Hello again,
I thank Hank de Wit, Brian Albinson and Jan Safar for replies (off list) offering suggestions for how to resolve my enquiry. Hank managed to find a journal article ( https://tinyurl.com/y7rmknpf ) that discusses sources of error in equatorial dials. The reference is: Garstang, R. H. (1997). /The errors of an equatorial sundial/, in The Observatory, v. 117, p. 344-351.
Unfortunately the article doesn't provide derivations, simply saying "a simple calculation by spherical geometry shows...", but, case (ii) of the analysis addresses the situation I encountered, stating
δH=a2(cos(ϕ)cos(H)tan(δ)-sin(ϕ))δH = {a}_{2}(\cos\left({ϕ}\right) \cos\left({H}\right) \tan\left({δ}\right) - \sin\left({ϕ}\right))
where δH is the time error, a2{a}_{2} is the angle by which the dial is twisted, ϕϕ is latitude, HH is hour angle, and δδ is solar declination.
Putting in the known values, gives a2{a}_{2} = 7° 18'. Using my cell phone compass I had guessed it as 10° or a little under.
Cheers, Steve
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.
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