a) Distribution
b) Intersections
c) Connectivity of two families
d) Connectivity of 4 discontinuities from 2 families
e) Connectivity net formed by 3 sets of discontinuities
This tool contains 3 dinstinct parts:
Following JABOYEDOFF ET AL. (1996), the three parts are calculated as follows:
and
3.
= 1 m
Using frequencies li of 2 sets of discontinuity, the angle j between these 2 sets and the angle q between the intersection line and the pole to the surface of observation, the total length of the intersection lines ltot in a given volume V and the number of intersection lines Ni on a given surface S are calculated.
Following JABOYEDOFF ET AL. (1996), the total length of the insterection lines ltot is given by:
The number of intersection lines Ni is calculated as follows:
l1 = 1 m
l2 = 5 m
j = 90° (the two sets are orthogonal)
q = 0° (the surface of observation is perpendicular
to the intersection line between the two sets)
This tool calculates the threshold distances d1 and d2 of sets 1 and 2, the probability of crossing Pc, the probability of continuity Ps and the probability of intersection Pt of 2 sets of discontinuity, using the mean persistences and the mean frequencies of the 2 sets (l1, l1) and (l2, l2), the angle (j) between the 2 sets, the probability of continuity threshold Pct and the number of successive intersections Ni (number of "stairs").
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For the calculation of the threshold distancies d1 and d2, see section III.1.3. The probability of crossing Pc of the 2 sets is:
The probability of continuity Ps is given by:
Finally, the probability of intersection Pt for Ni successive intersections is given by:
l1 = 10 m
l1 = 0.1 m-1
l2 = 20 m
l2 = 0.2 m-1
j = 60°
Pct = 97.5 %
Ni = 1 (2 discontinuities) or 2 (4 discontinuities)
d1 = 36.89 m
d2 = 18.44 m
Pc = 5.51 %
Ps = 95.06 %
Pt = 5.24 % (2 disc.) or 0.275 % (4 disc.)
Using the mean persistences and the mean frequencies of 2 sets of discontinuity (l1, l1) and (l2, l2), the angle (j) between the 2 sets, this tool calculates:
Two solutions are used for this problem. The different probabilities for the first one are given by:
where lapp1 and lapp2 are the apparent frequencies l1 and l2 along the discontunities of the sets 2 and 1 respectively:
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and
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The second solution is a simplified version of the first. In this case, the probabilities P1 and P2 are given by:
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and
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l1 = 14 m
l1 = 0.111 m-1
l2 = 37 m
l2 = 0.036 m-1
j = 76°
Solution 1
P1 = 90.7 %
P2 = 8.7 %
P = 7.9 %
Solution 2
P1 = 86.4 %
P2 = 21.7 %
P = 18.7 %
This tool is still under testing and is not activated in this Beata release.