Hi again,
It seems that the equation that I typed out in my message below does not
show correctly on all email software, so here it is again but written
using ordinary letters:
dH = a2 ( cos(lat) x cos(H) x tan(dec) – sin(lat) )
where
dH : delta H, error in hour angle
H: hour angle (0 at noon)
a2: azimuth error (assumed small)
dec: solar declination
I visited the dial in question again yesterday at solar noon. I
estimated the time error as equivalent to 5.28° of hour angle. For the
latitude of 49.24°N, the equation yields -7.83° for a2, the azimuth error.
That's reassuringly close to the previous day's estimate of -7.31°, and
consistent with a guess of -7 ± 1° that I obtained by sighting a
landmark across the arms of the dial and then measuring its bearing on
Google Maps.
Steve
On 2018-09-26 10:05 AM, Steve Lelievre wrote:
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
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