RE: Reverse-engineering oldest SGS

[Roger Bailey]

Thanks Chris.   This looks like a simple direct solution. Measure SH and SD. Calculate the design latitude from Sin Lat = Cos SH x Cos SD, and then the design wall declination from Sin Dec = Tan SD * Tan Lat. This works when the gnomon is still in place and you can measure SD and SH   I was not aware of your first equation  for Sin Lat but it is easily derived from the triangles in the vertical declining tetragon. The small pdf sketch is attached. At 5kb it should get through the size filter. In the meridian plan triangle Sin Lat = CM/CS. In the wall plane triangle Cos SD = CM/CH. In the inclined declined plane Cos SH = CM/CH so Sin Lat = Cos SD x Cos SH. QED.   Fred Sawyer also suggested a different direct solution but no one commented on it. I lacked skill in math to derive or prove the relationship outlined in his concise description.   Roger Bailey -----Original Message----- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]On Behalf Of Chris Lusby Taylor Sent: May 25, 2004 3:09 AM To: sundial@rrz.uni-koeln.de Subject: Re: Reverse-engineering oldest SGS I've only just got around to looking into this. The question, if you remember, was how to find the latitude and wall declination for which a vertical sundial (such as an SGS) had been made, given that you can measure the style height (SH) and angle between the style and vertical (SD). Nobody posted a correct analytical solution to Roger Bailey's simultaneous equations, and several posters thought it had to be done by trial and error. The thread then turned to using the hour line angles, which is a valid approach. But there is an easy solution. It is:   Sin Lat = Cos SH x Cos SD      Sin Dec = Tan SD * Tan Lat   Best wishes to all Chris Lusby Taylor 51.4N 1.3W Attachment converted: Macintosh HD:VD Tetra.pdf (PDF /CARO) (000C75F4)