Comment by Nevermark

Comment by Nevermark 3 days ago

Things scale so differently, we don’t need a parts list to make a general tradeoff relationship.

Of the moon and orbital, orbit is much closer and will be cheaper to start with.

But a lunar site would scale to much greater mean density and unlimited total capacities. And be much cheaper for reasons I gave, at some threshold scale.

Neither is easy, and it’s not at all clear that either is actually better than down here. Especially with nuclear efforts and funding rising quickly.

godelski 3 days ago

I'm not asking for a parts list, I'm asking for math

Reply View 4 replies

Nevermark 2 days ago

if you want some math beyond identifying different scaling issues, go for it.

Reply View | 3 replies
- godelski 2 days ago
  
  I have. That's why I'm certain you're wrong. But if you're too lazy to do it yourself there's plenty of papers written on this stuff
  
  Reply View | 2 replies
  
  Nevermark 2 days ago
  
  Well then, use your math to address a point. Just one.
  In a vacuum, radiative heat loss per time hyper scales with temperature to the 4th power.
  In orbit large and complex heat transfer systems are not going to be practical. On a surface, specialized heat pumps can localize heat energy to very high intensity. With critical reliability advantages of stability, vibration control, complete sun shading, weaker size constraints, etc.
  That is a tremendous advantage that will overwhelm most other details and tradeoffs, because the two main constraints, and operating costs, are energy production and heat dispersion. The latter imposing a limit on the former.
  (You have no knowledge of how effectively I use my time. If you have a valid point, make it, instead of - whatever you are doing. Claiming you know things without sharing your reasoning and aspersive language are for the posers. Just communicate why you think, what you think.)
  
  Reply View | 1 reply
  
  godelski 18 hours ago
  
  > In a vacuum, radiative heat loss per time hyper scales with temperature to the 4th power.
  Correct, but you do know that the Stefan–Boltzmann Constant is 5.67e-8 W/(m^2K^4), right? And that emissivity <1 in all real world applications? It is only 1 when a blackbody is radiating in a vacuum.
  This is how I know you don't understand the math. Because you didn't take the time to understand it. Plug in dummy numbers. I'll make it easy, 100^4=1e8.
  You didn't think about how T works and the domain we're operating in. T's power isn't a huge factor when we're trying to dump lower levels of heat.
  Let's say we're trying to discharge our power draw of 300W at 100C, that's still going to take a 0.5m^2 black body radiator sitting in perfect darkness PER CPU!!! A data center has hundreds of thousands!
  You realize how much fucking surface area that is?
  And this is before we consider all the other heat generated from the datacenter and the fact that you're dumping heat back to the moon's surface which will radiate it right back at your radiator making it much less efficient. Add the sun and you're fucked.
  > You have no knowledge of how effectively I use my time.
  Of course I do. You made a statement so preposterous I know you don't use it to do math or physics. Maybe you watch some math YouTube but that's not the same. I don't have to know everything you do do to know what you don't do.
  You know how I know this stuff? It's because I've put things into space. Yet you were even too arrogant to check NASA's website
  
  Reply View | 0 replies