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    Re: Bygrave sight reduction by slide rule
    From: Paul Hirose
    Date: 2018 Jan 7, 20:40 -0800

    For a very short distance (high altitude) test of the Bygrave formulas
    on a slide rule, solve for the separation and position angle of two
    components of a double star. To make the test more extreme, let the star
    be Polaris, which is a triple system. Coordinates of the two main
    components from Wikipedia:
    
    02h 31m 49.09s  +89° 15′ 50.8″  α UMi Aa
    02h 30m 41.63s  +89° 15′ 38.1″  α UMi B
    
    If Aa (the star visible to the naked eye) is the "observer" and B is the
    "celestial body", then
    
    89.26411 lat
    89.26058 dec
    0.28108° LHA
    
    W = arctan (tan 89.26058 / cos .28108)
    W = 89.26058
    
    At slide rule accuracy the cosine of such a small angle is virtually 1,
    so W equals the declination.
    
    X = 90 - 89.26411 + 89.26058
    X = 89.99647
    
    X is in the range 0 - 180, so "altitude" is positive.
    
    Z = arc tan(tan 0.28108 * cos 89.26058 / cos 89.99647)
    
    That formula is a good lesson on techniques to extend the range of the
    ST scale.
    
    1. Tan 0.28108 is not on ST, so set it as tan 2.81.
    
    2. Cos 89.26058 (red 89.26058) may be set directly. Since red 89 = black
    1, begin at black 1 and move left .26°. (A few rules include the red
    numbers on ST. But usually they're omitted since the numbers are obvious.)
    
    3. The method in step 2 won't work for cos 89.99647. Therefore, delete
    nines to get a value you can set: 86.47. Begin at black 4 (= red 86) and
    move left .47° to get cos 86.47.
    
    Though it's not immediately obvious, the same principle is at work in
    step 1 and step 3. I.e., cos 89.99647 (red 89.99647) is equivalent to
    black .00353, which is outside the range of ST. By shifting the decimal
    point three places you get black 3.53, alias red 86.47, which is on the
    scale. This flexibility with the decimal point is possible on ST because
    the sine or tangent of a small angle is practically proportional to the
    angle. That's not true on the S and T scales!
    
    If the angles were literally equal to the numbers on ST, the
    characteristic of the tangent would be -2, i.e., the result is between
    .01 and .1. But if you include the decimal point adjustments in steps 1
    and 3, the characteristic is 0 and so the arc tan is read on T. With the
    slide in its final position (index at the tangent on D), set cursor to
    the D index.
    
    Z = 134.26° on T at the cursor. Correct angle is 134.13.
    
    Hc = arc tan (cos 134.26 * tan 89.99647)
    Hc = 89.99494
    
    The cosine can be set directly. Change the multiplication to division by
    cot 89.99647, which you set as red 86.47 as in the Z computation. Read
    84.94, adjust to 89.99494 to account for the deleted 9s. Separation
    angle is the complement of that angle: .00506°. Correct angle is .005054
    (18.19″) and error is .02″.
    
    It wasn't easy, but a 10 inch slide rule and the Bygrave formulas
    performed well in this test.
    

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