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?)
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.)
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.
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.
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.
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.
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.
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:
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.
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.
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.)
Refraction and reflection. Most non specialty (consumer) mirrors have low quality and standards, so are affected by these more than other specialty mirrors.
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.
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.
Yeah this is completely wrong. Square law, the atmosphere absorption, non-coherence are all right on. You need a laser based in space to deliver real usable power.
The ruzzians are probably going to fly a space based sealed CO2 laser powered by radioactive generator.
Even if you took the entire face of the earth and converted it to mirrors, you would probably not have enough power to burn satellites. Simple demonstration…get a lens that is good for burning ants outside…now go inside your house and turn on your powerful TV. Focus it’s light on paper and see if you can burn it. Repeat with fire and with a projector.
get a lens that is good for burning ants outside…now go inside your house and turn on your powerful TV. Focus it’s light on paper and see if you can burn it. Repeat with fire and with a projector.
The light from the sun has traveled millions of km already, the fraction it has to travel from the earth back up the satellite is mathematically insignificant.
Your tv isn’t powerful… use something that has a similar function, like an actual light and it can work, but it will be affected by the inverse square law since the source is right there, unlike the sun.
What’s that got to do with your comment? A lightbulb is hot enough to start a fire, a tv isn’t… so using that instead of the other asinine examples you provided would absolutely work…
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?)
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.)
Filtered through the atmosphere twice*
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.
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.
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.
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.
the mirrors are flat
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.
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.
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.
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
The hell even is it that you think you are representing by these numbers? What is it you think you are saying?
fourth
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.
The distance you need to account for is Sun to earth plus back to the satellite. Which I think is what they were using.
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.
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.)
Refraction and reflection. Most non specialty (consumer) mirrors have low quality and standards, so are affected by these more than other specialty mirrors.
I’m gonna leave the source of that and the other number and where the pi went as an exercise for the reader
So you are like, an actual idiot then. I guess I picked the right meme for you.
Will you bet me money on how signal mirrors work and how bright the flash is at a certain distance from the mirror?
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.
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.
Yeah this is completely wrong. Square law, the atmosphere absorption, non-coherence are all right on. You need a laser based in space to deliver real usable power.
The ruzzians are probably going to fly a space based sealed CO2 laser powered by radioactive generator.
Even if you took the entire face of the earth and converted it to mirrors, you would probably not have enough power to burn satellites. Simple demonstration…get a lens that is good for burning ants outside…now go inside your house and turn on your powerful TV. Focus it’s light on paper and see if you can burn it. Repeat with fire and with a projector.
Bet you $100 that the inverse square law doesn’t apply in the way you think it does to sunlight reflected off a flat mirror
Step up, step up, you can’t win if you don’t play
The light from the sun has traveled millions of km already, the fraction it has to travel from the earth back up the satellite is mathematically insignificant.
Your tv isn’t powerful… use something that has a similar function, like an actual light and it can work, but it will be affected by the inverse square law since the source is right there, unlike the sun.
https://what-if.xkcd.com/145/
Take a look and let me know if you got any questions lol. Hey it’s xkcd right? Something’s gotta be right. Better source of info than me.
What’s that got to do with your comment? A lightbulb is hot enough to start a fire, a tv isn’t… so using that instead of the other asinine examples you provided would absolutely work…