Questions? Review Consider the following two-layer problem -
Revisiting the problem discussed in class on Tuesday Z RV RH .000 1.000000 1.000000 .200 .9284767 .6770329 .400 .7808688 .4806249 .600 .6401844 .3620499 .800 .5299989 .2867962 1.000 .4472136 .2360680 1.200 .3846154 .2000000 1.400 .3363364 .1732137 1.600 .2982750 .1526108 1.800 .2676438 .1363084 2.000 .2425356 .1231055 2.200 .2216211 .1122055 2.400 .2039542 .1030602 2.600 .1888474 .0952811 2.800 .1757906 .0885849 3.000 .1643990 .0827627 3.200 .1543768 .0776539 3.400 .1454940 .0731363 3.600 .1375683 .0691128 3.800 .1304545 .0655074 4.000 .1240347 .0622578 4.200 .1182129 .0593147 4.400 .1129097 .0566359 4.600 .1080592 .0541887 4.800 .1036061 .0519428 5.000 .0995037 .0498762 5.200 .0957124 .0479660 5.400 .0921982 .0461979 5.600 .0889320 .0445547 5.800 .0858884 .0430231 1=20 mmhos/m 2=2 mmhos/m 3=20 mmhos/m Z1 = 0.5 Z2 = 1 Given the above diagram could you set up the equation below?
How many different conductivity layers will you actually have to consider? - 3 layer problem Does it matter whether d (depth) and s (intercoil spacing) are in feet or meters? - No Set up your equation following the example presented by McNeill and reviewed in class, and solve for the apparent conductivity recorded by the EM31 over this area of the spoil.
The equation you solved should have looked like this. Z RV RH .000 1.000000 1.000000 .200 .9284767 .6770329 .400 .7808688 .4806249 .600 .6401844 .3620499 .800 .5299989 .2867962 1.000 .4472136 .2360680 1.200 .3846154 .2000000 1.400 .3363364 .1732137 1.600 .2982750 .1526108 1.800 .2676438 .1363084 2.000 .2425356 .1231055 2.200 .2216211 .1122055 2.400 .2039542 .1030602 2.600 .1888474 .0952811 2.800 .1757906 .0885849 3.000 .1643990 .0827627 3.200 .1543768 .0776539 3.400 .1454940 .0731363 3.600 .1375683 .0691128 3.800 .1304545 .0655074 4.000 .1240347 .0622578 4.200 .1182129 .0593147 4.400 .1129097 .0566359 4.600 .1080592 .0541887 4.800 .1036061 .0519428 5.000 .0995037 .0498762 5.200 .0957124 .0479660 5.400 .0921982 .0461979 5.600 .0889320 .0445547 5.800 .0858884 .0430231 The equation you solved should have looked like this. where - 1 = 3 = 4 mmhos/m 2 = 100 mmhos/m z1 = (30/12) = 2.5 z2 = (40/12) = 3.33
The EM31 has a 12 foot intercoil spacing hence - z1 = (30 feet/12 feet) = 2.5 z2 = (40 feet/12 feet) = 3.33 Z RV RH .000 1.000000 1.000000 .200 .9284767 .6770329 .400 .7808688 .4806249 .600 .6401844 .3620499 .800 .5299989 .2867962 1.000 .4472136 .2360680 1.200 .3846154 .2000000 1.400 .3363364 .1732137 1.600 .2982750 .1526108 1.800 .2676438 .1363084 2.000 .2425356 .1231055 2.200 .2216211 .1122055 2.400 .2039542 .1030602 2.600 .1888474 .0952811 2.800 .1757906 .0885849 3.000 .1643990 .0827627 3.200 .1543768 .0776539 3.400 .1454940 .0731363 3.600 .1375683 .0691128 3.800 .1304545 .0655074 4.000 .1240347 .0622578 4.200 .1182129 .0593147 4.400 .1129097 .0566359 4.600 .1080592 .0541887 4.800 .1036061 .0519428 5.000 .0995037 .0498762 5.200 .0957124 .0479660 5.400 .0921982 .0461979 5.600 .0889320 .0445547 5.800 .0858884 .0430231 Given the tables of R values at right RV(2.5) ~ 0.197 (average of R’s for z = 2.4 and 2.6) RV(3.33) ~ 0.149 (2/3rds the way from 3.2 to 3.4) Given also that 1 = 3 = 4 mmhos/m 2 = 100 mmhos/m
Recall those “rules of thumb” regarding the optimal sensing depth or exploration depth. For the EM31 operated in the vertical dipole mode the “ROT” says exploration depth is 18feet. Examining the terms on the equation we just computed - you can see that the middle term - which arises from an average depth of 35 feet - contributes significantly to the apparent conductivity measured at this location. More than 50% of the value of ground conductivity comes from the layer centered at depths well beyond (almost twice) the optimal exploration depth. This is a point to keep in mind especially when trying to locate contamination zones which may have abnormally high conductivity. We might normally exclude use of the EM31 in attempts to detect something at depths greater than 20 feet or so.