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An Optimization Technique for EstimatingVelocities and Fracture Orientation in
Orthorombic Media
by
Debora Cores
José G. Meza
Universidad Simón Boĺıvar
ISPM2000 - ATLANTA
August 2000
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An Optimization Technique for EstimatingVelocities and Fracture Orientation in
Orthorombic Media
by
Debora Cores
José G. Meza
Universidad Simón Boĺıvar
III Jornadas IBO
Intevep, Noviembre 2000
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The Influence of the Seismic ParameterAcquisition on the Optimization Problem for
Estimating Velocities and Fracture Orientation
by
Debora Cores
José G. Meza
Universidad Simón Boĺıvar
Optimization 2001
Aveiro-Portugal, July 2001
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OUTLINE
An orthorombic media (OM)
The Reflection Tomography Problem in OM
Historical Overview
Discretized Problem
Numerical Approach
Numerical Results
Conclusions
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Anisotropy
The velocity does not changes
with the wave propagation direc-
tion.
V
1
6 =
V
2
The velocity changes with the wave propa-
gation direction.
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Anisotropy: An Orthorombic Media (OM)
An orthorombic media is ananisotropic stratified medium with
vertical fractures.
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PROBLEM: Reflection Tomography (SRT) Problem in OM
M i n i m i z e
1
2
k T
r
; T (
V )
k
2
2
T : I R
m
! I R
n
travel time function, T = (
T
1
(
V ) T
2
(
V ) : : : T
n
(
V ) )
where,
T
i
( V ) =
Z
R a y
i
1
V ( x y z )
d l
i
T
r
2 I R
n
real travel time
vector.
V 2 I R
m
is the velocity
vector in OM.
n is the number of layers.
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Historical Overview
ISOTROPIC MEDIA :
Normal Equations:
Gauss Seidel with Successive Over-
relaxation: T. Bishop et al, 1985
Levenberg and Marquardt Method
with SVD descomposition : Lines and
Treitel, 1985: S. Chiu el al, 1986; T.
Zhu and L. Brown, 1987; Farra and
Madariaga, 1998.
Low Storage Opt. Techniques:
Spectral Gradient Method: Castillo,
Cores and Raydan , 2000.
ANISOTROPIC MEDIA:
In 2D elliptical anisotropic medium
Michelena et al. 1994.
In 2D medium with a sinusoidal aproxi-
mation of the velocity: Toshiki et al 1995.
In a 3D Transversally anisotropic
medium : Grechka, 1995.
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DISCRETIZED PROBLEM: Ellipsoidal Aproximation
Contreras et al, 1997
1
V
2
j
=
1
V
2
z
j
c o s
2
(
1
) +
1
V
2
x
j
c o s
2
(
2
) s i n
2
(
1
) +
1
V
2
y
j
s i n
2
(
2
) s i n
2
(
1
)
where j = P S V S H
correspond to the different
wave propagation modes.
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DISCRETIZED PROBLEM: Travel tiem Function
The travel time function for a ray corresponding to the ( i j
) pair reflecting in
the layer k ,
T
i j k
( X Y ) =
P
2 k + 1
h = 2
r
( x
i j k
h
; x
i j k
h ; 1
)
2
v
2
x
h
+
( y
i j k
h
; y
i j k
h ; 1
)
2
v
2
y
h
+
( z
i j k
h
; z
i j k
h ; 1
)
2
v
2
z
h
X = (
x
1
x
2
: : : x
2 n + 1
)
Y = (
y
1
y
2
: : : y
2 n + 1
)
Z = (
z
1
z
2
: : : z
2 n + 1
)
V
x
= (
v
x
1
v
x
2
: : : v
x
2 n + 1
)
V
y
= (
v
y
1
v
y
2
: : : v
y
2 n + 1
)
V
x
= (
v
z
1
v
z
2
: : : v
z
2 n + 1
)
z
i
= f
i
( x
i
y
i
) :
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DISCRETIZED PROBLEM
Consider any symmetry axes, (Group angle 6
=
Ray angle)
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DISCRETIZED PROBLEM
Azimutal Rotation
0
@
c o s ( ) s i n ( ) 0
; s i n ( ) c o s ( ) 0
0 0 1
1
A
Polar Rotation
0
@
c o s ( ) 0 ; s i n ( )
0 1 0
s i n ( ) 0 c o s ( )
1
A
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GENERAL DISCRETIZED PROBLEM
T
i j k
( S ) =
k + 1
X
h = 2
v
u
u
t
i j k
x
h
v
x
h
!
2
+
i j k
y
h
v
y
h
!
2
+
i j k
z
h
v
z
h
!
2
+
2 n + 1
X
h = 2 n + 2 ; k
v
u
u
t
i j k
x
h
v
x
h
!
