• mozz@mbin.grits.dev
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    5 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?)

    • lurker2718@lemmings.world
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      5 months ago

      The problem is the size of the sun. If you could look at the sun (don’t, try the moon its approximately the same size in the sky), you see it has a relatively large angular size. Its not just a point in the sky.

      So the problem, the rays from one point of the sun are almost parallel. But the rays from the different points of the sun are not. So they also aren’t parallel after your mirror. They spread in an angle similar to the size of the sun on the sky. And this is much larger than a satellite. So you cannot focus all energy on a satellite.

      • mozz@mbin.grits.dev
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        5 months ago

        Yep this is 100% accurate. I got so carried away disagreeing with the idea that it’ll spread out again in inverse-square fashion like from a point source, that I completely missed the people telling me that it’ll spread slightly because of the size of the sun. Absolutely true.

    • TropicalDingdong@lemmy.world
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      5 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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        5 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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          5 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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            5 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.

            • TropicalDingdong@lemmy.world
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              5 months ago

              149,597,971

              Where are you getting this number from? The number you need to be using is the distance from the earth to the ISS, 408,000 meters.

              Second, your formulation of the inverse square law is incorrect, in that you are missing the 4pi component, but in the grand scheme of distances we’re looking at, its negligible. It also looks like you may have gotten the order of operations wrong.

              Third

              149,597,971 km^2 * 149,597,871 km^2

              The hell even is it that you think you are representing by these numbers? What is it you think you are saying?

              fourth

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

              I can provide sources for all my claims, and I’m pretty sure I got all my math correct, within a ROM. I can’t say the same for your work.

              • mozz@mbin.grits.dev
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                5 months ago

                Where are you getting this number from?

                I’m gonna leave the source of that and the other number and where the pi went as an exercise for the reader

                  • mozz@mbin.grits.dev
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                    5 months ago

                    Will you bet me money on how signal mirrors work and how bright the flash is at a certain distance from the mirror?