"0something0" wrote in message
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Its also going to need to have YUUGE radiators and be able to operate at
high temperatures. The radiator needs to stay inside the shadow of the solar
array, which will limit how big it can be.

Assuming that in that distance, you get 55800 watts/square meter(from a
figure on Wikipedia and Space.com), and we use the equation

A = P / (ε * σ * T^4)

From He http://www.projectrho.com/public_htm...asicdesign.php

Few assumptions: We use the alloy with the highest known melting point with
a max emissivity. That is 4488 degrees kelvin. We also assume a 1km^2 square
array. So, it collects 44,800,000,000 watts, or 44 gigawatts of heat. So,
the equation substituted is

A = 44,800,000,000 watts/(1*5.670367(13)*10^−8*W* 1 m^−2*1 K^−4*4488 k) =

https://www.wolframalpha.com/input/?...224488.15+K%22

Umm... did I do it right? If I did, Its not that bad!
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What's the area you conclude is required for the radiators?

In any case, what you want is most of the power to be reflected away by the
mirrors.

BTW, I estimate the power per square meter is 862,000 watts/square meter
based on the fact the light power goes inversely by the square of the
distance and 3.7 million miles is 25 times closer than the Earth's distance
of 93 million miles.

Bob Clark

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