2
+
i j k
y
h
v
y
h
!
2
+
i j k
z
h
v
z
h
!
2
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GENERAL DISCRETIZED PROBLEM
i j k
x
h
= D
x
h
c o s (
h
) c o s (
h
) + D
y
h
c o s (
h
) s i n (
h
) ;
D
z
h
s i n (
h
)
i j k
y
h
= ;
D
x
h
s i n (
h
) + D
y
h
c o s (
h
)
i j k
z
h
= D
x
h
s i n (
h
) c o s (
h
) ;
D
y
h
s i n (
h
) s i n (
h
) + D
z
h
c o s (
h
)
D
x
h
= x
i j k
h
;
x
i j k
h ; 1
D
y
h
= y
i j k
h
;
y
i j k
h ; 1
D
z
h
= z
i j k
h
;
z
i j k
h ; 1
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NUMERICAL APPROACH
An optimization scheme for solving
M i n i m i z e k T
r
; T ( S ) k
2
s : t : L S
U
that satisfies the following conditions:
Only function and gradient evaluations (first order information) arerequired
r
f ( S ) =
J
T
f
( S ) (
T ( S ) ;
T
r
)
Global convergence
Fast local convergence
Low computational cost and storage
Box constraints
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NUMERICAL APROACH
Low cost and storage orthorombic ray tracing algorithm (Loreto and
Cores, 1998-1999)
M i n i m i z e T (
X Y Z )
The Spectral Proyected Gradient Method (SPG) (Birgin, Martinez and
Raydan, 1999) to solve,
M i n i m i z e k T
r
;
T ( S ) k
2
s : t : L S
U
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NUMERICAL APPROACH
Spectral Projected Gradient Method (SPG)
Step 1: If k P ( S k
; r f
(
S
k
) )
; S
k
k t o l , then Stop.
Step 2: Nonmonotone Line-search
Step 2.1: Set =
k
Step 2.2: Set S +
=
P (
S
k
; r f (
S
k
) )
Step 2.3: If f (
S
+
)
m a x
0 j k M ; 1
f (
S
k ; j
) +
(
S
+
; S
k
)
T
r f (
S
k
)
then
k
=
, S k + 1
=
S
+
, W k
=
S
k + 1
; S
k
, y k
=
r f (
S
k + 1
) ; r f
(
S
k
)
, go
to Step 3.
else, 2
1
2
]
go to Step 2.2
Step 3: b k
=
W
T
k
y
k
If b k
0 , k + 1
=
m a x
,
else a k
=
W
T
k
W
k
and k + 1
= m i n
f
m a x
m a x
f
m i n
a
k
b
k
g g
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NUMERICAL RESULTS
We consider the following two synthetic models.
−4
−2
0
2
4
−4 −3 −2 −1 0 1 2 3 4
0
1
2
3
4
5
6
MODEL 1
Model 1
−4
−2
0
2
4−4 −3
−2 −10 1
2 34
0
1
2
3
4
5
6
MODEL 2
Model 2
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NUMERICAL RESULTS
The distribution of the sources and recievers was made in:Squared Mesh: a a
b squared Radial Mesh: a circle of radious r .
0 0.5 1 1.5 2 2.5 3 3.5 40
0.5
1
1.5
2
2.5
3
3.5
4Squared Mesh
−4 −3 −2 −1 0 1 2 3 4−4
−3
−2
−1
0
1
2
3
4Radial Mesh
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NUMERICAL RESULTS
Let the vector ( L U
)
T be the lower and upper bounds of the velocities and
fracture orientation angles respectively.
Unconstrained Case: For i
= 1 : : : 2 n
+ 1
l (
i ) = 0
: 0 0 2
V
x
(
i )
5 0 0 =
u (
i )
l (
i ) = 0
: 0 0 2
V
y
(
i )
5 0 0 =
u (
i )
l ( i ) = 0 : 0 0 2 V
z
( i ) 5 0 0 = u ( i )
l (
i ) = 0
(
i )
9 0 =
u (
i )
l (
i ) = 0
(
i )
9 0 =
u (
i )
Constrained Case: For i = 1 : : : 2
n + 1
l (
i ) = 0
: 2
V
x
(
i )
5 =
u (
i )
l (
i ) = 0
: 2
V
y
(
i )
5 =
u (
i )
l (
i ) = 0
: 2
V
z
(
i )
5 =
u (
i )
l (
i ) = 1 0
(
i )
3 0 =
u (
i )
l ( i ) = 2 ( i ) 9 = u ( i )
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NUMERICAL RESULTS
Initial Iterates:
V
0
x
= V
0
y
= V
0
z
= ( 3 4 5 5 4
3 )
T
0
=
0
= ( 1 0
1 0
1 0
1 0
1 0
1 0 )
T
Note: The velocities are measuared in Km
Angles are measuared in degrees.
