Is this curve also described as a catenary curve, the hyperbolic cosine function which defines the shape of a uniform cable hanging between two points?
My axes are different but a catenary is usually described by y = (H/?g) Cosh (?gx/H)+C where ? is the mass per unit length of the cable, g the acceleration due to gravity and H the tension in the cable and C a constant. Roger Bailey -----Original Message----- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of Willy Leenders Sent: September 12, 2002 4:24 AM To: [email protected] 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. > > 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 > > > > > > > > > > > > > > > - > > > > > > > - > > > > - > > - - -
