Calculation of true thickness of the bed

 

12. Apparent resistivity sounding data for Schlumberger array is theoretically generated by the teacher for the following 4-layer model as $\rho_1$=100Ωm, $\rho_2$=20Ωm, $\rho_3$=500Ωm, $\rho_4$=10Ωm and the layer thicknesses $h_1$=50m, $h_2$=20m, $h_3$=50m. If the student interprets this theoretical sounding data for $\rho_3$ as 750Ωm, then according to the Principle of Equivalence, the thickness $h_3$ would be __________m (round off to 2 decimal places)

 

Solution:

 

 

Principle of equivalence: It is impossible to distinguish between two highly resistive beds of different "h" and "$\rho$" values if the product "$h*\rho$" is the same.

 

So,

Apparent resistivity sounding data for Schlumberger array is theoretically generated by Teacher

 $h_3=50m$

 

and $\rho_3 =500$ ohm m

 

If the student interprets this theoretical sounding data,

 

 $\rho_3 =750$ ohm m

 

And what is the $h_3$ = ?

 

$h_3*\rho_3 =h_3*\rho_3$

 

$50*500 =h_3*750$

 

$h_3*750=50*500$

 

$h_3=\frac{50*500}{750}$

 

$h_3=33.33 m$

 

 

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