Stopping Criterium :
k
P ( S
k
; r
f ( S
k
) ) ; S
k
k
2
1 0
; 6
Sun Station Ultra 10
M=8 in the SPG Method
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Model 1 (P-Wave), Unconstrained Case
Squared Mesh, ns=2, nr=15 Radial Mesh, ns=5, nr=16
V
x
V
a
x
V
y
V
a
y
V
z
V
a
z
V
a
x
V
a
y
V
a
z
1.5 1.35 1.7 1.69 1.9 2.12 1.28 1.7 2.23
2 2.97 2.3 2.29 2.5 1.68 2.56 2.3 1.95
3 3 2.8 2.79 3.3 3.3 3.45 2.8 2.86
3 3 2.8 2.8 3.3 3.3 3.45 2.8 2.86
2 2.97 2.3 2.3 2.5 1.68 2.65 2.29 1.94
1.5 1.35 1.7 1.7 1.9 2.1 1.28 1.69 2.23
Squared Mesh, ns=2, nr=15 Radial Mesh, ns=5, nr=16
(Aprox.) (Aprox.) (Aprox.) (Aprox.)
5 34.43 20 20 39.7 19.99
7 48.88 15 14.91 68.52 14.993 4.47 25 23.62 58.25 24.98
3 4.38 25 23.28 58.25 24.98
7 48.84 15 14.91 68.52 14.99
5 35.24 20 20 39.71 19.99
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Model 1 (P-Wave), Constrained Case
Squared Mesh, ns=2, nr=15 Radial Mesh, ns=5, nr=16
V
x
V
a
x
V
y
V
a
y
V
z
V
a
z
V
a
x
V
a
y
V
a
z
1.5 1.49 1.7 1.7 1.9 1.91 1.49 1.69 1.91
2 1.99 2.3 2.29 2.5 2.5 1.99 2.3 2.5
3 3 2.8 2.8 3.3 3.3 2.99 2.8 3.3
3 3 2.8 2.8 3.3 3.3 2.99 2.79 3.3
2 1.99 2.3 2.29 2.5 2.5 1.99 2.29 2.51
1.5 1.49 1.7 1.7 1.9 1.91 1.49 1.7 1.91
Squared Mesh, ns=2, nr=15 Radial Mesh, ns=5, nr=16
(Aprox.) (Aprox.) (Aprox.) (Aprox.)
5 8.95 20 19.99 8.97 19.99
7 8.68 15 14.91 8.79 14.993 5.81 25 23.42 5.78 24.95
3 5.81 25 23.42 5.78 24.98
7 8.68 15 14.91 8.79 14.99
5 8.95 20 19.99 8.97 19.99
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Model 1 (P-S Wave), Unconstrained Case
Squared Mesh, ns=2, nr=15 Radial Mesh, ns=5, nr=16
V
x
V
a
x
V
y
V
a
y
V
z
V
a
z
V
a
x
V
a
y
V
a
z
1.5 1.92 1.7 1.66 1.9 1.24 1.14 1.59 2.12
2 1.95 2.3 3.13 2.5 2.37 2.77 2.11 1.58
3 2.85 2.8 2.85 3.3 3.19 3.7 2.85 2.46
2.7 2.83 2.9 2.85 3.1 3.21 3.69 2.84 2.48
1.8 1.81 2 2.17 2.3 2.45 2.76 2.2 1.69
1.3 1.9 1.6 1.64 1.8 1.46 1.28 1.69 2.15
Squared Mesh, ns=2, nr=15 Radial Mesh, ns=5, nr=16
(Aprox.) (Aprox.) (Aprox.) (Aprox.)
5 71.82 20 19.99 34.49 19.99
7 14.64 15 14.89 53.22 14.993 6.78 25 0 48.34 18.61
3 6.78 25 1.74 46.94 18.55
7 10.44 15 15.03 48.78 14.99
5 64.52 20 19.97 40.57 19.99
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Model 1 (P-S Wave), Constrained Case
Squared Mesh, ns=2, nr=15 Radial Mesh, ns=5, nr=16
V
x
V
a
x
V
y
V
a
y
V
z
V
a
z
V
a
x
V
a
y
V
a
z
1.5 1.42 1.7 1.64 1.9 1.84 1.29 1.59 1.83
2 1.89 2.3 2.14 2.5 2.38 1.98 2.22 2.39
3 2.86 2.8 2.84 3.3 3.17 2.85 2.85 3.19
2.7 2.84 2.9 2.86 3.1 3.23 2.85 2.84 3.21
1.8 1.89 2 2.16 2.3 2.42 1.81 2.09 2.4
1.3 1.38 1.6 1.66 1.8 1.87 1.49 1.7 1.87
Squared Mesh, ns=2, nr=15 Radial Mesh, ns=5, nr=16
(Aprox.) (Aprox.) (Aprox.) (Aprox.)
