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

CSIR-NET JUNE 2018

99) An n-fold 12 channel CDP survey was carried out with the source spacing (S)equal to the receiver spacing(R).The source spacing required in units of receiver spacing, to obtain a ‘2n’ foldage is

a)S=0.5R 

b)S=0.75R 

c)S=R 

d)S=2R

(Thanks to Suresh sir ANU)

(Thanks to Neeraja AU)

sol:

$Foldage=\frac{1}{2}[number of Geophones (or)channels\times[\frac{Geophone spacing(or)receiver spacing(R)}{Source spacing(S)}]]$

$Foldage=\frac{1}{2}[number of channels\times[\frac{(R)}{(S)}]]------(1)$

$n fold\rightarrow12channel$

S=R

$n=\frac{1}{2}[12\times\frac{R}{S}]$

R=S

$n=\frac{1}{2}[12\times1]$

$n=6            ------(2)$

2n foldage

$2n=\frac{1}{2}[12\times\frac{R}{S}]-(3)$

n=6

$2\times6=\frac{1}{2}[12\times\frac{R}{S}]$

$12=\frac{1}{2}[12\times\frac{R}{S}]$

$24=[12\times\frac{R}{S}]$

$\frac{1}{24}=\frac{S}{12R}$

$\frac{S}{R}=\frac{12}{24}$

$\frac{S}{R}=0.5$

S=0.5R