Karney
Thu, 01 Jul 2010 14:06:38 -0700
[acos(-tan{latitude} x tan{declination})] / 7.5
Declination = 23.5 at midsummer , 0 at equinox -23.5 at midwinter
Kevin Karney Freedom Cottage, Llandogo, Monmouth, NP25 4TP Phone 01594 539 595. Mobile 07595 024 960 On 1 Jul 2010, at 18:12, John Goodman <johngood...@mac.com> wrote: > Hello all, > > I usually shudder when I see equations in messages posted to the list, but > now I have a problem than needs a mathematical solution. > > I'm looking for a simple formula which calculates the number of hours that > the sun will be above the horizon on the summer solstice for any given > latitude. I don't need to worry about refraction, or take into account the > sun's diameter. I'm just interested in the simple geometric case, using the > center of the sun's disk for determining the start and end times. > > For my purposes, these simple relationships will be true: If the answer for > summer solstice hours is SSH, then winter solstice hours will be 24 - SSH, > and equinox hours will be 12, regardless of the latitude. > > Thanks very much, > John > > --------------------------------------------------- > https://lists.uni-koeln.de/mailman/listinfo/sundial > --------------------------------------------------- https://lists.uni-koeln.de/mailman/listinfo/sundial