# NavList:

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

**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.