Isostasy and basin analysis powerpoint
-
Upload
serc-at-carleton-college -
Category
Technology
-
view
599 -
download
3
description
Transcript of Isostasy and basin analysis powerpoint
Isostasy in Geology andBasin Analysis
INCOMPLETE DRAFT
This exercise is drawn from Angevine, Heller and Paola (1990), with inspiration and essential planning by R. Dorsey. A. Martin-Barajas generously provided material used in this exercise.
Archimedes Principle: When a body is immersed in a fluid, the fluid exerts an upward force on the body that is equal to the weight of the fluid that is displaced by the body.”
This rule applies to all mountain belts and basins under conditions of local (Airy) isostatic compensation: the lithosphere has no lateral strength, and thus each lithospheric column is independent of neighboring columns (e.g. rift basins).
To work isostasy problems, we assume that the lithosphere (crust + upper mantle) is “floating” in the fluid asthenosphere. A simple, nongeologic example looks like this -
Solid
rs
1 2
h2h1Fluid (rf)
depth of equal compensation
Because fluid has no shear strength (yield stress =0),it cannot maintain lateral pressure differences. It will flow to eliminate the pressure gradient.
Solid
rs
1 2
h2h1Fluid (rf)
depth of equal compensation
Solid
rs
1 2
h2h1Fluid (rf)
depth of equal compensation
To calculate equilibrium forces, set forces of two columns equal to each other: F1 = F2 (f=ma)
m1xa = m2xa
m1 = m2
(gravitational acceleration
cancels out)
Because m=rxv (density x volume),convert to m=rh, and: rfh1=rsh2
Solid
rs
1 2
h2h1Fluid (rf)
depth of equal compensation
This equation correctly describes equilibrium isostatic balance in the diagram.
Onward to geology –
EXAMPLE 1 Estimate thickness of lithosphere:In this example, we’ve measured the depth to the moho (hc) using seismic refraction.Elevation (e) is known, and standard densities for thecrust, mantle, and asthenosphere are used:rc=2800 kg/m3
rm=3400 kg/m3
ra=3300 kg/m3
rc
elevation=3km
rm
hc=35km
hm=?
asthenosphere (rc)
Z=?
How deep to the base of the lithosphere?
Solve for Z:ra(Z) = rc(hc+e) + rm(Z-hc)
ra(Z) - rmZ = rc(hc+e) - rmhc
rc(hc+e) - rmhc
(ra-rm)
rc
elevation=3km
rm
hc=35km
hm=?
asthenosphere (ra)
Z=?
Z=
How deep to the base of the lithosphere?
Solve for Z:ra(Z) = rc(hc+e) + rm(Z-hc)
ra(Z) - rmZ = rm(Z-hc) - rmhc
rc(hc+e) - rmhc
(ra-rm)
2800(35+3) – 3400(35)(3300-
3400) -12,600
-100
rc
elevation=3km
rm
hc=35km
hm=?
asthenosphere (rc)
Z=?
Z=
=
= Z = 126 km
EXAMPLE 2 What is the effect of filling a basin with sediment?
Consider a basin 1km deep that is filled only with water. How much sediment would it take to fill the basin up to sea level?
rC=2800
rm=3400
ra= 3300
water
crust
mantlelithosphere
rs= 2300
sediment
crust
mantlelithosphere
depth of equal compensation
Let rw= 1000 kg/m3
Let rs= 2300 kg/m3 ho= 1km hc
hm
hs=?
rC=2800
rm=3400
1 2
rC=2800
rm=3400
ra= 3300
water
crust
mantlelithosphere
rs= 2300
sediment
crust
mantlelithosphere
depth of equal compensation
ho= 1km hc
hm
hs=?
rC=2800
rm=3400
Remember, force balance must be calculated for entire column down to depth of compensation (=depth below which there is no density difference between columns). Also, thickness of crust and mantle lithosphere does not change, so they cancel out on both sides of the equation.
1 2
rC=2800
rm=3400
ra= 3300
water
crust
mantlelithosphere
rs= 2300
sediment
crust
mantlelithosphere
depth of equal compensation
ho= 1km hc
hm
(hs-ho)
hs=?
rC=2800
rm=3400
rwho + rchc + rmhm + ra(hs-ho) = rshs + rchc + rmhmrwho + ra(hs-ho) = rshsrwho + rahs – raho = rshshs(ra- rs) = raho – rwhohs = ho(ra- rw)
(ra- rs)
rC=2800
rm=3400
ra= 3300
water
crust
mantlelithosphere
rs= 2300
sediment
crust
mantlelithosphere
depth of equal compensation
ho= 1km hc
hm
(hs-ho)
hs=?
rC=2800
rm=3400
hs = ho(ra- rw) (ra- rs)
= 1.0 (3.3-1.0)
(3.3-2.3)= 2.3 km
rC=2800
rm=3400
ra= 3300
water
crust
mantlelithosphere
rs= 2300
sediment
crust
mantlelithosphere
depth of equal compensation
ho= 1km hc
hm
(hs-ho)
hs=?
rC=2800
rm=3400
“rule of thumb”: the thickness of sediment needed to fill
a basin is ~ 2.3 times the depth of water that the sediment replaces
EXAMPLE 2B Sediment filling – Alarcon Basin example
Determine the maximum water depth in the Alarcon Basin from your profile or spreadsheet.
