Re: Brief explanation/derivation for Horizontal-Dial's declination-lines?
Oh alright, here's the derivation by plane trig at the dial: . .First, of course this is what gives the natural and easily-explained derivation for a Horizontal-Dial's hour-lines, and so I'll show that first: . I'll designate lines by the letter-names of their two endpoints. . When I state a sequence of three points' letter-names, I'm referring to the angle that they define. . If I want to refer to the triangle that they define, then I'll precede them with the word "triangle". . The following letters will stand for the following points: . N is the nodus. . P is the point where the shortest line from N, perpendicular to ON intersects the ground. . E is the nodus's shadow at the equinox, at the hour of interest (h/15). . D is the nodus's shadow at the declination on the date of interest (dec) at h/15. . O is the origin and intersection of the hour-lines. . Say that, at N, on the polar axis ON, there is mounted a Disk-Equatorial Dial. . NP, that dial's noon-line, has a length of OP sin lat. If that dial's hour-line for hour h/15 is extended to the ground, it extends to E. . PE = NP tan h = OP sin lat tan h. . PE/OP = tan POE (the angle on the dial for the hour-line for hour h/15). . That straighforwardly demonstrates the construction of an hour-line of a Horizontal-Dial, based on a disk-equatorial dial. . Now, for the position of D: . Regarding the triangle NEO: . OE can be gotten as OP/cos POE. . ...or as sqr( OP^2 + PE^2). . NEO = asin(ON/OE). . ...or, if you prefer, = acos(NE/OE) ...or ATAN(ON/NE) . NED = 180 - NEO, because D and O are points on the same line, on opposite sides of E. . So you have NE, NED, and END (the declination, a given quantity). . So you have ASA (angle-side-angle) . So you can solve triangle NDE for DE, by the Law of Sines. . So, measure distance DE, along the hour-line for h, to mark the point on the declination-line for the date of interest. . --- . But I'd rather give the Horizontal-Dial declination-line derivation that uses the Sun's altitude. It seems to me that the orrery derivation of the altitude-formula is easier and clearer to explain than the above trig-at-the-dial derivation for a point on the declination-line. . ...and that someone who'd like a derivation for the Horizontal-Dial's declination-lines, is going to later more likely have use or need for altitude (or terrestrial-distance) or azimuth calculations, than for other trig problems or other 3-dimensional analytic-geometry problems. . 47 Su November 17th 2140 UTC . Michael Ossipoff --- https://lists.uni-koeln.de/mailman/listinfo/sundial
Re: Brief explanation/derivation for Horizontal-Dial's declination-lines?
When I listed trig, in my previous message, I was referring to the fact that there's a 3rd derivation-approach that applies plane trig at the dial itself (but of course not just on the horizontal-plane). With any one of those 3 derivation-approaches, it would be necessary to explain either some 3-dimensional analytic-geometry; or the derivation of at least the formula for altitude (or maybe azimuth too) from h and dec; or trig for the solving of triangles. The trig needed for deriving the altitude and azimuth formulas consists only of direct use of the definitions of the trig functions, whereas the trig-at-the-dial derivation involves several plane triangles, and the solution of a non-right triangle...meaning that the person you're explaining to would have to hear about more trig than that required for deriving the altitude-formula. The altitude and azimuth formulas can be directly and straighforwardly derived by applying the definitions of the trig functions to an orrery. ...for a brief and straightforward derivation that would make sense to anyone with no prior experience in the subject. On Sun, Nov 17, 2019 at 1:42 PM Michael Ossipoff wrote: > I mean, when you're choosing which declination-lines > construction-explanation to use, there's the matter of: Which of the > following is that person more likely to have occasion to use? Or which is > more likely to be of interest and use to someone interested in sundials or > astonomical matters?: > > Altitudes (or terrestrial distances) and azimuths? > > 3-dimensional analytic geometry > > trig > > 3-dimensional analytic geometry and trig are of course useful for many > things. But celestial altitudes and azimuths, and terrestrial distances, > are of frequent and direct use and interest to people interested in > sundials or astronomical matters. > > So that's why I'd give the altitude &/or azimuth explanation for > declination-line construction. > > --- https://lists.uni-koeln.de/mailman/listinfo/sundial
Re: Brief explanation/derivation for Horizontal-Dial's declination-lines?
Frank-- . As you described, a Horizontal-Dials' (or any Flat-Dial's) declination-lines can be constructed, by 3-dimensional analytic geometry, as the intersection of a cone with a plane. . Here's another way: . 1. For a particular day, and at an hour shown on the dial, calclate the Sun's altitude. . 2. From that altitude can be calculated the nodus-shadow's distance of the nodus's shadow from the sub-nodus point. ...by dividing the height of the nodus by the tangent of the Sun's altitude. . 3. Measuring from the sub-nodus point, mark the point on that hour's hour-line at the above-calculated distance. . - . But maybe you'd rather just measure the distance from the hour-lines' intersection-point. In that case: . 1. Calculate both the Sun's altitude and azimuth at a particular date and time. . 2. As above, from that altitude can be calculated the nodus-shadow's distance from the sub-nodus point. That and the azimuth give you polar co-ordinates of the nodus-shadow with respect to the sub-nodus point. . 3. Convert the polar co-ordinates to rectangular co-ordinates with respect to the sub-noduc point. . 4. Add to the north-south co-ordinate the distance between the sub-nodus point and the hour-lines' intersection point. Then you have the rectangular co-ordinates of the nodus-shadow with respect to the hour-lines' intersection point. . 5. Convert the rectangular co-ordinates to polar co-ordinates, and mark the appropriate hour-line at the distance in those polar co-ordinates. . Or, alternatively: . 4. Divide the east-west co-ordinate by the north-south co-ordinate. . 5. Multiply that result by the distance between the sub-nodus point and the north edge of the dial or the construction-page. Mark that distance along that north-edge, from the dial's north-south axis. . 6. From the rectangular-co-ordinates, calculate the nodus-shadow's distance from the sub-nodus point. Measure that distance along the line to the point that you marked on the north-edge. . . So, there's the analytical-geometry solution that you described, and these various ways of doing it via calculation of the Sun's altitude, and maybe azimuth. . And those aren't the only derivations either. . Of all the derivations that I'm aware of, I prefer the altitude or altitude & azimuth, approach, because those calculations are of interest and use for various other matters, for sundials and all sorts of things. . . 47 Su November 17th 1830 UTC . Michael Ossipoff --- https://lists.uni-koeln.de/mailman/listinfo/sundial
