Microterraforming

noosfractal · 2007-08-08 03:45:15

I was quite inspired by this paper ...

Rigel Woida, The Road to Mars: A computer modeled analysis of the feasibility of using Large Deployable Reflectors to redirecting solar radiation to the Martian surface, May 2007
http://www.niac.usra.edu/files/students … _Rigel.pdf

... recently discussed here ...

Methods of terraforming - How to go from bone dry & lifeless
http://www.newmars.com/forums/viewtopic … 2&start=20

The paper is big on ideas but a little short on detail. I wanted to get a better handle on the physics of the situation so I could examine new scenarios (heating bigger areas, enhanced solar power, melting large quantities of water ice, etc). I thought NewMarsers might be interested as well. My calcs are a bit at odds with the paper, so you might want to compare and contrast.

The paper uses convex mirrors. That seemed wrong to me. You've got this orbital mirror with a certain area (A) collecting sunlight and you're going to reflect that to an area (A') on the Martian surface. Say A' = 10A, then the intensity of your reflected sunlight at the Martian surface is going to be 1/10th (on average) that collected at the mirror (ignoring complications like imperfect reflectors for the moment). Convex mirrors spread light out over a greater area, so you get lower intensity at the surface or you need more mirrors.

So my first step was to try and work out A' for a given A - just with flat mirrors, 'cause you can build convex and concave mirrors out of those.

I read some old papers on Solares (the mirror only version of solar power satellites) like ...

Billman et al., "Space reflector technology and its system implications," Feb 1979
http://pdf.aiaa.org/GetFileGoogle.cfm?g … e=mtgpaper

... and heliostat technology for solar towers (which has a lot in common), but the best paper I read for this purpose was ...

Mills et al., Reflections of the 'Burning mirrors of Archimedes', June 1992
http://www.iop.org/EJ/abstract/0143-0807/13/6/004/

... which is a really fun read.

A' doesn't equal A even for plane mirrors because of the finite angular size (alpha) of the Sun. Usually this factor isn't so important, but at orbital distances it is dominant. Working with circular mirrors ...

r' = r + h.tan(alpha/2)

... where:
r is the radius of the mirror,
r' is the radius of the reflected spot on the surface, and
h is the height of the orbit above the surface.

The tan(alpha/2) factor is small ( = radius of the Sun / Sun-Mars distance ) but h is big. The Sun-Mars distance varies (1.4-1.7 AU), so I'll do the examples with the worst case (1.4 AU) which gives

h.tan(alpha/2) ~ h/300

i.e., if you're orbiting at 1200 km, your spot on the surface has a radius of 4 km (plus mirror radius).

So then the ratio A'/A = (pi.r'^2)/(pi.r^2) = [1 + (1/300).(h/r)]^2

The orbital stuff is quite hard, so I'll leave it for the next post, but the geostationary orbit provides a simple example ...

h_mars_geo ~ 17000 km
assume r_mirror = 1 km
then r' = 58 km and
A'/A ~ 3300

So you would need (ignoring complications for the moment) 3300 flat 1 km radius mirrors, each aimed at the same spot on the surface, to yield Mars full sunlight (say 600 W/m^2) throughout the 58 km radius spot. Or, equivalently, a 58 km radius concave mirror with its focus point at the center of the surface spot (the 3300 flat mirrors can be considered a Fresnel implementation of the concave mirror - each has the same reflecting area).

Actually, because you are effectively creating an image of the Sun, and the edges of the Sun are darker than the center ("limb darkening"), you only get full sunlight (say, 90% of peak) in an inner circle of 1/2 the radius, but the point of the example is that, from Mars GEO, you can't make the spot smaller and just concentrate the sunlight on, say, 1 square kilometer. To do that you have to go closer.

But when you go closer, you lose the nice properties of geosynchronous orbit and you have to think about satellite ground tracks, scan rates and avoiding the shadow cast by Mars. It turns out to be worth it though.

To be continued ...

*** EDIT

Corrected ideal concave mirror radius.

RickSmith · 2007-08-08 15:09:39

Hi Everyone.

