A Tough One?
Dear Friends, I have a vertical east declining sundial that I believe was once fixed to a church building. Its gnomon is missing. What I wish to calculate is the latitude of the dial, and its declination. I am sure that it is a very easy problem to solve, but so far I have failed. When I have this information, I hope to be able to locate its original site, (I already have a rough idea of the area of England), and maybe find some evidence of where it was fixed to the building. The only real clue to its location is given by the angles made by its hour lines - assuming them to be accurate. Thanks in advance to anyone who may be able to help. Regards, Mike Cowham. Cambridge, England.
Re: A Tough One?
Mike Cowham wrote: > > Dear Friends, > I have a vertical east declining sundial that I believe was once > fixed to a church building. Its gnomon is missing. > What I wish to calculate is the latitude of the dial, and its > declination. I am sure that it is a very easy problem to solve, but so > far I have failed. > When I have this information, I hope to be able to locate its > original site, (I already have a rough idea of the area of England), and > maybe find some evidence of where it was fixed to the building. > The only real clue to its location is given by the angles made > by its hour lines - assuming them to be accurate. > Thanks in advance to anyone who may be able to help. > Regards, > Mike Cowham. > Cambridge, England. Dear Mike, Here I give just a number of formulae by which the latitude phi and the declination d of a vertical sundial can be calculated, assuming the pattern is well drawn. Measure the angles of the following 2 hourlines : - for east decliner : hour 6 and 9 - for west decliner : hour 18 and 15 Name these angles t45 and t90 and use positive signs for the angles. Calculate : P = cot(t45) - cot(t90) Q = cot(t90) X = P*P Y = Q*Q a = Y b = X + Y - 1 c= -1 Z = (-b + sqrt(b.b - 4.a.c)) / (2.a) or Z = (-b - sqrt(b.b - 4.a.c)) / (2.a) Take the positive answer for Z Then phi = atn(1/sqrt(Z)) d = asin(Q/tan(phi)) ( sqrt is square root out of... ) Example : t45 = 29 degrees t90 = 68.78 degrees X = 2.0044 Y = 0,1508 a = 0,1508 b = 1.1552 c = -1 ( of course ) Z = 0,7852 phi = 48.4552 degrees d = 20.1244 degrees ( east or west ) I hope I didn't make any typing error. Otherwise have a look in bulletin of De Zonnewijzerkring, 88.3, page 31. Best wishes, Fer. -- Fer J. de Vries [EMAIL PROTECTED] http://www.iaehv.nl/users/ferdv/ lat. 51:30 Nlong. 5:30 E
Re: A Tough One?
Mike, I encountered a similar problem in analysing the stained glass dial from Wendon Lofts. Assuming that there are no signs of the gnomon support to define the sub-style angle on your dial, I would get a value of the declination from the asymmetry of the "earliest" and "latest" hour lines on the dial. Then, having measured all the hour (and half- and quarter-hours if they exist), I would plot y = hourline angle - LHA vs x = LHA. This should give a curve which starts at the origin, rises to a peak, then returns to zero at LHA = 90 degrees (this for the case of a direct south dial - the curve will be offset for the am and pm halves for a decliner). This experimental plot can then be compared to a series of theoretical ones calculated for different latitudes eg by the BSS Sundial Constructor program. The closest fit tells you the latitude of your dial. The advantage of the rather strange-looking x-y parameters that I have suggested is that it (a) it makes use of all the hour line information you have, (b) it is most sensitive to lines around 45 degrees, and (c) plotting the difference between hourline angle and LHA maximises the differences between curves for different latitudes. You may have to iterate around the above loop a couple of times if there is significant uncertainty as to the declination. I hope this helps - I have the Excel plots from the Wendon Lofts dial if they are of any use. John Davis