RE: Question About Measuring Wall Declination
You can build one of these. https://www.mysundial.ca/sdu/sdu_wall_declinometer.html From: sundial [mailto:sundial-boun...@uni-koeln.de] On Behalf Of Michael Ossipoff Sent: July 1, 2023 1:07 PM To: Jeffery Brewer; sundial list Subject: Re: Question About Measuring Wall Declination 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-trigonometr
Re: Question About Measuring Wall Declination
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
Re: Question About Measuring Wall Declination
Hi Jeffery you are actually calculating the horizontal angle indicated as 'angolo' on the diagram below ie. deviation of the Sun from the wall under consideration. Hope this helps, Alexei [image: image.png] On Mon, 26 Jun 2023 at 16:37, Jeffery Brewer wrote: > I'm attempting to measure the declination of a wall using a method > described on this web page of The Sundial Primer > https://www.mysundial.ca/tsp/wall_declination.html (also described in > "Sundials: Their Theory and Construction" by Albert E Waugh Chapter 10). > > Referring to Figure 1 of The Sundial Primer reference, "The direction of > the sun relative to the wall, θ, can be determined as follows: θ = arctan( > AB / Nail Length)°" If I label the ends of the nail with points C and D > (see figure below) then the formula can be understood as θ = arctan( AB / > CD )° > > [image: Figure1Modified.jpg] > > With my very rudimentary understanding of basic trigonometry, I understand > how the formula would work for a simple right triangle existing in a single > plane, but not how it works here. It seems to me that AB lies in an XY > plane parallel to the wall, but CD lies along the Z axis, perpendicular to > the XY plane. The shape described by ABCD is a sort of twisted rectangle > and I don't understand how the formula applies. > > I'm almost certainly thinking about this wrong (it feels like an optical > illusion where I can only see the vase and not the faces). > > [image: image.png] > > If anyone can help me "see the light" I would appreciate it. > > Jeff Brewer > > > > > > --- > https://lists.uni-koeln.de/mailman/listinfo/sundial > > --- https://lists.uni-koeln.de/mailman/listinfo/sundial
Question About Measuring Wall Declination
I'm attempting to measure the declination of a wall using a method described on this web page of The Sundial Primer https://www.mysundial.ca/tsp/wall_declination.html (also described in "Sundials: Their Theory and Construction" by Albert E Waugh Chapter 10). Referring to Figure 1 of The Sundial Primer reference, "The direction of the sun relative to the wall, θ, can be determined as follows: θ = arctan( AB / Nail Length)°" If I label the ends of the nail with points C and D (see figure below) then the formula can be understood as θ = arctan( AB / CD )° [image: Figure1Modified.jpg] With my very rudimentary understanding of basic trigonometry, I understand how the formula would work for a simple right triangle existing in a single plane, but not how it works here. It seems to me that AB lies in an XY plane parallel to the wall, but CD lies along the Z axis, perpendicular to the XY plane. The shape described by ABCD is a sort of twisted rectangle and I don't understand how the formula applies. I'm almost certainly thinking about this wrong (it feels like an optical illusion where I can only see the vase and not the faces). [image: image.png] If anyone can help me "see the light" I would appreciate it. Jeff Brewer --- https://lists.uni-koeln.de/mailman/listinfo/sundial