I read this article and the main thing that struck me is that the chance of NASA increasing the cost of the mission many fold to make the area around their astronauts a bit warmer just won't happen. But for the terraforming forum this data is great.

I was really surprised that they found that 9% of the sunlight was scattered by the atmosphere. With Mars air pressure being a near vacuum, I would have guessed a much lower number.

The reason they use convex is because they want to inflate them with a gas charge. They make up for the spreading of light with additional mirrors. I think that they are trying to make the mirror as flat as possible given how they hope to deploy it.

Thanks noosfractal for bring this to our attention!

Warm regards, Rick

noosfractal · 2007-08-09 01:36:02

I read this article and the main thing that struck me is that the chance of NASA increasing the cost of the mission many fold to make the area around their astronauts a bit warmer just won't happen.

You could possibly justify a small system on safety grounds. What if two or three 80 ton payloads to Mars orbit could keep temperatures in a 1/2 km radius circle above freezing day and night? What if the heating caused a 15 mbar "high pressure" zone - enough to prevent your skin from blistering in case of a suit tear? What if the same system tripled your solar power production and eliminated the need for high mass power storage?

I was really surprised that they found that 9% of the sunlight was scattered by the atmosphere. With Mars air pressure being a near vacuum, I would have guessed a much lower number.

Well there is still ~100 km of atmosphere which includes layers and clouds and dust. They said the rovers received just 1% of normal daylight during the worst of the current dust storm.

The reason they use convex is because they want to inflate them with a gas charge.

I'm all for inflatables where it makes sense, but I'm sure I saw inflatable concave dishes out there. There was an inflatable ring with the reflective film stretched across it. I can't remember how they made it concave. I'll look into it.

I remember they complained that you needed to continually reinflate due to gas loss to the vacuum - which limited the lifetime of the dish. They were working on something that would let you harden the support structure in place once you inflated it.

I think the biggest lifetime issue for space solar mirrors is the control momentum gyroscopes (CGMs). CGMs are perfect for the pointing needs (high mass efficiency for the performance), and are electrically powered, but they have bearings that wear out in ~10 years. That's not too bad for the smaller systems ( say < 500 tonnes ), but if you invest in a bigger system, replacing gyros or relaunching mirrors is going to be a pain.

They make up for the spreading of light with additional mirrors.

Yeah, but not enough by a long shot by my calculations.

The paper says that the convexity results in an angular magnification of 22. So the h/300 factor goes to ~h/16. Even at h = 200 km, that gives a spot size of radius ~13 km. You'd need 30000 convex 75 meter radius mirrors to build up the center of that spot to Mars normal daylight, let alone Earth normal daylight, and that isn't even taking into account imperfect reflection, atmosphere reduction or orbital considerations.

I may be misunderstanding the paper though.

I think that they are trying to make the mirror as flat as possible given how they hope to deploy it.

In another place in the paper (the radiative transfer section), the mirrors seem to be treated as though they were flat. So maybe they are substantially flat. It's a little unclear.

SpaceNut · 2007-08-09 10:19:39

One issue with a light mass large diameter structures of this type is that we are creating a solar sail

noosfractal · 2007-08-09 18:50:57

One issue with a light mass large diameter structures of this type is that we are creating a solar sail

Yes, orbital adjustments have to take that effect into account. It isn't a first order issue though for a couple of reasons:

- For Mars, the acceleration due to light is less than the acceleration due to Mars' gravity until out around the 200,000 km mark (it depends upon the specific mass of the mirror, for example, at 7 g/m^2, the balance point is closer to 300,000 km). This is the basis for Forward's statite idea. The statite effectively hovers at the balance point. But the acceleration due to light is basically constant in this region, whereas the acceleration due to gravity increases in proportion to 1/r^2 as you come in closer to Mars. For close orbits, e.g., h < 1000 km, the acceleration due to light will be, not negligible, but quite small.

