Dear Anselmo and all, With respect to the "bifilar polar dial" in Appingedam (NL), my site does not specify the shape of the curved gnomon. It is certainly no ellipse, nor a hyperbola. Do you like some math? Have a look at the article by Fer de Vries in the NASS Compendium 8 (4), in particular fig. 4.
Point E (the center of the dial face, for those who don't have the Compendium at hand) has been taken as the origin of the coordinate system, EB (the east-west line) as the x-axis and EF (perpendicular to the dial face, intersecting the pole-style) as the z-axis. The coordinates of a point Q on the curved gnomon were derived as: x = EC = g.tan(t) - g.sin(t), and z = CQ = g.cos(t), in which t is the hour angle of the sun and g the height of the pole-style above the dial face. Scaling x and z in units of g, the shape of the curved gnomon is given by the parametric equations: x(t) = tan(t) - sin(t) and z(t) = cos(t). Your question actually is to convert this pair of equations into an analytic expression z(x). This can be done by making the usual substitutions: tan(t) = sin(t) / cos(t) and cos(t) = sqrt[1-sin(t)^2], but it is not going to look very nice. In case you would like to probe this route, it is perhaps easier to swap the axes and calculate x(z). My result is (please check): x^2 = (1 - 2z + z^2 + 2z^3 - z^4) / z^2. Definitely not the equation of a conic section! Some properties of the curve can be obtained from looking at the parametric equations. For t->90 degrees (6 hr local time), x->infinity and z->0. The curve thus approaches the dial face asymptotically when moving out. In the center, the curved gnomon touches the pole-style. The slope dz/dx of the curve at this point (x=0) is infinite. There are several ways to arrive at this result. You love calculus, do you? 1) For t->0 degrees (local noon), x->0 and z->1, as expected for a polar dial. The slope of the curve at x=0 is: dz/dx = (dz/dt) / (dx/dt) = -sin(t) / [1/cos(t)^2 - cos(t)]. For t=0, this unfortunately gives 0/0, an indeterminate value. According to the rule of Bernoulli (or De l'Hopital) one may take the derivatives of the numerator and the denominator, which at t=0 leads to 1/0, or infinite. 2) Make a Taylor series expansion of the quotient: x(t) = tan(t) - sin(t) = t^3/2 + (t^5)/8 + ..., and: z(t) = cos(t) = 1 - t^2/2 + (t^4)/24 + ... Hence dz/dx (t->0) = 2/(t^3), which approaches infinity for t->0. 3) Differentiate the analytic expression given above (your homework for today ;-). 4) The intuitive approach: if the slope were finite, the pole-style and the initial part of the curved gnomon would span an inclined plane. As long as the sun would be below this plane, an intersection point of the two shadow edges would be formed, which would fall on the straight, perpendicular date line. When the sun would rise above this plane, the intersection point would disappear and the initial part of the curved gnomon would cast an oblique shadow, which is incompatible with the existence of a perpendicular date line. Hence, the slope should be infinite. Kind regards, Frans Maes 53.1 N, 6.5 E www.biol.rug.nl/maes/sundials/ ----- Original Message ----- From: "Anselmo PÈrez Serrada" <[EMAIL PROTECTED]> To: <sundial@rrz.uni-koeln.de> Sent: Sunday, September 08, 2002 10:42 PM Subject: On bifilar polar sundial > Hoi, Frans! > > I have been playing a bit with the equations for a bifilar dial trying to > reproduce the bifilar polar dial that I saw in your web. There you say > that the transversal gnomon is a piece of hyperbola, but I have found > that it is really a piece of ellipse (there are more solutions, but none > is an hyperbolic arc). I suppose my calculations are wrong but I can't > find the mistake in them. Can you please provide me more > information about this topic? Do you know the exact equation for this > curve? > > Hartelijke bedankt, > > Anselmo P. Serrada > > > > > - > -
