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[-] TropicalDingdong@lemmy.world 7 points 7 months ago* (last edited 7 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.)

[-] Donjuanme@lemmy.world 6 points 7 months ago

Filtered through the atmosphere twice*

[-] SchmidtGenetics@lemmy.world 5 points 7 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 -2 points 7 months ago* (last edited 7 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 6 points 7 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.

[-] TropicalDingdong@lemmy.world -3 points 7 months ago

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.

That depends on the mirror, but I get the point you are making. However the light reflected off the mirror going to be subject to the angle of incident of the mirror. These are concave mirrors with specific focal lengths, not in the range of kilometers, but the range of meters. The efficiency of these mirrors is going to be far far far far far lower than 1% at a distance of 400,000 meters.

You'll get far more energy to the ISS if you use a laser pointer than if you use a mirror, even if thousands of times as much energy is being reflected by the mirror.

[-] einfach_orangensaft@feddit.de 6 points 7 months ago

the mirrors are flat

[-] mozz@mbin.grits.dev 4 points 7 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 -1 points 7 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/m^2^

A 5 milliwatt, off the shelf laser pointer with a beam divergence of 1.5 millirads would deliver approximately 4.25x10^-9^ watts/m^2^, 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 8.9x10^6^ 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 7 points 7 months ago* (last edited 7 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 -1 points 7 months ago* (last edited 7 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.

[-] SchmidtGenetics@lemmy.world 4 points 7 months ago

The distance you need to account for is Sun to earth plus back to the satellite. Which I think is what they were using.

[-] TropicalDingdong@lemmy.world -1 points 7 months ago

Yeah but what they are not accounting for is that the light actually has to come off the mirror. You can demonstrate this with a hand mirror that the illumination spot gets larger quickly at a distance. Take a mirror and find the sun. Send a reflection to the wall nearest you. Then send the reflection to a wall further away. The reflection on the wall further away is larger and therefore, the energy more spread out. The light coming off the mirror is not perfectly parallel as it had to pass through the atmosphere, then interact with the surface of the mirror. We do use mirrors for calibration in satellite remote sensing, but you will get far far far far more power arriving at something like the ISS coming from a much less powerful laser over such a distance. If we controlled by wattage, a laser will absolute crush a mirror in its ability to transmit energy over a distance.

[-] mozz@mbin.grits.dev 3 points 7 months ago* (last edited 7 months ago)

Send a reflection to the wall nearest you. Then send the reflection to a wall further away. The reflection on the wall further away is larger and therefore, the energy more spread out.

I am very confident that you have not tried this for yourself. I want you to show me pictures of this happening (with sunlight); I think you will find the experience educational.

(Edit: The mirror and the wall must both remain at the same rotational angle -- if you angle the mirror to move the spot, and the spot becomes elongated because it's now coming in at a more acute angle, it doesn't count. Shining a sunbeam on the edge of a doorframe and then through the door to a faraway wall that's through the doorway would be a good way to do it.)

[-] SchmidtGenetics@lemmy.world 1 points 7 months ago

Refraction and reflection. Most non specialty (consumer) mirrors have low quality and standards, so are affected by these more than other specialty mirrors.

[-] TropicalDingdong@lemmy.world 0 points 7 months ago

Yeah but any mirror that isn't imaginary has some kind of surface coating that's a different refractive index than the atmosphere, and then the light has to interact with the surface of the mirror, which is not a perfect reflector and has imperfections, not to mention the light just had to pass through 400km of fluid atmosphere of varying density and composition.

Its a bet I'm very willing to take.

[-] SchmidtGenetics@lemmy.world 1 points 7 months ago

Just give up dude , inverse square law doesn’t apply here, you were incorrect and are now making an idiot of yourself trying to incorrectly explain it.

[-] mozz@mbin.grits.dev 2 points 7 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

[-] TropicalDingdong@lemmy.world -1 points 7 months ago

So you are like, an actual idiot then. I guess I picked the right meme for you.

[-] mozz@mbin.grits.dev 2 points 7 months ago* (last edited 7 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?

[-] TropicalDingdong@lemmy.world -1 points 7 months ago

I will take a bet from you that the energy arriving at the ISS from a laser pointer you or I could purchase off Amazon, so consumer grade, is more than the energy that would arrive at the ISS from a concave mirror that would be used at a solar generating station.

