from lowish orbit satellites might work, it’s a totally different issue to power generation, which would be looking for 50,000 lux – or more.

31. George Turner says:

Yes it is. My stadium thought was that once you develop a power system, people will use the electricity to light a stadium, so why not skip all the middle steps? ^_^

Of course the direct lighting would cause all kinds of other difficulties, such as migrating birds circling the beam, the problem that you have to swing the beam to the target, leaving a mile-wide path of shock and alarm, etc.

Another thought is that with a tri-color laser system to make white light, you might easily achieve the lower illumination levels required for a stadium in a far more focused and controlled manner than a simple mirror, skipping the steps where you ground convert the light into electricity and the electricity back into light.

The maximum theoretical efficiency of lighting is 683 lumens/watt, whereas a top range LED or metal halide is going to top out at about 120 lumens/watt (17.5% efficient), combined with about 30 percent efficient photovoltaic cells and you’re looking at about 5 percent efficiency on the ground end compared to using the illumination directly as illumination.

The laser system would also have the advantage that you don’t have to steer the whole mirror to switch targets, you just have to steer the laser, and of course you can turn it on and off.

32. Andrew_W says:

The reflectors in orbit to extend daylight hours is an old idea (I can remember it getting publicity in the early ’80’s).

http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19820025545.pdf

“A constellation of 16 solar reflectors in geosynchronous Earth orbit (GEO) can illuminate to 8 lux (56 full Moons) an area approximately 333 km in diameter.”

Guess where they get that 333 km figure from.

33. I think a few of you folks have already understood the problem of etendue: https://en.wikipedia.org/wiki/Etendue.

Concentrating solar power systems often refer to the number of suns. This is because, from the POV of the receiver, the radiant intensity (power per solid angle per receiver area) of incoming reflected solar light can never be larger than the sun. What these reflectors do is make many virtual images of the sun from the POV of the receiver. If you have 10 virtual images, and each is 99% as radiantly intense as the sun, you have 9.9x the usual amount of incoming sunlight, etc.

For a reflector to make an image of the sun, it has to be the same apparent angular size as the sun. If it’s smaller, then it just makes an image of part of the sun.

If the reflector has a larger angular size than the sun, then it makes sense to curve it a bit. This makes a magnified image of the sun, but the intensity stays the same.

A reflector 400 km away would have to be about 4 km diameter to produce a full sun image at one point. As you travel away from that point, the image of the sun will be partially vignetted, so that intensity will drop off as you get away from the point. The best possible focus would be a 8 km diameter spot at full intensity in the center and zero intensity at the edge.

A larger reflector would have the same best possible focus size on the ground, but would be brighter.

34. A geosynchronous mirror lighting up a photovoltaic array on the ground is a terrible idea, for the reasons specified in this post. You cannot fix the etendue problem with a scattering or hot receiver/emitter at an intermediate focus.

The easy way to explain this is that the receiver/emitter receives light from some fraction of the hemisphere, but emits in all directions. It makes etendue larger, probably a lot larger, when what is needed is to make etendue smaller.

It seems to me that a laser can make etendue smaller. If I look at a pump chamber for a Nd:glass rod amplifier, I see that the lasing medium receives energy from many different directions over a large area (side of the rod), but emits some useful fraction of that energy with small beam dispersion over a small area (end of the rod). I googled a bit for an explanation of why lasers get to violate the etendue rule, and found nothing.

I’m still suspicious. This smells distinctly like entropy going down. My guess is that the laser efficiency is bounded by etendue in some subtle and inescapable way.

Even if the laser avoids the etendue problem, it will still have large conversion inefficiency because it emits monochromatic light but is pumped by broad-spectrum light. Most of the photons going in will have different energy than the required pump photons. If they have less, they’ll not pump. If they have more, they might pump, but any excess energy will be converted into heat. This same problem affects the efficiency of photovoltaic panels lit by incandescent light.

35. Peterh says:

I suspect the narrow beam possible from a laser relates to efficiency. Similarly, with an aperture mask and secondary optics you could get a narrower beam from sunlight, with loss of efficiency.