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Aubrey,

Try "Geometry of Navigation" by Roy Williams. Section 2.10 is titled
"Distance along the Arc of a Meridian on the Surface of an ellipsoid".
Equation 2.18 is the L(phi) you show below and derivation of that equation
is shown in that section. Williams calls that function Latitude Part similar
to the Meridional Parts. William's book also gives the derivation and
computation of the Meridional Parts. I have implemented the algorithm using
a spread sheet and generated a table of Meridional parts as well as Latitude
Parts.

Going back to your question, the function L(phi) is not the equation for a
general elliptic curve. It is strictly along a meridian.

Sam
----- Original Message -----
From  "Aubrey O'Callaghan" 
To: 
Sent: Sunday, August 12, 2001 2:54 PM
Subject: Meridional parts


> 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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