The USNO Webpage http://aa.usno.navy.mil/data/docs/AltAz.php will also compute elevation angle (altitude) and azimuth of the sun for a given date and location at specified intervals.
On Saturday, January 31, 2015, 31, at 12:31 PM, Bill Gottesman wrote: > You can download a free excel spreadsheet, sunpositioncalculator at > http://precisionsundials.com/sunpositioncalculator.xls. The Azimuth page > allows you to input date, latitude, longitude, and azimuth, and it gives you > the civil time, eot, declination, and altitude. When opening, you must allow > macros to run if the computer asks. > > -Bill > > On Sat, Jan 31, 2015 at 10:49 AM, Richard B. Langley <[email protected]> wrote: > If you know the zenith distance, z, of the sun (90° - elevation angle) as > well as the azimuth (A) then you could use: > > sin(h) = -sin(z)*sin(A)/cos(delta) > > where delta is the sun's declination. The latitude of the site, phi, is not > needed. > > Computing the hour angle when the zenith distance is not known is a little > trickier. In principle, this equation could be used: > > sin(h) = tan(A)*(sin(phi)*cos(h) - cos(phi)*tan(delta)) > > but you'll notice that h appears on both sides of the equation. Possibly this > can be solved in an iterative fashion by selecting an approximate trial value > for h and using it on the r.h.s. to compute a new value of h. You would then > use this new value on the r.h.s. and continue the iterative procedure until > the new value does not change significantly from the previous value. I've not > actually tried this myself so proceed with caution. > > -- Richard Langley > > On Saturday, January 31, 2015, 31, at 11:05 AM, John Goodman wrote: > > > Dear dialists, > > > > Does anyone know a formula for calculating the hour angle given the > > azimuth, declination, and latitude? > > > > I’d like to know the time of day, throughout the year, when the sun will be > > positioned at a particular angle. This will allow me to determine when > > sunshine will stream squarely through a window on any (sunny) day. > > > > I’ve seen several formulae for calculating azimuth. I suspect that one of > > them could be rewritten to solve for the hour angle given the azimuth > > instead of the finding the azimuth using the hour angle (plus the > > declination and latitude). Unfortunately, I don’t have the math skills for > > this conversion. > > > > Thanks for any suggestions. > > --------------------------------------------------- > > https://lists.uni-koeln.de/mailman/listinfo/sundial > > > > ----------------------------------------------------------------------------- > | Richard B. Langley E-mail: [email protected] | > | Geodetic Research Laboratory Web: http://gge.unb.ca/ | > | Dept. of Geodesy and Geomatics Engineering Phone: +1 506 453-5142 | > | University of New Brunswick Fax: +1 506 453-4943 | > | Fredericton, N.B., Canada E3B 5A3 | > | Fredericton? Where's that? See: http://www.fredericton.ca/ | > ----------------------------------------------------------------------------- > > --------------------------------------------------- > https://lists.uni-koeln.de/mailman/listinfo/sundial > > ----------------------------------------------------------------------------- | Richard B. Langley E-mail: [email protected] | | Geodetic Research Laboratory Web: http://gge.unb.ca/ | | Dept. of Geodesy and Geomatics Engineering Phone: +1 506 453-5142 | | University of New Brunswick Fax: +1 506 453-4943 | | Fredericton, N.B., Canada E3B 5A3 | | Fredericton? Where's that? See: http://www.fredericton.ca/ | ----------------------------------------------------------------------------- --------------------------------------------------- https://lists.uni-koeln.de/mailman/listinfo/sundial
