Frans, Jose, Anselmo, Roger and all, We now have two formulae for the east-west thread. x = (1 - z) * ((1 - z^2)^0.5) / z and x^2 = (1 - 2z + 2z^3 - z^4) / z^2
The second is the first^2 and both formulae are correct. I varied z from 1 to 0 where also z = cos(hourangle) The output, multiplied for a gnomon of 40, is equal with the table in the Dutch article so I have no doubt about the two formulae. These formulae are based on x = g.tan(t) - g.sin(t) z = g.cos(t) However, there are more solutions for this dial. It isn't necessary that the curved east-west thread starts at the top of the north-south thread ( = polar style ). Assume the polar style has an height of g1 and the east-west thread starts at g2, where g2 =< g1. Then I get for the x and y coordinates: x = g1.tan(t) - g2.sin(t) z = g2.cos(t) What than is the analytical formula? I quess it is wise to substitute g1/g2 by a factor k or so. I am not good enough in analytical mathematics to figure it out myself. >From long ago I remeber it is even possible that g2 > g1 and this can be solved with the general formula I asked for. Then a part of the east-west thread around noon is bend backwards and a ( small) loop in the thread is formed. By that this solution isn't practical but mathematical it may be interesting. Somewhere in the n cubic meters of my sundial papers the solution will be in an letter by Thijs de Vries, but I can't find it back so your help to get the answer again would be appreciated. It was Thijs de Vries who started with the bifilar dial with one curved thread in the late 1970's. It's nice that this dial gets all the attention now and I am glad I rewrote the article for NASS. As I was told it will also be published in a French translation for the dialists in France. To Roger: it certainly isn't a catenary curve. Best wishes, Fer. Fer J. de Vries mailto:[EMAIL PROTECTED] http://www.iae.nl/users/ferdv/ Eindhoven, Netherlands lat. 51:30 N long. 5:30 E ----- Original Message ----- From: "Willy Leenders" <[EMAIL PROTECTED]> To: <[email protected]> Sent: Thursday, September 12, 2002 12:23 PM Subject: Re: On bifilar polar sundial > Dear Frans, Jose, Anselmo and all, > > There is an other analytical equation for the curved gnomon of the > Appingedam-sundial:: > > x = (1 - z) * ((1 - z^2)^0.5) / z > > Willy Leenders > Hasselt, Flanders in Belgium > > > > "Frans W. Maes" wrote: > > > You are right, Jose. Thanks for checking! > > Frans > > > > ----- Original Message ----- > > From: "Jose Luis Diaz" <[EMAIL PROTECTED]> > > To: <[email protected]> > > Sent: Tuesday, September 10, 2002 2:10 PM > > Subject: RE: On bifilar polar sundial > > > > I think the result is x^2 = (1 - 2z + 2z^3 - z^4) / z^2. > > > > Kind regards, > > > > ----- Original Message ----- > > From: Frans W. Maes <[EMAIL PROTECTED]> > > To: <[email protected]> > > Sent: Tuesday, September 10, 2002 9:13 AM > > Subject: Re: On bifilar polar sundial > > > > > 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. ( note: see correction by Jose ) > > > 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 Pérez Serrada" <[EMAIL PROTECTED]> > > > To: <[email protected]> > > > 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 -
