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    Re: Bygrave sight reduction by slide rule
    From: Paul Hirose
    Date: 2018 Jan 2, 14:44 -0800

    On 2017-12-13 14:38, I wrote:
    > W = arctan(tan dec / cos LHA)
    > If LHA is between 90 and 270, replace W with its supplement: W = 180 - W
    If LHA = 90 its cosine is zero and therefore the equation is undefined.
    In that case, alter LHA by an insignificant amount, say .001°.
    For example, lat = 50, dec = 40, LHA = 90. After you change LHA to
    89.999, W = 89.99881
    > If latitude and declination have same name
    >      X = 90 - lat + W
    X = 130 to practical accuracy
    > Z = arc tan(tan LHA * cos W / cos X)
    Z = arc tan(tan 89.999 * cos 89.99881 / cos 130)
    At first that looks impossible on a slide rule, but tan 89.999 = 1 / tan
    .001, and cos 89.99881 = sin .00119. Thus the part in parentheses is
    equivalent to (sin .00119 / tan .001 / cos 130). The first quotient is
    easily and accurately computed on ST: 1) align indices on slide and
    body, 2) set cursor to 1.19 on ST, 3) set 1 on ST to cursor.
    The quotient is on D at the C index. Its characteristic is 0 (i.e., the
    value is in the range 0 - 10) if the angles are literally as set. The
    actual angles are both a thousand times less, so there's no net effect
    on the characteristic. The quotient is 1.19. Finish the equation: 4) set
    cursor to C index, 5) set 130 (alias black 40) on S to cursor, 6) set
    cursor to D index.
    Z = 61.65 at the cursor on T
    > Hc = arc tan(cos Z * tan X)
    Hc = 29.50
    The errors in Z and Hc are respectively -.6 and .08 minutes, with
    respect to a solution computed for LHA exactly 90.

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