Earth section paths


Earth section paths are paths on the earth defined by the intersection of a reference ellipsoid and a plane. Common examples of earth sections include the great ellipse and normal sections. This page provides a unifying approach to all earth sections and their associated geodetic problems.

The Indirect Problem

The indirect problem for earth sections is: given two points, and on the surface of the reference ellipsoid, find the length,, of the short arc of a spheroid section from to and also find the departure and arrival azimuths of that curve, and. Let have geodetic latitude and longitude . This problem is best solved using analytic geometry in ECEF coordinates.
Let and be the ECEF coordinates of the two points, computed using the geodetic to ECEF transformations discussed here.

Section plane

To define the section plane select any third point not on the line from to. Choosing to be on the surface normal at will define the normal section at. If is the origin then the earth section is the great ellipse.. Since there are infinitely many choices for, the above problem is really a class of problems, where is the unit vector in the direction of. The orientation convention used here is that points to the left of the path. If this is not the case then redefine = -. Finally, the parameter d for the plane may be computed using the dot product of with a vector from the origin to any point on the plane, such as, i.e. d =. The equation of the plane is thus ⋅ = d, where is the position vector of.

Azimuth

Examination of the ENU to ECEF transformation reveals that the ECEF coordinates of a unit vector pointing east at any point on the ellipsoid is: =, a unit vector pointing north is =, and a unit vector pointing up is =. A vector tangent to the path is:
so the east component of is, and the north component is. Therefore, the azimuth may be obtained from a two-argument arctangent function, =. Use this method at both and to get and.

Section Ellipse

The intersection of a plane and ellipsoid is an ellipse. Therefore, the arc length,, on the section path from to is an elliptic integral that may be computed to any desired accuracy using a truncated series. Before this can be done the ellipse must be defined and the limits of integration computed.
Let the ellipsoid given by, and let.
If p=0 then the section is a horizontal circle of radius, which has no solution if.
If p>0 then Gilbertson showed that the ECEF coordinates of the center of the ellipse is, where,
the semi-major axis is, in the direction, and
the semi-minor axis is, in the direction, which has no solution if.

Arc Length

The polar form relative to center for the equation of an ellipse is, where, relates to the ellipse eccentricity, not the spheroid eccentricity. Let P be a point on the ellipse and, then the vector from to has components. Using an argument similar to the one for azimuth above, let, then, and, and. In this way we obtain the central angles and corresponding to and respectively. Care must be taken to ensure that ≤ ≤. Then the arc length along the ellipse is given by = Substituting above into this formula, performing the indicated operations, using one more term than Gilbertson's expression and regrouping, results in
, where
Alternatively, expansions for the Meridian arc may be used here by replacing the spheroid eccentricity with the section ellipse eccentricity.

The Direct Problem

The direct problem is given, the distance,, and departure azimuth,, find and the arrival azimuth,.

Section plane

Construct the tangent vector at,, where and are unit vectors pointing north and east at. Pick a vector, , to define the section plane, paying attention to orientation. Observe that must not be in span, together with defines the plane.

Locate {P_2}

This is a 2-d problem in span, which will be solved with the help of the arc length formula above. The basic approach is to use Newton-Raphson iteration to arrive at. The basis of the estimate is that the position vector of any point on the section ellipse may be expressed in terms of the position vector of the center and the central angle as
To get an initial estimate of, let, =Central_Angle,
,.
Now initialize =, and iterate the following steps:
exit when
No more than three iterations are usually necessary, although nearly antipodal cases can be problematic.
Finally, let, and = ECEF_to_Geo using Bowring's 1985 algorithm, or the algorithm here.
Alternatively, inversion of the arc length series may be used to avoid iterations.

Azimuth

Azimuth may be obtained by the same method as the indirect problem: =, where the subscript 2 indicates evaluation of the associated quantity at.

Examples

The great ellipse

Let be the origin, so that = the position vector of. The above approach provides an alternative to that of others, such as Bowring.

Normal sections

The normal section at is determined by letting = . The above approach provides an alternative to that of others, such as Bowring.

The mean normal section

The mean normal section from to is determined by letting = . This is a good approximation to the geodesic from to for aviation or sailing.

A class of sections

A class of sections may be imagined by rotating about the chord connecting and All of these may be solved with the single approach above.

Intersections

Let two section planes be given: ⋅ =, and ⋅ =. Assuming that the two planes are not parallel, the line of intersection is on both planes. Hence orthogonal to both normals, i.e. in the direction of.
Since and are not colinear,, is a basis for. Therefore, there exist constants and such that the line of intersection of the 2 planes is given by = + + t, where t is an independent parameter.
Since this line is on both section planes, it satisfies both:
+ =, and
+ =.
Solving these equations for and gives
[1 -, and
[1 -.
Define the "dihedral angle",, by = ·.
Then = , and =.
On the intersection line we have = + t, where = +.
Hence: = + t, = + t, and = + t, where
= +, = +, and = +.
and =, for i=1,2,3.
To find the intersection of this line with the earth, plug the line equations into, to get
, where =,
=,
=.
Therefore, the line intersects the earth at. If, then there is no intersection. If, then the line is tangent to the earth at .
Observe that since and are not colinear. Plugging t into
= + t, gives the points of intersection of the earth sections.

Examples

Maximum or Minimum Latitude

on an earth section path may be found by dropping the subscripts on the given section;, , and setting, so that. Then solve for such that.
Since, and, we must have. Plugging t into =, gives the points of intersection of the earth sections. Alternatively, just set.

Maximum or Minimum Longitude

on an earth section path may be found by dropping the subscripts on the given section;, , and setting, where is the longitude to be solved for such that.
Alternatively, just set.