Effect of a hidden layer
Assume that you wish to map the 5.75 and 6.40 km/s formations in the Illinois basin. Given the velocity information shown in Figure 11.3a, what difficulties would you expect to encounter? The shale at 420–620 m and the lower velocity at 790–960 m form “hidden layers”; how much error will neglect of the hidden layers involve?
The velocity-depth data are summarized in Table 11.3a. Each of the three high-velocity layers will produce a head wave whose apparent velocity is that of the layers if the layering is all horizontal (which we assume, knowing that dips are generally gentle). We calculate the intercept times in order to plot the time-distance curves.
Because the layers are assumed to be horizontal, equation (3.1a) gives the angles of incidence for the ray that produces the head waves. For the 5150 m/s head wave,
|0–300 m||2650 km/s|
We use equation (4.18d) to calculate the intercept time :
For the 5750 m/s head wave we have
Its intercept time will be
To complete the time-distance curve, we have for the 6400 m/s head wave, allowing for 170 m of 5000 m/s layer that interrupts the 5750 medium (note ray direction is the same in both parts at 5750 m/s),
Its intercept time will be
The crossover between the 5150 and 5750 m/s head waves is given by
and the crossover between the 5750 and 6400 m/s head waves is given by
The 5750 m/s curve will be responsible for first breaks for only 1.30 km.
The data are plotted in Figure 11.3b. Interpretation of this time-distance plot will be difficult because the slopes of the three head-wave curves are nearly the same. The ratios of the successive head-wave velocities in this situation are only 1.12 and 1.11; generally ratios should be 1.25 or larger to be interpreted unambiguously.
Failure to recognize a hidden layer means that the time spent in that layer will be interpreted as spent in a layer with higher velocity, which will make the depth appear too large. The shallow refraction event should be interpreted correctly because there are no hidden layers, but the depth calculated for the deeper interfaces will be too great because of the hidden layers.
If we recognize only the 5150 m/s and 6400 m/s head waves (the most probable situation unless additional information is available), that is, the 5750 m/s layer is a hidden layer, then we would calculate the thickness of the 5150 m/s layer as
This gives m, which, when added to the 300 m thickness of the top layer, gives the depth of the 6400 layer as 1257 m. Comparing with the correct value of 1200 m, the error is 60 m or 5%.
If we should recognize the 5750 m/s head wave, but are not aware of the 3650 m/s layer, we would calculate the thickness of the 5150 m/s layer as
This gives m, which, when added to the 300-m thickness of the top layer, gives a depth of 900 m. Comparing with the correct value of 620 m, the error is 280 m or 45%.
The travel through the 170 m thick 5000 m/s layer, if it is not recognized, would probably be assumed to be at the velocity of 5750 m/s, producing a time error of only 4 ms:
The error is small because the difference in assumed velocities is small.
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Also in this chapter
- Salt lead time as a function of depth
- Effect of assumptions on refraction interpretation
- Effect of a hidden layer
- Proof of the ABC refraction equation
- Adachi’s method
- Refraction interpretation by stripping
- Proof of a generalized reciprocal method relation
- Delay time
- Barry’s delay-time refraction interpretation method
- Parallelism of half-intercept and delay-time curves
- Wyrobek’s refraction interpretation method
- Properties of a coincident-time curve
- Interpretation by the plus-minus method
- Comparison of refraction interpretation methods
- Feasibility of mapping a horizon using head waves
- Refraction blind spot
- Interpreting marine refraction data