Head waves (refractions) and effect of hidden layer

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Problem 4.18a

Derive traveltime equations for head waves [equations (4.18a,b)]. What is the significance of intercept times $ t_{1} $ and $ t_{2} $ in these equations?

Background

When a P-wave is refracted at a horizontal interface between two layers of different velocities, equation (3.1a) shows that $ \sin \theta _{2}/V_{2}=\sin \theta _{1}/V_{1} $. When $ V_{2}>V_{1} $, $ \theta _{2}>\theta _{1} $, so if $ \theta _{1} $ increases sufficiently, $ \theta _{2} $ will eventually reach its maximum value of $ \pi /2 $, at which point the refracted wave will be traveling parallel to the interface as a head wave or a “refraction.” As the head wave progresses along the interface, it radiates energy upward into the first medium at the same angle as the angle of incidence on that interface. From equation (3.1a) we see that the angle of incidence at which the refraction is generated is $ \sin \theta _{c}=V_{1}/V_{2} $ (see Figure 4.18a); $ \theta _{c} $ is called the critical angle. In addition to the head wave, a direct wave travels directly from the source to the receiver, that is, along the line $ OR'' $ in Figure 4.18a. Near the source the direct wave arrives before the refraction, but eventually the higher-velocity path of the refraction allows it to overtake the direct wave at point $ R' $, the crossover point. The refraction exists at offsets greater than the critical distance $ OQ $, but between $ Q $ and $ R' $ it is often obscured by the earlier direct wave.

The traveltime curve of the direct wave is given by $ t=x/V_{1} $, where $ x $ is the offset. For a two-layer situation the traveltime curve for the refraction from a horizontal refractor at

Figure 4.18a.  Two-layer raypaths and traveltime curves, horizontal layering.
Figure 4.18b.  Three-layer head-wave and travel-time curve.

depth $ h $ (Figure 4.18a) is (Sheriff and Geldart, 1995, 95)


$ {\begin{aligned}t={\frac {x}{V_{2}}}+{\frac {2h\cos \theta _{c}}{V_{1}}}={\frac {x}{V_{2}}}+t_{i}.\end{aligned}} $ (4.18a)

The traveltime curve is a straight line of slope $ 1/V_{2} $ with a projected intercept on the $ t $-axis equal to $ t_{i}=2h\cos \theta _{c}/V_{1} $. The slope of the direct-wave traveltime curve gives $ V_{1} $, while that of the refraction curve gives $ V_{2} $. Although the refraction curve does not extend to the $ t $-axis, we can project it to get the intercept $ t_{1} $. Thus we can find $ \theta _{c} $ and get the depth $ h $ from $ t_{i} $.

For a three-layer situation (Figure 4.18b) such that $ V_{1}<V_{2}<V_{3} $, two refractions exist. The shallow one has the critical angle $ \theta _{c1} $, where $ \sin \theta _{c1}=V_{1}/V_{2} $, whereas the deeper refraction has angles of incidence given by $ \sin \theta _{1}/V_{1}=\sin \theta _{c2}/V_{2}=1/V_{3} $. The traveltime curve for the shallow refraction is given by equation (4.18a) while that for the deeper refractor is (Sheriff and Geldart, 1995, 96)


$ {\begin{aligned}t={\frac {x}{V_{3}}}+\left({\frac {2h_{1}}{V_{1}}}\right)\cos \theta _{1}+\left({\frac {2h_{2}}{V_{2}}}\right)\cos \theta _{c2}={\frac {x}{V_{3}}}+t_{2},\end{aligned}} $ (4.18b)


$ {\begin{aligned}{\hbox{where}}\qquad \qquad t_{2}=\left({\frac {2h_{1}\cos \theta _{1}}{V_{1}}}\right)+\left({\frac {2h_{2}\cos \theta _{c2}}{V_{2}}}\right).\end{aligned}} $ (4.18c)

Note that the first term in equation (4.18c) is not $ t_{i} $ in equation (4.18a) because $ \theta _{1}<\theta _{c1} $.

Equation (4.18c) can be generalized for $ n $ layers in the form


$ {\begin{aligned}t_{n}=\sum _{i-1}^{n-1}{\frac {2h_{i}\cos \theta _{i}}{V_{i}}}+{\frac {2h_{n}\cos \theta _{cn}}{V_{n}}}.\end{aligned}} $ (4.18d)

Solution

A typical refraction path in Figure 4.18a is OMPR. Taking $ OR=x $, the traveltime $ t $ is

$ {\begin{aligned}t&={\frac {(OM+PR)}{V_{1}}}+{\frac {MP}{V_{2}}}={\frac {2h}{V_{1}\cos \theta _{c}}}+{\frac {(x-2h\tan \theta _{c})}{V_{2}}}\\&={\frac {x}{V_{2}}}+{\frac {2h}{\cos \theta _{c}}}\left({\frac {1}{V_{1}}}-{\frac {\sin \theta _{c}}{V_{2}}}\right)={\frac {x}{V_{2}}}+{\frac {2h}{V_{1}\cos \theta _{c}}}(1-\sin ^{2}\theta _{c})\\&={\frac {x}{V_{2}}}+{\frac {2h\cos \theta _{c}}{V_{1}}}={\frac {x}{V_{2}}}+t_{i}.\end{aligned}} $

Equation (4.18b) can be derived in a similar way; from Figure 4.18b,

$ {\begin{aligned}t&={\frac {2OM'}{V_{1}}}+{\frac {2M'M''}{V_{2}}}+{\frac {M''P''}{V_{3}}}\\&={\frac {2h_{1}}{V_{1}\cos \theta _{1}}}+{\frac {2h_{2}}{V_{2}\cos \theta _{c2}}}+{\frac {(x-2h_{1}\tan \theta _{1}-2h_{2}\tan \theta _{c2})}{V_{3}}}\\&={\frac {x}{V_{3}}}+{\frac {2h_{1}}{V_{1}\cos \theta _{1}}}\left(1-{\frac {V_{1}\sin \theta _{1}}{V_{3}}}\right)+{\frac {2h_{2}}{V_{2}\cos \theta _{c2}}}\left(1-{\frac {V_{2}\sin \theta _{c}2}{V_{3}}}\right)\\&={\frac {x}{V_{3}}}+{\frac {2h_{1}\cos \theta _{1}}{V_{1}}}+{\frac {2h_{2}\cos \theta _{c2}}{V_{2}}}={\frac {x}{V_{3}}}+t_{2}.\end{aligned}} $

The traveltime curve is again a straight line with slope $ 1/V_{3} $ and intercept $ t_{2} $.

