# NavList:

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

**Re: Lecky's mistake**

**From:**David Pike

**Date:**2020 Oct 31, 12:55 -0700

I’ve been looking at this and perhaps someone will confirm or contradict my understanding. I think what Lecky is saying is for a given latitude close to your own, and at a particular astronomical time, there will be a longitude which corresponds to a particular altitude of Sun. Once you know that, you can dot-in that lat and long. There are also tables which will give you the Sun’s azimuth at that place and time. Add 90 degrees and draw a line though the dotted-in lat & long. This is your Sumner line. Your true position is somewhere on this line.

The mathematics seems to rely upon solving the side, side, side, PZX triangle of co-dec, co-assumed lat, and co-alt in order to find the angle at the pole between the Sun’s and the observer’s meridian. If you know the Sun’s meridian and the angle at the Pole, you add the two to get your longitude. Being 19^{th} Century, he works in hours minutes and seconds, not degrees, for hour angles. Well, they are called hour angles after all.

It’s the actual maths that I’m struggling to follow. I’ve taken the 11.00 AM example, which seems more general than the 07.08 AM example. The 07.08 example is a special case, because the Sun is on the Prime Vertical.

First, he finds X, the Sun’s sub stellar point at the time of the observation. He must get declination and hence co-dec (Polar Distance) from tables. He get’s the Sun’s astronomical meridian by adding an equation of time value from the nautical almanac to GMT from his chronometer.

He seems to multiply Cosecant polar distance x secant assumed latitude x cosine of ½ (Polar Distance + Latitude + the Sun’s Altitude) x Sine of ‘the remainder’ using 9’s convention six figure log tables. What on Earth is ‘the remainder’, and which spherical triangle formula is he using? Like in ‘A Hitch Hiker’s Guide to the Galaxy’, the answer comes out as 59minutes, 58 seconds of time. Which antilog of 8.230947 is 59minutes 58 seconds?

He then flicks the 59min 58s the other side of 12.00 hours, probably to get the sense of the hour angle correct and gets his actual astronomical time. Then he subtracts that from astronomical time at Greenwich to get the longitude in hours minutes and seconds corresponding to his assumed latitude.

After that he uses tables to get the sun’s azimuth for that position and time and draws his next Sumner line at 90 degrees to it. Phew; it’s taken me all day to work that out, and those new taps in the Utility Room still need fitting. Overall, I think I prefer using AP3270. DaveP