For every location on earth, there are parts of the sky that the sun can never be in (e.g., directly overhead anywhere outside the tropics).

I think the sun can be in about 40% of the sky (a hemisphere).

I'll withhold my reasoning/calculations for now, because I'd like the opinion(s) of others more-expert (most are) than me in this subject.

Please weigh in. I believe this "simple" geometric question will have been answered centuries ago, but I have not found it as an answer, nor even as a subject.

  • $\begingroup$ You shouldn't withhold your calculations on this. Either your math is correct or it isn't. When asking a question, it is important that you give as much information as you can. $\endgroup$ – Phiteros Jan 10 '17 at 0:56
  • $\begingroup$ OK - I used sin(23.5) (degrees, the inclination of the axis of the rotation of the earth vis-a-vis the plane of its orbit around the sun). The sun's angle at any given time of day varies by twice that much across the year, but half of that angle is occluded by the earth itself (at night). $\endgroup$ – N. Joseph Potts Jan 10 '17 at 0:59
  • $\begingroup$ My first thought is "all of it". Consider an observer at the South Pole. Over the course of a six month long summer, the Sun will have painted every bit of the sky at an elevation below 23.5 degrees. That's 40% of the sky right there. Add in strings of observers along the Arctic and Antarctic Circles and the Sun will have painted every bit of the sky at an elevation below 47 degrees. That's 73 percent of the sky. A couple of more strings of observers along the Tropics should get that remaining 27 percent. $\endgroup$ – David Hammen Jan 10 '17 at 12:24

Yes, you're right. The current inclination of the earth's axis of rotation to the plane of orbit around sun is 23.43708° degrees. The sun can effectively cover the area from latitudes +23.43708° to -23.43708° on a sphere. Ignoring the night side of the celestial sphere gives the required area as Sin(23.43708°) = 39.774% ~40%.

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  • $\begingroup$ Do you mean it covers from the Tropic of Cancer to the Tropic of Capricorn? ;) $\endgroup$ – Dean Jan 10 '17 at 10:33
  • $\begingroup$ No, I mean the percentage of the sky as viewed from ANY ONE PLACE. The view is of a HEMIsphere, not of the notional SPHERE encompassing the entire earth. My current thinking is, 73% (47°) everywhere between the x-arctic circles. WITHIN the x-arctic circles, the percentage falls gradually to 40% (23.5°) at the poles. I BELIEVE the 47° figure holds good throughout the tropics, as well as the temperate latitudes. $\endgroup$ – N. Joseph Potts Jan 10 '17 at 14:50
  • $\begingroup$ I don't understand your last step 'sin(23.4)`. $\endgroup$ – usernumber Jan 15 at 12:54

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