Problem 4.18b

Show that the two geological sections illustrated in Figure 4.18c produce the same time-distance refraction curves.

Solution

Although no refraction will be generated at the intermediate interface because the velocity decreases at the interface, equation (4.18b) is still valid. Therefore, for the three-layer case in Figure 4.18c, Snell’s law gives

$ {\begin{aligned}{\frac {\sin \theta _{1}}{3.00}}={\frac {\sin \theta _{c2}}{1.50}}={\frac {1}{6.00}};\end{aligned}} $

thus, $ \theta _{1}=30.0^{\circ } $, $ \theta _{c2}=14.5^{\circ } $, and equation (4.18b) gives

$ {\begin{aligned}t&={\frac {x}{6.00}}+{\frac {2\times 0.300\cos 30^{\circ }}{1.50}}+{\frac {2\times 0.300\cos \,14.5^{\circ }}{3.00}}\\&=0.167x+0.5040.\end{aligned}} $

Figure 4.18c.  Sections giving the same traveltime curves.

For the two-layer case in Figure 4.18c, we use equation (4.18a); since $ \theta _{c}=\sin ^{-1}(3.00/6.00)=30.0^{\circ } $,

$ {\begin{aligned}t={\frac {x}{6.00}}+(2\times 0.935/3.00)\cos \,30.0^{\circ }={\frac {x}{6.00}}+0.540.\end{aligned}} $

The effect of the 1.5 km/s hidden layer is to shorten the critical distance ($ OQ $ in Figure 4.18a), but this would probably not be detected and hence the depth to $ V_{3} $ would probably be overestimated, for example,

$ {\begin{aligned}h={\frac {1}{2}}Vt_{i}/\cos \theta _{c}={\frac {1}{2}}\times 3.0\times 0.54/\cos \,14.5^{\circ }=837\,\mathrm {m} .\end{aligned}} $

Problem 4.18c

What would be the depth to the top of the 6 km/s layer in the three-layer case of Figure 4.18c if $ V_{2}=4.00 $ km/s instead of 1.5 km/s?

Solution

When $ V_{2} $ is 4.00 km/s, we have for the shallow headwave:

$ {\begin{aligned}&\sin \theta _{c1}/3.00=1/4.00,\quad \theta _{c1}=48.6^{\circ },\quad \cos \theta _{c1}=0.661,\\&t_{1}=x/4.00+(2\times 0.300/3.00)\cos \,48.6^{\circ }=x/4.00+0.132\,\mathrm {s} ,\end{aligned}} $

For the headwave from the 6.00 km/s layer,

$ {\begin{aligned}\sin \theta _{1}/3.00&=\sin \theta _{c2}/4.00=1/6.00,\quad \theta _{1}=30.0^{\circ },\theta _{c2}=41.8^{\circ },\\\cos \theta _{1}&=0.866,\quad \cos \theta _{c2}=0.745,\\t_{2}&=x/6.00+(2\times 0.300/3.00)\cos \,30.0^{\circ }\\&\quad +(2\times 0.300/4.00)\cos \,41.8^{\circ },\\&={x}/{6.00}+0.173+0.112=x/6.00+0.285\,\mathrm {s} .\end{aligned}} $

Figure 4.18d shows the plot of x versus $ t $

While the 4.00 km/s layer does generate a refraction event, it might not be evident that it is a separate alignment. We might interpret the results as a two-layer problem missing the 4.00 km/s event, thus underestimating the average velocity to the 6 km/s layer and under-estimating its depth. If we should see the intermediate layer clearly enough, we could obtain the depths correctly.

Figure 4.18d.  Plot of $ t $ versus $ x $ for Figure 4.18c(i).

Problem 4.18d

What would be the apparent depth to the refractor in problem 4.18b if $ V_{3}=4 $ km/s instead of 6.00 km/s?

Solution

In the three-layer case, we have

$ {\begin{aligned}\sin \theta _{1}/3.00&=\sin \theta _{c2}/1.50=1/4.00,\\\theta _{1}&=48.6^{\circ },\theta _{c2}=22.0^{\circ },\cos \theta _{c2}=0.927,\\t&=x/4.00+(2\times 0.300/3.00)\cos \,48.6^{\circ }+(2+0.300/1.50)\cos \,22.0^{\circ }\\&=x/4.00+0.503\,\mathrm {s} .\end{aligned}} $

If we do not know that the 1.50 Km/s layer exists, we would take $ V_{1}=3.00 $, $ V_{2}=4.00 $, $ \theta _{c}=48.6 $, $ t_{i}=0.540 $, $ h=(1/2)V_{1}t_{1}/\cos \theta _{c}=(1/2)3.00\times 0.540/\cos \,48.6^{\circ }=1.22 $ km. For the two-layer case we would calculate the depth correctly since there is no hidden layer. However, we might not recognize the headwave since its slope (1/4.00) is not much different from that of the direct wave (1/3.00) (see Figure 4.18d).

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