GATE-2018 (24) :Wyllie – Time average equation

GATE-2012

27. A small scale seismic reflection survey was conducted with a shot point located at the middle of a 500m long geophone spread. The NMO correction travel times at the end of the spread were found to be 1.227s and 1.255s. If the average seismic wave velocity above the reflector is 2500 m/s , what is the dip of the reflector? (Give the value in degrees in nearest integer)

A) 4                               

B) 6                              

C) 8                                  

D)10

Solution:

A dipping reflector leads to asymmetric arrival times for shots recorded in different directions. The dip moveout is defined as the difference in travel  Times at –X and X, knowing the velocity V, the angle θ can be estimated.

∆T_d= t_x- t_-x

$\triangle T_{d}=\frac{2xsin\theta}{V}$ ----------(1)

Travel times are

t_x= 1.255 s

t_-x= 1.227 s

∆T_d= 1.255 – 1.227

       = 0.028s

Form equation (1)

$\triangle t_{d}=\frac{2xsin\theta}{V}$


$\sin \theta=\frac{V \triangle T_d}{2 x}$

Velocity = 2500 m/s

x=250 m ( offset distance means the distance between Source and Receiver )

sinθ= (2500* 0.028)/ 2* 250

        = 5 * 0.028

        = 0.14

     Θ= sin^-1( 0.14)

     Θ= 8.047

Therefore Dip (θ) = 8⁰


Reference:  An introduction to Geophysical Exploration by Philip Kearey, Michael Brooks and Ian Hill