Hi Arthur, You wrote about the sundial in Appingedam, Netherlands : (Since the second shadow is not needed to tell the time, I would hesitate to classify this as a bifilar dial at all.)
This remark has more to do with 'classification of sundials' and that's a different but also interesting discussion. To my opinion this dial surely is a bifilar dial. You are right by stating 'the second shadow is not needed to tell time', but a sundial is able to show us many more phenomenae of the sun then just the dayly time. It also may show the date or the sun's altitude or other phenomenae and still it is a sundial. To read the date in this example we need the second shadow too. And for that part this dial is a bifilar dial, however in stead of filars 2 edges of planes are used, but the principle is the same. Best wishes, 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: Arthur Carlson <[EMAIL PROTECTED]> To: sundial list <sundial@rrz.uni-koeln.de> Sent: Thursday, October 19, 2000 10:05 AM Subject: Re: Bifilar dial in Genk Sundial Park > Chris Lusby Taylor <[EMAIL PROTECTED]> writes: > > The principle is relatively straightforward. As the description says, > the pole style and base plate together constitute a polar dial. > (Since the second shadow is not needed to tell the time, I would > hesitate to classify this as a bifilar dial at all.) At any given > time of day, the shadow plane will always cut the edge of the yellow > glass at the same point. For different dates/declinations, the shadow > of this point will move up and down by the distance L*tan(D), where D > is the declination and L is the distance from the edge of the gnomon > to the edge of the shadow. At noon, L must equal the height of the > style, H. The trick is to make L = H for every time of day. If x is > the distance from the base of the style, measured in units of H, and y > is the distance above the base plate in the same units, then the > equation for the necessary curve is this: x = (1-y)*sqrt(1-y^2)/y > > Have fun proving this! > > --Art Carlson >
