NavList:

Bill Ritchie, you wrote:
"In a marine setting, I wonder how the solo competitors in the (astro only) 2018 Golden Globe non-stop round the world race will use their "ellipse error proportions" knowledge. I hope that if they are avoiding an isolated reef they ignore the math and assume that they are at the nearest edge of such ellipse to the reef, plus their planned offing."

Huh? Please do tell how this would be "ignoring the math". How would they assume they are near some edge of the error ellipse unless, in fact, they employ some of that darn math to draw that error ellipse?

To answer your scenario directly, if any of the GGR sailors reached a point where he or she believed that some isolated reef might be a clear danger, and anything about the accuracy of their celestial work was suspect, then I would hope that the sailor in question would do the prudent thing: throw in the towel and pull out the GPS (which they will carry for exactly that possibility). Or at least the sailor would recognize the need to heave to, deploy a sea anchor, and wait until there are better conditions to reset the traditional navigation.

Now suppose instead one of those GGR racers decides to use a little more than "Celestial Navigation for Weekend Sailors" as the basis for the work. That navigator might decide to plot a fix on the chart and might regularly draw an error ellipse around it. What would be the shape of that error ellipse for a standard two-body fix? Obvious symmetry tells you that it's a circle when the lines of position cross near 90°. Most of us would have no problem drawing a circle perhaps two miles across centered on the fix, and most of us would understand the meaning of that circle (it's not an absolute guarantee!). No problem so far. But what would you draw if those two lines of position were separated by 11 or 12°? You know the ellipse is going to be longer and its long axis will fall between the lines of position. But how long? Well, now you know the real answer. If you would normarlly draw a circle two miles across, then your error ellipse in this case would have an aspect ratio of 10-to-1. That is, you would draw an ellipse roughly two miles across and twenty miles long. If instead the crossing angle was double that, then the ratio would be 5-to-1. These are the implications, nicely distilled to a simple rule that anyone --and I do mean anyone-- can learn, of some very powerful math. Isn't that a good thing??

To repeat the rule, since it was originally in another thread:
If two LOPs are separated by some angle dZ, the "aspect ratio" of the error ellipse, short dimension compared to long dimension, is given by:
  ratio = tan(dZ/2).
That's the exact ratio. And there is a very nice fast approximation for this: the ratio is dZ/114 when dZ is in degrees. If you prefer, you can think of the ratio as pi·(dZ/360) with dZ in degrees. This latter approximation is most useful for cases where the angle dZ is considerably less than 90°, but cases near 90 are obvious anyway --more or less a circle. I'll post a quick version of the derivation under the original thread.

Frank Reed