A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: recommendation for slide rule ?
From: Gary LaPook
Date: 2009 May 24, 14:48 -0700
From: Gary LaPook
Date: 2009 May 24, 14:48 -0700
After reading Dave Walden's post on this thread on May 20 about using the Bygrave formulas on the ten inch slide rule, it occured to me that that may be the better method since it uses tans with the advantages that you point out. I haven't tried this out yet but it makes sense to me. BTW your matrix math left me in the dust, I don't remember that part of my math education. gl On May 23, 9:00 pm, Paul Hirose
wrote: > I decided to work Gary LaPook's sight reduction problem on a slide rule > by means of rectangular coordinates. > > The inputs are lat. = +34°, LHA = 14°, dec. = +20°. > > First, convert spherical coordinates LHA and dec. into rectangular > coordinates in a system whose +z axis coincides with the north pole and > +y axis intersects Earth's axis. This requires 1) negating LHA then > subtracting 90° to obtain an angle "theta", -104°, which conforms to the > usual spherical convention, then 2) converting theta and dec. into xyz > coordinates via these formulas: > > x = cos dec. cos theta = -.2274 > y = cos dec. sin theta = -.911 > z = sin dec. = .342 > > Form a 3x3 rotation matrix to convert the above vector to the observer's > horizontal orientation. This requires an x rotation by (90° - lat). The > matrix is: > > 0 0 1 > 0 sin lat. cos lat. > 0 -cos lat. sin lat. > > Multiply the rotation matrix and the vector. That yields a new vector > which represents the body in a system whose +x axis is directed east, +y > north, and +z to the zenith: > > x = -.2274 > y = -.226 > z = .946 > > It's easy to blunder when doing this by hand; a good check is to compute > the sum of the squares of x, y, and z. It should be near 1.000: > > .0518 = x squared > .0511 = y squared > .895 = z squared > ----- > .9979 > > Also take the square root of the sum of x squared and y squared. This > value, which I'll label d, is the distance of the body from the +z > (zenith) axis: > > .321 = d = √(.0518 + .0511) > > Altitude is the arc tangent of z/d. Set the C scale left index to d > (.321), the hairline to z (.946) on scale D, and read 71.28° (71°17') at > the hairline on scale T. That is, if your slide rule has a double T scale. > > With a single T you must compute the arc cotangent of d/z. Set C left > index to .946, hairline to .321 on D, read 71.28° on the T scale *red* > numbers. Had the altitude been less than 45°, the computation would have > been the same as with a double T scale, and you'd read the angle on the > *black* numbers. This is an example of how a double T scale simplifies > the slide rule. The physical moves are no easier, and the result is no > more accurate, but you don't have to think as hard. The rule itself > isn't any more "powerful". (Having said that, I'm surprised so many high > end rules have a single T. This includes the Keuffel & Esser "Deci-Lon" > and Post "Versalog", both flagships for their companies to the end of > the slide rule era.) > > To obtain azimuth, compute the arc tangent of y/x. Disregard the signs > at first. Set the C right index to .2274, the hairline to .226 on D, > read 44.9° at the hairline on T. Since x and y are both negative, the > actual angle is in the third quadrant of the Cartesian system, and its > value is 44.9° - 180°, or -135.1°. > > But remember the orientation of the coordinate axes. Zero degrees is > east, and increases north. To change this to the normal convention, > negate the angle and add 90°, resulting in 225.1° for azimuth. > > Compared to HO 229, the slide rule solution is off .4' in altitude, > perfect in azimuth. That's partly good luck. (In case you're wondering, > nothing was fudged to make it come out right, and those numbers are my > one and only attempt.) However, accuracy also comes from arranging the > calculation so the results come out as arc tangents. Look at the T scale > on a slide rule. The most compressed portion is at 45°. Even there, it's > graduated every .2° on a 10 inch rule. > > Then look at the S scale. At small angles it has practically the same > resolution as T. But at 30° it's clearly more compressed, and at 70° the > graduations are single degrees, tightly packed. Clearly, accuracy is > going to be highly variable if you read results as arc sines. > > With the non-inverse trig functions (e.g., sine rather than arc sine) > things are different. Basically, you can obtain these to constant > accuracy for all angles. Say, to .1% on a 10 inch rule. And > multiplication and division also have nearly constant relative error > everywhere on the scales. It's these things, plus avoidance of arc sine > and arc cosine, that give the vector method its accuracy. > > Of course the disadvantage is the greater quantity of computation. > However, the pencil and paper work is simple addition of numbers at > slide rule accuracy. For instance, multiplying the rotation matrix times > the vector requires four sums from a total of nine numbers, including > the check I suggested. > > The choice may come down to where you're more comfortable. To do the > above computation with spherical trig I'd have to look in a book! > > -- > I filter out messages with attachments or HTML. --~--~---------~--~----~------------~-------~--~----~ Navigation List archive: www.fer3.com/arc To post, email NavList@fer3.com To unsubscribe, email NavListemail@example.com -~----------~----~----~----~------~----~------~--~---