[math-fun] Celestial Geometry
I have a question in celestial geometry whose answer is surely well-known -- but not to me. Do any of you know this? I want to know the function whose inputs are my latitude on the earth (eg 0=equator, +-90=poles) and the latitude the sun is over today (0 on equinox, +-23.5 on the solstices), and whose output is what direction I need to look in to see the sun at sunrise/sunset (0=due east/west, -90=due south -- which would be the case if it's the winter solstice and I'm at the arctic circle). Surely all almanac-writers have a button on their pocket calculators which does this... --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
Thanks to all. cos(azimuth) = sin(declination) / cos(latitude). Declination = what latitude the sun is over, azimuth = angle on the horizon with 0 = due north, 90 = due east. Googling "azimuth at sunrise" reveals all. --Michael On Fri, Jul 11, 2008 at 9:27 AM, Michael Kleber <michael.kleber@gmail.com> wrote:
I have a question in celestial geometry whose answer is surely well-known -- but not to me. Do any of you know this?
I want to know the function whose inputs are my latitude on the earth (eg 0=equator, +-90=poles) and the latitude the sun is over today (0 on equinox, +-23.5 on the solstices), and whose output is what direction I need to look in to see the sun at sunrise/sunset (0=due east/west, -90=due south -- which would be the case if it's the winter solstice and I'm at the arctic circle).
Surely all almanac-writers have a button on their pocket calculators which does this...
--Michael Kleber
-- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
-- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
participants (1)
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Michael Kleber