# NavList:

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

**Re: Celestial Navigation in 1866**

**From:**Frank Reed

**Date:**2019 Feb 16, 22:55 -0800

Hello Lloyd Browne,

* Oysina Point* is simply an obsolete placename. It was an old name for the southwest point of the island of Timor. How can you know that? Google is your friend. If you search on Google for "Oysina Point" (being careful to search with it in double-quotes like this, so that Google looks for the exact phrase), you'll get a few hits which answer the question. We can confirm it's the right place by doing the math. And that's the answer to your primary question. Your explorer/navigator/surveyor was keeping track of the direction and distance to this arbitrary location, his "ground zero" for imagining where he was in his travels, by simple calculation. And this calculation would have been greatly facilitated by standard tables. It's not complicated.

On a flat sheet of paper where points are plotted as x and y coordinates, the distance from one point to another is given by the Pythagorean Theorem:

dist = sqrt(dx^{2} + dy^{2}),

where dx is the difference in the horizontal or x coordinates of the two points and dy the difference in the vertical or y coordinates. We can use this equation with latitudes and longitudes if dy is the difference in latitude between the two places, and dx is the difference in longitude, but scaled by a factor of cos(lat) to account for the fact that longitude degrees start out exactly equal to latitude degrees at the equator, but the lines converge as we approach the poles. Go visit Google Maps, right-click on the west point of Timor and select "What's Here". That will give you a latitude of about 10.30° S and a longitude of about 123.54° E. Work out dy (=diff in lat) and dx (=diff in lon, reduced by 2% for that scaling of longitude degrees...) and then calculate square root of the sum of the squares of differences... and what do you know, you get almost exactly the differences you find in that logbook: 338 nautical miles from the first point and 245 miles from the second point. The bearings work out, too.

Why pick that point? Why compare locations in his explorations to the southwest point of Timor? As I noted above, it's nothing more than a "ground zero" for his work. A place to compare back to as the exploration continues. He could just as easily have computed his bearing and distance from Singapore's town hall or from the Eiffel Tower in Paris. The numbers are perhaps more meaningful relative to a relatively nearby anchor location, but really any point will do.

You said also that "Edmunds was a gifted navigator."

Ok... How do you know? You mentioned that his latitudes were often within 200m of modern GPS positions. How often? In the two examples you have posted, the apparent precision in the latitudes is half a minute of arc, which is not hard work for any navigator on land. Were the latitudes also that accurate? Half a minute of arc is about 900m. Does Edmunds really get latitudes four times better than that? And how do you know?

Frank Reed

Clockwork Mapping / ReedNavigation.com

Conanicut Island USA