1. Caracas, Marzo 2006 Algorithm for 3D-Modeling H.-J. Götze IfG, Christian-Albrechts-Universität Kiel Interpretation Interpretation

2. Gravity Modeling Caracas, Marzo 2006

3. Geological and geophysical model Caracas, Marzo 2006

4. Potential Integral Caracas, Marzo 2006 Homogenous density, only attraction – no centrifugal force with f: gravity constant 6,67*10 -11  :  = constant for polyhedron V: volume of the attracting body The direction of the coordinate axis of a rectangular system are defined by a basis of unit vectors The three components of the gravity vector with respect to this basis are given by: and the components of the three vertical second derivatives of which may be of special interest in geophysics: !

5. Caracas, Marzo 2006 Find the gravitational potential U, the gravity components and the vertical derivatives of a homogenous polyhedron with respect to a point P. Determine the following integrals: The problem will be solved by transforming the volume integrals first into surface integrals and then into line integrals.

6. Random Polyhedron Caracas, Marzo 2006

7. one has to find a vector field such that Caracas, Marzo 2006 By applying the divergence theorem: with or and However, this problem has no unique solution. Among all possible solutions a ‘simple’ one holds: with:(for the potential) with: (for the gravity vector) and (for the vertical gradient of gravity)

8. Using these functions one finds: Caracas, Marzo 2006 for the potential U: for the components of the gravity vector: and the three components of the vertical gravity gradient: where the other components can easily be written by replacing the index 3 by 2 and 1 respectively. Because the surface S is required to consist of t polygons Pt the outer normal of (t=1,…,n) each polygonal plane Pt is constant.

9. Caracas, Marzo 2006 We express the surface plane in the Hession normal form: D t : positive distance between a point P s and the plane P t SIGN(D) t : denotes the sign of D t. It holds: Hence, we get: and The problem of determining the potential U and its derivatives is reduced to evaluate two integrals of that kind: We aim to transfer these two surface integrals into four line integrals which may surround the surfaces P t. with j=1,2,3 and

10. Caracas, Marzo 2006 Transformation Procedure Let us introduce a new orthogonal coordinate system for each single surface with a base such that:, and ! Clearly the choice of depends on the derivation in the integral which should be transformed. To transform any direction in the polygon plane can be chosen for ! Firstly, and. To apply the divergence theorem (now for a plane) again, we have Now denotes the outer normal to the closed boundary line of the line polygon. Note that lies IN the polygonal surface ! To transform integrals and we have to look for a vector field such that Again, the problem has no unique solution.

11. contains of n straight linesEach polygonal surface Caracas, Marzo 2006 However, one possible solution to evaluate is: We express to be a gradient: and One gets:. Therefore we get which holds generally. Due to later usage of these formulas in an interactive computing environment we restrict n to be n=3. This is to approximate further ‘normal surfaces’ by triangles whose surface normals can be determined unique always. Thus, In each plane we have: Similar to the procedure as used in the transition from volume to surface integrals we conclude that is the distance between the Point and the straight line.

12. Caracas, Marzo 2006 Due to the chosen new coordinate system is the orthogonal projection of onto the surface. Next step is to rewrite the expression for : and therefore The transformation of the second integral is more complicated. Its derivation is not given here. As a final result we get the expression: Note, that in the final expressions for and no quantities depend on the transformed coordinate system !

13. Caracas, Marzo 2006 Evaluation of the remaining two line-integrals and are necessary to compute the potential, the gravity vector components and the vertical gravity gradient. Later, we will see, that also the components of the magnetic field of a polyhedron can be computed. No further integrals are necessary! Let the point be the orthogonal projection of point onto the line. By taking this point as the origin of a 1-dimensional ‘local’ coordinate system at we can use the distance L from it as a coordinate. Then and is: and The integrals can be solved by applying to standard tables of integrals (e.g. Gröbner and Hofreiter, 1961). One finds: and

14. Geometry for potential field calculus Caracas, Marzo 2006

15. Note that and are the special distances between the first and second end point of the line and the point (which is the site, where potential, gravity, gradient and magnetic field will be determined). Then we have: putandand write: For arctan-terms in brackets use the following indentities:

16. Caracas, Marzo 2006 Finally we find Coming to the end we receive seven equations, which contain the lowest possible number of transcendent functions: which depends on the mentioned special occasions. +++… !

