Algorithm for 3D-Modeling
description
Transcript of Algorithm for 3D-Modeling
Caracas, Marzo 2006
Algorithm for 3D-ModelingAlgorithm for 3D-Modeling
H.-J. GötzeIfG, Christian-Albrechts-Universität Kiel
• InterpretationInterpretation
Gravity ModelingGravity Modeling
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Geological and geophysical modelGeological and geophysical model
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Potential IntegralPotential Integral
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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
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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
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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
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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
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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
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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 '
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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
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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
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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