Mercator Sailing: Course and Distance

[Dieter]

(08-26-2018 12:30 AM)Gene222 Wrote:  "a bit of calculus"??? Your math skills are far better than mine.

It's just the limit for the distance as lat1 → lat2.
Re. math skills: the result includes the derivative M'(lat). I was lazy and had it calculated by Wolfram Alpha. ;-)

(08-26-2018 12:30 AM)Gene222 Wrote:  It looks like you are calculating the distance directly from the ellipsoid and not from the map projection.

No, these are the Mercator sailing equations. The meridional parts calculation is just a bit more accurate in that the eccentricity e is not hidden in other constants so that it can be adjusted. Compare the results with other sources, e.g. online calculators.

Hint: the expression ln(tan(45°+x/2)) in the "book" formula is equivalent to artanh(sin(x)).

For a course on the same parallel the given formula simply calculates the limit of the original formula as the two latitudes approach each other. Try it: use the original formula and make the two latitudes almost equal, e.g. 40° and 40,0001°. You will get the same result.

(08-26-2018 12:30 AM)Gene222 Wrote:  I did remember of another way to solve the course and distance on an East-West line for pt 1 and pt 2. This was on a old forum. This person said to create a third point mid-longitude between the two points, but with a latitude well above the two points. Then calculate the course and distance from pt 1 to pt 3 and from pt 3 to pt 2. He then added the vectors to give him the course and distance from pt 1 to pt 2. I need to think about this East-West issue some more.

Way too complicated. The above formula works.

(08-26-2018 12:30 AM)Gene222 Wrote:  Regarding the WGS84 ellipsoid, it certainly makes sense to use a more modern ellipsoid that is used by GPS, but this is a Mercator Sailing program. The WGS84 ellipsoid seem more like Rhumb-line sailing. Regarding the accuracy of the author's equation, I did find a more accurate meridian part equation from an on-line Mercator calculator's site. I don't know if it is Clark's ellipsoid of 1866, but it closely resembles the author's equation.

That looks like a series expansion of the true formula.
Of course you can also use this approximation. But why do so if you can have the exact result?

(08-26-2018 12:30 AM)Gene222 Wrote:  I think I need to keep this program simple and stick with the Mercator sailing equations.

Again: these are the Mercator sailing equations.
And the artanh method makes them even more compact.

Gene, it looks like you think that the formulas I gave calculate something different from the Mercator sailing course and distance. They don't. The Meridional parts formula is just more accurate and can handle different ellipsoids. The results match exactly what is shown in other online sources on this topic. And the East-West-formula is exactly what you get with the original formula if the two latitudes approach each other.

Edit:
Maybe a few examples are more convincing. Please take a look at this PDF with some calculations including the meridional parts. I assume their values were taken from nautical tables, so they may serve as a reference. Here are the values from the PDF, then those calculated with the artanh formula (assuming the Clarke ellipsoid) and finally using the simplified formula due to "The Calculator Afloat".

Code:

 given                     Meridional Parts
latitude      due to PDF    artanh formula    simple formula
------------------------------------------------------------
 42°45'        2826,73          2826,73          2826,83
 27°30'        1706,46          1706,46          1706,52
 47°14,3'      3206,54          3206,54          3206,64
 43°55'        2922,63          2922,63          2922,73

You see that the artanh formula exactly matches the values in the PDF (which I assume are correct) while the results from the formula in the book are slightly high (~ +0,1). The latter doesn't matter much as in our case only the difference between the meridional parts of the two positions is considered. Also an accuracy of one decimal seems to be adequate here.

BTW, you can minimize the error of the simple formula if you replace the constant 23,2689 with 23,4286 for the Clarke ellipsoid or with 23,0522 for WGS84/GRS80. This way the deviation is less than ±0,014 meridional parts for any latitude. On the other hand 23,2689 is the best possible value for an eccentricity of 0,0822... which is roughly in the middle between the two mentioned ellipsoids.

But instead of all this you can also simply use the more accurate formula. Either the artanh version or the other one. ;-)

Edit 2:
If you want to stick to the simplified formula you can determine the distance on the same parallel (lat) as follows:

Code:

                                                     1
distance  =  120 · |long2 - long1| / [ ----------------------------- – 23,2689·pi/5400·cos(lat) ]
                                       sin(45°+lat/2)·cos(45°+lat/2)

The constant in the last term can be simplified to 0,0135373. Or, even shorter, 1/73,87.

For your example this evaluates to 230,73 nmi.

Dieter