Caracas, Marzo 2006

Algorithm for 3D-ModelingAlgorithm for 3D-Modeling

H.-J. GötzeIfG, Christian-Albrechts-Universität Kiel

• InterpretationInterpretation

Gravity ModelingGravity Modeling

Caracas, Marzo 2006

Geological and geophysical modelGeological and geophysical model

Caracas, Marzo 2006

Potential IntegralPotential Integral

Caracas, Marzo 2006

Homogenous density, only attraction – no centrifugal force

Rrr '

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

zyxeee,,

V

dVrr

fU '

1

The three components of the gravity vector with respect to this basis are given by:T

z

U

y

U

x

Ug

,,

and the components of the three vertical second derivatives of g

z

g

y

g

x

gzzz

,, which may be of special interest in geophysics:

zz

U

z

g

yz

U

y

g

xz

U

x

gzzz

222

;; !

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:

Vx

Vz

Vy

Vx

V

dVrrxz

fVG

dVrrz

fg

dVrry

fg

dVrrx

fg

dVrr

fU

'

1

'

1

'

1

'

1

'

1

2

The problem will be solved by transforming the volume integrals first into surface integrals and then into line integrals.

Random PolyhedronRandom Polyhedron

Caracas, Marzo 2006

one has to find a vector field such that

Caracas, Marzo 2006

By applying the divergence theorem: S

dSnUdVudiv

with

normal surface

function vector

:

:

n

u

u

integral-potential thefor rr

udiv

'

1or

integralgravity thefor

rrxudiv

i

'

1and

gradient verticalthefor

rrxzudiv

i

'

1

However, this problem has no unique solution. Among all possible solutions a ‘simple’ one holds:

2

1

2

1321

Rgradxxx

Ru

T

Ru

1

Rx

ui

1

with: R

udiv1

(for the potential)

with:

with:

Rx

udivi

1

Rxz

udivi

12

(for the gravity vector) and

(for the vertical gradient of gravity)

Using these functions one finds:

Caracas, Marzo 2006

u

for the potential U:

SdenxenxenxR

fU

SSSS

332211

,cos,cos,cos1

2

for the components of the gravity vector:

dSenR

fUS

iSi

,cos1

and the three components of the vertical gravity gradient:

S

S

j

jdSen

RxfU

33,cos

1

where the other components can easily be written by replacing the index 3 by 2 and 1 respectively.

jjUU

12 and

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.

tn

Caracas, Marzo 2006

We express the surface plane in the Hession normal form:

332211

,cos,cos,cos enxenxenxDDSIGNttttt

Dt: positive distance between a point Ps and the plane Pt

SIGN(D)t: denotes the sign of Dt. It holds:

Stt PnDSIGN

containnot does

contains which space-half the to points if

1

1

Hence, we get:

m

tt

Ptt

dPR

DDSIGNK

Ut

1

1

2

t

P

m

titi

dPR

enKUt

1,cos

1

dPRx

enKUtP j

m

ttj

1,cos

133

and

The problem of determining the potential U and its derivatives is reduced to evaluate two integrals of that kind:

tP

t

t R

dPi

tP j

tjdP

Rxii

t

1

We aim to transfer these two surface integrals into four line integrals which may surround thesurfaces Pt.

with j=1,2,3and

Caracas, Marzo 2006

Transformation ProcedureLet us introduce a new orthogonal coordinate system for each single surface with a base such that:

tP

'ie

,tPne '

3

t

t

Pj

Pj

ne

nee

'

2 321' eee

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 !

je

jx

tjii

ti t

P '2e

Firstly, and .