5 8.08 20 19.89 5.28 20
7 8.96 15 14.97 8.62 14.98
3 5.48 25 10 5.42 13.02
3 5.45 25 10.25 5.94 14.91
7 8.99 15 15.28 6.62 14.99
5 8.41 20 19.91 8.76 19.99
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Model 2 (P Wave), Unconstrained Case
Squared Mesh, ns=2, nr=15 Radial Mesh, ns=5, nr=16
V
x
V
a
x
V
y
V
a
y
V
z
V
a
z
V
a
x
V
a
y
V
a
z
1.5 1.9 1.7 1.7 1.9 1.5 1.49 1.7 1.91
2 2.6 2.3 2.08 2.5 2.07 2.01 2.29 2.48
3 3.96 2.8 2.82 3.3 2.62 3.46 2.81 2.96
3 3.97 2.8 2.89 3.3 2.69 3.36 2.8 2.95
2 2.82 2.3 2.49 2.5 1.47 2.01 2.31 2.5
1.5 1.88 1.7 1.69 1.9 1.49 1.49 1.69 1.9
Squared Mesh, ns=2, nr=15 Radial Mesh, ns=5, nr=16
(Aprox.) (Aprox.) (Aprox.) (Aprox.)
5 90 20 19.8 6.7 19.99
7 38.69 15 0 3.3 12.83
3 48.46 25 45.68 62.77 36.31
3 36.77 25 49.29 57.28 34.31
7 71.88 15 0 3.13 12.88
5 90 20 20.15 6.73 20
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Model 2 (P-S Wave), Unconstrained Case
Squared Mesh, ns=2, nr=15 Radial Mesh, ns=5, nr=16
V
x
V
a
x
V
y
V
a
y
V
z
V
a
z
V
a
x
V
a
y
V
a
z
1.5 1.92 1.7 1.71 1.9 1.503 1.502 1.69 1.89
2 2.39 2.3 1.87 2.5 1.92 1.76 2.02 2.25
3 3.73 2.8 3.02 3.3 2.29 3.51 2.89 2.96
2.7 3.63 2.9 2.75 3.1 2.71 2.91 2.81 2.76
1.8 2.89 2 2.44 2.3 1.35 1.96 2.28 2.69
1.3 1.77 1.6 1.59 1.8 1.303 1.302 1.59 1.79
Squared Mesh, ns=2, nr=15 Radial Mesh, ns=5, nr=16
(Aprox.) (Aprox.) (Aprox.) (Aprox.)
5 88.92 20 18.4 0 20
7 32.38 15 0 18.36 22.73
3 66.92 25 22.55 89.16 13.05
3 28.39 25 36.24 86.88 40.32
7 74.75 15 0 14.57 21.16
5 90 20 21.22 1.07 19.98
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Model 2 (P-S Wave), Constrained Case
Squared Mesh, ns=2, nr=15 Radial Mesh, ns=5, nr=16
V
x
V
a
x
V
y
V
a
y
V
z
V
a
z
V
a
x
V
a
y
V
a
z
1.5 1.5 1.7 1.69 1.9 1.86 1.49 1.69 1.903
2 1.91 2.3 2.22 2.5 2.37 1.81 2.06 2.21
3 2.86 2.8 2.85 3.3 3.21 3.01 2.81 3.55
2.7 2.83 2.9 2.85 3.1 3.19 2.73 2.88 2.91
1.8 1.89 2 2.07 2.3 2.42 2.01 2.27 2.59
1.3 1.29 1.6 1.61 1.8 1.84 1.29 1.6 1.81
Squared Mesh, ns=2, nr=15 Radial Mesh, ns=5, nr=16
(Aprox.) (Aprox.) (Aprox.) (Aprox.)
5 5.84 20 21.49 8.46 20.1
7 7.36 15 15.46 4.93 13.46
3 6 25 20.31 2 25.13
3 5.78 25 19.14 6 20.43
7 7.47 15 14.29 4.47 12
5 5.01 20 18.94 7.64 19.86
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NUMERICAL RESULTS
0 1000 2000
0
0.02
0.04
0.06
0.08
0.1# rays vs. velocity error
S
R
0 1000 2000
4
4.5
5
5.5
6
6.5
7
7.5# rays vs. polar error
S
R
0 1000 2000
0
1
2
3
4# rays vs. azimutal error
S
R
0 1000 2000
0
20
40
60
80
100# rays vs. cpu−time
S
R
0 1000 2000
500
1000
1500
2000
2500
3000
3500
4000# rays vs. # iteration
S
R
0 1000 2000
0
100
200
300
400
500
600
700# rays vs. # back−tracking
S
R
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CONCLUSIONS
We solve the velocity and fracture orientation inversion problem in OM
using the Spectral Projected Gradient Method (SPG) and an ellipsoidal
approximation of the velocity.
