This gives the correct area
S = 2pc (c+ar)
for prolate spheroids.
On 2004-05-17, we received the first attempt at optimizing a symmetrical
formula by Knud Thomsen, who investigated the following expression, featuring
a second parameter (k) generalizing his earlier formula (the case k = 0):
S » 4p
[ ( apbp +
apcp +
bpcp ) /
( 3 - k
{ 1 - 27abc /
(a+b+c)3 } ) ] 1/p
The formula is designed to be correct in the case of a sphere for all values of p and k.
Selecting p = ln(2) / ln(p/2), Thomsen claims optimal results
when k is around 0.0942 and suggests the
convergent 3/32 (0.9375)
which is good enough to yield a relative error between
-0.204 % and +0.187 %
[the next convergents would be 5/53 and 8/85].
This represents an improvement of one order of magnitude over his
original formula (k=0).
Earlier work of Achim Flammenkamp
(of Universität Bielefeld, Germany)
along similar lines was brought to our attention on 2004-06-14 by his colleague
Torsten Sillke:
Building on an article of Klamkin, Flammenkamp
investigated several approximations to the surface area of an ellipsoid around 1990,
including the following two expressions.
The first one has only a 10% accuracy,
whereas a worst relative error of 2.09% is claimed for the second formula.
S » 2p
[ ( ab + ac + bc )
- 3 abc / (a+b+c) ]
S » p
[ (1-1/Ö3)
( ab + ac + bc )
+
(1+1/Ö3)
( a2b2 +
a2c2 +
b2c2 )½ ]
Originally, one nautical mile was meant to correspond to the distance between
two points on the Ocean separated by a geocentric
angle of one minute (1°/60).
Taking into account the oblateness of the Earth, the nautical mile should be defined,
more precisely, as an "average" minute of latitude.
Following a 1929 resolution of the International Hydrographic Conference,
the international nautical mile (NM)
is now officially defined as exactly equal to 1852 m.
(This conventional definition had been legal in France since 1906.)
The so-called "British Admiralty" nautical mile also has a conventional value:
6080 ft, or exactly 1853.184 m.
If the meter was still defined in reference to the Meridian, there would be
exactly 10 000 000 meters in a quarter meridian, corresponding to a change
in latitude of 5400 minutes.
A "typical" single minute of latitude would
then be equal to 100000/54 or about 1851.852 m.
To the nearest whole number of meters,
this gives the above definition of the international nautical mile.
The nautical mile formerly used in the United States had a different definition:
It was
1/21600 of the meridian of the so-called Clark spheroid
(which is a sphere having the same surface as the Earth).
Modern values for the Reference Ellipsoid (see below)
would put the radius of Clark's spheroid at about 6371007.181 m and
the "conventional" value of the
US nautical mile at 1853.250866 m (6080.219 ft, or 6080.207 "US Survey" ft).
The true shape of the Earth (the so-called "geoid") is
defined as an equipotential surface
(i.e., a surface which is everywhere perpendicular to
the local vertical indicated by a plumb line).
Such a shape turns out to be quite complicated,
but its irregularities are best charted in comparison
with a close "regular" approximation, the "Reference Ellipsoid", whose exact
shape and size have been defined in 1979, in Canberra (IUGG 1980).
The Reference Ellipsoid is defined to have the following characteristics:
- Equatorial radius (a) exactly equal to 6378137 m.
- Oblateness f = 1 / 298.257222101 =
1 - b/a
The polar radius b = a (1- f )
is the distance of either pole to the center.
The Meridian is an ellipse of eccentricity
e = Ö(2f-f 2 ) = 0.0818191910428.
The length of the reference Meridian is thus:
40007862.91692186(12) m
The "uncertainty" stated (between parentheses) is in units of the last significant digit shown
and corresponds to an "uncertainty" of half a unit in the least significant digit of the
value for the oblateness f specified above.
That's a ludicrous precision of 120 nm.
