
Latitudes again!
Scott Zelenak posted a problem not long ago asking “What latitude is half the distance from the equator to the pole”
I would like to give some information to this question and answer Bill93 question “How hard is the math to calculate
this without using those tools”.First you must know the ellipsoid to calculate the problem. This was NOT given; Bill93 used GRS80 and then Zelenak used
Clarke 1866 Ellipsoid.Second you must know the quadrant distance, which of course is different for each ellipsoid. No one asked how to calculate this distance.
I gave the distance by calculating it in Mathematica V8.0 as I posted. Most people don’t have this program so I will show how to get this
by other means (and not using NGS tools).A very brilliant man from India, Srinivasa Ramanujan in 1914 gave a very simple approximate equation for the perimeter of an ellipse.
p~ pi(a+b)[1+3h/(10+sqrt(43h))]
here pi is of course 3.14159265358979………..
a= radius of major axis of ellipsoid
b= radius of minor axis of ellipsoid
h=(ab)^2/(a+b)^2
sqrt= square rootNow to get maximum accuracy you must carry out all calculations to 15 or more places(pi, a, b,). Easy to find these numbers so I will not post them. See a paper by A. Weintrit “So, What is Actually the Distance from the Equator to the Pole?–Overview of the Meridian Distance
Approximations.Oh I forgot to tell you to divide the perimeter by 4 for the quadrant( just incase somebody might say something).
OK now we have the quadrant distance and we divide it in half for Zelenak problem.
Now in the old days we had tables that were computed for the different ellipsoids. The Army used 5 ellipsoids , International,Bessel,
Everest, Clarke 1880 and the Clarke 1866. Tables were made and you could open the table to the page that came close to your distance.
They were calculated to every minute and also gave the difference per second. SO Bill93 it was very easy to find the Latitude given any distance
from the equator.Now in modern times with NGS Geodetic Tools you can do as Bill93 did. Get the ellipsoid, do an inverse between zero Lat. and 90 degree Lat.
This gives the quadrant distance. Divide it in half for Zelenak problem. Then do a forward solution from zero Lat , azimuth zero, and the 1/2 quadrant distance an you have your answer. I do not know why Zelenak said he had an answer “after three iterations”If one looks in most books on map projections you will find that they develop series for this type of problem.
NGS Special Pub. #241 Natural Tables for the Computation of Geodetic Positions, Clarke Spheroid 1866 has the following.
(here for the Clarke ellipsoid and Zelenak problem we want a distance of 5 000 944.02149+) The table only gives you distance to 0.001 m.
We see on page 46 for Lat. of 45d 08m a Meridional arcs(meters) of 4 999 544.727 and for 45d 09m 5 001 396.955 and a 1 second diff. of
30.870467 meters. By interpolation ones gets the answer of 45d 08m 45.32795000944.021494999544.727 (the table value) = 1399.29449
1399.29449/30.870467 (table value Diff. 1 sec) = 45.32793 seconds , add this to the entry value of 45d 08m and you get a Lat. of
45d 08m 45.32793 seconds which is the same in NGS Tools and the answer to the problem for Clarke 1866 ellipsoid.JOHN NOLTON
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