RE: A question for the mathematically inclined
OK, I would also like to take a turn and ask a question to the mathematically inclined: I have been trying to figure out how to plot the duration of daylight over the course of the year as a function of latitude. (I would generate a curve for each latitude I am interested in.) I believe the result should be a sine curve which looks comparatively flat at the equator, growing increasingly steeper until the polar circle, where it would turn into a binary step curve and the six month day turns to six month night -- leaving aside physical effects like refraction. I am particularly interested in the slope of the curve around the equinoxes at northern latitudes, when the transition from long summer days to short winter days is quite abrupt. Jack Aubert -Original Message- From: sundial [mailto:sundial-boun...@uni-koeln.de] On Behalf Of John Goodman Sent: Sunday, February 01, 2015 12:37 PM To: Sundial List Subject: Re: A question for the mathematically inclined Thanks to everyone who replied with suggestions, both on and off the list. When I asked my question, I assumed there was a trivial solution that could be simply explained. I realize now that the calculations are not straightforward. Roger Bailey has given me an approach that I believe will work for me. I’m now trying to understand how the math represents the spatial geometry of the problem. The variety of solutions I received are an indication of the broad experience and wisdom embodied in this group. I'm always grateful for that asset. On Jan 31, 2015, at 10:05 AM, John Goodman johngood...@mac.com 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 -- This message has been scanned for viruses and dangerous content by MailScanner, and is believed to be clean. --- https://lists.uni-koeln.de/mailman/listinfo/sundial
Re: A question for the mathematically inclined
Hi Jack, Let me offer the solution to a related question that came up while hiking with friends around the time of the solstice. One friend asked about the changes he had noticed in the times of sunrise and sunset near the solstice. Sunrise kept getting later after the solstice but sunset minimum was before the solstice. Why? I responded saying it was due to the difference between clock time and solar time. This difference, called the Equation of Time, is due to the tilt of the earth's axis and the eccentricity of the earth's orbit. As an engineer I follow the dictum Don't speculate! Calculate. To fully answer the question, I developed a spreadsheet to calculate sunrise and sunset times for specified dates at a specified location, Sidney at 48.66 N, 123.4 W and specified atmospheric refraction (50 arc min). The spreadsheet with all the details, chart and data tabs is attached. Anything in the tables in blue type, like location, refraction and start date you can change to see the effect. Anything in ghost letters is part of the internal calculations for solar declination and the equation of time using Meeus Astronomical Algorithms as well as sunrise and sunset times by spherical trigonometry. Jack, the math is all there for to answer your question. Just change the data in cells with blue printing. I calculated but did not plot the duration. It is easy to do. These calculations covered a two month period around the winter solstice but changing the start date changes the whole period of interest. It is easy to change the increment from 1 to any other increment like 7 for each week and copy this down through the date column. Copy the last full row to extend the calculation to a full year. It is all there for you to hack to answer the question on the effect of latitude. Regards, Roger Bailey Walking Shadow Designs -- From: Jack Aubert j...@chezaubert.net Sent: Tuesday, February 03, 2015 12:23 PM To: 'Sundial List' sundial@uni-koeln.de Subject: RE: A question for the mathematically inclined OK, I would also like to take a turn and ask a question to the mathematically inclined: I have been trying to figure out how to plot the duration of daylight over the course of the year as a function of latitude. (I would generate a curve for each latitude I am interested in.) I believe the result should be a sine curve which looks comparatively flat at the equator, growing increasingly steeper until the polar circle, where it would turn into a binary step curve and the six month day turns to six month night -- leaving aside physical effects like refraction. I am particularly interested in the slope of the curve around the equinoxes at northern latitudes, when the transition from long summer days to short winter days is quite abrupt. Jack Aubert SolsticeRiseSet.xls Description: MS-Excel spreadsheet --- https://lists.uni-koeln.de/mailman/listinfo/sundial
RE: A question for the mathematically inclined (Jack Aubert)
I find this website very helpful for visualizing the changes in daylight over time and latitude. Daylight Hours Explorer http://astro.unl.edu/classaction/animations/coordsmotion/daylighthoursexplorer.html On Feb 3, 2015, at 3:36 PM, sundial-requ...@uni-koeln.de wrote: OK, I would also like to take a turn and ask a question to the mathematically inclined: I have been trying to figure out how to plot the duration of daylight over the course of the year as a function of latitude. (I would generate a curve for each latitude I am interested in.) I believe the result should be a sine curve which looks comparatively flat at the equator, growing increasingly steeper until the polar circle, where it would turn into a binary step curve and the six month day turns to six month night -- leaving aside physical effects like refraction. I am particularly interested in the slope of the curve around the equinoxes at northern latitudes, when the transition from long summer days to short winter days is quite abrupt. Jack Aubert --- https://lists.uni-koeln.de/mailman/listinfo/sundial
Re: A question for the mathematically inclined
Thanks to everyone who replied with suggestions, both on and off the list. When I asked my question, I assumed there was a trivial solution that could be simply explained. I realize now that the calculations are not straightforward. Roger Bailey has given me an approach that I believe will work for me. I’m now trying to understand how the math represents the spatial geometry of the problem. The variety of solutions I received are an indication of the broad experience and wisdom embodied in this group. I'm always grateful for that asset. On Jan 31, 2015, at 10:05 AM, John Goodman johngood...@mac.com 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
Re: A question for the mathematically inclined
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 l...@unb.ca 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. LangleyE-mail: l...@unb.ca | | Geodetic Research Laboratory Web: http://gge.unb.ca/ | | Dept. of Geodesy and Geomatics EngineeringPhone:+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 --- https://lists.uni-koeln.de/mailman/listinfo/sundial
A question for the mathematically inclined
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
Re: A question for the mathematically inclined
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. LangleyE-mail: l...@unb.ca | | Geodetic Research Laboratory Web: http://gge.unb.ca/ | | Dept. of Geodesy and Geomatics EngineeringPhone:+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
Re: A question for the mathematically inclined
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 l...@unb.ca 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. LangleyE-mail: l...@unb.ca | | Geodetic Research Laboratory Web: http://gge.unb.ca/ | | Dept. of Geodesy and Geomatics EngineeringPhone:+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. LangleyE-mail: l...@unb.ca | | Geodetic Research Laboratory Web: http://gge.unb.ca/ | | Dept. of Geodesy and Geomatics EngineeringPhone:+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
Re: A question for the mathematically inclined
Hello John, I routinely use Napier's Analogue as suggested by Fred Sawyer when I asked this question several years ago. This involves an intermediate step involving an angle B. Here are the formulae. Napier's Analogues: Knowing Latitude, Declination and Azimuth, Solve for Altitude and TimeFindangle B : Sin B =Sin Az Cos Lat/ Cos Dec. Then Tan .5(90-Alt)=Tan .5(Lat-Dec)Cos.5(B-Az)/Cos .5(B+Az), Then the Sine Rule for t: Sin Az=CosDec Sint/CosAlt or Sint=SinAzCosAlt/CosDec These are fairly easy to program into a spreadsheet. Regards, Roger Bailey -- From: John Goodman johngood...@mac.com Sent: Saturday, January 31, 2015 7:05 AM To: Sundial List sundial@uni-koeln.de Subject: A question for the mathematically inclined 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 - No virus found in this message. Checked by AVG - www.avg.com Version: 2015.0.5646 / Virus Database: 4273/9032 - Release Date: 01/31/15 --- https://lists.uni-koeln.de/mailman/listinfo/sundial
