THE SMOOTH SOUNDING GRAPH. A Manual for Field Work ... - BGR
THE SMOOTH SOUNDING GRAPH. A Manual for Field Work ... - BGR THE SMOOTH SOUNDING GRAPH. A Manual for Field Work ... - BGR
and formula (13) will then be 49 U K K ρa = K = jρ1 + j'ρ 1 I aI aI For our two cases I and II this means K ( I ) K I. ρ a = j × 10 000 + j' × 10 000 aI aI K ( II ) K II. ρ a = j × 10 + j' × 10 aI aI Comparing the current densities j (I) and j (II) and keeping in mind that only here the information from the underground with respect to its resis- tivity distribution is concentrated we see that j (I) is only 1‰ (promille!) of j (II) because the ratio ρ1 (II) :ρ1 (I) is 1:1000. From this simple calculation we learn that measuring the same voltage U between M and N the information from the underground is 1000 times weaker in the case of a highly resistive surface (case I) than in the case of a well conducting surface (case II). Now we look on the second term in the last two formulas containing the disturbing "quasi-constant" current density j’. Perhaps the reader was a bit surprised on the form of these last formulas I. and II. because the two terms on the right side are written in a different way. But in the second j' "disturbing" term the quotient is constant in both cases I. and II. The aI remaining K × ρ1 is 1000 times larger in case I. than in case II. The result of comparing both cases will be: In case I. the first term shows a 1000 times weaker j (I) - that is the underground information - than in case II. On the other hand the "disturb- ing" second term with j’ is 1000 times larger in case I. than in case II. The conclusion will be that in case I. the first term can be neglected as during continuing the measurement we get into the situation j (I)
50 Fig.31. They are ascending with an angle of ~63.5° (i.e. arctan(2) from formula (10) in chapter 1.3), drawn in bi-log. scale. The result: On a highly resistive surface a disturbing leakage current will suppress the underground information (j (I) ) and finally the sounding graph will run into a 63.5° ascending rear branch. This will happen if j' is positive, i.e. really added to j (I) .
- Page 1 and 2: THE SMOOTH SOUNDING GRAPH A Manual
- Page 3 and 4: Preface 2 This manual shall be a pr
- Page 5 and 6: 1. Basic rules 4 The first chapter
- Page 7 and 8: This current density is marked as j
- Page 9 and 10: 8 Fig.2 Fig.3 Fig.4 Fig.5
- Page 11 and 12: 10 1.3. The four-electrode arrangem
- Page 13 and 14: 12 Fig.6 Fig.7 Fig.8
- Page 15 and 16: 14 If we compare equations (10) and
- Page 17 and 18: Case 2 (Fig. 10) 16 Now we observe
- Page 19 and 20: 1.5. The fundamental principle for
- Page 21 and 22: Case 3 (Fig.14) 20 The electrode di
- Page 23 and 24: with just the same factor K. 22 Aft
- Page 25 and 26: 24 Fig17 layers with different resi
- Page 27 and 28: 26 Simulating this zooming by enlar
- Page 29 and 30: 28 1.6. Shifting of potential elect
- Page 31 and 32: 30 Fig.21 Fig.22 Fig23
- Page 33 and 34: 32 2.1. How to carry out a field me
- Page 35 and 36: L/2 a/2 1,5 6,28 2 11,8 2,5 18,9 34
- Page 37 and 38: 36 Fig.27 Fig.28 Before we start th
- Page 39 and 40: 38 fence, ditch) especially, if the
- Page 41 and 42: 40 by experience. The measurement i
- Page 43 and 44: 42 2.2. Possible errors influencing
- Page 45 and 46: 2.2.4. Crossing a ditch (Fig.26/29)
- Page 47 and 48: 46 accuracy is not so important the
- Page 49: 2.2.8. Insulation and leakage curre
- Page 53 and 54: 52 If j' is negative, i.e. the dist
and <strong>for</strong>mula (13) will then be<br />
49<br />
U K K<br />
ρa = K = jρ1<br />
+ j'ρ<br />
1<br />
I aI aI<br />
For our two cases I and II this means<br />
K ( I ) K<br />
I. ρ a = j × 10 000 + j'<br />
× 10 000<br />
aI<br />
aI<br />
K ( II ) K<br />
II. ρ a = j × 10 + j'<br />
× 10<br />
aI aI<br />
Comparing the current densities j (I) and j (II) and keeping in mind that<br />
only here the in<strong>for</strong>mation from the underground with respect to its resis-<br />
tivity distribution is concentrated we see that j (I) is only 1‰ (promille!)<br />
of j (II) because the ratio ρ1 (II) :ρ1 (I) is 1:1000.<br />
From this simple calculation we learn that measuring the same voltage U<br />
between M and N the in<strong>for</strong>mation from the underground is 1000 times<br />
weaker in the case of a highly resistive surface (case I) than in the case<br />
of a well conducting surface (case II).<br />
Now we look on the second term in the last two <strong>for</strong>mulas containing the<br />
disturbing "quasi-constant" current density j’. Perhaps the reader was a<br />
bit surprised on the <strong>for</strong>m of these last <strong>for</strong>mulas I. and II. because the two<br />
terms on the right side are written in a different way. But in the second<br />
j'<br />
"disturbing" term the quotient is constant in both cases I. and II. The<br />
aI<br />
remaining K × ρ1<br />
is 1000 times larger in case I. than in case II. The result<br />
of comparing both cases will be:<br />
In case I. the first term shows a 1000 times weaker j (I) - that is the underground<br />
in<strong>for</strong>mation - than in case II. On the other hand the "disturb-<br />
ing" second term with j’ is 1000 times larger in case I. than in case II.<br />
The conclusion will be that in case I. the first term can be neglected as<br />
during continuing the measurement we get into the situation<br />
j (I)
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