You didnt say what the solar declination was when the dial-reading and clock-time had the values you stated.
Suppose that the dial is rotated about a vertical axis, away from the meridian. Say the sun is on a certain meridian, but at a higher declination. That will affect the dial-reading, because the dial's gnomon is no longer parallel to the Earth's axis. Therefore, to say how much the dial is rotated about a vertical axis, from its desired orientation, you'd need to know the solar declination. ------------------------------------- Spherical trig isn't needed. Spherical co-ordinate transformations can solve that problem, and probably any sundial problem that spherical trig can solve. The formulas for spherical co-ordinate-transformations can be derived by plane geometry, without need for spherical trigonometry, which looks at spherical triangles. Maybe there are sundial problems that spherical trig can solve more elegantly or briefly, or "easily" (if you have the formulas and theorems of spherical trig). But, as I said, I'd expect that any sundial problem that spherical trig can solve can be solved by co-ordinate transformations. For one thing, the method that I'd use would probably result, overall, in a problem that could only be solved iteratively. Maybe a spherical trig solution wouldn't need that. (For iterative solution of a big equation like that, I'd use Regula Falsi. Newton's method is faster, but it sometimes diverges. Regula Falsi always converges. ...sometimes (but not often) too slowly. But when it converges too slowly Newton's method likely would diverge anyway. Newtons' method requires calculus, and Regula Falsi, an ancient method, doesn't. When Regular Falsi converges too slowly, there are methods to speed it up. They can be found at a web article or paper from South America, by googling "Regula Falsi Methods". Of course if those, too, don't improve convergence enough, then there's always Bisection, which reliably converges at a known adequate but not spectactular rate.) Some months ago, in a thread at this forum, about Analemmatic Dials, I outlined some topics by which the construction of an Analemmatic Dial or a Reclining-Declining Dial can be explained in several discussions, on several successive days. I'll fill in those topics*, describing the derivation in more detail, in postings here, if you'd like, but it would be best if I do so over a series of days, with a day for each topic, as I described in that outline at the Anelammatic Dial thread a few months ago. *for the problem of finding how azimuth-misaligned the dial must be, to have a given wrong time-reading, at a given local true solar time and solar declination. Let me know if I should post that. ...and the detailed derivation-discussion for the Analemmatic and Reclining-Declining too? By the way, I should add that of course all that's necessary is that the dial be azimuth-rotated so that it tells the time that it should tell, and so it isn't necessary to solve the problem that we're talking about in order to azimuth-align the dial. But that problem could be important if you intend to use a portable equatorial dial as a sun-compass. Michael Ossipoff Michael Ossipoff On Tue, Sep 25, 2018 at 9:01 PM Steve Lelievre < [email protected]> wrote: > Hi folks, > > The equatorial sundial at Vandusen Gardens in Vancouver BC is a lovely > piece but I think it is set up wrong - I suspect the dial's axis is not > aligned to the meridian. This suspicion is based on the measurements > described below, but also because when I viewed the dial today I > concluded it is not aligned north-south (but then again, my idea of n-s > was based on my potentially unreliable cell phone compass). > > I want to figure out the angle that the dial is twisted by so I have > some kind of spherical geometry problem to solve. Unfortunately, I've > never been able to get my head around spherical trig. I've tried to > learn about it a few times, but it's still a dark art for me. > > The dial is an equatorial, latitude 49.2N, 123.2W. As far as I can tell, > the slope angle of the axis of the equatorial is correct for the > latitude. As well, the dial base is flush to the plinth, which appears > to be a properly flat (horizontal) surface. Today at 12:27 pm Pacific > Daylight Time, the dial showed 12:45 pm (note, although the dial shows > local solar hours, the hour labels are advanced by one hour - like a > Daylight Saving shift). > > My thinking: The site is 3.2 degrees west of the timezone meridian, > which is 12.8 minutes of time, so 12:27 PDT is like 12:14.2 local mean > time, or 11:14.2 if we take out the Daylight hour. The Equation of Time > is 8.3 minutes today (dial is fast), so the actual reading of 12:45 is > like 12:36.7 local mean time, or 11:36.7 if we take out the Daylight > hour. Hence it seems to me that the dial was off by 22.5 minutes of > time, which is the difference between 11:36.7 and 11:14.2. > > The thing I want to know: assuming all the error is due to rotation > about a vertical axis, what is the angle that the dial is twisted by? > > Please could some kind soul help me out by explaining the steps involved > in the calculation - my problem is in knowing which equations to use, > and why. > > Steve > > > --------------------------------------------------- > https://lists.uni-koeln.de/mailman/listinfo/sundial > >
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