# User:Ageary/Chapter 14

Problems-in-Exploration-Seismology-and-their-Solutions.jpg | |

Series | Geophysical References Series |
---|---|

Title | Problems in Exploration Seismology and their Solutions |

Author | Lloyd P. Geldart and Robert E. Sheriff |

DOI | http://dx.doi.org/10.1190/1.9781560801733 |

ISBN | ISBN 9781560801153 |

Store | SEG Online Store |

## Chapter 14 Specialized applications

## 14.1 Using refraction method to find depth to bedrock

**To find the depth to bedrock in a damsite survey, 12 geophones were laid out at 15-m intervals from 15 to 180 m. Determine the overburden depth from the data in Table 14.1a assuming a single layer above the refractor. By how much does the depth differ if we assume two layers above the refractor?**

**Background**

The refraction method is discussed in problem 4.18.

15 | 19 |

30 | 29 |

45 | 39 |

60 | 50 |

75 | 59 |

90 | 62 |

105 | 65 |

120 | 68 |

135 | 72 |

150 | 76 |

165 | 78 |

180 | 83 |

**Solution**

Figure 14.1a shows the plotted data. The single layer interpretation (fine lines) gives , , . The critical angle . Equation (4.18a) gives the depth to bedrock as

The three-layer solution (heavy lines) gives , , , . Note that is not reliable; it could be any smaller value. If it were smaller, the first-layer values would change slightly, but it would not significantly change the values for the other layers.

For the first interface, ,

For the second interface, , so

Thus we get from equation (4.18d)

so ,

The total depth is .

The difference between the two interpretations is 9 m or about 17%.

## 14.2 Interpreting engineering refraction profiles

**14.2a What can you conclude from the time-distance data in Figure 14.2a?**

**Background**

Refraction interpretation is discussed in problems 4.18 and 4.24.

**Solution**

While the alignments for are short, we have four of them, giving the velocities 780, 700, 820, 800 m/s. Because three values agreed so closely, we discarded the 700 value and took . The other velocities were obtained using the intercept times.

From the reversed profiles with sources and , we measure

We ignore the short alignment at the left end of with intercept at of 16 ms, because it depends entirely on one point on the vertical axis at , and there is no matching event elsewhere.

Using equation (4.24f), , so

From equation (4.24b), we get

The reversed profiles with sources and yield the following values:

Equation (4.24f) now gives

**14.2b Apply equation (4.18c) to get approximate thicknesses of the second layer.**

**Solution**

From the profiles with sources and extending to offsets of 180 m we observe a high-velocity event giving the following measurements:

We need also the value of for the reversed profiles and :

Assuming 800 m/s for (since it is not determined), we have

We get the total thickness at using data from part (a) and equation (4.18c):

**14.2c What is the dip of the deeper interface?**

To get the dip of the deeper interface, we get the total depths at and . Using data from (a) and (b), the depth at is

The thickness of the lower bed at is obtained as in part (b):

The total thickness at ,

down from to .

**14.2d Why are the answers in parts (b) and (c) approximate?**

**Solution**

Some of the intercept times have only one significant figure so that the calculated depths in part (a) are not accurate. Also we used equation (4.18c) in both (b) and (c) to get depths. Since equation (4.18c) is valid only for zero dip, these values are approximate for this reason also.

## 14.3 Interpretation of four-shot refraction data

**Find the velocities and depths of refractors in Figure 14.3b.**

**Background**

The four-shot method uses a refraction spread with a source at each end to provide “short-shot” data and two sources offset inline beyond the critical distance (see problem 4.18) to provide “long-shot” data that permit accurate measurement of the refraction velocities. A refraction event appearing on both profiles can be identified by the slopes of the - curves. Geophones at the shot points are offset when shots at or are used.

**Solution**

Figure 14.3b shows east-west reversed profiles with sources and providing the short-shot (SS) data and sources and (not shown) that are offset inline to provide the long-shot (LS) data. We have labeled the different segments from to and fitted each with a straight line whose slope gives the apparent velocity. The SS segments were used to get intercept times as shown in the figure. Note that fitting straight lines to the data assumes that the refractors are planar, clearly an approximation.

The data suggest that this is a three-layer situation with dip down to the east. We note that the LS data segments , , and show a time-offset between geophones 6 and 7, as if indicating a fault downthrown to the east. However, evidences for features at depth (such as a fault cutting the refractor) should be displaced away from the source (because the head-wave path from the refractor to a geophone is inclined, as in Figure 14.3a). However profiles and show the same anomaly at the same location even though shot from opposite directions. Because our geophones spacing is only 20 m, this suggests that they are caused by something shallower than 20 m, and the most probable cause is a near-surface anomaly—perhaps a statics error in near-surface velocity or elevation corrections. We could make an empirical surface correction, but have not done so.

The SS data show a shallow refraction event (segments and ) that probably is a refraction from the top of an intermediate layer. The measured velocities are , , (note that neither nor is well determined, both being based on two points only), with intercept times , . Using equation (4.24f) we get

Because of the small dip, we can take and as vertical depths.

Comparison of velocities shows that segments and are the LS equivalent of SS segment , while and are equivalent to . To interpret these events we shall use the LS velocities and the SS intercept times.

Based on our previous discussion we ignore the displacements of segments and and and and draw the best-fit straight lines as shown in Figure 14.3b, obtaining , . We use these values plus the intercept times and to get the depth and dip.

We use the method of problem 4.24b to get and :

so , , .

Now , , (the fact that this value of agrees so closely with the above value is because both values depend upon the velocities).

The SS intercept times of 40 and 165 ms give the following depths:

Note that this value of is based on intercept times plus velocities, whereas the value is based on the velocities only.