Re: Reverse Engineer Oldest SGS

Fri, 9 Apr 2004 13:05:51 -0400 (EDT)

Hi Anselmo,
 
This is the next step in the problem of John.
I made a small spreadsheet with the fomulas you mention but I got for the substyle angle a value four times the real value.
I changed:
tan(zSS / 2) = 2B / (A^2 + B^2 - 1)
into
tan(zSS * 2) = 2B / (A^2 + B^2 - 1)
and now it runs well.
A typo error I think.
 
I have the book of Soler but perhaps an older version because it has less then 395 pages and I don't see the formulas in that book so I  can't check what he writes.
 
To John I sent a small basic program to use the method I published on the sundial list.
So John is able to start with his job to recalculate vertical dials that show sun time.
 
Martin Bernhardt also published a procedure to recalculate a vertical dial with longitude correction. So for vertical dials we have the solutions complete I think.
 
Best wishes, Fer.
 
 
Fer J. de Vries
 
 
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----- Original Message -----
Sent: Friday, April 09, 2004 12:06 PM
Subject: Re: Reverse Engineer Oldest SGS

Fer J. de Vries wrote:
Hi John,
 
Assuming the sundial is vertical (because the XII hourline is vertical) and assuming the sundial is well made, just measure two angles in the pattern and it is possible to recalculate the latitude and declination of the dial.
 
The angles you need are the hourlineangles for hour 6 and 9 for a morning dial or 15 and 18 for an afternoon dial.
 
There is a more general formula in case you can't measure these hours. I got it from Soler's book, page 395:

Suppose that you have the angle z1 for hour H1 and z2 for hour H2,  let's define

P = cotg(H1)   Q = cotg(H2)
p = cotg(z1)     q = cotg(z2)
A = (p-q) / (P-Q)
B = (Pq - pQ) / (P-Q)

Then you can get (and check!) the substylar angle from

tan(zSS / 2) = 2B / (A^2 + B^2 - 1)

and from it

tan(LAT) = sqrt( B/tan(zSS) )
sin(DEC) = B/tan(LAT) = sqrt( B*tan(zSS) )

These formulae are complete general so it's worthwhile taking the effort to type them into
a spreadsheet.

Regards

Anselmo

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