• mozz@mbin.grits.dev
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      8 months ago

      My scientific research of squinting at the poster says a spy satellite is probably about as long as a pickup truck which is probably about 20 feet long.

      xkcd says space is 100 km away and I’m sure there’s nothing else I need to understand about that.

      At 100 km away, the change of angle that will move your beam by 20 feet (enough to make the difference between hitting or not, if the thing and the flat mirror are both about 20 feet long I guess) is (20 feet / 100 km / pi) radians or 0.0000194 radians, meaning you raised or lowered one edge of the mirror by 0.004 inches or around the width of pretty-thick hair. I would be a little surprised if the mirrors even stayed within that tolerance just from flexing around in the wind for as big as they are.

      On the other hand, you wouldn’t have to hit the spy satellite with every mirror; you could probably heat it up significantly just by hitting it with a bunch of the beams as they were swinging wildly around and mostly missing it. And if it was specifically a spy satellite, you could probably fry its optics with not really a lot of mirrors for not a long time actually managing to hit it.

      On the other other hand the thing would be flying along at around 8 km/s, so you’d have to get your mirrors positioned accurately enough, and then start moving them at a relatively insane speed while still keeping their absolute positioning dead accurate when their motors and overall construction clearly weren’t designed for either of those tasks at the required level of precision.

      TL;DR Let’s try it

      Also there’s this

      • TropicalDingdong@lemmy.world
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        8 months ago

        You still have a crap-ton of atmosphere you have to get through, and the beams being reflected aren’t coherent. So the light reflected is subject to the inverse square law, which means that the energy diminishes as the inverse square of the distance. So the actually energy reaching the satellite would be minuscule. If you want to effectively use light to punch all the way through the atmosphere, you’ll need beam coherence.

        • mozz@mbin.grits.dev
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          8 months ago

          The difference in the angles of the beams is the angle difference of a beam that came from an object 149,597,871 km away at a separation of 20 feet i.e. basically fuck-all. For this purpose I think they’re effectively (edit: coherent) parallel. And I think the atmospheric reduction would be significant but not defeating-to-the-purpose; I mean the sunbeam on its way in still had plenty of effectiveness after getting through the same atmosphere. If you did it on a cloudy day or something then yeah it wouldn’t work at all.

          (Edit: Wait, I don’t understand optics; I mean parallel, not coherent. I don’t think coherence enters into it?)

          • TropicalDingdong@lemmy.world
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            8 months ago

            The losses due to beam angle is nothing compared to the losses due to the inverse square law. This is why coherence is so critical for getting substantial quantity of photons from point A to point B. Lasers are defined by this difference, in that the light they produce is coherent. Because of this lasers are detraction limited, and have very low divergence at distance. Incoherent light sources like the sun have random amplitudes and phases in regards to time and space, so have very short coherence distances.

            You could buy and build what this guy did, and probably get a few photons all the way through the atmosphere. The GEDI space laser fires with a power of 10mJ, and still results in a beam footprint of 25m. Granted the laser has to make a two way trip, but only a couple of hundred thousand photons are making it back to the sensor. So you would probably be able to see the glittering object using a high resolution camera, but there is no way that incoherent light could make any meaningful difference to something in space (considering, you know, its also being hit by radiation from the sun, you know radiation that hasn’t been filtered trough the atmosphere.)

            • mozz@mbin.grits.dev
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              8 months ago

              Divergence and lack of coherence are two very different things (as I fully realized only after I typed up my message, I guess).

              Divergence is a result of the angle. If you’re producing light from a local point-source, you have to work very very hard to make sure the angle of the departing rays is as close as you can make it, and you’re still not going to get anywhere even remotely close to 20 feet divided by 149,597,871 km. That’s where all the insane dropoff in the examples you’re talking about is coming from. The rays from the sun, though, are effectively parallel by the time they reach the earth to points 20 feet separated.

              The inverse-square law is a result of the power in the beam spreading out over a larger area and spreading out its energy output over a wider area. It’s just a way of expressing that if the beam has spread itself out from hitting 1’x1’ into hitting 10’x10’ at a distance 10 times greater, each square foot of the target will now only get 1/100 of the energy. It won’t get weaker in total, without being absorbed by something along the way; that would violate conservation of energy. In this case the beams are parallel, the target is still 20’x20’ plus some tiny tiny fraction, there is a little bit of absorption by the atmosphere but not enough to make it not bright. The sun’s light goes through the atmosphere and it’s still bright (somewhat brighter if you’re on a mountain or in space, with a lot more UV, but not like night and day.)

