a) 2 discontinuity sets
b) 2 discontinuity set and a direction of fracturation
c) 3 discontinuity sets
This tool converts the mean spacings (L) / mean persistences (l) of 2 discontinuity sets (L1, l1) and (L2, l2) to a compartment mean volume estimate, using the angle (j) between the 2 sets.
This calculation method is adequate for fresh and homogeneous compartments, comporting as a block
Following JABOYEDOFF ET AL. (1996), the mean volume Vm of the compartment is defined as follows:
T is the thickness of the compartment. It is choosen as the half of the mean persistence of the less persistent set, and is given by:
S^ is the surface of the compartment in a direction perpendicular to the intersection of the 2 sets of discontinuity:
where q is the acute angle between the two sets of discontinuity. The distancies d1 and d2 are taken as the mean spacing of the discontinuities or may be calculated using a probability threshold (see Spacing distribution and density of intersection tool for more details). Thus, the volume of the compartment Vm is given by:
(4) Example
L1 = 10 m
l1 = 10 m
L2 = 5 m
l2 = 20 m
j = 60°
Vm = 367.6 m³
This tool converts the mean spacings of 2 discontinuity sets L1 and L2, the angle j between the 2 sets and the apparent frequency l^12 ( in a direction perpendicular to the intersection of the two discontinuity sets) to a compartment mean volume estimate.
This calculation method is adequate for highly fractured rocks presenting a high apparent fractures frequency in a direction close to the intersection of the two sets.
Sets 1 and 2 form a potentially instable wedge.
In this case, the thickness T of the compartment is given by:
So Vm is given by:
L1 = 10 m
L2 = 5 m
j = 60°
l^12 = 4 m-¹
Vm = 14.4 m³
This tool converts the mean spacings, dip direction and dip angle of 3 discontinuity sets (L1, a1, b1), (L2, a2, b2) and (L3, a3, b3) to a compartment mean volume estimate.
For 3 discontinuity sets delimiting prisms.
The estimated mean volume Vm corresponds to the volume of the (1, 2, 3) parallelepiped, which is given by:
where the vectors are
normal to the sets 1, 2 and 3.
L1 = 10 m
a1 = 352°
b1 = 0°
L2 = 5 m
a2 = 352°
b2 = 60°
L3 = 3 m
a3 = 82°
b3 = 90°
Vm = 173.2 m³