Microblog Memes

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is this what it sounds like!? (infosec.pub)

submitted 2 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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[–] hobovision@mander.xyz 5 points 2 months ago (1 children)

What is the physics or math behind that? Light from the sun is essentially aligned by the time it reaches earth. If the mirror is perfectly reflective, a 10 m^2 mirror should light up a patch of Earth roughly 10 m^2 times the cosine of the angle of the mirror. So unless the angle is close to 90°, most of the losses would be from poor reflectivity.

I totally agree it's a stupid idea. But maybe it's even worse than I am thinking of?

[+] LostXOR@fedia.io 12 points 2 months ago* (last edited 2 months ago) (2 children)

[deleted]

[–] davad@lemmy.world 5 points 2 months ago* (last edited 2 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:

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.
We want 1000 W/m^2 on the ground to qualify as daylight [1]
Collimated light
No attenuation or scatter from the atmosphere, but we will assume the beam diameter spreads 0.5 degrees [2]
Perfectly reflective mirror
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

[–] exasperation@lemmy.dbzer0.com 2 points 2 months ago (1 children)

You can't get away with less because a mirror can't appear brighter than what it's reflecting; this is a fundamental property of optical systems.

I can understand that a single flat mirror cannot ever appear brighter than whatever is being reflected. But why can't multiple mirrors pointed at one spot have a total intensity greater than that of any one of the mirrors (or a curved dish that focuses the light)?