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    Re: recommendation for slide rule ?
    From: Paul Hirose
    Date: 2009 May 23, 21:00 -0700

    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
    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.
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