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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...@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 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!
>
> > --
> > 
>
>
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