This is a highly nonlinear problem that has many solutions, so
regularization of the problem is required.
The SPG method obtain good precision for the velocities estimates using
a relative small number of rays and no regularization.
To obtain a good estimate of the azimuthal and polar angle vectors
regularity is essential.
To get a constrained region (regularity ) is not a difficult task in seismic
since the maximum and minimum values of the velocities in the medium
is know a priori.
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$
%
A better estimate of the azimuthal angle vector can be obtained if there
are rays in all different azimuths (For example, using Radial Mesh).
None of the Mesh distributions used in this work give enough information
for obtaining a good estimate of the polar angle vector.
The problem in obtaining a better estimate of the polar angle vector is not
the optimization scheme used, but depends on the seismic data
acquisition.
Increasing the number of rays, the error in the velocity vector and in the
azimuthal angle vector can be reduced, obviously this imply an increase
in the cpu-time.
Also, increasing the number of rays, the number of iterations and number
of back-trackings may be reduced.
On the other hand, the error in the polar angle vector increases even if
the number of rays increase, since the seismic data distribution is not the
adequated for estimating the polar angle vector.
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NUMERICAL RESULTS
TEST 1: Corresponding to Model 1
L = ( l
1
: : : l
3 0
)
T
and U = ( u
1
: : : u
3 0
)
T
where l
i
= 0 : 0 0 2 i = 1 : : : 1 8 l
i
= 0 i = 1 9 : : : 3 0
u
i
= 5 0 0 i = 1 : : : 1 8 u
i
= 5 i = 1 9 : : : 3 0
n l s = 3 , n s = 6 , n l r = 4 and n r = 2 8
Initial Velocities Real Velocities Approximated Velocities
V
0
x
V
0
y
V
0
z
V
R
x
V
R
y
V
R
z
V
a
x
V
a
y
V
a
x
2 2 2 1.5 1.7 1.9 1.4999973 1.7019851 1.9019638
3 3 3 2 2.3 2.5 1.9984656 2.3011494 2.5005751
4 4 4 3 2.8 3.3 2.9998747 2.7991592 3.3009248
4 4 4 3 2.8 3.3 2.9997968 2.8008410 3.2994352
3 3 3 2 2.3 2.5 2.0011699 2.2988378 2.4999144
2 2 2 1.5 1.7 1.9 1.5062839 1.6980178 1.8985726
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NUMERICAL RESULTS
TEST 1: Corresponding to Model 1
L = ( l
1
: : : l
3 0
)
T
and U = ( u 1
: : : u
3 0
)
T
where
l
i
= 0 : 0 0 2 i = 1 : : : 1 8 l
i
= 0 i = 1 9 : : : 3 0
u
i
= 5 0 0 i = 1 : : : 1 8 u
i
= 5 i = 1 9 : : : 3 0
n l s = 3
, n s = 6
, n l r = 4
and n r = 2 8
Initial Angles Real Angles Approximated Angles
0
0
R
R
a
a
1 1 0 0 1.5322834 0.0033316
1 1 0 0 1.3463286 0.0049373
2 2 0 0 1.4478746 0.0000000
2 2 0 0 1.4356391 0.3929935
2 2 0 0 1.3585243 0.0341731
2 2 0 0 1.5861549 0.0000000
C P U ; t i m e = 2 2 : 8 2 m i n , i t e r = 5 0 7 and l i n e ; s e a r c h e s = 1 2 6
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NUMERICAL RESULTS
TEST 2: Corresponding to Model 1
L = ( l
1
: : : l
3 0
)
T
and U = ( u
1
: : : u
3 0