[-] mozz@mbin.grits.dev 3 points 7 months ago

A perfectly flat mirror, exactly like one of the ones in the OP generating station picture. Both are aimed perfectly on target, and the mirror is reflecting light from the sun on a sunny day. With those caveats I'll bet $1,000. I'm happy with any university physicist or physics professor to be the judge, or Randall Munroe, or make a proposal of some other person if neither of those are acceptable to you. A lower amount of money is also fine if you're not comfortable with $1,000.

[-] TropicalDingdong@lemmy.world 0 points 7 months ago

I think the mirror should be a standard one from the Ivanpah Solar power facility. Wikipedia puts them at 7 meters in area. Wikipedia also puts it at 7.4 kWh/m2/day. I think those would be acceptable parameters for you? I think this fits in with the spirit of the bet because these would be the specific parameters taken from the mirrors mentioned in the meme.

Do you have any suggestions for me on parameters to constrain my shopping on Amazon for laser parameters? I said a laser I could purchase on Amazon, so I'm ok with sticking with them as a source. I can buy some pretty damn beefy lasers off Amazon. For example, I can buy a 2000 watt laser on prime right now. Do you want constraints here?

Also, $1000 is out of my price range. I can afford to lose $100. Are you ok with that?

[-] mozz@mbin.grits.dev 3 points 7 months ago* (last edited 7 months ago)

Well but the actual mirrors would not work at all because of any number of reasons (among other reasons they can't track fast enough or precisely enough to actually hit a satellite, and they're going to have little imperfections in their flatness which will distort the reflected beam away from what the laws of optics say would happen for an idealized situation). The whole actually-shooting-down-satellites thing is clearly a joke; my point was more disagreeing with your description of how the laws work in the idealized situation.

Hmm... I am confident that optically, an idealized flat mirror will reflect a patch of sunbeam that's more collimated than any human-produced laser at any price. I'm less sure about the actual Ivanpah mirrors but I would guess that they are flat enough to produce a beam that's more collimated than a standard consumer laser. The thing is that that's more or less impossible to test... we can ask a physicist about the physical laws, but my guess is that they would be as clueless as I am about how precisely flat the mirrors actually are and of course there's a lot of wiggle room in how fancy a laser we want to say you can get.

We could ask one of those Youtubers like the ones I linked to if they want to fly a helicopter into the beam from one mirror at a great distance and do measurements of how much the beam had attenuated in practice, but that seems like a great setup to a "what could go wrong" disaster video in the making...

[-] TropicalDingdong@lemmy.world 0 points 7 months ago

So bet or not dude? Cus you seem like you are backing out.

[-] mozz@mbin.grits.dev 3 points 7 months ago

I will bet $1,000 (or $100) that this is wrong:

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/m^2^

A 5 milliwatt, off the shelf laser pointer with a beam divergence of 1.5 millirads would deliver approximately 4.25x10^-9^ watts/m^2^, 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 8.9x10^6^ times as power effecient at getting light from earth to the ISS as a mirror would be.

And this is right:

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

I will also bet on the behavior of an idealized flat mirror. I won't bet on whether you can actually shoot down satellites with the Ivanpah solar plant, because there are real-engineering issues that interfere with it aside from the physics of how mirrors and sunlight work.

If you want to try to chart a middle ground, I think it'd be better to talk about something actually testable than trying to argue about the real-world behavior of the Ivanpah mirrors. I'd be happy to bet $100 that:

Take a mirror and find the sun. Send a reflection to the wall nearest you. Then send the reflection to a wall further away. The reflection on the wall further away is larger and therefore, the energy more spread out.

... is wrong, as long as the mirror is flat. This one is easy to test so this might be a better bet.

[-] blanketswithsmallpox@lemmy.world 3 points 7 months ago
[-] mozz@mbin.grits.dev 4 points 7 months ago

I care so much that I was outside earlier today, messing with my little reflective discs. It's actually really hard to get the angles right and I couldn't find a wall that was at the right type of angle to be able to test it without the reflection skewing from where the sun was, and a couple of people came near me, and clearly looked at me sort of wondering "what the hell is this guy doing."

I made no attempt to explain. Y'all can think I'm insane or whatever. I'm doing science. Get the fuck away.

[-] TropicalDingdong@lemmy.world -1 points 2 months ago* (last edited 2 months ago)

… is wrong, as long as the mirror is flat. This one is easy to test so this might be a better bet.

But that was never the bet. The bet was about transmitting light through the atmosphere. This is just some weird little aside you got yourself tangled in. More than happy to take on a bet about the transmission power of mirrors versus lasers from space, which is what were actually discussing.

(Dug this one out of the grave because I'm trying to find another bet I just won, so was searching for "bet")

this post was submitted on 15 Feb 2024
451 points (95.4% liked)

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