There are six special curves connecting two given points P and Q on an ellipsoid:
A geodesic is the shortest curve inside a surface, which connects two given points. In our case the surface is always an ellipsoid of revolution approximating the figure of the earth.
In essence, there are six quantities describing the course of a geodesic:
Besides this, the following quantities have some importance:
Instead of the ellipsoidal latitudes φ for P,Q,m we may also use the reduced latitudes β or the geocentric latitudes ψ .
| P | point on ellipsoid |
| P′ | point on circumscribed sphere |
| O | geocentre |
| a | semimajor axis |
| b | semiminor axis |
| m | length of meridian arc |
| ρ | radius of parallel |
| N | radius of curvature in the prime vertical |
| φ | ellipsoidal latitude |
| ψ | geocentric latitude |
| β | reduced latitude |
| μ | rectified latitude |
Out of the 15 possible quantities, three must be given, to compute the remaining 12 quantities. The two classical primary problems of the geodesic on the ellipsoid of revolution are:
| given quantities | |
|---|---|
| 1st primary problem | φP, αP, s |
| 2nd primary problem | φP, φQ, Δλ |
But also almost all other combinations of known quantities are computed. In few cases a computation is practically impossible, either because the problem is in principle theoretically unsolvable or because a solution has not yet been implemented.
In some cases the solution is non-unique. If two solutions exist, both are computed.
If one has computed the ellipsoidal difference of longitudes Δλ , one may use the ellipsoidal longitude λP to compute λQ or vice versa. This can be computed manually by the function Create coordinate list.
Die computation is partly iterative. The extended setting ''Criterion of convergence'' controls, how many iteration steps are executed.
If both points are close to the equator or one point is near a pole, the computation may be inaccurate, while a warning is issued, or even fail. A solution is pending.
This symbol shows the rough course of the geodesic PQ:
| ↗ | from southwest to northeast | → | on the equator from west to east |
| ↙ | from northeast to southwest | ← | on the equator from east to west |
| ↘ | from northwest to southeast | ↑ | on the meridian from south to north |
| ↖ | from southeast to northwest | ↓ | on the meridian from north to south |
| ↷ | from southwest to southeast by touching the northern bounding parallel | ||
| ↶ | from southeast to southwest by touching the northern bounding parallel | ||
| ↺ | from northwest to northeast by touching the southern bounding parallel | ||
| ↻ | from northeast to northwest by touching the southern bounding parallel | ||
We consider the following points in ellipsoidal coordinates referring to :
The second primary problem should be solved on the WGS84 ellipsoid. In the following, the heights are set to zero.
Result: The length of the geodesic is s = 6848049 m. The forward azimuth at the start point is αP = 303.521454° The reverse azimuth at the end point is αQ = 45.4313890°.
Problem: Use Loxodrome from Dresden (Saxony) to Dresden (Ontario), to determine the arc length between these points on the loxodrome. (Solution: It is sLox = 7373778 m and is therefore almost 8% larger than the arc length on the geodesic.
Now the midpoint M on the geodesic is desired. For this purpose wi start at the start point P (Saxony) with the same forward azimuth αP = 303.521454°, but we go only half the distance s/2 = 6848049 m/2. This is a first primary problem.
Result: The midpoint M in called Q in this computation. The ellipsoidal latitude of M is φM = 57.7794406°The ellipsoidal longitude of M may be computed as follows: λM = 13.735186° - 52.931461° = -39.196275° = 320.803725°. The azimuth at point M looking into the direction of Ontario amounts to αM = 259.3707113°. This corresponds to a heading from southeast to northwest touching the northern bounding parallel. Point M is reached after the geodesic PM reaches the bounding latitude φm = 58.3981055° .
Zur Kontrolle gehen wir von Q (Ontario) mit dem Rückwärtsazimut To check this, we go from Q (Ontario) with reverse azimuth αQ again half the distance into the direction of P.
Result: The midpoint M is called P in this computation. The ellipsoid latitude and the azimuth at point M agree with high accuracy. The ellipsoid longitude of M is computed as follows: λM = -82.181667° + 42.985389° = -39.196278° = 320.803722°. Also this value agrees very well with the previous computation.