)
T
where l
i
= 0 : 0 0 2 i = 1 : : : 1 8 l
i
= 0 i = 1 9 : : : 3 0
u
i
= 6 i = 1 : : : 1 8 u
i
= 4 0 i = 1 9 : : : 3 0
n l s = 3 , n s = 6 , n l r = 4 and n r = 2 8
Initial Velocities Real Velocities Approximated Velocities
V
0
x
V
0
y
V
0
z
V
R
x
V
R
y
V
R
z
V
a
x
V
a
y
V
a
x
3 3 3 1.5 1.7 1.9 1.4867472 1.6086736 1.7609429
3.5 3.5 3.5 2.7 2.5 2.9 2.7073193 2.4508121 2.7925423
4 4 4 3 2.8 3.3 3.0203938 2.7488306 3.1746943
4 4 4 2.9 2.7 3.2 3.0147983 2.7494752 3.1800087
3.5 3.5 3.5 2.6 2.4 2.8 2.6841448 2.4514979 2.8105226
3 3 3 1.4 1.6 1.8 1.5616237 1.6898496 1.7545015
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NUMERICAL RESULTS
TEST 2: Corresponding to Model 1
L = ( l
1
: : : l
3 0
)
T
and U = ( u 1
: : : u
3 0
)
T
where
l
i
= 0 : 0 0 2 i = 1 : : : 1 8 l
i
= 0 i = 1 9 : : : 3 0
u
i
= 6 i = 1 : : : 1 8 u
i
= 4 0 i = 1 9 : : : 3 0
n l s = 3
, n s = 6
, n l r = 4
and n r = 2 8
Initial Angles Real Angles Approximated Angles
0
0
R
R
a
a
0 0 30 10 1.4881100 9.9471689
0 0 30 10 5.3222734 10.2593893
0 0 30 10 0.3525908 9.8410988
0 0 30 10 1.2738337 9.7745561
0 0 30 10 5.3366231 9.8324231
0 0 30 10 4.7458023 10.0201751
C P U ; t i m e = 4 1 : 9 m i n , i t e r = 3 7 9 and l i n e ; s e a r c h e s = 9 3
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NUMERICAL RESULTS
TEST 3: Corresponding to Model 1
L = ( l
1
: : : l
3 0
)
T
and U = ( u 1
: : : u
3 0
)
T
where
l
i
= 0 : 2 i = 1 : : : 1 8 l
i
= 2 0 i = 1 9 : : : 2 4 l
i
= 5 i = 2 5 : : : 3 0
u
i
= 6 i = 1 : : : 1 8 u
i
= 4 0 i = 1 9 : : : 2 4 u
i
= 2 0 i = 2 5 : : : 3 0
n l s = 4 , n s = 2 0 , n l r = 6 and n r = 6 6
Initial Velocities Real Velocitie s Approximated Velocities
V
0
x
V
0
y
V
0
z
V
R
x
V
R
y
V
R
z
V
a
x
V
a
y
V
a
x
3 3 3 1.5 1.7 1.9 1.4376400 1.7011965 1.9290055
3.5 3.5 3.5 2.7 2.5 2.9 2.6773830 2.4507081 2.8187457
4 4 4 3 2.8 3.3 2.9590742 2.7493184 3.2417599
4 4 4 2.9 2.7 3.2 2.9503655 2.7497884 3.2461665
3.5 3.5 3.5 2.6 2.4 2.8 2.6758592 2.4530501 2.8238717
3 3 3 1.4 1.6 1.8 1.3586700 1.5958847 1.9018278
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NUMERICAL RESULTS
TEST 3: Corresponding to Model 1
L = ( l
1
: : : l
3 0
)
T
and U = ( u 1
: : : u
3 0
)
T
where
l
i
= 0 : 2 i = 1 : : : 1 8 l
i
= 2 0 i = 1 9 : : : 2 4 l
i
= 5 i = 2 5 : : : 3 0
u
i
= 6 i = 1 : : : 1 8 u
i
= 4 0 i = 1 9 : : : 2 4 u
i
= 2 0 i = 2 5 : : : 3 0
n l s = 4
, n s = 2 0
, n l r = 6
and n r = 6 6
Initial Angles Real Angles Approximated Angles
0
0
R
R
a
a
0 0 30 10 37.5290024 9.9980061
0 0 30 10 22.9054416 10.0189242
0 0 30 10 29.4423860 9.9919281
0 0 30 10 28.9710453 9.9914708
0 0 30 10 23.3144158 10.0187249
0 0 30 10 33.8245728 9.9980831
C P U ; t i m e = 1 4 h o u r s , i t e r = 2 6 3 3 and l i n e ; s e a r c h e s = 4 9 1
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NUMERICAL RESULTS
TEST 4: Corresponding to Model 2 L = ( l
1
: : : l
5 0
)
T
and U = ( u 1
: : : u
5 0
)
T
where
l
i
= 0 : 2 i = 1 : : : 3 0 l
i
= 0 i = 3 1 : : : 5 0
u
i
