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
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