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

98) In a Seismic reflection survey over a Dipping interface separating two media with the velocity in the Upper medium being 2000 m/sec, the travel times of the reflected waves at offset distances of -100m and +400 m are 981.25 ms and 1013 ms respectively. If the slant depth two way travel time at the shot point is 1 sec. then dip move out(ms/meter) is_________________

 

(Thank to Chandrasekhar, ANU)

Solution:

Given that the offset distances 

$X_{1}$ = -100 m

$X_{2}$=+400m

The travel times 

$T_{1}$ = 981.25 ms

$T_{2}$=1013 ms

 

Velocity of the layer (V) = 2000 m/sec

 

The two way travel time($T_{0}$)=1sec

Dip move out=?

 

For calculating the dip move out we have to calculate normal move out first and then subtract that from the travel times we get the corrected travel times

 

Normal move out correction $\triangledown t_{d}=\frac{X^{2}}{2 V^{2}t_{0}}$

For offset $x_1=-100m$ normal move out  is

$\triangledown t_{d_{1}}=\frac{(100)^{2}}{2 (2000)^{2}(1)}$

 

$\triangledown t_{d_{1}}=\frac{10^{4}}{8 \times10^{6}}$

 

$\triangledown t_{d_{1}}=0.00125 seconds$

 

$\triangledown t_{d_{1}}=1.25 milli seconds$

 

similarly for the for offset $X_2=400m$ normal move out is

$\triangledown t_{d_{2}}=\frac{(400)^{2}}{2 (2000)^{2}(1)}$

 

$\triangledown t_{d_{2}}=\frac{16\times10^{4}}{8 \times10^{6}}$

 

$\triangledown t_{d_{2}}=0.02 seconds$

 

$\triangledown t_{d_{2}}=20 milli seconds$

Hence the corrected Travel times for the two offset distances are as follows

 

$T_{1}$=981.25-1.25=980 milli seconds

 

$T_{2}$=1013-20 = 993 milli seconds

 

Now the Dip move out =$\frac{\triangledown t_{d}}{\triangledown x}$
 

=$\frac{t_{2}-t_{1}}{x_{2}-x_{1}}$

 

=$\frac{993-980}{400-100}$

 

=$\frac{13}{300}$

 

=0.043 ms/m

Note: We may not have a symmetrical spread and we find the dip moveout by removing the effect of normal moveout.

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(Special Thanks to Rohit Jha)

(Thanks to Ashok, AKNU)

Solution:

Formula

$t=t\circ(1+\frac{x^2+4hxsin\theta}{8h^2})---(1)$

Offset distance $(x_1)$ =-100 , $(x_2)$ =400

Travel time $t_1$=981.25 ms , $t_2$=1013ms

Velocity (v)=2000m/sec

$t_0$ = 1 sec

$h=\frac{v t_0}{2}=\frac{2000\times1}{2}= 1000 m/s$

 

Substitute the above values in equation 1, then

$981.25=(1+\frac{(-100^2)+4(1000)(-100)sin\theta}{8(1000)^2})$

$981.25=(1+\frac{(10^4)-4(10^5)sin\theta}{8(10)^6})$

$981.25=(1+\frac{1}{800}-\frac{1}{20}){sin\theta}---(2)$

$1013=(1+\frac{(400^2)+4(1000)(400)sin\theta}{8(1000)^2})$

$1013=(1+\frac{(16*10^4)+(16*10^5)sin\theta}{8(10)^6})$

$1013=(1+\frac{16}{800}+\frac{16}{80}){sin\theta}$

$1013=(1+\frac{1}{50}+\frac{1}{5}){sin\theta}---(3)$

subtract equation (3) from equation (2)

$1013-981.25=(1+\frac{1}{50}+\frac{1}{5}){sin\theta}-(1+\frac{1}{800}-\frac{1}{20}){sin\theta}$

$1013-981.25=(\frac{1}{50}-\frac{1}{800})+(\frac{1}{5}+\frac{1}{20}){sin\theta}$

$31.75=(0.02-0.00125)+(0.2+0.05){sin\theta}$

$31.75=(0.01875+0.25){sin\theta}$

$31.75=(18.75+0.25){sin\theta}$

$31.75-18.75=0.25{sin\theta}$

$13=0.25{sin\theta}$

${sin\theta}=\frac{13}{0.25}$

${sin\theta}=52^0$

Now dip moveout $=\frac{2xsin\theta}{v}$

For required moveout (ms/ meter) is

$=\frac{2*1000*52}{2000}$

$=52$

$=0.052ms/m$

Reference: Applied Geophysics by W.M.Telford, L.P.Geldart and R.E.Sheriff.