this post was submitted on 08 Aug 2025
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Microblog Memes

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is this what it sounds like!? (infosec.pub)
submitted 7 months ago by not_IO@lemmy.blahaj.zone to c/microblogmemes@lemmy.world
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https://mastodon.social/@sundogplanets/114994624601006489

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[–] davad@lemmy.world 5 points 7 months ago* (last edited 7 months ago) (1 children)

To light up a location as brightly as the Sun would, you need to cover a half-degree circle in the sky (viewed from that location) with mirrors that reflect the Sun directly at the location.

That's the best, simplest example I've seen for why this doesn't work. But...I wanted to look at it from the perspective of irradiance losses from the beam spreading. It's been a long time since I did any optics, so I could be way off-base with my approach. Feel free to correct anything I screw up.

Here are my assumptions:

  1. Near space irradiance from the sun is 1,367 W/m^2 [0]. Let's round up and assume the mirror gets 1400 W/m^2 from the sun.
  2. We want 1000 W/m^2 on the ground to qualify as daylight [1]
  3. Collimated light
  4. No attenuation or scatter from the atmosphere, but we will assume the beam diameter spreads 0.5 degrees [2]
  5. Perfectly reflective mirror
  6. Mirror 600 km away from the earth

Beam spreading loss is a function of distance. So however large the beam width (mirror diameter) starts, it'll be this much bigger when it reaches the ground:

600km * tan (0.5 degree) = 5.24km

That means if we have a 1m diameter mirror, we get a beam 5.24km + 1m on the ground. If we have a 5km diameter mirror, we get a 10.24km beam on the ground.

To get our target of 1000 W/m^2, we need at least 1000/1400 = 0.71 of what hits the mirror to hit our target.

mirror/(mirror+spread) >= 0.71 mirror >= 12.83km

[0] https://en.wikipedia.org/wiki/Sunlight#Measurement
[1] Wikipedia says that we actually get more like 1100 W/m^2 when the sun is at its zenith.
[2] https://en.wikipedia.org/wiki/Collimated_beam#Distant_sources