- If you imagine Mars half lit by the Sun, and half in shadow, with the mirror orbiting, you can see that the mirror can't reflect anything at noon (surface time), and that it reflects best at midnight. In between, the mirror reflects with varying efficiency depending on the surface target. For example, if a flat circular mirror above the terminator between day and night wants to light up a spot directly beneath it on the terminator, the efficiency is ~70% (ignoring all other complications) because the surface "sees" an oval rather than a circle. The efficiency quickly drops off towards noon (0% efficient) and gets better towards midnight (100% efficient), so one strategy could be to use the mirrors to light up the surface during the night half of the orbit, but during the day half of the orbit, to use them as solar sails for orbital adjustment. Orbital height isn't the only useful adjustment either, in particular, you probably want a sun-synchronous orbit which requires control of the orbital precession.

So the solar sail effect will probably be more helpful than anything. It might not be enough though. We might need some sort of solar-electric propulsion (SEP) for stationkeeping.

noosfractal · 2007-08-10 03:05:23

Again imagining Mars half in day and half in night, and, say, an equatorial orbit (pretending Mars has no axial tilt for the moment), you can see that the orbiting mirror would spend half its time in sunlight (on the inefficient reflecting side of the orbit) and half its time in shadow (on the efficient reflecting side of the orbit). If the Sun is eclipsed by Mars from the point of view of the orbiting mirror, the mirror can't reflect any sunlight to the surface, so the orbit has to be inclined ...

In both figures north pole is up, in the top figure we're looking from the side, in the bottom figure we're looking from behind (inside the shadow and back towards the Sun).

In the top figure, the orbit is inclined just enough so that it exits the shadow at its "highest" point. The bottom figure shows that you don't have to do any more than that to make sure the mirror is always in sunlight.

The required inclination (phi) depends on the orbital height. From the geometry ...

sin(phi) = r_mars / ( r_mars + h )

Some inclination angles for some orbital heights ...

h (km) -- min r' (km) -- phi (degrees, approx)

214,000 (statite) -- 713 -- 1
17,000 (geo) -- 57 -- 10
9,500 (geo/2) -- 32 -- 15
6,400 (geo/3) -- 21 -- 20
2,800 (geo/6) -- 9.3 -- 33
1,700 (geo/8) -- 5.7 -- 42
1,000 (geo/10) -- 3.3 -- 50
500 (geo/12) -- 1.7 -- 60
300 (geo/13) -- 1.0 -- 67

Actually these angles should be slightly smaller to avoid only the umbra, and slightly larger to avoid the penumbra, but they are fine for this purpose. I've also included the (ideal, unachievable in practice) spot size radius (r') for reference.

The "geo/n" in brackets means the mirror orbits Mars "n" times in the amount of time it takes for Mars to rotate once. I stopped at 300 km because that seems to be the minimum h for efficient stationkeeping. Below that height, the atmosphere will be dense enough to deorbit the mirror without constant and significant orbital height adjustments. Here's some heights at which Martian atmospheric densities are equivalent to Earth's atmospheric densities ...

Earth height from surface (km) -- Mars height for same pressure

400 -- 200
500 -- 220
1000 -- 300
2000 -- 375

Many would recommend the 2000 km equivalent height (i.e., 375 km on Mars) to guard against density inflation during solar storms.

Other than just constraining the set of useful orbits for the mirrors, the minimum inclination angle also constrains the surface latitudes that the mirrors can efficiently light up. For example, a mirror at 300 km can only contribute 90% of peak reflected light to latitudes within +/- 4 degrees of the latitude directly below it (again because of mirror tilting so that the target sees an oval).

But I've probably been abusing the term "inclination" angle because I've been ignoring axial tilt. Mars has an axial tilt of 25 degrees (just a little more than Earth), so the latitude "directly beneath" the mirror in, say, the northern summer, will be 50 degrees to the south in the northern winter. On this basis alone, you might set the minimum orbital height to 1700 km (geo/8) if you want to be able to continuously light up a particular patch of ground throughout the Martian year.

