John Richard: h from Az, dec, Lat
John:
(I don't know if each paragraph will be enclosed in asterisks. That
happened in one of my posts, but not in the other. I use an asterisk for
multiplaction and any spurious asterisk at the beginning and end of a
paragraph shouldn't be confuse with that multiplying asterisk.)
Here is the formula for h, for a given Az, Lat, and dec. ...which I said
that I'd soon post.
From the size of this formula, it isn't surprising that others have
recommended something briefer. But this formula comes directly from the
formula for Az, from Lat, dec, h.
This is in the form of a quadratic-formula solution. It can give 2 answers,
and I'll say something about how to choose which one is right.
What follows will be correct if I didn't make any algebraic errors.
I use these abbreviations:
Az = azimuth
Lat = latitude
dec = declination
h = hour angle (The sun's hour angle is the sundial equal-hours before or
after solar.noon).
(The Greek letters save space, when they're available, but not when they
have to be written out in Latin characters. And the above abbreviations are
clearer to people who aren't familiar with the Greek-letter symbols.)
h =
-2(tan dec * cos Lat)/tan Az plus-or-minus the square root of:
{4( tan^2 dec * cos^2 lat/tan^2 Az)
- 4(sin^2 Lat +1/tan^2 Az) * (tan^2 dec * cos^2 Lat -sin^2 Lat) }
The result of evaluating the above is divided by:
2 * (sin^2 Lat + 1/tan^2 Az)
-
Because the quadratic formula often gives 2 answers, then here are some
suggestions for choosing the right one:
If your input azimuth is east of south, then h and its sine must be
positive.
If your input azimuth is west of south, then h and its sine must be
negative.
If your input azimuth is south of the east-west line, then the cosine of h
must be greater than:
tan dec/tan Lat.
.if your input azimuth is north of the east-west lilne, the cosine of h
must be less than tan dec/tan Lat.
Maybe a briefer solution-fomula can be gotten by setting equal to zero, the
sun's altitude in a co-ordinate system whose equator is the azimuth
circle passing through your desired azimuth, and solving for the h that
would achieve that.
If that's workable, then it could have the advantage that h only appears
once in the altitude formula.
I apologize in advance for any algebraic or copying errors.
Michael Ossipoff
---
https://lists.uni-koeln.de/mailman/listinfo/sundial
John Richard: h from Az, dec, Lat
Sorry, I accidentally wrote h = , when I meant sin h = , at the beginning of the formula for h, given Az, Lat, and dec. Michael Ossipoff --- https://lists.uni-koeln.de/mailman/listinfo/sundial
John Richard: h from Az, dec, Lat
I'm sorry--another typo: I meant to say: If your input azimuth is west of south, then h and its sine must be positive. If your input azimuth is east of south, then h and its sine must be negative I really didn't mean to post so much, but I wanted to correct those typos. Michael Ossipoff --- https://lists.uni-koeln.de/mailman/listinfo/sundial
John Richard: h from Az, dec, Lat
Another omission: By Az, I ;mean your desired azimuth, expressed in angular distance-in-azimuth from north or south, whichever of those distances is less (depends on whether your intended azimuth is closer to north or to south).. The appeal of the briefer solution that someone else posted is clear. Ok, I don't think there will need to be any more fixes or corrections by me. Again, sorry about the omissions and typos, and the necessary corrections and fixes. Michael Ossipoff --- https://lists.uni-koeln.de/mailman/listinfo/sundial
Fw: Jack: Duration of sunlight on a particular day
From: Roger Bailey Sent: Wednesday, March 04, 2015 5:28 PM To: Michael Ossipoff Subject: Re: Jack: Duration of sunlight on a particular day Hello Michael, I responded to Jack and provided a spreadsheet that calculated sunrise and sunset times that included the option for refraction. This spreadsheet also included Meeus Algorithms to calculate the equation of time and convert solar time to clock time. This spreadsheet was developed to show the sunrise and sunset shifts near the winter solstice but the start date is optional. All the options are in blue type. The detailed calculations are in a faint type. I included the spreadsheet in my reply to Jack with a cc. to the list but this response was never posted. The spreadsheet is attached to this reply to you and anyone on the list who is interested can send me a request. Jack found spreadsheet and the chart tab quite useful. Regards, Roger Bailey From: Michael Ossipoff Sent: Wednesday, March 04, 2015 6:51 AM To: sundial@uni-koeln.de Subject: Jack: Duration of sunlight on a particular day Jack: You wrote: 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.) [endquote] You said