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    Re: E-6B sight reduction trig
    From: Paul Hirose
    Date: 2012 Dec 28, 20:41 -0800

    I will demonstrate another Bygrave sight reduction on the E-6B. The
    correct values are:
    
    LHA = 295 51.2
    lat = -36 29.6
    dec = -17 54.1,
    
    az = 90 01.3
    alt = 31 05.5
    
    It's no accident azimuth is almost 90. Let's see if the solution is
    inaccurate at that angle.
    
    First convert sexagesimal angles to decimal degrees:
    
    lat = -36 29.6 = -36.49
    LHA = 295 51.2 = 295.85
    dec. = -17 54.1 = -17.90
    
    t = 360 - LHA = 64.15
    
    The Bygrave formula requires the cosine or secant of t. Either will do,
    since the cosine and secant are simply reciprocals of each other.
    However, the E-6B has only a tangent scale, so to obtain a secant we use
    the trigonometric identity sec² t = 1 + tan² t. This relationship 
    between tangent and secant will be used several times. For angle t:
    
    tan t = 2.076
    tan² t = 4.33
    sec² t = 5.33
    sec t = 2.304
    
    Now that sec t is known, solve for y:
    
    y = arctan(tan dec * sec t) = arctan(.742) = 36.6
    
    The cosine of y is needed later, so compute it from tan y:
    sec² y = 1 + tan² y = 1.550
    sec y = 1.245
    cos y = .803
    
    Compute upper case Y:
    Y = 90 - lat + y = 91.1
    
    With Y known, compute its secant, needed later. Start with the tangent.
    Unfortunately tan 91.1° is not on the scale. But it's equal to 1 / tan
    1.1°. To practical accuracy, tan 1.1° is simply 1.1° converted to
    radians. So opposite 1.1° on the outer scale set 57.3 on the inner. Read
    tangent at the inner index. But we really wanted its reciprocal, which
    is at the outer index: 52.0.
    
    With tan Y known, find its secant via the secant and tanget identity 
    again: sec² Y = 1 + tan² Y. This time it's easy. The tangent of Y is so 
    big, adding 1 to its square makes hardly any difference. To E-6B 
    accuracy, tan Y = sec Y = 52.0.
    
    Next solve for az = arctan(tan t * cos y * sec Y) = arctan(2.076 * .803
    * 52.0) = arctan(86.6). The arctangent of 86.6 is way outside the range
    of the scale. However, arctan 86.6 = 90 - arctan(1/86.6) = 90° -
    arctan(.01155). Again use the fact that the tan of a small angle is
    practically equal to the angle itself, expressed in radians. Therefore,
    arctan(.01155) = .01155 * 57.3 = .662°, and 90° - .662° = az 89.34°.
    
    According to the Bygrave rules, that value must be subtracted from 180
    because Y is greater than 90. So az = 90.66°.
    
    We're almost done. To compute altitude we need the cosine or secant of
    azimuth. Again the tangent is large, so to practical accuracy sec az =
    tan az = 86.6. (Formally, the value is negative, but the sign is ignored.)
    
    Finally Hc = arctan(cos az * tan Y) = arctan(tan Y / sec az) =
    arctan(52.0 / 86.6). At 52.0 on the outer scale, set 86.6 on the inner.
    Read arctan = 31.0° on the drift scale.
    
    The correct values are az = 90 01.3 (vs. 90.66), Hc = 31 05.5 (vs.
    31.0). Not bad, especially the azimuth was intentionally chosen to make
    the solution difficult. And it was difficult! Trig on any slide rule
    demands knowledge of identities. (As a schoolboy I disliked trig 
    identities, but have become more comfortable with them over the years.) 
    The minimum facilities for trig on the E-6B, plus the parameters of this 
    problem, made the solution even harder.
    
    -- 
    I filter out messages with attachments or HTML.
    
    
    
    
    

       
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