To be continued ...

noosfractal · 2007-08-12 01:41:50

In addition to the latitude constraints in the previous post, the mirrors will have a similar longitude constraint. For example, mirrors at 300 km can also only contribute 90% of peak reflected light to longitudes within +/- 4 degrees of the longitude directly below - and, actually, it isn't even as good as that since that is only for the most efficient midnight position, but mitigating factors such as having multiple mirror sets in orbit means that it will do for a first cut.

With increasing orbital height this becomes less of an issue, since the mirror doesn't have to tilt as much to move its reflected spot on Mars' surface. For example, at h = 6400 km (geo/3), a mirror could deliver 90% of peak to anywhere in the nightside hemisphere ... if it were a flat plane. But of course the planet's surface curves to make a sphere, which sets an upper limit of +/- 30 degrees as a longitude constraint, no matter what the orbital height, with geosynchronous being a special case because the mirror doesn't need to tilt for latitude.

The longitude constraint is important because, below GEO, the surface and the mirror move through their "orbits" at different rates. For example, at h = 9500 km (geo/2), a mirror will move through 180 degrees of orbit in the time it takes for the target to move through 90 degrees of longitude. Because of the longitude constraint, and the desire for continuous illumination, naively, the number of degrees of nighttime surface longitude a single mirror set can cover is 60/(n-1) where n is the number of complete orbits per Maritan day. That is, you need 3 sets of mirrors (180/60) for geo/2 and 6 sets of mirrors (180/30) for geo/3.

I can't find better solutions for geo/2 or geo/3, but there are better solutions for lower orbits. I use orbits like the one in the second figure above that skim the shadow cast by Mars. In the geo/2 case I use 3 orbits - one for each mirror set - but with one rotated clockwise 45 degrees, and one rotated counter-clockwise 45 degrees with respect to the pictured orbit. One orbit covers from twilight to twilight + 60 degrees, the pictured orbit covers from twilight + 60 degrees to dawn - 60 degrees, and the final orbit covers from dawn - 60 degrees to dawn. In other cases I use the single orbit pictured.

You can use multiple orbits in every case, but you save by not doing so. For example, you could use 6 separate orbits in the geo/3 case, each covering 30 degrees of longitude, but once you get to 6 separate mirror sets, you can just place them equidistant in the same orbit to provide continuous coverage. Then you only need to add more sets once the longitude constraint drops below 60 degrees (which happens somewhere around the 2000 km mark). In fact, at geo/3 and below, you can think in terms of a continuous ring of mirrors in the pictured orbit having a certain reflector density.

Once the longitude constraint ( = [(h.tan(60))/(pi.r_Mars)].180 ) drops below 60 degrees, the number of sets required is just 360/longitude_constraint (to give continuous illumination). Fractional values are fine it you are just going to use lots of smaller mirrors (it just increases the reflector density), but if you are going with single large concave mirrors, then you'll need to round up ...

Orbital height (km) -- number of sets required

17,000 (geo) -- 1
9,500 (geo/2) -- 3
6,400 (geo/3) -- 6
2,800 (geo/6) -- 6
1,700 (geo/8) -- 7.3
1,000 (geo/10) -- 12.4
500 (geo/12) -- 25
300 (geo/13) -- 42

The number of required sets is independent of the number of mirrors per set, but the number of mirrors per set is much lower at lower orbital heights, which more than makes up for the number of required sets in terms of total reflector area. However, GEO is most efficient in terms of how much illuminated surface area you get for your total reflector area. I'll leave reflector area calculations to next post

I'll be the first to admit that these may not be the optimum orbits, and they have quite a large effect on the total launch mass of the system, so if you think of better orbits I'd love to hear about them. I considered the HEPOs mentioned in the paper, but orbits with h = 300 km can only efficiently reflect light 4 or 5 degrees shadow-ward of their (necessarily high) inclination (at 18 degrees the light per unit area would drop to 10% of peak), so they wouldn't be able to reach the target for most of its nighttime hours. This seems a serious flaw to me (and occurs with convex mirrors as well as flat), but I'll be happy to hear why it isn't.

To be continued ...

noosfractal · 2007-08-13 02:27:57

I've mentioned some imperfections and inefficiencies, and cumulatively they have a significant effect, so they are worth looking at a bit more closely.