The point M lies southeast of Cape Farewell (southernmost piont of Greenland) near the borderline between the Labrador basin and the Irminger sea.
The problem is to generate equidistant points on the computed geodesic and to collect them in a coordinate list . The first point A of this list must not necessarily be the point P and the last point E of this list must not necessarily be the point Q. Instead of this, these points are specified by there arc lengths sA, sE along the geodesic referring to sP = 0 . For example, the point A=P would be addressed by sA = 0 and the point E=Q would be addressed by sE = s. These are also the default values, if no arc lengths are specified. For example, the midpoint of the geodesic PQ would by addressed by s/2 . A or E may well be outside PQ, but not extremely far away of P and Q.
Since up to now we only worked with ellipsoidal longitude differences Δλ , we need an absolute ellipsoidal longitude λof one point out of P or Q or A or E.
The equidistancy of the points and the belonging to the desired geodesic can be checked by means of . Side lengths and equatorial azimuths must be identical for all polyline sides. Equatorial azimuths must also be identical with the specification α0 . The equidistancy has a high, but not the highest accuracy.
Example: If you want only the points P and Q in the coordinate list, chose n = 2 as point number and leave the areas for the arc lengths blank.
The coordinate list consists of ellipsoidal latitudes and longitudes. All output of ellipsoidal heights is zero. These coordinates can be converted to other system types by .
The coordinate list may have one of the following column formats :
On the geodesic from Dresden (Saxony) to Dresden (Ontario) we compute nine equidistant points, the first point is Dresden (Saxony) and the last point is Dresden (Ontario):
The fifth point is the midpoint and is obtained by φ = 57.77944707385° λ = -39.1962115331° = 320.8037884669° . The last digits differ slightly from the midpoint previously obtained, because, as has been stated, the equidistancy is not always realised by the highest accuracy. If these points are computed by , a correct and constant equatorial azimuth of α0 = 328.3121600° is obtained, but the polyline side lengths differ by at most 7 m.
Now, we want to generate additional points between the two northernmost points, and the distance should be 1000 m. For this purpose we chose the arc lengths sA = 2568018.4 m; sE = 3424024.5 m and the point spacing Δ = 1000 m.
We obtain 856 points with a true distance between Δmin = 1001.1757 m and Δmax = 1001.1768 m. The difference to the desired value Δ = 1000 m is due to the fact that the arc length is not an integer multiple of the selected point distance, such that the point distance has been adapted slightly. The equatorial azimuths all amount to α0 = 328.3121600. The northernmost point has an ellipsoidal latitude of φm = 58.39810555 at an arc length of φm = 2684154.9. These results agree exactly with the values for the bounding parallel computed above.
If the computational task makes it likely that internal rounding errors falsify the output, a warning is issued.
If given quantities are not known errorfree, you can compute, how much these errors propagate to the computed quantities. See .
Using the computation tool also meridian arc lengths can be computed by setting Δλ = 0 . In this way for the meridian arc of the Bessel-Ellipsoid between the parallels φP = 59° and φQ = 60° you obtain the arc length s = 111390.561 m. Using the computation tool you obtain the same value, but with more digits: s = 111390.5612847 m
If reversely you want to compute from φP = 59°, φQ = 60°, s = 111390.561 m back to Δλ = 0 , you do not obtain a result. The reason is that s has been rounded down. A tiny extension to s = 111390.562 m is working and yields two solutions Δλ = ±0.00025° ≈ ±0.9''. Geometrically we intersect a parallel by an ellipsoidal circular arc, which yields two intersection points. They are about 27 m distant from the desired meridian. The reason for this big true deviation is the unfavourable error propagation of this inappropriate section.
One could have grasped this situation without knowing the true solution by propagation the error in s . In the following example we assume a maximum error of Δs = 0.001 .
This yields a deviation in Δλ of 0.00011° ≈ 0.4''.
Problem: If you also specify error values for φP and φQ , you do not obtain a result. Please find out, why.
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