Page 271 - Physical Principles of Sedimentary Basin Analysis

7.16 Stretching and thinning of the sediments 253

are the basin thickness from burial history modeling or back-stripping, and they do not
involve stretching of the sedimentary basin. We will also assume four rift phases in the
following. A generalization to more or less rift phases is straightforward. The cumulative
β-factor at the beginning of the first rift phase is
β max,1 = 1 (7.198)

because the crust is unstretched until then. At the beginning of the second rift phase (at
time t 2 ) the cumulative β-factor becomes
c 0
β max,2 = (7.199)
c 0 − f w w 2 − β 2 β 3 β 4 f 2 s 2
where f i = ( m − i )/( m − c ) and i is the average basin density at time t i .The
thickness of the basin is now increased by the product β 2 β 3 β 4 . The sediments deposited
to the beginning of the second rift phase, with the thickness s 2 , must have been a factor
β 2 β 3 β 4 thicker, because they have experienced rift phases 2, 3, and 4. Note 7.18 shows how
porous sediments can be stretched by a β-factor in a mass-conservative manner. Similarly,
the cumulative β-factor for the third rift phase is
c 0
β max,3 = (7.200)
c 0 − f w w 3 − β 3 β 4 f 3 s 3
because the sediments deposited until the beginning of the third rift phase, with the thick-
ness s 3 , have gone through rift phases 3 and 4. Therefore, this thickness was a factor β 3 β 4
thicker at the beginning of rift phase 3. At the beginning of the fourth rift phase we have
c 0
β max,4 = (7.201)
c 0 − f w w 4 − β 4 f 4 s 4
and at the present time
c 0
β max,5 = . (7.202)
c 0 − f w w 5 − f 5 s 5
At the same time the β-factor for each rift phase is given by the cumulative β-factor by
expression (7.190). These equations for the β-factors can be solved by the following pro-
cedure, as shown in Note 7.17, where the last β-factor is found first, which is then used to
obtain the next last β-factor and so forth. The β-factors in reverse order become

c 0 − f w w 4
β 4 = (7.203)
c 0 − f w w 5 − ( f 5 s 5 − f 4 s 4 )
c 0 − f w w 3
β 3 = (7.204)
c 0 − f w w 4 − β 4 ( f 4 s 4 − f 3 s 3 )
c 0 − f w w 2
β 2 = (7.205)
c 0 − f w w 3 − β 4 β 3 ( f 3 s 3 − f 2 s 2 )
c 0 − f w w 1
β 1 = . (7.206)
c 0 − f w w 2 − β 4 β 3 β 2 ( f 2 s 2 − f 1 s 2 )