The first is imperfect mirror reflection. For solar sail type reflectors, reflectivity is typically in the range 0.8-0.9. I used 0.9. This can go higher, but if you need 20% extra mass to get a 10% improvement then it is not worth it.

Next is counter-reflection by the Martian atmosphere. The paper quoted 9%. I rounded that up to 10%. This will likely only get worse as terraforming takes place.

Next is latitude inefficiency. Most of the analysis above assumes that the mirror is reflecting at least 90% of peak light to its target. I conservatively use a 0.9 multiplier here to give some wiggle room. As noted this rules out orbit heights below 1,700 km for year-round illumination, I include lower heights in the table below for reference anyway.

Next is longitude inefficiency. Again, the analysis assumes at least 90% of peak light is reaching the target. However, a 0.9 multiplier is less conservative here because the longitude constraint should really take into account orbital position, so I use 0.85 based on some quick modeling. These last two factors are dependent on the orbit scheme, and it is quite possible that they (and/or the number of sets) can be improved.

That gives a total inefficiency of (0.9)(0.9)(0.9)(0.85) = 0.62. I use 0.6 for every case except GEO and the statite where I use 0.7 to highlight that longitude inefficiency doesn't apply. The final reduction is a nod to pointing inaccuracy, out of service time, and everything else I haven't thought of. I should probably have used 0.5.

The following table shows the reflector area required to send Mars full daylight to a spot with radius r' throughout the Martian night from orbital height h, taking into account the total inefficiency ...

h (km) -- r' (km) -- reflector area (km^2) -- concave mirrors (radius km)

214,000 (statite) -- 714 -- 2.3 million -- 1 x 850
17,000 (geo) -- 58 -- 15,000 -- 1 x 69
9,500 (geo/2) -- 33 -- 17,000 -- 3 x 42
6,400 (geo/3) -- 22 -- 16,000 -- 6 x 29
2,800 (geo/6) -- 10 -- 3,000 -- 6 x 13
1,700 (geo/8) -- 6 -- 1,300 -- 8 x 7.7
1,000 (geo/10) -- 3.6 -- 840 -- 13 x 4.6
500 (geo/12) -- 1.8 -- 400 -- 25 x 2.3
300 (geo/13) -- 1.1 -- 260 -- 42 x 1.4
200 -- 0.74 -- 180 -- 62 x 0.96

The concave mirrors column shows the minimum number and size of concave mirrors required to provide the reflector area. However, as mentioned above, at geo/3 and below, you would likely use smaller mirrors in a single orbiting ring. For example, to provide the 1,300 km^2 reflector area for the geo/8 case, you could use 430 x 1 km radius flat mirrors, or 43,000 x 100 meter radius flat mirrors, in a circular orbit inclined ~42 degrees to the ecliptic.

Current solar sail projections use mass estimates that correspond to betwen 5 and 10 tonnes per km^2. So conservatively, multiply reflector area by 10 to get the mass in tonnes, e.g., 13,000 tonnes for geo/8. There was an interesting paper that talked about using dusty plasmas for solar sails ...

Sheldon et al., Dynamic and optical characterization of dusty plasmas for use as solar sails, Jan 2002
http://adsabs.harvard.edu/abs/2002stai.conf..425S

... that yeilded the equivalent of 1 tonne per km^2 (1% reflectivity, 10 kg/km^2), but that tech has a way to go before we even know if it is possible.

1,700 km seems to be the minimum height to allow for year round illumination (and the target would need to be between latitudes 35 and 45), so that sets our launch mass at 13,000 tonnes - not something we're going to see being launched from Earth any time soon. It does get you 30 km^2 of continuous full Martian daylight (and up to 75 km^2 of partial daylight). That could give you a pretty interesting microclimate, but it seems a stretch that resources would be dedicated to such a project any time soon.

noosfractal · 2007-08-14 04:04:25

The reason they use convex is because they want to inflate them with a gas charge.