Calculate how much sediment would be needed to fill the Alarcon Basin up to sea level.
rC=2800
rm=3400
ra= 3300
water
crust
mantlelithosphere
rs= 2300
sediment
crust
mantlelithosphere
depth of equal compensation
ho= 1km hc
hm
(hs-ho)
hs=?
rC=2800
rm=3400
hs = ho(ra- rw) (ra- rs)
= 3 (3.3-1.0)
(3.3-2.3)= 6.9 km
crust
mantlelitho-
sphere
crust
mantlelitho-
sphere
EXAMPLE 3 How does crustal thinning effect the depth of sedimentary basins?
1 2
hc1=30km hc2
hm1=90km
hm2
ha
Newly-formed basin
Thin crust and mantle lithosphere to half of original.How deep a basin forms in response to thinning?
Solve for Z.
Note: ha = hc1 + hm1 - hc2 - hm2 - Z
Z
crust
mantlelitho-
sphere
crust
mantlelitho-
sphere
1 2
hc1=30km hc2
hm1=90km
hm2
ha
Note: ha = hc1 + hm1 - hc2 - hm2 - Z
Z
rc(hc1) + rm(hm1) = rw(Z) + rc(hc2) + rm(hm2) + ra(ha) 30rc + 90rm = rw(Z) + 15rc + 45rm + ra(120-15-45-Z)30rc + 90rm = Zrw + 15rc + 45rm + 60ra- Zra
Z(ra-rw) = 60ra-45rm-15rc
crust
mantlelitho-
sphere
crust
mantlelitho-
sphere
1 2
hc1=30km hc2
hm1=90km
hm2
ha
Z
Z(ra-rw) = 60ra-45rm-15rc
If the new basin is filled by water,what is its depth (Z)? -
A: Fill with water(rw = 1.01 g/cm2)
Z=60ra-45rm-15rc
(ra – rw)
crust
mantlelitho-
sphere
crust
mantlelitho-
sphere
1 2
hc1=30km hc2
hm1=90km
hm2
ha
Z
Z(ra-rw) = 60ra-45rm-15rc
If the new basin is filled by water -
A: Fill with water(rw = 1.01 g/cm2)
Z=60ra-45rm-15rc
(ra – rw)
60(3.3)- 45(3.4)-15(2.8) (3.3-1.01)=
=3.002.29
= 1.31 kmfor water
crust
mantlelitho-
sphere
crust
mantlelitho-
sphere
1 2
hc1=30km hc2
hm1=90km
hm2
ha
Z
Z(ra-rw) = 60ra-45rm-15rc
If the new basin is filled by sediment,what is its depth? -
B: Fill with sediment(rw = 2.3 g/cm2)
Z=60ra-45rm-15rc
(ra – rw)
crust
mantlelitho-
sphere
crust
mantlelitho-
sphere
1 2
hc1=30km hc2
hm1=90km
hm2
ha
Z
Z(ra-rw) = 60ra-45rm-15rc
Basin is formed:A: fill with waterB: fill with sediment
B: Fill with sediment(rs = 2.3 g/cm2)
Z=60ra-45rm-15rc
(ra – rs)
60(3.3)- 45(3.4)-15(2.8) (3.3-2.3)=
=3.001.0
= 3.0 kmfor sediment
EXAMPLE 3B Use Upper Delfin Basin sediment fill and crustal thickness to estimate amount of thinning of mantle lithosphere.
crust
mantlelitho-
sphere
crust
mantlelitho-
sphere
1 2
hc1 hc2 =
hm1 hm2=
Sedimentthickness
Mantle lithosphereremoved
Delfin Basin
Tiburon Basin
Location map, Northern Gulf of California
Low-angle detachment fault
intrusions
Interpreted seismic line, Delfin Basin and Tiburon Basin
Thickness of sediments and crust are interpreted from seismic lines
Minimum thicknesses, Delfin Basin Sediments 4 km
Metasediments 4 kmIntrusions 0.4 kmOther 1 km
from Dorsey, 2010, Table 1
EXAMPLE 3B Use Upper Delfin Basin sediment fill and crustal thickness to estimate amount of thinning of mantle lithosphere.
crust
mantlelitho-
sphere
crust
mantlelitho-
sphere
1 2
hc1=35km hc2=
~10km?
hm1=?km
hm2=?km
Sedimentthickness
Mantle lithosphereremoved
hs=4km
Notes, 9-20-13Example 3b - another G of CA example – how much mantle lithosphere has been removed? [“inversion” of the question. First solve for amt of sed fill, 2nd measure crust and sed. Thickness and solve for amt of mantle lithosphere removed]
Dorsey 2010 Geology paper – table of approximate sediment & metasediment fill in basin (Salton Trough and other northern G of CA basins)
Need 1) initial (pre-rifting crustal thickness – Martin-Barajas paper in press, or Couch et al. ‘91??) [~35 km]2) Thinned crust (either in Couch, or in an existing Martin-Barajas lead or co-author) [~ 14ish, of which 7ish is sed.s + metased.s]