3 ms·

I don't really follow this argument, and I would like to. I think one thing which would help me develop an intuition for it would be to see the calculation of

I don't really follow this argument, and I would like to. I think one thing which would help me develop an intuition for it would be to see the calculation of the lens size for heating a one square-centimeter area on the earth to as high a temperature as possible by the light of the moon, and what that optimal temperature is. Anyone reading for whom this is straightforward? Even a description of how to do the calculation would go a long way.

Dylan16807 8y ago
You get the highest temperature by reflecting moonlight from every angle. For a single lens you just want something really big. Make it take up 90+% of the sky from your target. For temperature, the no-calculation way is to measure a rock on the moon (article says 100C) and use that number for how hot you should be able to get. The calculation way goes as follows: Near earth you get 1400 watts per square meter of sunlight, so if that bounces perfectly off the center of a full moon and gets through the atmosphere with no losses, your target will get 1400 watts per square meter. That's equal to a black body at 122C. After taking into account the spherical shape and atmospheric losses you might get less than half of that, so ambient heat might drown out your results.
- deleted 8y ago
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- AlexCoventry 8y ago
  > You get the highest temperature by reflecting moonlight from every angle. I see the thermodynamic principle at work, here, but I don't really understand how it's operating at a mechanistic level. Is it possible to demonstrate that assertion optically?
  - Dylan16807 8y ago
    Take your target and trace a ray in every direction away from every point on its surface. The more of these that hit the energy source, the more energy you're getting. And 100% is obviously the best you can do.
    - AlexCoventry 8y ago
      From an optical perspective, how does the argument dismissed at the start of the OP break down? Is there some reason the moon's light, gathered from a lens covering hundreds of acres, carries insufficient energy to light a fire in concentrated form?
      - Dylan16807 8y ago
        The optical argument goes like this: Measure the brightness (in energy per square meter) right at the moon's surface. No matter what you do with lenses, you can't concentrate moonlight beyond this level. You can make an enormous lens where all the energy coming off a particular acre of moon goes through it. But you can't focus all of it onto a spot smaller than an acre, no matter what you do.
        AlexCoventry 8y ago
        Thanks for the explanations. The current top comment here seems to agree with my original intuition, though. https://news.ycombinator.com/item?id=18739120 https://news.ycombinator.com/item?id=18739120
        Dylan16807 8y ago
        As a few people pointed out, the reflection off the moon ruins the étendue, so there's no concentrating it beyond what all the rocks on the surface already experience.
        AlexCoventry 8y ago
        Thanks again for your help. I think that led me in the right direction. This is the optical argument for maximum concentration given conservation of étendue (the same page has an optical argument for the conservation): https://en.wikipedia.org/wiki/Etendue#Maximum_concentration https://en.wikipedia.org/wiki/Etendue#Maximum_concentration The angle subtended by the moon is approximately 0.54 degrees, or about 0.01 radians, so the maximum concentration factor is about 10,000. The moon provides about 0.1 lux of illumination, so the maximum illumination you can achieve by optics is about 1000 lux. The sun subtends about the same angle, and we receive 30,000-100,000 lux from it, and the maximum illumination you can achieve from concentrating it optically is about 1B lux. I'm willing to believe that you can't light a fire from the moon, if the intensity's a million times lower.

Dylan16807 8y ago

You get the highest temperature by reflecting moonlight from every angle. For a single lens you just want something really big. Make it take up 90+% of the sky from your target. For temperature, the no-calculation way is to measure a rock on the moon (article says 100C) and use that number for how hot you should be able to get. The calculation way goes as follows: Near earth you get 1400 watts per square meter of sunlight, so if that bounces perfectly off the center of a full moon and gets through the atmosphere with no losses, your target will get 1400 watts per square meter. That's equal to a black body at 122C. After taking into account the spherical shape and atmospheric losses you might get less than half of that, so ambient heat might drown out your results.

deleted 8y ago
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AlexCoventry 8y ago
> You get the highest temperature by reflecting moonlight from every angle. I see the thermodynamic principle at work, here, but I don't really understand how it's operating at a mechanistic level. Is it possible to demonstrate that assertion optically?
- Dylan16807 8y ago
  Take your target and trace a ray in every direction away from every point on its surface. The more of these that hit the energy source, the more energy you're getting. And 100% is obviously the best you can do.
  - AlexCoventry 8y ago
    From an optical perspective, how does the argument dismissed at the start of the OP break down? Is there some reason the moon's light, gathered from a lens covering hundreds of acres, carries insufficient energy to light a fire in concentrated form?
    - Dylan16807 8y ago
      The optical argument goes like this: Measure the brightness (in energy per square meter) right at the moon's surface. No matter what you do with lenses, you can't concentrate moonlight beyond this level. You can make an enormous lens where all the energy coming off a particular acre of moon goes through it. But you can't focus all of it onto a spot smaller than an acre, no matter what you do.
      - AlexCoventry 8y ago
        Thanks for the explanations. The current top comment here seems to agree with my original intuition, though. https://news.ycombinator.com/item?id=18739120 https://news.ycombinator.com/item?id=18739120
        Dylan16807 8y ago
        As a few people pointed out, the reflection off the moon ruins the étendue, so there's no concentrating it beyond what all the rocks on the surface already experience.
        AlexCoventry 8y ago
        Thanks again for your help. I think that led me in the right direction. This is the optical argument for maximum concentration given conservation of étendue (the same page has an optical argument for the conservation): https://en.wikipedia.org/wiki/Etendue#Maximum_concentration https://en.wikipedia.org/wiki/Etendue#Maximum_concentration The angle subtended by the moon is approximately 0.54 degrees, or about 0.01 radians, so the maximum concentration factor is about 10,000. The moon provides about 0.1 lux of illumination, so the maximum illumination you can achieve by optics is about 1000 lux. The sun subtends about the same angle, and we receive 30,000-100,000 lux from it, and the maximum illumination you can achieve from concentrating it optically is about 1B lux. I'm willing to believe that you can't light a fire from the moon, if the intensity's a million times lower.