Hello Arthur .
In the last cloudy weekend I' ve studied also the inverse problem : given a hyperbola or at least 5 points find the foci and asymptotes. I've got a simple solution in the particular case that among these points there is a vertex . The solution is based on this nice theorem valid for all conics: " the middle points of all parallel diameters of a conic lie on a straight line ." A diameter is defined to be that segment generated by two intersections of a straight line with the conic, therefore the parallel diameters are those generated by a set of parallel straight lines. " if the conic is a hyperbola or an ellipse the straight line joining the middle points of the diameters crosses its simmetry centre ", which is , for the hyperbola , the intersection of the two asymptotes half way between the two vertexes. That's all about the theory . Hereafter I'll apply all that to find foci and asymptotes given at least 5 points got by shadow observation along several hours . The procedure is very easy to be applied and self-explaining. 1) Draw a however directed set of parallel straight lines on the wall ( two would be sufficient, but we don't know the exact position of the final hyperbola . ). 2) In the morning , when the point of the gnomon's shadow crosses the straight lines mark these points A, B, C on the wall . 3) Mark the point V of the shadow in the instant of your local noon : this is a vertex of the hyperbola. 4 ) Repeat instructions 2) in the afternoon , so that we mark the corresponding points ., C', B', A' 5) Find the middle points on the segments AA' , BB' , CC' . and draw the straight line joining them. 6) Draw the straight line joining the base of the gnomon and the vertex of the hyperbola : this is a simmetry axis of the hyperbola , containing also the foci. The other axis is perpendicular in V. 7) The intersection S of the two straight lines drawn in 5) and 6) is the simmetry centre of the hyperbola 8) Measure the distance SV : this is the value of the parameter a of the hyperbola 9) Draw the perpendicular segment from a whatever marked point , say A , to the simmetry axis and mark the intersection point P. 10) Measure the length Y of the segment AP. 11) Measure the length X of the segment SP. 12 Calculate the parameter b from the hyperbola equation (X^2/ a^2)-(Y^2/ b^2) = 1 13) Calculate c = (a^2+b^2)^1/2, which is the distance of the two foci from S on the simmetry axis. 14)Draw the segment HK of length 2b perpendicular to the simmetry axis in V, in such way that V results its middle point. 15)The two straight lines containing the segments SH and SK are the asymptotes of the hyperbola. That's all. I believe that this method could be useful. Does it fit your needs ? Best Regards Alberto Nicelli Italy (45,5 N ; 7,8 E) > ---------- > From: Arthur Carlson[SMTP:[EMAIL PROTECTED] > Sent: lunedì 27 aprile 1998 15.53 > To: sundial@rrz.uni-koeln.de > Subject: Re: latitude with pegs and strings > > Arthur Carlson <[EMAIL PROTECTED]> writes: > > > I've been playing in my garden, ... > > > > *** Does anybody know a relatively simple method for finding the > > latitude from observations of the sun over the course of several hours > > without recourse to tables and calculations? *** > > Thanks to all for the suggestions. I admit that the problem sounds > somewhat contrived, but it seems to me to be in the spirit of dialing > in the age of quartz watches. David Higgon's answer came closest to > what I was looking for. It is nearly as transparent as the shortest > shadow method of finding North. I may point out that one can find > North and the current declination if one observes the length of the > shadow, rather than just timing the direction. > > In the present case, I actually want to mark the local noon at the > solstices and the equinox as well as verify my latitude, so I prefer > to use no other shadow-caster than the corner of the roof. > Furthermore, I don't have easy access to all the ground North of the > corner (Walls and such get in the way.), which rules out some > possibilities. The recent posts on astrolabes led me to devise a > method involving stereographic projections. The idea is to take three > shadow positions (in the minimalist spirit of a mathematician) and > project them stereographicly (using a separate diagram, also > constructed with pegs and strings). I can then construct the circle > through the new points, find the closest and farthest points to my > origin, and convert these back to either points on the ground or > angles in my diagram. The key is that it is easy to construct a > circle from three points, but hard to construct a hyperbola from 5. > Furthermore, I know how to utilize the knowledge of the position and > height of the gnomon in this scheme (which is why three point are > enough). Alberto Nicelli described how to construct a hyperbola given > the foci, but the inverse problem is harder. > > Anyway, I find the geometry of the method interesting. If I get a > chance this weekend I'll see if it works in the field. I suspect the > biggest errors will come from the fact that the ground is not exactly > flat and level. > > Art Carlson > [EMAIL PROTECTED] >
