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principles and applications of microearthquake networks

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92 4. Seismic Ruy Tracing for Minimum Time Path<br />

(4.68)<br />

a = arccos(dx/ds).<br />

y = arccos(dz/ds)<br />

p = arccos(dy/ds)<br />

In the usual Cartesian coordinate system, the positive z axis is pointing<br />

upward. However, in seismological <strong>applications</strong>, it is more convenient to<br />

let the positive z axis point downward. The take-<strong>of</strong>f angle I#J at the source<br />

(with respect to the downward vertical) is just the direction angle y at the<br />

source, i.e., + = yIA, or<br />

(4.69) $ = arccos(dz/ds)/A<br />

If we solve the ray equation in the form <strong>of</strong> Eq. (4.56), then ws(0) is the<br />

value <strong>of</strong> dz/ds at the source. Therefore, Eq. (4.69) may be written as<br />

(4.70) $ = arccos[w6(0)]<br />

Equation (4.66) is very important because it relates the spatial derivatives<br />

<strong>of</strong> the travel time to the direction cosines <strong>of</strong> the seismic ray <strong>and</strong> the<br />

slowness (Le., the reciprocal <strong>of</strong> the velocity) <strong>of</strong> the medium at the earthquake<br />

source. We will use this equation frequently in the following<br />

subsections.<br />

4.4.2. Constant Velocity Model<br />

This is the simplest velocity model; the velocity v is a constant <strong>and</strong> is<br />

independent <strong>of</strong> the spatial coordinates. The minimum time path between<br />

any two points A <strong>and</strong> B is a straight line (i.e., AT). If the earthquake<br />

source is at point A with coordinates (xA,<br />

zA), <strong>and</strong> the receiving station<br />

is at point B with coordinates (xs, Ve, zB), then the path length S between A<br />

<strong>and</strong> B is [(xB - xA)2 + (ys - yA)2 + (za - zA)*]”*. The travel time T is<br />

simply given by<br />

(4.71) T = S /E<br />

The direction cosines for the ray are constant <strong>and</strong> are given by (xE-<br />

xA)/S, (Ys - yA)/S, <strong>and</strong> (zB- zA)/S. Using Eq. (4.661, the spatial derivatives<br />

<strong>of</strong> the travel time at the source for this case are<br />

aT/ax/A = -(xB - xA)/(ns), aT/ay(A = -(YE - 4)A)/(uS)<br />

(4.72)<br />

aT/azl, = -(zB - zA)/(us)<br />

Using Eq. (4.69), the take-<strong>of</strong>f angle at the source is<br />

(4.73) 4 = arccos[(zB - zA)/S]<br />

In this simple model, we do not really need to apply the formulas for the<br />

general three-dimensional case. Eq. (4.72) for the spatial derivatives can

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