->  DW> Certainly, the ISS is is darkness for a significant fraction of 
->  DW> the time, but that's no reason to waste the sunlight it does 
->  DW> get. I don't follow your "as much logic" argument.  
  
-> where do you get "significant fraction" from? it takes 90 minutes
-> for the IS to go around the earth... seems to me that 45 minutes of 
-> that is in the light and the other half in the dark ;) 
  
 DW> So 1/2 is not a significant fraction? 

i dunno... i hear "significant fraction" as "more than half" ;)

 DW> Actually, the ISS is in darkness for less than half the time. It's
 DW> a  few hundred kilometres above the surface fo the earth, so it
 DW> receives  sunlight when the ground below it is just in darkness.

right... that's what enables us to see it when it flys over :)

 DW> Suppose it's 400  km up. 

right now, as of the writting of this message, it is 357.64km in altitude
and "falling" as it crosses the tip of south america... before i
complete this message, it will be down in the lower 345km range as it
crosses the southern 20th parallel ;)

 DW> The earth's diameter is about 13,000 km.

mmm... across the poles, around the equator or an average of them?

dia{at}equator - 12756km
dia{at}poles   - 12713km

avg dia     - 12734km

yeah, yeah... i know... 400 and 13000 are nice round numbers and make for
easy math :P

 DW> So the length of the  tangent from the surface to the station would
 DW> be sqr(400 * 13400),  

where do you get 13400 from? the diameter of the earth plus the station's
altitude above the earth? wouldn't that be 13800 then? if the station is in
a perfectly circular orbit, that is... not that its going to make that much
difference in the math... what's a few kms, here and there out of several
hundred ;)

 DW> which is about 2300 km. That's about 1/20 of the circumference of 
 DW> the orbit. If my brain is working right, that means the station 
 DW> is in sunlight for about 60% of the time, when the sun is in the 
 DW> plane of the orbit. When it isn't, the fraction would be even 
 DW> higher.    

kinda makes sense... at least without sitting down and drawing it out ;)

mmmmm... i think you may be on to something, too... i'm seeing a SUN LOS of
21:00 and a SUN AOS of 52:12... subtract the LOS from the AOS and we get a
nice 31 minutes, 12 seconds of "no sun"... since i was figuring
45 minutes and i/we see that it is actually 31 minutes, that gives us 59
minutes in sunlight... so figure 60/30 ... 2/3rds of the time is spent in
the sunlight...

interesting...

now to continue on with the quest as to why the station rotates from
"pushing" the shuttle when it docks to "pulling" it and
then back to "pushing" it when the time comes to undock ;)

)\/(ark