Hello Frank, I hope that our Emails, interesting for us, don't bore other readers :-)
Your answers put always in evidence some details that we have to consider but that often slip our mind (at least mine !) and remain hidden. I agree completely with your LESS GOOD FEATURES : I have to admit that I had thought only to horizontal sundials, also remembering one made by a friend in a square in Varese, a the town in northern Italy. Obviously a hole with the axis pointing at the position of the equinoctial sun at 12 noon cannot work for a vertical plane facing East or West! For a declining sundial I think that the direction of the axis of the hole should coincide with the intersection of the plane of the equator with the substyle plane (hour plane normal to the dial and on which the polar style lies) This direction is given by the extreme point of the polar style and the point in which the substyle line crosses the equinoctial one; it belongs to the plane of the polar style and makes with the dial plane an angle = (90-style height). If the sundial faces East or West the axis of the hole is normal to the plane, that is the plane of the hole is parallel to the dial . In the figure : Lat.=45° , wall-decl=50° W GG' = 100 ; CG = 220.01 ; SG' = 51.03 ; CS = 247.0 Style height GCS = 27.03° ; SCM = 37.45° ; CSG = 62.97° CAMERAOBSCURA MERIDIANA Note- This is the Latin name but we can call it in different other modes (pin-hole sundial, dark room sundial, etc. ) In an old article (presented in one of our meetings in 2003 ) I made the same considerations (even if using different formulas ) on the uniformity of the brightness, along the meridian line, of the image in a dark room sundial . I had considered different positions of the hole (also a vertical hole) but I had not thought to the King angle and to the possibility to lengthen the hole: this is a good idea . However I have to note that in my formula I have found sin(45+dec) x sin(45+dec) x sin(45+dec) x sin(45+dec+ang) that is sin(h) is cubed . I try to explain this in the note at the end , so the readers not interested can jump it :-) Obviously our considerations are valid only if the "dark room" condition is satisfied, that is if the ratio (distance from the hole to the image) / (least dimension of the hole-as seen by the image) is much greater than 108 (at least 300-500 ) This is not true in the meridian lines that have to work in open air. The brightness of the image in room sundials is too low and cannot be perceived in open air but only in .. dark rooms :-) Gianni Ferrari NOTE D = hole diameter ; H = hole height from floor ; Fi = sun Diameter (rad) ; J = illumination constant ; h= Sun altitude ; ang = angle of hole to the horizontal If the Sun has altitude = 90° and ang = 0° then : Light through the hole = J x D x D x pi / 4 Image area = (H x Fi) x (H x Fi) x pi / 4 Brightness of the image = (J x D x D) / [ (H x Fi) x(H x Fi) ] = K When the sun has altitude h we have : Light through the hole = J x D x D x sin(h+ang) x pi / 4 Distance hole-image = H / sin(h) Image area = [H x Fi /sin(h)] x [H x Fi /sin(h)] x [1/sin(h)] x pi / 4 = (H x Fi) x (H x Fi) x [1/ sin(h)]^3 Brightness of the image = K x sin(h+ang) x sin(h) x sin(h) x sin(h) Then in my opinion the function to use is sa(dec,ang) = sin(h) x sin(h) x sin(hc) x sin(h+ang) or sa(dec,ang) = sin(45+dec) x sin(45+dec) x sin(45+dec) x sin(45+dec+ang) G.F.
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