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Subject: Meridional parts
From: Aubrey O'Callaghan (ocallaghan@XXX.XXX)
Date: Sun Aug 12 2001 - 17:54:46 EDT
I have been doing some simple research on meridional parts as related to
rhumb line naviagation.
As we know the meridional parts helps us to get a better course than the
other plane sailing methods.
There are a couple of interesting Javascript programs available on the web
to make the computation. I have compared these with a simple program I
wrote using HP49G
http://www.info.gov.hk/mardep/javascpt/mp.htm
http://pollux.nss.nima.mil/calc/parts.html
The first site gives a result which compares well with the meridional parts
in Nories Nautical Tables - the computations here use Clarkes projection
1880 and my calculator.
The second site does not give the same results. However if I input the
ellipicity or compression - C, (C=(A-B)/A, where A is major axis and B is
minor axis) of 1/298.2572 of the WGS 84 spheroid (Clarke spheroid
1/293.465) then I get pretty much the same answers. It would be nice if the
authors advised which spheroid is being used. As most charts today use the
WGS 84 datum, perhaps this is what we should be using for navigation ?
The meridional parts does not help us for computing distances. We have to
resort to distance= d.lat x sec.Course.
However, owing to the non- spherical shape of the earth the linear value
of a nautical mile varies from 1843 m at the equator to 1862 m at the
poles. (A nautical mile's definition is the length of an arc subtending
1'). To overcome this inconvenience of having a varying unit of distance a
standard nautical mile of 1852 m was adopted. Therefore the equation of
distance shown above is somewhat deficient.
Apparently a Mr. D.H. Sadler presented a paper to the Royal Institute of
Navigation in 1956 (I have not been able to find a clear reference to it on
the RIN Journal web page). This calculated the linear distance of a
parallel of latitude from the equator in "standard nautical miles". He
computes a fn. L(phi) which is the distance in standard nautical miles from
the equator.
We can then compute d.L(phi) and feed this into our distance equation
distance = d.L(phi) x sec.Course.
The equation that I have gleaned, with no derivation, from some old
mercator sailing notes is:
L(phi) = integral [limits: 0, phi] a(1-e^2) d(phi)/(1 - e^2.(sin(phi))^2)^3/2
I have plugged in the values of a (major axis) and e (eccentricity)
(sqrt[(a^2-b^2)/a^2]) and run the integration on my HP, but do not get the
same results as are tabulated in my notes for the Clarke Spheroid .
Does anybody know of Sadler's work and might be able to comment on it ? I
wonder if it is a mathematical construction to find the length along an
ellipic curve ?
Aubrey.
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