Divide this by 21600 and you obtain:
1852.215875783419(5)m
Forsaking most of the ludicrous precision, we may state this result in terms
of the aforementioned international nautical mile (NM) henceforth equal to 1852 m.
1852.216 m = 1.000117 NM
Now, none of the above computations
yield the proper "average minute of latitude" on that spheroid.
Actually, proper averaging yields subtler and simpler results
with expressions involving only elementary functions and shunning
all the arcana related to the
perimeter of an ellipse.
Like many questions about averages,
this one can only be properly addressed if probabilistic assumptions are made explicit:
Let's do our "averaging" by giving equal weight to any
square inch on the surface of the Globe.
This is the simplest natural approach which refers only
to the spheroid's geometry, shunning ad hoc alternatives for the
real Earth (like considering only open oceans, or even weighing maritime
routes according to their estimated traffic).
Thus, a "random" point on the Earth is more likely to be at a low latitude
than a high one. Loosely speaking, there are "more points" in the vicinity
of the equator, at latitude 0°, than in the North Pole's vicinity, at
latitude +90°.
On a sphere, for any infinitesimal positive quantity
d,
a random
latitude would be between q and
q+d with probability
½ cos(q) d.
The correct "weighing" for an oblate spheroid is somewhat more complicated and
yields the following (exact) formula, in terms of the radius of the equator
(a) and the eccentricity of the meridian (e):
Average Minute of Geocentric Latitude
For the Earth, this is : 1853.256 m = 1.000678 NM
| 2p a |
|
2 [ 1 - e2 ( 1 - 2 e2 / 15) ] |
 |
|
| 21600 |
Ö(1-e2 )
(1 + (1-e2 ) atanh(e) / e )
|
|
The second factor in the right-hand side is close to unity for small values
of e, but it can become large when e approches one.
Technically, the value of an average minute of latitude is indeed infinite
for a flat disk (e = 1).
The difference with the "British Admiralty nautical mile" of 6080 ft
is less than 3 inches, and the former definition (6080.2 ft) of the US
nautical mile is only a centimeter off the mark.
These two are indeed good
approximations to the "average minute of latitude", while the more naive
"new" nautical mile of 1852 m is over a meter short!
This is not the end of the story, though.
So far, the "latitude" we spoke
about has been the "geocentric" one: the angle between the plane of the
equator and a line going through the center of the Earth.
This is not the "geodetic" latitude which is really seen by navigators who measure the angle
between the horizon (or a vertical line, which does not go through the
center of the Earth) and a fixed direction, like the celestial pole.
If the Earth was a sphere, there would not be any difference whatsoever.
Also, since there are still 21600 "geodetic" minutes of latitude in a full
meridian; this yields the same "naive" average of 1852.216 m as for the
"geocentric" minute.
Things change when we average "correctly" over the
surface of the globe: First, we observe that distance changes along a
meridian with geodetic latitude at a rate (per radian) equal to the
meridian's radius of curvature —this is one way the radius of curvature
can be defined, for any planar curve.
Thus, at a point where the
radius of curvature of the meridian is R, the geodetic minute of latitude
is equal to Rp/10800.
The average (over the ellipsoid's surface) of R is
6356828.89 m, which gives a low value (1849.12657 m) for the average
"geodetic" mile, almost 3 meters short of the international nautical mile.
The corresponding formula is very similar to the "geocentric" one:
Average Minute of Geodetic Latitude
For the Earth, this is : 1849.127 m = 0.998448 NM
| 2p a |
|
2 [ 1 - e2 ( 4/3 - 8 e2 / 15) ] |
|
|
| 21600 |
Ö(1-e2 )
(1 + (1-e2 ) atanh(e) / e )
|
|
This is smaller than the geocentric expression for roundish spheroids
but becomes higher when the polar radius is below
a/Ö6,
or about 40.825% of the equatorial radius.
The geocentric value can be as much as 14.564% higher or 33.333%
lower than the geodetic one
(respectively, for
e = Ö(9-Ö21)/Ö8)
or e near 1).
where
