I realize that you’ve already gotten good answers, but I’d like to say a few things too.
… I’m really late replying, because I’ve been trying to figure out how to word answers to a few long assertion-posts from the usual confused self-sure kids at a philosophical forum. After this time, I’m going to, one way or another, in the forum-options, or my inbox-settings, do a setting that stops topic-announcements from those forums from appearing at my inbox. … First, are you sure that a nail in the wall is the best way? It’s very unlikely to go in perpendicular to the wall. Best would be a block or box that’s reliably rectangular-prism in shape. Lacking that, why not use the short cardboard tube from inside a bathroom-tissue roll? … Assume that the plane of its edge at the ends is perpendicular to its axis & cylindrical-surface. … Stand it on a flat surface, & use a carpenter’s square, a right-triangle drafting square, or a protractor, to mark a vertical line on the tube…or at least the two endpoints of a vertical line. … At the top end of the line, make a small notch, & let that be the shadow-casting point, using the line as the nail. … You’ve got the formula for the declination of a vertical wall, in terms of the measurements of the shadow of a perpendicular object, but you’re interested in the derivation of the solution, & you’ve already gotten good answers about that. But I’d like to make a few comments. … I’m going to refer to the declining-ness of a declining wall, its distance from due-south, as its “facing”, because the word “declination” of course already has a meaning in dialing & astronomy—altitude with respect to the equatorial-plane. … Referring to the spherical coordinate-system whose equatorial-plane is the surface of the declining-wall, I’ll call it the “declining-wall system”. To refer to the spherical coordinate-system whose equatorial plane is the surface of a south-facing wall, I’ll call it the “south-face system”. … This is one of those problems in which, it seems to me, the most computationally-efficient derivation isn’t the most straightforward, obvious, natural easiest one. ...where, in particular, the computationally-efficient derivation uses plane-trigonometry, & the more straightforward easy natural one uses a spherical-coordinate transformation. … Formulas for the length & direction of the nail’s shadow, from the Sun’s position in the coordinate-system with its equator parallel to the wall, can be gotten by coordinate transformations from the Sun’s position in the equatorial co-ordinate-system. … Determine the Sun’s equatorial-coordinates: … The Sun’s hour-angle, its longitude in the equatorial-system, is given by the sundial-time (French hours, equal-hours), the True-Solar Time, gotten from the clock-time by the usual use of the Equation-of-Time & the longitude correction. Hour angle is reckoned clockwise (westward) from the meridian. … The Sun’s declination (altitude in the equatorial-system) for a particular day can be looked up, & interpolated for a particular hour. … It seems to me that the most straightforward solution is to transform the Sun’s equatorial coordinates to the south-face system. … Then transform the Sun’s south-face coordinates to the declining-wall system. … The Sun’s altitude in the declining-wall system gives the length of the shadow, Its longitude in the declining-wall system gives the direction of the shadow on the wall. … You could use the shadow’s length or its direction. The shadow’s length, from the Sun’s altitude in the declining-wall system, has a briefer formula, & the length of the shadow is easier to measure than its direction. …& so I’ll speak of using the length of the shadow. … Resuming: When you’ve transformed the Sun’s south-face coordinates to declining-wall coordinates, the resulting formula for the Sun’s altitude in the declining-wall system will include a variable consisting of the angle between one system’s pole & the other system’s equatorial-plane. (That’s the latitude when you’re converting between the horizontal & equatorial systems, & so I call it the “latitude” for any coordinate transformation. That’s what I mean by “latitude”, in quotes, here) … Solve that formula for the “latitude”. Evaluate the “latitude”. Subtract that from 90 degrees, to get the wall’s facing. …thje amount by which it declines. … This assumes that the wall declines by less than 90 degrees. … Incidentally, this isn’t the only problem in which coordinate-transformations seem more straightforward than the plane-trigonometry solution: … I once noticed that a vertical-declining dial can be marked by plane trigonometry, but spherical coordinate-transformations seem more straightforward. … Likewise, it seems to me that the marking of the declination-lines for a Horizontal-Dial can be done most computationally-efficiently by plane trigonometry at the dial., …but calculating the Sun’s altitude & azimuth for each hour, for each of the several Solar declinations, seems much more straightforward, obvious & natural.
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