In my previous e-mail I forgot to mention that there is a sundial that incorporates a catenary.
You can see it in the Sundial Park in Genk (Flanders in Belgium) and on the website of Frans Maes http://www.biol.rug.nl/maes/genk/welcome-e.htm go to R.I.C. Quadrant Willy Leenders Hasselt, Flanders in Belgium Willy Leenders wrote: > Roger, > > The curve (of the gnomon of the Appingedam-sundial) is NOT a catenary curve. > > For the catenary curve there is an equation independent from the mass per unit > length of the cabel, the acceleration due to gravity and the tension in the > cable. > > It is: > > y = a * cosh (x/a) > > or > > y = a/2 * (e^(x/a) + e^(-x/a)) > > where a = the distance between the x-axis and the lowest point of the > catenary. > > Willy Leenders > Hasselt, Flanders in Belgium > > Roger Bailey wrote: > > > 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: sundial@rrz.uni-koeln.de > > 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: <sundial@rrz.uni-koeln.de> > > > 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: <sundial@rrz.uni-koeln.de> > > > 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: <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 > > > > > > > > > > > > > > > > > > > > > > > > > - > > > > > > > > > > > > > - > > > > > > > > > > - > > > > > > - > > > > - > > > > - > > - -
