From: nuny@bid.nes   
      
   On Tuesday, June 20, 2017 at 10:31:48 AM UTC-7, Robert Clark wrote:   
      
   > At a solar radius of 700,000 km away from the Sun, based on the light    
   > intensity going inversely by the square of the distance, and with 1,360    
   > watts per sq. meter (in space) at the Earth’s distance, or 1.36 gigawatts    
   > per sq. km., I estimate this should give 60 terawatts per sq. km. at only a    
   > solar radius away from the Sun.   
   >    
   > But in the post above,  I had estimated that fully *on* the Sun’s surface   
   we    
   > could collect 60 terawatts per sq. km. of power. Anyone have an explanation    
   > of this discrepancy?   
      
     I'm not going to go through the math for you, but think about how much of   
   the Sun your hypothetical power-collector can "see" from a radius away as   
   opposed to "on the surface" (which I take to mean just above the photosphere).   
      
     Also, how do you plan to keep it there? It can't orbit, and it can't float.   
   I'm pretty sure radiation/solar wind pressure won't cut it.   
      
   >   In any case if this research team succeeds in producing this ultra high    
   > reflective, high temperature material, then a mirror smaller than a    
   > kilometer across a solar radius away from the Sun could collect enough    
   > energy for the total energy usage for the entire human population of the    
   > Earth.   
      
     I still want to know how the refrigeration system is going to get rid of the   
   heat. At the photosphere's surface the radiator is going to "see" the corona   
   no matter how the photosphere is shaded from it.   
      
     At a radius out it will "see" much colder interplanetary space.   
      
      
     Mark L. Fergerson   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)