= 6 i = 1 : : : 3 0 u
i
= 4 0 i = 3 1 : : : 4 0 u
i
= 5 i = 4 1 : : : 5 0
n l s = 1 , n s = 2 , n l r = 2 and n r = 1 5
V
0
x
V
0
y
V
0
z
V
R
x
V
R
y
V
R
z
V
a
x
V
a
y
V
a
x
2.5 2.6 3 1.5 1.7 1.9 1.5161705 1.7044085 1.8036717
3.5 3.2 3.6 2.7 2.5 2.9 2.6627301 2.4650854 2.8372655
4 3.8 4.1 3 2.8 3.3 2.9680257 2.7636329 3.2310795
4.1 4.3 4.5 3.3 3.5 3.6 3.1759822 3.4724621 3.6971851
4.3 4.5 4.5 3.5 3.6 3.8 3.7108556 3.7538805 3.5634800
4.3 4.5 4.5 3.4 3.5 3.7 3.8196233 3.5490186 3.5673048
4.1 4.3 4.5 3.2 3.4 3.5 3.1410118 3.4148884 3.5903418
4 3.8 4.1 2.9 2.7 3.2 2.9575074 2.7362643 3.2367196
3.5 3.2 3.6 2.6 2.4 2.8 2.6577356 2.4378346 2.8389060
2.5 2.6 3 1.4 1.6 1.8 1.4447071 1.5913248 1.8147302
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NUMERICAL RESULTS
TEST 4: Corresponding to Model 2 L = ( l
1
: : : l
5 0
)
T
and U = ( u 1
: : : u
5 0
)
T
where
l
i
= 0 : 2 i = 1 : : : 3 0 l
i
= 0 i = 3 1 : : : 5 0
u
i
= 6 i = 1 : : : 3 0 u
i
= 4 0 i = 3 1 : : : 4 0 u
i
= 5 i = 4 1 : : : 5 0
n l s = 1 , n s = 2 , n l r = 2 and n r = 1 5
0
0
R
R
a
a
23 3 30 0 26.9826548 0.0023920
23 3 30 0 27.8955924 0.0056462
23 3 30 0 28.1072325 0.0169888
23 3 30 0 32.2204377 0.9686877
23 3 30 0 36.8106643 3.5928424
23 3 30 0 6.5430180 0.9112516
23 3 30 0 38.4896379 1.0487732
23 3 30 0 27.2661348 0.0000001
23 3 30 0 28.0762195 0.0000000
23 3 30 0 22.3459052 0.0065288
C P U ; t i m e = 2 : 1 7 h o u r s , i t e r = 2 2 6 4 and l i n e ; s e a r c h e s = 5 2 9
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NUMERICAL RESULTS
TEST 5: Corresponding to Model 2 L = ( l
1
: : : l
5 0
)
T
and U = ( u 1
: : : u
5 0
)
T
where
l
i
= 0 : 2 i = 1 : : : 3 0 l
i
= 0 i = 3 1 : : : 5 0
u
i
= 6 i = 1 : : : 3 0 u
i
= 4 0 i = 3 1 : : : 4 0 u
i
= 5 i = 4 1 : : : 5 0
n l s = 2 , n s = 6 , n l r = 4 and n r = 2 8
V
0
x
V
0
y
V
0
z
V
R
x
V
R
y
V
R
z
V
a
x
V
a
y
V
a
x
2.5 2.6 3 1.5 1.7 1.9 1.5146239 1.6834034 1.8664965
3.5 3.2 3.6 2.7 2.5 2.9 2.6520520 2.4716170 2.8821383
4 3.8 4.1 3 2.8 3.3 2.9540754 2.7565136 3.2766908
4.1 4.3 4.5 3.3 3.5 3.6 3.2934095 3.5050076 3.6028292
4.3 4.5 4.5 3.5 3.6 3.8 3.7115094 3.6230909 3.5623813
4.3 4.5 4.5 3.4 3.5 3.7 3.7822892 3.5025558 3.5610915
4.1 4.3 4.5 3.2 3.4 3.5 3.2050763 3.3926163 3.4969652
4 3.8 4.1 2.9 2.7 3.2 2.9172908 2.7421486 3.2543635
3.5 3.2 3.6 2.6 2.4 2.8 2.6143580 2.4293056 2.8551777
2.5 2.6 3 1.4 1.6 1.8 1.3901383 1.6163345 1.8183832
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NUMERICAL RESULTS
TEST 5: Corresponding to Model 2 L = ( l
1
: : : l
5 0
)
T
and U = ( u 1
: : : u
5 0
)
T
where
l
i
= 0 : 2 i = 1 : : : 3 0 l
i
= 0 i = 3 1 : : : 5 0
u
i
= 6 i = 1 : : : 3 0 u
i
= 4 0 i = 3 1 : : : 4 0 u
i
= 5 i = 4 1 : : : 5 0
n l s = 2 , n s = 6 , n l r = 4 and n r = 2 8
0
0
R
R
a
a
23 3 30 0 32.3391389 0.0000489
23 3 30 0 34.1106580 0.0000030
23 3 30 0 32.3994191 0.0060426
23 3 30 0 28.2734365 0.2313614
23 3 30 0 33.7757315 1.8283798
23 3 30 0 14.0161269 1.4759419
23 3 30 0 31.3560557 0.2281124