2

2

2

1'' xxr 2

1221

rDR t

To apply the divergence theorem (now for a plane) again, we have

t t

t

P LLtdLnudPudiv

Now denotes the outer normal to the closed boundary line of the line polygon . Note that lies IN the polygonal surface !

tLn

tL

tLn

tP

To transform integrals and we have to look for a vector field such that ti

tjii u

R

udiv1

Again, the problem has no unique solution.

contains of n straight linesEach polygonal surface

Caracas, Marzo 2006

However, one possible solution to evaluate is:ti Txx

r

Ru

212','

We express to be a gradient:u

t

tt

DR

DRDRgradu ln

2

and TtT

xxRr

Dxx

Ru

212

2

21','','

1

One gets: dLnxxRr

DdLnxxR

it

tt

t L

T

LLtL

T

t

212

2

21','

1','

1

tP

tL. Therefore we get

n

vtvtii

1which

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. n

3

1vtvtii Thus, dLnxx

RrDdLnxx

Ri

tv

T

LLttv

T

tv

ttv

212

2

21','

1','

1

In each plane we have: tP

221121',cos'',cos'',' enxenxnxx

tvtvtv

T

Similar to the procedure as used in the transition from volume to surface integrals we conclude that

2211',cos'',cos' enxenxddsign

tvtvtvtv

is the distance between the Point and the straight line . tvd *

SP

tvL

Caracas, Marzo 2006

Due to the chosen new coordinate system is the orthogonal projection of onto the surface .

i

e '

*SP

SP

tP

.*1

1Stv Ppoint the

containnot does

contains which plane-half to points n if is

tvdsign

Next step is to rewrite the expression for :tvi

tv

Ltvtvttv

Ltvtvtv

dLRr

ddsignDdLR

ddsignitvtv

2

2 11

and therefore

tv

Ltvtvttv

Ltv

m

ttvt

dLRr

ddsignDdLR

ddsignitvtv

2

2

1

11

The transformation of the second integral is more complicated. Its derivation is not given here. As a final result we get the expression:

tjii

tv

Ltv

vtvjttttv

v Ljtvtj

dLRr

ddsignenDDsigndLR

eniitvtv

2

3

1

3

1

1,cos

1,cos

Note, that in the final expressions for and no quantities depend on the transformed coordinate system !

ti

tjii

j

e '

Caracas, Marzo 2006

Evaluation of the remaining two line-integrals

tvL

tvdL

RA

tv

1

and tvL

tvdL

RrB

tv

2

1 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:

**SP *

SP

tvL

tvL

tvA

tvB

dLLdDAtv

tv

BN

ANtvttv

21222

dLLdDLdBtv

tv

BN

ANtvttvtv

2122222

and

The integrals can be solved by applying to standard tables of integrals (e.g. Gröbner and Hofreiter, 1961). One finds:

tv

tv

BN

ANtvttv

LdDsA 21222ln and

tv

tv

BN

ANtvttv

t

tvttv

LdDd

LDdDB

21222

1arctan

Geometry for potential field calculusGeometry for potential field calculus

Caracas, Marzo 2006

Caracas, Marzo 2006

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).

tvRR1

tvRR2

tvL

SP

Then we have: tv

tvtv

tvtv

tvtvtvtvtvLN

RRAN

RRBNRRANRRBNA

1

2ln1ln2ln

tvtv

tvt

tvtv

tvt

tvttv RRd

AND

RRd

BNDdDB

1arctan

2arctan

1

XRRd

BND

tvtv

tvt 2

YRRd

AND

tvtv

tvt 1

put and and write:

YXdDBtvttv

arctanarctan1

For arctan-terms in brackets use the following indentities:

XY

YXYXARC

tv

1

arctanarctanarctan 1XY if

XY

YXARC

tv 1arctan 01 XXY and if

XY

YXARC

tv 1arctan

01 XXY and if

Caracas, Marzo 2006

Finally we find

XY

YXdDB

tvttv 1arctan

1

Coming to the end we receive seven equations, which contain the lowest possible number

of transcendent functions:

which depends on the mentioned special occasions.

3

1

3

112 v vtvtvttvtvtvt

m

tt ARCdsignDLNddsignDDsign

fU

3

1

3

11

,cosv v

tvtvttvtvtv

m

titi

ARCdsignDLNddsignenfU

3

1

3

1133

,cos,cos,cosv v

tvtvjtttvjtv

m

ttj

ARCdsignenDsignLNenenfU

+++…

2

2s

s

m

U

gN !

