NavList:

   

Reply

Guy Schwartz asked-

This has probably been discussed here before, but hopefully someone
will help me out. I'm trying to find the formula for caculating your
DR after traveling from a know position.
For instance:
Last Fix
40 deg 07 min N
44 deg 37 min W
Course 090
distance traveled 56.88 Nm
New DR?
It would be a big help to me if someone that knows the formula could
incorparate the above example so I could follow along.
Thank you,
Guy
===========================

Pierre Boucher provided a helpful response, though his statement -
"To find DLo (difference of longitude in minutes of longitude)
distance travel / cos(mid latitude)"
omits reference to course, and is true only in the special
circumstances of Guy's Easterly-course example
Not sure about that from Yanni Nikopoulos. But I think theres a bit
more to add. We will assume that for our purposes the Earth is a
sphere, which is usually good enough.

If you are steering a constant course (i.e. a rhumb line), then it's
always exactly true that your change in latitude in arc-minutes,
measured positive Northwards, is distance (in miles) x cos course,
where course is an angle measured clockwise from North, 0 to 360
degrees. It will be a positive change, if your course has a Northerly
component, and a negative change if it's at all Southerly. Add that
change to your initial latitude, allowing for sign, and you have the
latitude of your destination. That part is simple, then, however long
your journey is.

Calculating change in longitude is a bit more complicated, because the
length of a degree of longitude varies, over the Earth's surface, as
the latitude changes. I always like to consider longitudes as positive
Westward (though several authorities differ) in order to keep
longitudes in line with hour-angles, which increase Westward. In which
case, when travelling small distances (of less than 60 miles or so)
it's nearly true to say that the change in longitude in arc-minutes,
measured positive Westward, is distance (in miles) x (- sin course) /
(cos lat). The problem with that simple approach is that the latitude
is changing slightly, from one end of the leg to the other, so what
value of lat do you use in that formula to calculate change of long?
If there's hardly any change in lat, from one end of the leg to the
other, either because the leg is so short, or because the course in
nearly due East or due West, then it just doesn't matter; you can take
the lat, in that expression, to be the latitude of your starting
point. Doing the calculation that way is often referred to as "plane
sailing", because it's treating the Earth locally as if it was a plane
surface rather than as part of a sphere. In those circumstances, this
is a good enough approximation.

The example that Guy has quoted is not really an informative one, in
that with a course of exactly 90 degrees (due East) it's trivially
simple. The latitude stays the same throughout, at 40 deg 07' N.
Because it does, our plane-sailing approximation becomes exact, at
56.88 x (- sin 90) / cos 40 deg 07', or - 74.4 arc-minutes, so the
initial long of 44 deg 37' W must be reduced by 74.4 minutes to 43 deg
22.6' W. Just as Pierre Boucher showed.

For a longer travel, which involves a change in latitude, you can make
a much better approximation, quite simply, in the following way.
First, calculate the change in latitude as before, to get the exact
latitude at the end of the leg. Now simply average those two values
for latitude, at the start of the leg and at the end of it. Now use
this average latitude, or "middle-latitude" in that formula for change
in long, which becomes-
Westing in minutes = distance x (- sin course) / cos (mid-lat).
Calculating in this way was known as "middle-latitude sailing", for
obvious reasons, and was used in medium-length sea voyages, from one
headland to another. Even that, however, is only an approximation, and
for long ocean passages, particularly those that cross the Equator, it
can be seriously in error.

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.

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

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!

Note that for very long legs of a passage, Mercator sailing becomes
somewhat academic. Great-circle sailing, in which the course is
regularly adjusted, is almost always significantly shorter, so in
practice long legs of Mercator sailing are seldom adopted.

I suspect that parts of this answer provide more detail than Guy
expects or requires,

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.

   

Reply