# NavList:

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

**Re: Sumner's Line (Navigation question)**

**From:**Bill B

**Date:**2006 Feb 11, 15:43 -0500

Frank You wrote: "So what's the math that converts a measured altitude into an LHA? This is an easy spherical triangle problem. The Sun's local hour angle is the angle measured at the elevated celestial pole between the observer's meridian and the arc from the pole to the Sun (draw it!). To calculate that angle, we can use the three sides of the spherical triangle made by the Sun, Zenith, and the Elevated Celestial Pole. I won't spoil the fun. You try it out for yourself. Draw that triangle and solve for LHA using the cosine formula." I believe I understand you, but want to check my reasoning. My derivation of the Hc formula is: cos LHA = (sine Hc - (sine lat sin dec)) / (cos lat cos dec) When I use my derivation of the Hc formula to solve for LHA, I do not get LHA in the case of a body east of the meridian. In your example, an angle of 45d (0 Eqn of Time) would yield a local time of 3 PM or 9 AM. Clearly the 9 AM LHA would be 315 if we measure from the observer west to the body. Of course my calculator does not know whether the body is pre or post meridian passage, and cos 45d = cos 315d. At least the way I am doing it for now, I believe I am getting angle t, the angle formed from the observer to the elevated pole to the the body, which is not the LHA when the body is east of the meridian. You also wrote: "...then the Sun's altitude, which can be measured accurately with a sextant, automatically yields the Sun's local hour angle. This hour angle is exactly the same thing as the observer's local apparent time." Perhaps a matter of semantics, but in the case of a body east of the meridian, to get local time I think I should subtract the arc-to-time conversion from 12 to get local time. Have I missed something? Thanks Bill