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On 24 June, in reply to a question by Guy Schwartz, I expounded about
Mercator Sailing. I'm sorry to say that there were serious errors in
that explanation, which I will try to correct here.

This is what I wrote-

| However, there is a way to calculate a rhumb-line destination
| precisely, for any journey A to B, in which a constant course is
| maintained, no matter how long it might be. This is known as
"Mercator
| sailing". First, knowing the latitude of A, work out the exact
| latitude change just as before, so now we have the two latitudes A
and
| B. Next, convert each of those latitudes into a value that's known
as
| "Meridional Parts", or MP. Tables to do this were available in the
| epitomes that were available to old mariners. But you can convert a
| lat to its equivalent in MP yourself dead easily with a calculator.
| Just halve the lat in degrees (keeping its sign carefully, positive
| North, negative South). Then add 45 degrees (remembering that sign),
| the resulting angle being always positive, between 0 and 90, and
find
| the tan of that angle. Now find the log (the ordinary log, to the
base
| 10) of that result, and multiply it by 7819. It sounds a handful,
but
| a calculator will do it with no trouble at all. The end result,
which
| you should be able to check out , is that a lat of zero, on the
| equator, results in a MP of zero; for a lat in the Northern
hemisphere
| of +45 deg, it's MP is 2992.9; and for a Southern lat of -45 deg,
| it's -2992.9.
|
| (Note that these values are for a spherical Earth, and to achieve an
| even more precise result, it's possible tinker slightly with the
| conversion of lat to MP to get an even better fit to the Earth's
true
| ellipsoidal figure. On that basis are meridional-parts tables, and
| Mercator charts made. We will ignore that tinkering here.)
|
| Now take the difference, by subtracting the initial MP (of A; call
it
| MPa) from the final MP (of B; call it MPb), taking account of their
| signs. The result will be positive if the course has any Northing,
and
| negative if it has any Southing.

=========================

OK so far, as far as I can tell. But then I continued-

| And now, finally,
| Westerly change in long (in minutes)= (MPb - MPa) / (- tan course)

And that was my first error. Instead of dividing by (- tan course), I
should have MULTIPLIED. So what I should have written was-

Westerly change in long (in minutes)= (MPb - MPa) x (- tan course).

Sorry about that.

Later, Guy Schwartz then posed another example, in which the course
was chosen as 45 degrees. It so happens that tan 45 deg is exactly 1,
so that whether you divide by tan 45, or multiply by tan 45, you get
exactly the same answer. In that special case my erroneous calculation
gave the right answer for the wrong reason!

=========================

I then went on-

| Note that this works precisely under even the most extreme
conditions,
| even when travelling between points on opposite sides of the
Equator.
| And even if you specify such a daft long distance, and a course
| suitably close to East or West, that the travellers rhumb-line
course
| takes him several times round the world before he gets there, it
will
| still give the right answer!

But that is not completely true, and I certainly ought to have said
so. Calculating difference of longitude, using meridional parts, gets
imprecise for rhumb-line courses very near to 90 or 270 degrees, due
East or West, and becomes impossible if the course is exactly East or
West. The problem is that with such courses, the difference of
latitude (and so the difference of Meridional Parts) approaches zero,
which is compensated for by the way (tan course) increases towards
infinity for courses near due East and West. So the result is a
decreasing precision, for courses near East and West, and an
meaninglessness, multiplying zero by infinity, for due East and West.

But if course, where the differences in latitude are small, that's
just the case where middle-latitude sailing works accurately, even
over a long distance; so in those cases, within a very few degrees of
East and West, it's perfectly appropriate to switch methods.

I have a vague memory that some guru, on this list, introduced a
method, perhaps using versines, that remained accurate for all course;
but if that's so, I can't recall any details.

George.

contact George Huxtable at george@huxtable.u-net.com
or at +44 1865 820222 (from UK, 01865 820222)
or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.

   

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