# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**360 degree Bygrave slide rule sight reduction**

**From:**Paul Hirose

**Date:**2017 Feb 20, 23:20 -0800

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.