Can you? I know how to determine the chord distance between two ECEF coordinates, but can you determine the direction of the chord - or only by converting to LLH first?
Just curious..
"But can you determine the?ÿdirection of the chord"
Please explain what?ÿ you mean by "Direction of the chord"
Do you mean normal section azimuth?
Do you mean geodetic azimuth?
Do you mean ____________?
JOHN NOLTON
This paper by Rod Deakin tells us how. The formula on page 9 will do it, but it requires both cartesian and geographic coordinates.
MERIDIAN DISTANCE (mygeodesy.id.au)
The paper is 59 pages of concise material, Reading it all is not onerous and the example MATLAB programs are instructive even without knowing MATLAB coding. I think that the section on the four major ellipsoidal curves beginning on page 30 is particularly interesting.
?ÿ
The direction along the line representing the mark to mark distance between 2 points.
Wowsers..thanks.
So you do need lat/lon as well. It's more complicated than I thought. I thought maybe you could get a direction from just the ECEF coordinates.
@cf-67?ÿ
Im sure GeoDD Mike can add the world of geodesy/geophysics into the equations too, add gravity, time Domain, expansion and contraction of the blob etc etc etc...
ECEF is the idealized simple model, or so i was learning in GIS land.?ÿ I'm back into surveying. GIS is my data management plan.
Not to put too fine a point on it, the ellipsoid is the model and ECEF is a coordinate system that allows calculations on the ellipsoid surface.
If you look at the definitions of the GRS80 and the WGS84 ellipsoids, they're virtually identical. Over the years, though, point (0, 0, 0) in the WGS84 ECEF coordinate system has been moved to more closely correspond to new measurements of the earth's center of gravity. No such adjustment has been made to the GRS80 center. When the two ellipsoids are placed in the same ECEF coordinate system, their centers will not both be (0, 0, 0), and different coordinates will be displayed for the same point on the earth's surface.
It's hard to make this stuff simple.
?ÿ
A simple inverse between the two coordinates would give you the a chord and a direction dimension. It would not be the distance between the two nor the bearing (either way) between the two. I have never worked with a hybrid plane/spherical system.?ÿ
Yes, I meant simple like the sun rises in the east, and sets in the west simple. Man, is everybody drinking decaf these days or what??ÿ
😉
?ÿ
?ÿ?ÿ
The "EASY" way is to use use the NGS Tool Kit Progran inverse3d
https://geodesy.noaa.gov/cgi-bin/Inv_Fwd/invers3d.prl
I don't know the "NOT-SO EASY" method, but I'll bet that John Nolton does.
I get it, but even sunrise isn't that simple. For about half the year, it's North of East and for the other half, it's South of East.
Working on my third cup of fully-caffeinated now.
Carry on!
I guess that as the distance can be calculated by expanding Pythagoras into 3D (Sqrt(X^2+Y^2+Z^2), I hoped maybe the azimuth could be similarly divined by some simple expansion of Atan(deltaX/deltaY)
But I am a tea drinker and clearly not thinking straight. ?????ÿ
Be aware the azimuth from A to B is not the same (nor the reciprocal) as the azimuth from B to A.
Paul in PA
@mathteacher I feel this is not quite right. NAD83 which uses GRS80 for its ellipsoid is offset from the 'true' center of mass, but other datums/geodetic coordinate reference systems that use GRS80 have not done that. ITRF realizations use GRS80--they're not offset. Also, it's not when two ellipsoids are in ECEF are they offset, it's when a datum, even though it is possibly using the same ellipsoid, does not coincide before converted to another datum.
