Re: Polar ceiling sundial

[fer j. de vries]

Hi John,   There are several limitations to concider but the principle still is true.   The problems you mention occur at latitudes smaller then 47 degrees.   The max. altitude of the sun h = 90 - phi + 23.5 degrees.   The inclination of the mirror i = 0.5 phi   We search for the latitude where the sun beam meets the mirror at 90 degrees.   With a simple drawing ( attached ) it is to see that:   0.5 phi + 90 + 90 - phi + 23.5 = 180.   Or phi = 47 degrees.   Also is seen that the mirror needs to be inward the room and the overhanging ceiling gives limits to the time the dial will work.   So pratical there are problems to solve, however the idea is worth to think about.   Best wishwes, Fer.     Fer J. de Vries [EMAIL PROTECTED] http://www.iae.nl/users/ferdv/ Eindhoven, Netherlands lat.  51:30 N      long.  5:30 E ----- Original Message ----- From: J Lynes To: fer j. de vries ; Anselmo Pérez Serrada ; Sundial, Mailinglist Sent: Sunday, January 06, 2002 10:34 AM Subject: Re: Polar ceiling sundial Thanks, Fer. Apologies all round, and especially to Anselmo.  I now think his proposal would work after all.  But only in the winter months.  At noon at the equinox the reflected beam would be vertically above the mirror.  During the summer months the beam would be reflected south of the east-west line, i.e. back through the window. Or am I wrong again? John Lynes   Fer de Vries wrote - The idea by Anselmo is correct.   The mirror is just between the polar axis and the horizontal ceilng and by this the pattern on the ceiling looks like a polar dial.   Attachment converted: Macintosh HD:ceiling.gif (GIFf/JVWR) (000398E2)