I'm all for inflatables where it makes sense, but I'm sure I saw inflatable concave dishes out there. There was an inflatable ring with the reflective film stretched across it. I can't remember how they made it concave. I'll look into it.

Here you go ...

Cassapakis et al., Inflatable structures technology development overview, Sept 1995
http://www.lgarde.com/papers/overview.pdf

... basically, you have 3 parts - an inflatable ring, a back film that is coated with something reflective, and a front film that is transparent. The two films are joined at the edges like a balloon, as well as to the ring. The reflective coating is on the inside of the "balloon". You inflate the ring and the space between the two films to form the concave reflective dish. You can even change the focal length of the dish by varying the gas pressure.

noosfractal · 2007-08-16 02:44:46

So the major problem is the h/300 spot size. The light from a single km^2 of reflecting surface gets spread out over a large area and requires a much larger collector area to compensate. The next thing to consider is putting a focusing lens in front of the mirror. The lens adds mass and reduces efficiency (transmissivity is always < 100%) but, with fairly moderate tech assumptions, it turns out to be quite an improvement.

Consider the orbiting flat circular mirror radius r again, right now it is acting as a light source that could be replaced by a point light source ~300.r further from the surface. So, if a convex lens the same size as the mirror, with a focal length of 300.r is placed in front of the mirror, it will compensate for the beam divergence caused by the finite angular size of the Sun, and the spot size on the surface will be r (ignoring atmospheric scattering).

That is, ignoring other complications for the moment, a 1 km radius mirror + 1 km radius lens with focal length 300 km = 1 km radius spot size on the surface. This is independent of orbital height, so the mirror can be conveniently placed at Martian GEO. With 50% total efficiency, this pretty much yields the illuminated 1 km^2 goal, and the mirror alone would have a conservative mass estimate of ~30 tonnes, which is fairly reasonable considering the tonnages mentioned in the posts above.

But what about the lens mass? I think you could use this technology ...

Inflatable Fresnel Lenses as Concentrators for Solar Power
http://www.techbriefs.com/content/view/1833/32/

... but designed as a collimating lens. So the design would look something like this ...

- vacuum
- front gas membrane (i.e., the planet facing part of the inflatable structure)
- collimating Fresnel lens
- inflating gas (Nitrogen seems to be the standard)
- reflector surface (microns thick coating of Aluminum)
- rear gas membrane (Mylar?)
- control elements (even GEO needs stationkeeping)
- vacuum

The 30 tonne estimate includes the gas membranes, the reflector surface, and the control elements.

The mass of the inflating gas depends on the structure volume and inflation pressure. At GEO distance, the ideal 1 km radius dish has a depth of 15 mm, which would create a volume of 120,000 m^3 for a symmetric structure. Typical inflation pressures are on the order of 100 Pa (i.e., 1/1000th of an atmosphere), so if temperatures within the structure can be kept reasonable (shouldn't be hard with constant solar illumination), then the gas mass ~ 150 kgs (i.e., negligible).

The cited article gives a figure of 2 kg/m^2 for a 250 micron thick Fresnel lens. That's a mass of 6000 tonnes, which dwarfs all the other figures, but it still compares favorably with the 13,000 tonnes previous best case. However, I don't think an order of magnitude improvement is overly optimistic (50 micron, using polycarbonate lens material), so that would put the total system mass at ~630 tonnes. Still not a done deal, but a huge improvement over previous estimates. A project that could conceivably be undertaken at some point.

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#1 2007-08-08 03:45:15

Re: Microterraforming

#2 2007-08-08 15:09:39

Re: Microterraforming

#3 2007-08-09 01:36:02

Re: Microterraforming

#4 2007-08-09 10:19:39

Re: Microterraforming

#5 2007-08-09 18:50:57

Re: Microterraforming

#6 2007-08-10 03:05:23

Re: Microterraforming

#7 2007-08-12 01:41:50

Re: Microterraforming

#8 2007-08-13 02:27:57

Re: Microterraforming

#9 2007-08-14 04:04:25

Re: Microterraforming

#10 2007-08-16 02:44:46

Re: Microterraforming

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