4. Seismic Ray Tracing for Minimum Time Path
4.4. Computing Travel Time and Derivatives 99 arbitrary depth in a model consisting of N multiple horizontal layers over
a half-space as shown in Fig. 23. Let the earthquake source be at point A with coordinates (xA, y A , Z A ) , and the station be at point B with coordinates
( X B , Y B , zB). We construct our model such that the station is at the top of the
first layer, and we use uL and hi to denote the velocity and the thickness of the ith layer, respectively. In Fig. 23, we let the source be inside thejth layer at depth
5
from the top of the jth layer. The difference between this case and the surface-source case is that the down-going travel path of the refracted wave starts at depth zA instead of at the surface. This means that the down-going seismic wave has the following less layers to travel: the first ( j - I ) layers from the surface, and an imaginary layer within thejth layer of thickness5.
We can therefore write down a set of equations for this case in a manner similar to Eq. (4.97)forj = 1 , 2 , .
.
. , N - l a n d k = 2 , 3 , ..
. , N , whereTjkisthetraveltime for the ray that is refracted along the top ofthekth layer from a source at theI
)---
jth layernth layer Fig. 23. Diagram of the refracted path along the kth layer for a source located in the j t h layer.
100 4. Seismic Ray Tracing for Minimum Time Path
jth layer, t j k is the corresponding critical distance, A is the epicentral distance given by A = [(x, - xA)'
+
( y B - Y ~ ) ' ] ' ~ ~ , the thickness5
is given by6
= zA - ( h ,+
h,+
*.
*+
hj...l), the symbol f l k l = (u$ - u;)"', and the symbolRw
= (u$ - u;L)"'.These equations allow us to calculate the travel times for various re- fracted paths and the corresponding critical distances beyond which re- fracted waves will exist. In many instances, one of the refracted paths is the minimum time path sought. However, there are cases in which the direct path is the minimum time path. This is considered next.
If the earthquake source is in the first layer (with velocity u,), then the travel time for the direct wave is the same as that in the constant velocity model (see Section 4.4.2), i.e.,
(4.99) T d = [(XB A X A ) ~
+
(Ye - YA)'+
(2, - 2~)*]"/2)1However, if the earthquake source is in the second or deeper layer, there is no explicit formula for computing the travel time of the direct path. In this case, we must use an iterative procedure to find an angle
4
such that its associated ray will reach the receiving station along a direct path. Let us consider a model where the earthquake source is in the second layer, as shown in Fig. 24. Let the first layer have a thickness hl, and let u1 and u2 be the velocity of the first and the second layer, respectively. In our model the layer velocity increases with depth so that u2 > v,. We choose a coordinate system such that thez
axis passes through the source at point A , and the 7 axis passes through the station at point B. Let the coordinatesT i i-
Fig. 24. Diagram to illustrate the computation of the travel path of a direct wave for a layered velocity model.
4.4. Computing Travel Time und Derivatives 101 of point A be (0, z A ) , and that for point B be (A, 0). For simplicity, we will label the points along the q axis by their q coordinates; for example, point At has the coordinates (A,, 0). We will also label the points along the line z
= hl by their q coordinates; for example, point q1 has the coordinates ( q l ,
h d . We define
5
= zA - hl. Let the 4’s be the angles measured from thenegative z direction to the lines joining point A with the appropriate points along the line z = h, (see Fig. 24).
To find by an iterative procedure an angle
4
whose associated ray path will reach the station, we first consider lower and upper bounds(41
and 42) for the angle4.
If we join points A and B by a straight line, thenAB
intersects the line z = hl at point ql such that its 7 coordinate is given by 77, = A(5/zA). By Snell’s law and the fact that u2 > u l . this ray A T will emerge to the surface at a point with q coordinate of A], which is always less than A. Thus the angle associated with the ray A< can be selected as the lower bound of
4.
To find the upper bound of4,
we let q2 be the point directly below the station and on the line z = h,, i.e., q2 = A. Again by Snell’s law and u2 > u,, the ray Axwill emerge to the surface at a point with q coordinate of A2, - which is always greater than A. Thus the angle 42 associated with the ray A q 2 can be selected as the upper bound of4.
and its associated trial angle To start the iterative procedure, we consider
4*
(see Fig. 24) such that q* a trial ray is approximated by (4.100)(v*
- q 1 ) / ( ~ 2 - q1) = (A - A1)/(A2 - A,)and
(4.101)
4*
= a r c t a n ( d 0By Snell’s law and knowing q*, we can determine the q coordinate, A,, of the point where the ray A X emerges to the surface. In order for the trial angle $* to be acceptable, the ray A X must emerge to the surface very close to the station, i.e., IA - A*[ < E , where the error limit E is typically a few tens of meters. If [A - A*[ > E , we select a new
+*
and repeat the same procedure until the trial ray emerges close to the station. To perform the iteration, we must consider the following two cases:( I ) If A* > A, the new
4*
is chosen between the two raysA7)1
and A T . To do this, in Eq. (4.100), the current values for A, and q*replace
A2
and q2 respectively, and a new value for q, is com- puted.If A, < A, the new C#J* is chosen between the rays A T a n d
G.