              I don’t see that coherence fits into this particular part of it in any way; as far as I know, we use lasers for this type of purpose because of their low divergence and the coherence has nothing to do with it. The rays originally from the sun have no coherence and they still manage to make it all the way out here.

              • TropicalDingdong@lemmy.world
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                8 months ago

                Coherence is the key here, I assure you. Incoherent light is subject to the inverse square law in a way that lasers, which demonstrate coherence, are not. Lasers are coherent and collimated, and as such don’t interfere with one another and are parallel contributing to the laser’s ability to remain focused over long distances without spreading out significantly. This collimated nature of laser beams is a direct result of their high degree of spatial coherence, allowing them to maintain intensity over distances where a non-coherent light source would have dispersed according to the inverse square law. You arent reflecting coherent, in-phase, collimated from mirror, even if the suns rays strike the mirror parallel.

                Lets assume each of the mirrors reflects 850 watts. The distance to the ISS is 408,000 meters.

                The energy reflected by one mirror as received by the ISS is subject to the inverse square law (because it is incoherent).

                E = (850 watts) / (4pi408000m)2,, or about 4.06x10 −10 watts/m2

                A 5 milliwatt, off the shelf laser pointer with a beam divergence of 1.5 millirads would deliver approximately 4.25x10-9 watts/m2, or about 10x as much energy as the 850 watt mirror.

                You can not melt a spy satellite with mirrors. You might be able to with lasers. A laser will be approximately 8.9x106 times as power effecient at getting light from earth to the ISS as a mirror would be. This is directly due to the properties of laser light, specifically coherence and collimation, which make it not subject to the inverse square law.

                • mozz@mbin.grits.dev
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                  8 months ago

                  You’re confused, sir. Light from the sun is collimated, yes, i.e. parallel rays. The correct equation if you’re going to apply the inverse square law is:

                  E = 850 watts / 149,597,971 km^2 * 149,597,871 km^2 = 849.998864 watts

                  Same reason a signal mirror can reflect a flash as bright as the sun even miles away off a surface a few inches square.

                  You can believe or not; I’ve explained it as clearly as I know how.

            • SchmidtGenetics@lemmy.world
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              8 months ago

              Inverse square law is negligible, it’s already traveled from the Sun to earth, from the earth back up is a fraction of what it’s already traveled.

              • TropicalDingdong@lemmy.world
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                8 months ago

                Well no its not because its also been filtered via the atmosphere, then it got reflected off a mirror, now it has to make the trip again, and for all intents and purposes is incoherent.

                Basically all of the energy reflecting from the mirror is lost before it hits the ISS.

                • SchmidtGenetics@lemmy.world
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                  8 months ago

                  The filtering the first time is marginal, same the second time.

                  The inverse square law is mathematically insignificant, why do you think you can still be blinded by a mirror? The source doesn’t become the mirror, the math is still calculated from the source, you need to account for the mirrors refraction in the calculation though.

                  It’s also thousands of sources, even at 1% (probably isn’t this low, but worst case here) is more than the direct energy hitting it from the sun.

    • Smuuthbrane@sh.itjust.works
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      8 months ago

      Uh… losses from transmitting through the atmosphere a second time?

      Damn. I wonder what its operational range would be.

      • einfach_orangensaft@feddit.deOP
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        8 months ago

        this thing is big enough to alter the average reflective index of a whole state if it swings around its mirrors

        the focus spot in theorie could be set to any range, just as u go more far the precision of each mirror angle will be the limiting factor amongst atmospheric losses distortions.

  • Donjuanme@lemmy.world
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    8 months ago

    Those are designed to focus on a large, stationary, object not far away, not a small hypersonic object very very far away.

  • FiskFisk33@startrek.website
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    8 months ago

    I suspect in order to stay focused on such distances you’d need extremely flat mirrors. Like, telescope grade stuff.

    I doubt the mirrors they have is even within an order of magnitude flat enough.

    • Liz@midwest.social
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      8 months ago

      You might even need adaptive mirrors to deal with atmospheric distortion. Also, they would have to move relatively quickly and very precisely (read: an impossible combination) to track satellites in low orbit. Plus, you could only hit satellites that crossed overhead at a relatively high angle.

      But yeah, one solar tower plant did a stunt where they reflected an image made of sunlight at the ISS and an astronaut took a picture. They didn’t melt.

      • einfach_orangensaft@feddit.deOP
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        8 months ago

        where they reflected an image made of sunlight at the ISS and an astronaut took a picture

        got a link to said picture? it may make for a good meme template. I saw that the chinese did that kind of ‘pixel art’ with there own near identical solar thermal plant