Spacing distribution and density of intersection tool

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


 

a) Distribution

(1) Description

This tool contains 3 dinstinct parts:

  1. discontinuities per volume: in this part, the probability of finding k or less than k discontinuities in a volume v are calculated using the mean spacing of the discontinuity set , a given number of discontinuities and a given volume
  2. spacing distribution: here, the probability of finding one or more discontinuities in a given interval of d meters is calculated using the mean spacing of the discontinuity set
  3. threshold distance: the interval distance dt is calculated for a given threshold Pt (probability of occurrence) using the mean spacing of the discontinuity set

(2) Solution

Following JABOYEDOFF ET AL. (1996), the three parts are calculated as follows:

  1. and

2.

3.

(3) Example

= 1 m

  1. k = 2
    v = 1 m3
    P(k, v) = 18.39 %
    P(<k, v) = 91.97 %
  2. d = 1
    P(³1, 1) = 63.21 %
  3. Pt = 99 %
    dt = 4.61 m

 

b) Intersections

(1) Description

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.

(2) Solution

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:

(3) Example

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)

  1. V = 1000 m³
    ltot = 5000 m
  2. S = 100 m²
    Ni = 500

 

c) Connectivity of 2 families

(1) Description

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").

(2) Solution

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:

(3) Example

l1 = 10 m
l1 = 0.1 m-1
l2 = 20 m
l
2 = 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.)

 

d) Connectivity of 4 discontinuities from 2 families

(1) Description

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:

(2) Solution

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:

and

The second solution is a simplified version of the first. In this case, the probabilities P1 and P2 are given by:

and

(3) Example

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 %

 

e) Connectivity net formed by 3 sets of discontinuities

This tool is still under testing and is not activated in this Beata release.