Questions and answers: fractures and hemispherical projection 101(b) How many pieces of core could be expected to have a length
greater than 0.2 m, and what is their mean length?A7.1 (a) We know the length of core, and so in order to calculate the
number of pieces we have to determine the fracture frequency (because
the number of pieces, N, is found from N = hL). As we know the RQD
of the core and the threshold value at which this was measured, we can
compute the frequency from the fundamental formula for RQD, i.e.
RQD = 100(ht + 1) exp(-At).
This formula for RQD is implicit in h (which means we cannot rearrange
it to leave A. by itself on one side of the equation), and so can only be
solved iteratively (or by trial and error). Thus, with RQD = 58% and
t = 0.1 m, we find that A = 14.344 m-'. Having found a value for the
frequency, we calculate the total number of fractures present in the core
as
Nt = hL = 14.344 * 134 = 1922.2
and, as a part of a piece cannot exist, this result should be rounded down
to 1922.
The definition of RQD is "the ratio of the total length of those pieces of
core longer than 0.1 m to the total length of the core recovered, expressed
as a percentage". In this case, the RQD is 58%, which directly represents
the proportion of core composed of pieces longer than 0.1 m. As the
overall length of the core is 134 m, the total length of these pieces is then
0.58 x 134 = 77.72 m.
(b) The number of pieces whose individual lengths are greater than
0.2 m is computed from the proportion of the total number that com-
prises pieces longer than 0.2 m. In turn, this proportion is found by
computing the probability of finding an interval of core with no fractures
over a length of 0.2 m.
If we sample the entire fractured core, the probability of finding a
fracture must be 1. This is stated mathematically as the area under the
probability density distribution, which is
Jdm exp(-h) d~ = 1.
If we are only interested in a particular range of spacing values, then this
integral is evaluated using limits that represent the range. So, for spacing
values between 0 and some value b we have
Pr(x I b) = hexp(-Ax) dx = 1 - exp(-Ab)
Ib
and, if we are only interested in spacing values larger than b, then
Pr(x > b) = 1 - Pr(x I b) = 1 - [1 - exp(-Ab)] = exp(-Ab).
In the case under consideration we have b = 0.2 m, giving
Pr(x > 0.2) = exp(-h - 0.02) = 0.057,