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With the slide rule techniques in my "360 degree slide rule trig" series

http://fer3.com/arc/m2.aspx/360-degree-slide-rule-trig-Hirose-nov-2016-g37154
http://fer3.com/arc/m2.aspx/360-degree-slide-rule-trig-Hirose-nov-2016-g37223
http://fer3.com/arc/m2.aspx/360-degree-slide-rule-trig-Hirose-feb-2017-g38237

a variant of the Bygrave sight reduction method is possible. It's a
universal solution, valid for all LHAs, latitudes, declinations, and
even negative altitudes. The three main formulas are the same, but the
rules and auxiliary formulas are different, and angle Y is unnecessary.
The solution is designed for a straight slide rule. I don't know if it's
practical on the Bygrave device.

In these formulas, ignore negative signs on trig functions. For
instance, tan -10° = -.176, but calculate as if it were +.176. This rule
simplifies the three arc tangents: they're all in the range 0 - 90.

Note that LHA doesn't require adjustment. Even the equivalent negative
angle will work, e.g., -10° instead of 350°.

W = arctan(tan dec / cos LHA)
If LHA is 90 to 270, W = 180 - W.

In the computation of X, south latitude is negative.

X = 90 - lat + W, if north dec.
X = 90 - lat - W, if south dec.

If X is not 0 to 180, add or subtract 180 to make it so. If that
adjustment is necessary, Hc will be negative.

A = arctan(tan LHA * cos W / cos X)
If X exceeds 90, A = 180 - A

Angle A is similar to azimuth angle, except that — regardless of assumed
latitude — south is zero when Hc is positive, north is zero when Hc is
negative. Go east or west from zero in the obvious way, according to
which side of the meridian the body lies.

Hc = arc tan(cos A * tan X)
Negate Hc if X required adjustment.


Example:
LHA = 213.712°
lat = 59.861° south
dec = 28.984° north

W = arctan(tan 28.984 / cos 213.712)
W = 33.66, adjusted to 146.34

X = 90 - -59.861 + 146.34
X = 296.20, adjusted to 116.20

A = arctan(tan 213.712 * cos 146.34 / cos 116.20)
A = 51.5, adjusted to 128.5 (+.8′ error)

Hc = arc tan(cos 128.5 * tan 116.20)
Hc = -51.70 (-2.0′ error)


In a Monte Carlo simulation of a 10 inch slide rule, with one million
sights randomly distributed between nadir and zenith, the root of the
mean squared altitude error was 1.76′, 1.07% of errors exceeded 5′, and
worst error was 10.7′. In azimuth, RMS error was 2.88′, .05% of errors
exceeded 30′, and worst error was 17.4 *degrees*. However, that occurred
within 15′ of the nadir. If test problems within 10° of the zenith or
nadir were excluded, the worst azimuth error was 30′.

My program has an accurate simulation of interpolation error, but it
assumes there's no error when a scale index and the hairline are brought
into coincidence. That's not true in the real world, so my statistics
are optimistic. But I believe they're not far from the truth.

   

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