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Peter Fogg wrote:

"I don't know where you can find them, and you certainly need them to
use them. Here is the provenance of the two sets of tables I have:
1. Bennett G, 1995. "Tables for the solution of problems associated with
rhumb line courses and distances using the World Geodetic System
Spheroid 1984" CN SYSTEMS.
2. Following a (much) earlier discussion of the topic here, Sam (Chinese
sounding name?) kindly sent me "Table of Latitude Parts (Meridian
Distance)". Sam was a list contributor in early days. Although they have
WGS
84 on them, as does the Bennett tables, they differ somewhat, as
mentioned earlier (while I was being coy). Eg; For 10d, B: 597.11, S:
596.04.
Bennett's require interpolation between whole degrees, Sam's have
factors for minutes of arc."

---
The tables mentioned above shows what is usually called "latitude
parts". Latitude parts are the number of length units, usually "minutes"
or "miles" of some kind, corresponding to the distance between the
equator and a certain latitude parallel, along a meridian. So the
expression Meridian Distance makes sense.

The reason that the values given for 10 degs latitude differ is that
they are expressed in different units of length.

Evaluating the integral provided by Andres Ruis (somewhat confusingly
called meridional parts) for 10d latitude (with e=0.081892 according to
the WGS84 earth model) yields a*0.17338, where a is the equatorial
radius.

If a is expressed in units of one minute of arc along the equator, then

a=(360*60')/(2*pi)=3437.747' and the resulting value of latitude parts
becomes 596.04', the value shown by Sam.

If a is expressed in units on one International nautical mile (of 1852
metres), then

a=6378137/1852=3443.918 nm, where the numerator is the WGS84 equatorial
radius in metres. Multiplied with 0.17338 this results in 597.11 nm,
corresponding to Bennett's tabulated value.

When calculating difference in latitude on the ellipsoid the answer is
somewhat difficult to interpret as the length of the arc minute (along a
meridian) varies due to the non-spherical shape of the ellipsoid. By
using latitude parts this is overcome because the length of the meridian
arc is expressed in a fixed unit of length.

Meridional parts, in its ususal interpretation, is the distance from the
equator to a parallel of latitude, along a meridian, expressed in
"minutes of longitude". The length of these minutes changes with
latitude because the meridians converge towards the poles. The
expression for meridional parts (on a sphere) is
a*integral(sec(lat)dlat), from 0 to lat. The solution to this integral
is a*ln(tan(45d+lat/2)) or a*ln(10)*log(tan(45d+lat/2)). With a
expressed in arc minutes of a great circle, then
a*ln(10)=3437.747'*2.30259=7915.7'. I haven't been able to figure out
what kind of minutes George uses with his factor 7819, maybe he meant to
use 7918?

To simplify (?) calculation it could be useful to know that
ln(tan(45d+lat/2))=artanh(sin(lat))=arsinh(tan(lat)).

For practical nautical navigation it doesn't matter much if you use a
sphere or an ellipsoid of some kind, or whether you use the
International nautical mile, the mile of 6080 feet or minutes of arc of
the equator. Uncertainties in log calibration, course accuracy, "set and
drift" are in most cases far larger than the small discrepancies found
between different ways to calculate sailings.

But it is always good to have some understanding of the theories behind
the calculations. The integral shown by Andres Ruis have been published
in various sources but I have never seen a complete derivation of it.
Many years ago I tried to derive it, and got a slightly different
result. My solution gave the same integrand but the upper limit became

arctan(sqrt(1-e^2)tan(lat))

Now, as sqrt(1-e^2) is very near unity it really doesn't matter - but
it's annoying not to get the same result!

   Lars 59N 18E

   

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