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:

SP

SP

SP

(1) P lies in the plane defined by . It follows that . U=0 and the -term of becomes zero. The ‘final” formulas still hold.

tP 0

tD

tvARC

iU

(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

*SP

tP Txx

Rr 212','

1*

SP

*SP

tP

*SP Fa

a

2lim0

2','1

212 dsnxx

Rr t

T

Lt

(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.

*SP

tL

tL

*25,0SP if

tL *

SP

Orientation & closed polygonOrientation & closed polygon

Caracas, Marzo 2006

Hints for programmingHints for programming

Caracas, Marzo 2006

Hints for programmingLet 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 :tP

-the length of each side: v = 1, 2, 3 vvv VVd

11

-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 .

v

vv

vv d

VVE

11

,1

14

VV

tP

1,2,1

1,2,1

vvvv

vvvv

tEE

EEn

321

,cos,,cos,,cos enenenttt

tP

tvvvv nEES

,1,1

321

21

,cos,cos,cos1

12

1

12

1

12

enenend

zz

d

yy

d

xxKJI

ES

ttt

rP

Hints for programmingHints 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

tP rP

33

22

11

,cos

,cos

,cos

3

2

1

1

Px

Px

Px

en

en

en

rPVnD

t

t

t

trt

33,cos22,cos11,cos321

PxenPxenPxenttt

rPVErAvvvtv

,1

vtvtv

drANrB 1

tvAN

tvBN

tvd

vL

rPVESrdvvvtv

,1

rd

21

JenATenCTLenCTenBTdtttt

32112122112121

,cos,cos,cos,cos(

PVKenBTenATtt

1121221

),cos,cos

Induced and remanent magneticsInduced 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

pzzpyypxxd 13121121

1tvdsign

1tvdsign

*rP

*rP tv

L

tvL

ijU

Sn

Vn

sdIR

dvIR

drR

IPV 111

nI

0nI

R

dsI

RPV

n

1

dsIsdInn

R

dsIPVPHnQP

Induced and remanent magneticsInduced 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:

T

PHzHyHxPH ,,

Sjj

m

jjn

m

j SjjjnPoly ds

RIds

RIPH

1111

dsIsdInn

RJZ

RJY

RJX

In

3

2

1

,cos

,cos

,cos

cos

cos

cos

en

en

en

s

t

t

t

3

2

1

,cos

,cos

,cos

enRJZ

enRJY

enRJX

sdI

t

t

t

n

PolyPH

m

ttttPoly dsen

RzRJZdsen

RyRJYdsen

RxRJXPH

1321

,cos1

,cos1

,cos1

Induced and remanent magneticsInduced 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

Fl

tt

vtvt

vttvtvtPoly ARCdsignenDsignLNenenRJXPHX

1

3

11

3

111

,cos,cos,cos

Fl

ttv

vtvt

vttvtvtPoly ARCdsignenDsignLNenenRJYPHY

1

3

12

3

122

,cos,cos,cos

Fl

ttv

vtvt

vttvtvtPoly ARCdsignenDsignLNenenRJZPHZ

1

3

13

3

133

,cos,cos,cos

Ti

REMINDTiii

INDi

REMi

Induced and remanent magneticsInduced 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 vectorI = inclination of field vector

In case remanent magnetization has to be considred:

with: Q = Königsbergfactor DR= declination of remanent field vectorIR = inclination of remanent field vector

yexe

21, ze

3

kRJZjRJYlRJXiIND

kRRJZjRRJYlRRJXiREM

IiRJZ

IDiRJY

IDiRJX

IND

IND

IND

sin

cossin

coscos

FiIND

F

F

F

RINDRREM

RRINDRRREM

RRINDRRREM

IiQIiRRJX

IDiQIDiRRJY

IDiQIDiRRJX

sinsin

cossincossin

coscoscoscos

FQiQiINDREM

Induced and remanent magneticsInduced 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=RJXRJTY=RJYRJTZ=RJZ

REMINDTiii

RRINDIND

RRRRINDIND

RRRRINDIND

IQIFIiQIiRJTZ

IDQIDFIDiQIDiRJTY

IDQIDFIDiQIDiRJTX

sinsinsinsin

cossincossincossincossin

coscoscoscoscoscoscoscos