23 3 30 0 31.2219247 0.0000007
23 3 30 0 32.5670208 0.0000002
23 3 30 0 26.4797392 0.0000259
C P U ; t i m e = 1 6 h o u r s , i t e r = 5 6 6 1 and l i n e ; s e a r c h e s = 1 2 4 7
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NUMERICAL RESULTS
TEST 6: Corresponding to Model 2 L = ( l
1
: : : l
5 0
)
T
and U = ( u 1
: : : u
5 0
)
T
where
l
i
= 0 : 2 i = 1 : : : 3 0 l
i
= 0 i = 3 1 : : : 5 0
u
i
= 6 i = 1 : : : 3 0 u
i
= 4 0 i = 3 1 : : : 4 0 u
i
= 5 i = 4 1 : : : 5 0
n l s = 3 , n s = 1 2 , n l r = 5 and n r = 4 5
V
0
x
V
0
y
V
0
z
V
R
x
V
R
y
V
R
z
V
a
x
V
a
y
V
a
x
2.5 2.6 3 1.5 1.7 1.9 1.5015321 1.6815115 1.8446112
3.5 3.2 3.6 2.7 2.5 2.9 2.6330562 2.4550646 2.8898359
4 3.8 4.1 3 2.8 3.3 2.9731147 2.7550248 3.2525118
4.1 4.3 4.5 3.3 3.5 3.6 3.3050384 3.453214 3.6168122
4.3 4.5 4.5 3.5 3.6 3.8 3.9870890 3.8343516 3.0960452
4.3 4.5 4.5 3.4 3.5 3.7 3.9948479 3.7634267 3.1293398
4.1 4.3 4.5 3.2 3.4 3.5 3.2320777 3.3985567 3.5229264
4 3.8 4.1 2.9 2.7 3.2 2.9321892 2.7441214 3.2395075
3.5 3.2 3.6 2.6 2.4 2.8 2.6083586 2.4465557 2.8750930
2.5 2.6 3 1.4 1.6 1.8 1.3929838 1.6172687 1.8505478
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NUMERICAL RESULTS
TEST 6: Corresponding to Model 2 L = ( l
1
: : : l
5 0
)
T
and U = ( u 1
: : : u
5 0
)
T
where
l
i
= 0 : 2 i = 1 : : : 3 0 l
i
= 0 i = 3 1 : : : 5 0
u
i
= 6 i = 1 : : : 3 0 u
i
= 4 0 i = 3 1 : : : 4 0 u
i
= 5 i = 4 1 : : : 5 0
n l s = 3 , n s = 1 2 , n l r = 5 and n r = 4 5
0
0
R
R
a
a
23 3 30 0 34.3991447 0.0043882
23 3 30 0 35.4810917 0.0013095
23 3 30 0 30.9784433 0.0881668
23 3 30 0 33.9740601 0.0000000
23 3 30 0 33.5488468 3.7197016
23 3 30 0 37.6389450 0.3137009
23 3 30 0 32.450822 0.0000000
23 3 30 0 28.059081 0.087736
23 3 30 0 34.2884020 0.0047640
23 3 30 0 25.9546206 0.0020617
C P U ; t i m e = 6 0 h o u r s , i t e r = 3 9 7 4 and l i n e ; s e a r c h e s = 9 6 2
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NUMERICAL RESULTS
150 rays 840 rays 2700 rays
k
S
r
;
S
k
k
2
1.749 0.622 1.27
k
V
r
;
V
k
k
2
0.631 0.54 1.25
k (
r
r
) ; (
k
k
) k
2
1.63 0.317 0.24
Iterations 2264 5661 3974
Line-searches 529 1247 962
CPU-time 2 16 60
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CONCLUSIONS
We solve the velocity and fracture orientation inversion problem in OM
using the Spectral Projected Gradient Method (SPG) and an ellipsoidal
approximation of the velocity.
This is a highly nonlinear problem that has many solutions, so
regularization of the problem is required..
The SPG method obtain good precision for the velocities estimates using
a relative small number of rays and no regularization.
To estimate de azimuthal and polar angles regularity is essential.
A better estimate of the azimuthal angle can be obtained if there are rays
in all different azimuths (For example using Radial Mesh).
The ray tracing takes most CPU time required for the inversion, so a
parallel low cost ray tracing will reduce the CPU time.
56