To do this, in Eq. (4.loo),
the current values of A, and q* replace Al and ql, respectively, and a new value for q* is computed. In both cases, a new value for4*
can be obtained from Eq. (4.101).(2)
102 4. Seismic Ray Tracing fw Minimum Time Path
The preceding procedure for a source in the second layer can be gener- alized to a source in a deeper layer, say, the jth layer. Once a trial angle
4*
is chosen, we may use Snell’s law to compute successive incident angles to each overlying layer until the trial ray reaches the surface at point A, with the T coordinate given by(4.102) A.+ = q*
+
i=$
-1 hf tane,
where 6, =
4*.
The incident angles are related by (4.103) - = - sin Of sin for 1 I i < jVI ui+1
where Oi is the incident angle for the ith layer with velocity vf and thick- ness hf. With the proper substitutions, Eq. (4.102) can also be used to calculate A1 and
A2.
In this iterative procedure, the trial angle
4*
converges rapidly to the angle4
whose associated ray path reaches the station within the error limit E . Since this ray path consists o f j segments of a straight line in each of thej layers, we can sum up the travel time in each layer to obtain the travel time from the source to the station.Knowing how to compute travel time for both the direct and the re- fracted paths, we can then select the minimum travel time path. The spatial derivatives and the take-off angle can be computed from the direc- tion cosines using results from Section 4.4.1 as follows.
Let us consider an earthquake source at point A with spatial coordi- nates (xA, yA, zA). and a station with spatial coordinates
be,
ye, ze). Let us choose a coordinate system such that points A and B lie on the qz plane, and A ’ is the projection of A on z = zg. In Fig. 25, let us consider an element of ray path ds with direction angles a,p,
and y , and components dx, dy, and dz (with respect to the x, y, and z axes, respectively). In Fig.25a, the projection of ds on the 71 axis is sin y ds. In Fig. 25b, this pro- jected element can be related to the components dx and dy of ds by
(4.104)
where a’ and
p‘
are the angles between the 7) axis and the x and y axes, respectively. Consequently, from Eq. (4.104) and the definition of direc- tion cosine for y , we have(4.105)
The angles a’ and
p’
can be determined from Fig. 25b as dx = cos a’ sin y ds, dy = cosp’
sin y dsdxlds = cos a’ sin y , dylds = cos /3’ sin y dzlds = cos y
4.4. Computing Travel Time and Derivatives (sin Y ds)
103
Fig. 25. Diagram of the projections of an element of ray path ds on theqz plane (a) and on the xy plane (b).
(4.106) COS a’ = ( X B - X,J/A, COSp’ = (Ye - YA)/A
whereh is the epicentral distance between points A ’ and B and is given by (4.107) h = [ ( X B - XA)’
+
( y ~ - yA)2]1r’For the refracted waves in a multilayer model (see Fig. 23), the direc- tion angle y at the source is the take-off angle for the refracted path. If the earthquake source is in thejth layer and the ray is refracted along the top of the k th layer, then by Eq. (4.96)
104 4. Seismic Rcig Trctcing .fir Minimum Time Puth (4.108)
Therefore Eq. (4.105) becomes
yIA = 61, = arcsin(uj/uk)
dx/ds = [ ( X B - x~)/A](uj/uk) (4.109) dY/ds = [ ( Y B - YA)/Al(Uj/uk)
dz/ds = [ I - (u]/uk)2]'12
Using Eq. (4.66), the spatial derivatives of the travel time evaluated at the source for this case are
dTjk/axlA = - ( x B - XA)/(A * uk)
(4.110) aTjk/aYlA = -(Ye - . Y A ) / ( A * uk)
dTjk/dzlA = -(u% - ~ ~ ) " ' / ( u ] ' uk)
and the take-off angle [according to Eq. (4. l08)] is (4. I 1 1 )
where the limits for j and k are given in Eq. (4.98).
For the direct wave with earthquake source in the first layer, the spatial derivatives of the travel time and the take-off angle are the same as those given in Section 4.4.2 for a constant velocity model.
For the direct wave with earthquake source in the jth layer (with velocity u j ) , we find the direct ray path and its associated angle
4
by an iterative procedure as described previously. The direction angle y at the source is the take-off angle J, for the direct path. Since the take-off angle is measured from the positive z direction whereas the angle4
is measured from the negative z direction (see Fig. 24), we have+jk = arcsin(uj/uk)
(4.112) Y I A = J, = 180" -
4
Thus, using Eqs. (4.103, (4.1061, and (4.66), and noting that sin( 180" -
4)
= sin 4, and cos( 180" -
4)
= -cos4,
the spatial derivatives of the travel time evaluated at the source for this case aredT/dxlA = -[(xB - xA)/(A. uj)] sin 4 (4.113) dT/aylA = - [ ( Y B - yA)/(A
. uj)l
sin4
d T/aZlA = COS