17. Caracas, Marzo 2006 How to treat possible singularities? We assumed the point to be located outside the polyhedron. For inside the polyhedron the formulas must be modified by excluding a small sphere around from the volume. However, additional singularities are introduces while performing transformations based on Gauß’s divergence theorem: (1) P lies in the plane defined by. It follows that. U=0 and the -term of becomes zero. The ‘final” formulas still hold. (2) Let be located inside. Since the vector field is not defined at point, we have to exclude from the to apply the divergence theorem. One has to take a circle around and let its radius (a) tend towards zero, resulting in the. Thus, we have replaced the corresponding (3) lies at the line or at the end points of. This singularity has to be solved in a similar way as it was done in (2). However F becomes lies between the end points of. In case lies at one of the end points a special term which depends on the angle of the two sides involved modify F.

18. Orientation & closed polygon Caracas, Marzo 2006

19. Hints for programming Caracas, Marzo 2006 Hints for programming Let us refer to the final set of equations which describe the potential, the gravity vector components and the vertical gravity gradient. The following quantities have to be calculated only once: 1) For each triangle surface : -the length of each side: v = 1, 2, 3 -the unit-vectors along sides v: with -the outer unit vector of surface : with its components: -the unit vectors orthogonal to the sides of each triangle : with v = 1, 2, 3 for example: Up to here, all quantities do NOT depend on the point.

20. Hints for programming Caracas, Marzo 2006 2) For each triangle surface and for each ‘station’ (point) we have to calculate the following items: ‘orthogonal distance Which are the lower and upper integral boundaries. and are quantities which consist of an orientation! -Finally we have to evaluate and its location ‘left’ or ‘right’ hand sited of the triangle side : with v = 1, 2, 3 e.g. for we have

21. Induced and remanent magnetics Caracas, Marzo 2006 If we count the terms, we get: We obtain: if lies left of and if lies right of. ------------------------------------------------------------------------------------------------------------------------ Trying to apply these equations to magnetic modelling, we can use the quantities which describe the gradients. Also included should be the possibility to deal with remanent magnetization. Following potential field theory we can write: Which holds for the magnetic potential if the field inclines at arbitary directions (not equal 90°). When = const, the dot product and because For the magnetic field components

22. Induced and remanent magnetics Caracas, Marzo 2006 In accordance with earlier developments (gravity gradient), for the field components of of a polyhedron with planes, Because one can write with and Where RJX, RJY, RJZ are the components of induced magnetization. If remanent magnetization has to be considered, please see next chapter. The is obtained:

23. Induced and remanent magnetics Caracas, Marzo 2006 The integrals were already solved for equation () and for the components of the induced magnetic field we have: HX, HY and HZ are all expressed in terms of quantities which have been calculated already. Performing RXS, RYS and RZS with and without the presence of remanent magnetization: The total magnetization vector can be described as: with: = vector of induced magnetization = vector of remanent magnetization

24. Induced and remanent magnetics Caracas, Marzo 2006 With respect to an orthogonal coordinate system and (which is used throughout this paper), we have: and with:, susceptibility  and inducing field D = declination of field vector I = inclination of field vector In case remanent magnetization has to be considred: with: Q = Königsbergfactor DR= declination of remanent field vector IR = inclination of remanent field vector

25. Induced and remanent magnetics Caracas, Marzo 2006 By vector addition, one gets for the components of : Finally, one is able to use RJTX, RJTY and RJTZ even if no remanent field is present. In that case, set Q=0 and we have: RJTX=RJX RJTY=RJY RJTZ=RJZ