# NavList:

## 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 25, 13:57 -0700

I have now worked some sample problems using Bygrave's formulas on a ten inch rule and it does seem to be an improvement over the sine- cosine method since the tan scale has better resolution. It is also obvious that a slide rule with two tan scales would be easier to use for this computation. gl On May 24, 2:48 pm, glap...{at}PACBELL.NET wrote: > 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 Hirosewrote: > > > 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 NavList-unsubscribe@fer3.com -~----------~----~----~----~------~----~------~--~---