That's pretty neat Gianni. The closest large German city at that latitude is
Frankfort. It would be interesting to see if Fer's equations produce the
same result.
----- Original Message -----
From: Gianni Ferrari
To: LISTA INGLESE
Sent: Saturday, April 10, 2004 9:07 AM
Subject: Re: Reverse Engineer Oldest SGS
Because in the ancient sundial that need restoration (very frequent in
Italy) it is not always possible to measure the angles between the hour
lines, some time ago I have devised a different method to find the sundial
parameters in which we need only the knowledge of the equinoctial line and
of some points where the hour line cross it.
Let be:
- E the point in which the meridian line crosses the equinoctial
one;
- P1 and P2 the points in which two hour lines cross the
equinoctial ;
- (P1-E) and (P2-E) the distances between the points P1 , E and
P2 , E (measured on the sundial). These values are positive if the point P
is in the afternoon, are negative if it is before noon.
- (N1*15) and (N2*15) the hour angles of the two hour lines
We get then :
(P1-E)/(P2-E) = [tan(N1*15-Ws)+tan(Ws)] / [tan(N2*15-Ws)+tan(Ws)]
where Ws is the hour angle of the sub-style line .
If k = [(P1-E)*tan(N2*15)] / [(P2-E)*tan(N1*15)] we obtain the
formula :
tan(Ws) = (k-1) / [tan(N2*15) - k*tan(N1*15)] from which Ws .
If Mu is the angle between the Equinoctial line and the horizontal one
and
A= tan(Ws) ; B = tan(Mu) we have
cos(Latitude)=[ B/A]*SQR {(1+A*A) / 1+ B*B)}
From sin(d) =tan(Mu)*tan(Latitude) we obtain the declination
d of the wall , and from
sin(Gam) = cos(Latitude)*cos(d) the angle Gam between
the style and the wall.
Finally from
(P1-E) = R*[tan(N1*15-Ws) + tan(Ws)] / cos(Gam)
we may calculate the length of the ortho-style R
--------------------------------
I have tried to do some measures from the photo of the "The Oldest Stained
Glass Sundial"
that is in John Carmichael's site and I have found the followings
results (obviously very approximate):
P1 on the equinoctial line = intersection with the 8h line; N1*15 = -60
P2 on the equinoctial line = intersection with the 14h line; N2*15 = 30
From the photo I have found : P1-E = -59 ; P2-E = +38.5 ; Mu = 13d
The results are :
Latitude = 44.9d
Wall_Decl = -13.3d (East)
Sub-style hour angle = -18.5d that is, sub-style hour = 10h 46m
Angle between the plane and the style = 43.6d
Ortho-style = 0.593 times the distance between the points E and P1
It would be interesting to compare these results with those gotten with
other methods.
Note 1 - in my opinion the sundial was probably calculated for a latitude
of 45d and not for a particular place in Germany or in Switwerland.
Note 2 - This method , that I have published some years ago on my volume,
is useful also for sundials with Italic or Babylonian hours.
Best
Gianni Ferrari
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