to disregard physical effects such as atmospheric refraction (and solar semi-diameter?). That simplifies the formula. Where h is the number of equal sundial-hours before or after solar noon at sunrise or sunset; dec is declination, and Lat is latitude: cos h = - tan Lat * tan dec. Double h, and that's the sunlight duration for that day, the day corresponding to some particular value of dec, at some particular latitude Lat. And yes, _lots_ of people at this list know that. I'm not posting something new. But I just wanted to mention it because I haven't seen it in the answers so far. Michael Ossipoff --- https://lists.uni-koeln.de/mailman/listinfo/sundial No virus found in this message. Checked by AVG - www.avg.com Version: 2015.0.5751 / Virus Database: 4299/9221 - Release Date: 03/03/15 SolsticeRiseSet.xls Description: MS-Excel spreadsheet --- https://lists.uni-koeln.de/mailman/listinfo/sundial
The BSS Horizontal Sundial and 3d printer
I used your splendid and accurate BSS Horizontal Sundial design pdf but with a couple of variations. I used a 6 square ceramic tile as the dial plate, rather than 6mm MDF board, but the making of the gnomon may you find more interesting. I drew the gnomon by using the free Sketchup Make 3D modelling program, (which I have used for some time in creating virtual dials), with the object of making it on a 3d printer. I decided that it would be better to go to a service bureau rather buy a printer myself, which would be expensive and maybe of limited use. There are a number of bureaux in the UK but in the end I chose Shapeways, a Dutch company who are helpful, who can make one-off smallish items at relatively reasonable prices, within two weeks and in a variety of materials from white, solid plastic to metals from brass, steel, etc.. The method is straightforward, my Sketchup model file, at actual size, is translated in STL format, a standard CAD software format. and the file is uploaded into Shapeways' website. Shapeways checks that my model can be printed without problems and quotes a price. I then decide whether to go ahead. I chose your plain gnomon in white Shapeway's Strong and Flexible Plastic material but I could have used a more complex gnomon in different colours or Here are my attachments, a photo of my BSS Horizontal Sundial with the 3d printed gnomon and Shapeway's image of my fancy gnomon model, should I want to replace the plain gnomon! Sunny days, Phil Walker BSS Member 52° 45' 46N 2° 22' 18W--- https://lists.uni-koeln.de/mailman/listinfo/sundial
Jack: Duration of sunlight on a particular day
Jack: You wrote: *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.)* *[endquote]* *You said to disregard physical effects such as atmospheric refraction (and solar semi-diameter?). That simplifies the formula.* *Where h is the number of equal sundial-hours before or after solar noon at sunrise or sunset; dec is declination, and Lat is latitude:* *cos h = - tan Lat * tan dec.* *Double h, and that's the sunlight duration for that day, the day corresponding to some particular value of dec, at some particular latitude Lat.* *And yes, _lots_ of people at this list know that. I'm not posting something new. But I just wanted to mention it because I haven't seen it in the answers so far.* *Michael Ossipoff* --- https://lists.uni-koeln.de/mailman/listinfo/sundial
John Richard: h from Az, dec, Lat
Richard: You 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. [endquote] Once, in an unfamiliar town, I wanted to find when the sun's direction is south, east, west, southeast, and southwest--for orientation in the new town. At first, I used the successive substitutions method that you describe above. As you suggest there, that method doesn't always work, but, when it does, it's convenient. But then I noticed that the equation can be solved as an equation quadratic in sin h. Just write cos h in terms of sin h, and put that term by itself on the left side of the equation. Square both sides. You get an equation quadratic in sin h. When dec is positive, then, to choose from the quadratic formula's 2 answers for sin h, it might be necessary to consider the requirement that the azimuth be north (or south) of the east-west line, as determined in the denominator of the azimuth formula. John: I'll soon post the resulting formula for h, given Lat, Az, and dec. Most likely someone else will post it before I do, though. I'm quite conscious of the fact that there are some people at is list, and in dialing, who are a lot more qualified than I am. Of course a frequent need to find h, given Lat, Az, and dec, is when it's necessary to find when a sundial won't be shaded by a building. Michael Ossipoff -- 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
