Geology Reference
In-Depth Information
Fig. 8.15
Schematic description of factors infl uencing the surface level, E, of a salt marsh located above an incompressible base of
consolidated sand
In this case, the salt marsh level change rate, D
E
, has to
be corrected for the eustatic level change rate, D
Eu
, in
order to give the rate of salt marsh level change in the
moving tidal frame D
E
rsl
:
Δ=Δ+Δ−Δ−Δ−Δ =Δ+Δ−Δ−Δ
ES
S
PI
US
S
PM
(8 .8)
rsl
sed
org
sed
org
It is here convenient to merge the isostatic and the
eustatic changes into one term describing the relative
sea-level change rate, D
M
. Eq.
8.8
is equivalent to the
equation suggested by Allan (Allan
1990
, p. 79) with
the exception that D
S
sed
in his version is exchanged
with D
S
min
being 'the time-rate of build up by mineral
sediment (m a
−1
)'. The choice of using D
S
sed
instead
of D
S
min
addresses the fact that most of the organic
material in silty salt marsh deposits is actually derived
from biogenic processes in the adjacent tidal area.
Here, more or less decomposed organic material com-
bines with mineral particles to form the resulting sedi-
ment source which feeds the salt marsh areas.
Separating organic and mineral material, therefore,
creates a problem as the measurable sediment supply
consists of both mineral and organic material. This
was also recognized by French (
1993
, p. 69) naming
this contribution 'externally derived (predominantly
clastic) material' and the organic one as 'that due to
internally derived plant detritus'. The latter, which
represents the organic below ground production, is
the same as D
S
org
and was estimated by French (
1993
)
to be on the order of 0.2 × 10
−3
m year
−1
for temperate
silty salt marshes.
In the following example, the salt marsh sediment
described in Figs.
8.13
and
8.14
is used as background
for practical solutions of the above stated variations of
the continuity equation for salt marsh sedimentation:
Given parameters: D
S
sed
= 4.18 × 10
−3
m year
−1
,
D
S
org
= 0.2 × 10
−3
m year
−1
, D
P
= 2.10 × 10
−3
m year
−1
,
D
I
= 0.5 × 10
−3
m year
−1
, D
Eu
= 1.78 × 10
−3
m year
−1
(
8.8
)
According to Eq.
8.7
, the absolute salt marsh level
is raised by:
−
3
−
3
−
3
−
3
−
1
−
3
−
1
Δ= ×+×−×−×
E
4.18
10
0.2
10
2.10
10
0.5
10
m year
=×
1.78
10
m year
Notice that if the isostatic term is neglected, we get:
−
3
−
3
−
3
−
3
−
1
Δ
E
=
4.18
×
10
+
0.2
×
10
−
2.10
×
10
=
2.28
×
10
m year
( without I )
Δ
