Long Line Inverse Position

ok, going with some presumptions:
i used a distance between two local CORS stations, USNO to GODE. computed in 'cg adjust' by charles ghilani. not sure if it is a 'long enough' line but this is output, meters, GRS80. my guess is the two different distances are arc and chord

Adjust for Windows
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DATUM: GRS80
First Station :
---------------
X = 1,112,190.4536 LAT = 38°55'08.23647" North
Y = -4,842,956.4805 LON = 77°03'58.39613" West
Z = 3,985,352.3622 EHT = 50.1511

Second Station :
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X = 1,130,774.4394 LAT = 39°01'18.18974" North
Y = -4,831,255.0703 LON = 76°49'36.57468" West
Z = 3,994,200.5587 EHT = 15.8450

Delta height = -34.3061
Ellipsoidal distance = 23,676.3971
Mark-to-Mark distance = 23,676.5308
Forward azimuth = 61°07'07.0"
Back azimuth = 241°16'09.0"

DX = 18,583.9858 DN = 11,435.6562
DY = 11,701.4102 DE = 20,731.5643
DZ = 8,848.1965 DU = -78.2343

Linear units are meters.

maybe i will compute something longer next...stay tuned

ok yellowknife5, alaska to richmond5, virginia:

DATUM: GRS80
First Station :
---------------
X = -1,224,451.7436 LAT = 62°28'51.18402" North
Y = -2,689,218.4030 LON = 114°28'50.40162" West
Z = 5,633,639.2316 EHT = 182.5234

Second Station :
----------------
X = 961,334.8644 LAT = 25°36'49.60933" North
Y = -5,674,076.9827 LON = 80°23'02.16425" West
Z = 2,740,536.6088 EHT = -12.4640

Delta height = -194.9874
Ellipsoidal distance = 4,809,797.1959
Mark-to-Mark distance = 4,696,497.2716
Forward azimuth = 132°22'28.5"
Back azimuth = 337°42'34.9"

DX = 2,185,786.6080 DN = -2,942,629.9727
DY = -2,984,858.5797 DE = 3,226,171.0343
DZ = -2,893,102.6228 DU = -1,729,114.2019

Linear units are meters.

An ellipsoidal distance is a distance along the surface of the ellipsoid, as if one laid a string on the surface of a ball.

A mark-to-mark distance is a straight line from point to point, cutting through the surface of the ellipsoid.

Think of a balloon. You could measure from A to B on the outside of the balloon, that would be like the ellipsoidal distance. It would not be possible to measure the mark to mark distance because it would puncture the balloon.

Of course, for nearby points the difference is very small.

At the risk of being cynical, "no job is difficult if you have the right tools . . . and use them properly."

My suggestion has already been implemented by several in the thread - perform the computations using ECEF coordinates. The mark-to-mark distance is obtained by the 3-D Pythagorean Theorem - square root sum of the squares of the components.

As Hamilton points out, we also need to know for sure which distance it is we want. The options, summarized from the GSDM book include:

1. Mark-to-mark (as above).
2. Mark-to-mark on the ellipsoid (compute X/Y/Z for each latitude/longitude position but hold ellipsoid height of each point to zero.
3. On the ellipsoid, mark-to-mark is identical to chord distance.
4. On a sphere, arc is computed using L=R*theta where R is the mean radius of curvature for the line and theta is computed from chord and R. For lines of intermediate length, ellipsoid arc and circular arc are nearly identical - say up to 20 kilometers.
5. To compute a really long geodetic distance on the ellipsoid precisely, look up and use the geodetic inverse equations used by physists and geodesists.

But, try it out. It is incredible what can be done using the solid geometry equations of the ECEF (as described and used in the GSDM).

Philosophical observation - traditional geodetic computations are done on the ellipsoid. A new approach is to perform the computations in 3-D space - using solid geometry equations and the ECEF values.

Source of Long Line Inverse Position form

It's from Spheroidal Geodesics, Reference Systems, and Local Geometry, published by US Naval Oceanographic Office and available in PDF from DTIC here. There's a note that it's the "best available copy." (which means it's fuzzy and unclear in some places, but is overall readable)

Long Line Inverse Position ?

Unfortunately my Geodesy text is not at hand.

The bearing from A to B is not equal to the bearing from B to A, that much I recall for sure. You imagine those geodesic lines to be "S" curves. The reason being that for points far apart we must account for the changes in curvature along the surface of an ellipsoid of revolution.

You are only on the ellipsoid for points due North or South.

I am unsure but I believe one must also measure the distance along the "S" curve.

Then there is the spherical trigonometry to consider and the spherical excess in figures. Consider that on a perfectly spherical earth one can lay out a triangle with 3 90° angles, any sum from 180° to 270°. Then complicate that with a perfect ellipse of revolution and we are only getting close to the correct answer.

Paul in PA

Long Line Inverse Position ?

Unfortunately my Geodesy text is not at hand. Found the text, Wolfgang Torge, "Geodesy" and corrected some of my terms. I have not checked out your computations but believe you have two meaurements along the two normal sections. The length along the geodesic would be shorter than both. When my edit time expired while I was editing.

The bearing from A to B is not equal to the bearing from B to A. There are two normal sections A-B & B-A. You imagine the geodesic line to be a "S" curve between the two. The reason being that for points far apart we must account for the changes in curvature along the surface of a rotational ellipsoid.

You are actually only on the ellipsoid for points due North or South.

The distance along the geodesic "S" curve is the shortest distance between A & B.

Then there is the spherical trigonometry to consider and the spherical excess in figures. Consider that on a perfectly spherical earth one can lay out a triangle with 3 90° angles, any sum from 180° to 270°. Then complicate that with a perfect ellipse of rotation and we are only getting close to the correct answer.

Paul in PA

This stuff makes a lot more sense if you have a small globe and a big rubber band to play with while you ponder.

Arrange the rubber band to start from a particular location and head in a particular direction or to another particular location some distance away. Then make it go straight around the globe in the great circle. Compare that to the lines of latitude and longitude.

All of the sudden, you see why planes fly near Alaska to get from the US to Japan, and other such things, and (much exaggerated) why lines of constant bearing aren't straight.

Source of Long Line Inverse Position form

Thank you very much! That's the book!

It's not easy reading, to say the least. I still haven't figured out why the two distances.