NavList:

Roger Errington, seven days ago, you asked:
"Is the Lat used when calculating the change in longitude (dLon = Dist x sin course / cos Lat) the original latitude (Lat0) or the adjusted latitude after adding the change in latitude (Lat = Lat0 + dLat/60)? "

Sorry to take a week replying to you, but I see you got a number of interesting replies already. The short answer is "take your pick; it makes little difference". And I agree with the reply from David McN which said nearly that.

First, let's talk about "needs". When would you need to apply this DDR or Digital Dead Reckoning, as I call it (for anyone else following, it's just "plane" DR run up on a scientific calculator)? There are two key scenarios. If you're attempting to do navigation by traditional methods, you may find you want to use it for an entire day of navigation now and then. And this is feasible so long as the distances on each leg of your "day's work" are not longer than 50-100 miles. And of course they would usually be shorter, possibly much shorter so that's probably fair. A second scenario, critical to our NavList discussions, covers the run between two celestial sights.

Suppose I have a pair of points on one celestial line of position, maybe a Sun sight taken in the morning. Each point on that LOP has a latitude and a longitude, call them Lat0, Lon0. Suppose I sail some distance, D, on some course, C, and then take another celestial sight, maybe an afternoon Sun sight. I want to bring the earlier celestial line of position forward in time. So I can use
   dy = D·cos(C),
   Lat1 = Lat0 + dy / 60
and
   dx = D·sin(C)
   Lon1 = Lon0 + dx / cos(Latx) / 60
to update each of the points on the first line of position. This drags the morning Sun line forward in time, into the afternoon, along with the vessel. But what should we use for the latitude in the longitude scaling factor? Here I've called it Latx to indicate that it might be something we should think about. Should we use Latx=Lat0? Or Latx=Lat1? Or maybe the simple average of those two, which would be equal to Lat0+dy/120? None of these are perfect for two reasons: these are plane approximations to the sphere, which is a valid approximation up to a couple of hundred miles for most purposes, and equally because the distance travelled and the course made good are not known in advance.

How should we decide? There's no absolute rules here. So experiment! Try different cases. So long as you stay away from high arctic latitudes you'll find that it makes very little difference. There is a small difference, and even without experimentation, you can guess that the average will be slightly better. But you don't need to worry about it.

What should you do in high arctic latitude? Bring your GPS. And your winter coat. :)

Frank Reed