powerpoint lecture with simple calculations

1. Isostasy in Geology and
Basin 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.

2. 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.”

3. 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).

4. 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 -
1
2
h1
h2
Solid
s
Fluid ( f)
depth of equal
compensation

5. Because fluid has no shear strength (yield stress =0),
it cannot maintain lateral pressure differences. It
will flow to eliminate the pressure gradient.
1
2
h1
h2
Solid
s
Fluid ( f)
depth of equal
compensation

6. To calculate equilibrium forces, set forces of two
columns equal to each other: F1 = F2
(f=ma)
m1xa = m2xa
m1 = m2
1
2
(gravitational acceleration
cancels out)
h1
h2
Solid
s
Fluid ( f)
depth of equal
compensation

7. Because m= xv (density x volume),
convert to m= h, and: fh1= sh2
1
This equation
correctly
describes
equilibrium
isostatic
balance in the
diagram.
2
h1
h2
Solid
s
Fluid ( f)
depth of equal
compensation

8. Onward to geology –

9. 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 the
crust, mantle, and
asthenosphere are used:
=2800 kg/m3
c
=3400 kg/m3
m
=3300 kg/m3
a
elevation=3km
c
hc=35km
Z=?
m
hm=?
asthenosphere ( c)

10. How deep to the base of the lithosphere?
Solve for Z:
a(Z) = c(hc+e) + m(Z-hc)
a(Z) - mZ = c(hc+e) - mhc Z=?
c(hc+e) - mhc
Z=
( a- m)
elevation=3km
c
m
hc=35km
hm=?
asthenosphere ( a)

11. How deep to the base of the lithosphere?
Solve for Z:
elevation=3km
c
hc=35km
a(Z) = c(hc+e) + m(Z-hc)
a(Z) - mZ = m(Z-hc) - mhc Z=?
m
hm=?
c(hc+e) - mhc
Z=
( a- m)
asthenosphere ( c)
2800(35+3) – 3400(35)
=
(3300-3400)
-12,600
Z = 126 km
=
-100

12. EXAMPLE 2 What is the effect of filling a basin with
sediment?

13. 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?
2
1
ho=
1km
hc
hm
water
C=
2800
m=
3400
a=
hs=?
crust
mantle
lithosphere
3300
s=
2300
C=
2800
m=
3400
Let w= 1000
kg/m3
sediment
Let s= 2300 kg/m3
crust
mantle
lithosphere
depth of equal
compensation

14. 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
water
ho=
s=
hs=? 2300 sediment
1km
=
C
hc
2800 crust
C=
crust
2800
m=
mantle
hm
mantle
3400 lithosphere
m=
3400 lithosphere
a= 3300
depth of equal
compensation

15. who
+ chc + mhm + a(hs-ho) = shs + chc + mhm
who + a(hs-ho) = shs
who + ahs – aho = shs
hs( a- s) = aho – who
hs = ho( a- w)
a- s)
water
ho=
sediment
s=
hs=?
1km
=
C
2300
hc
2800 crust
C=
crust
2800
m=
mantle
hm
mantle
3400 lithosphere
m=
3400 lithosphere
(hs-ho) a= 3300
depth of equal
compensation

16. ho( a- w)
hs =
a- s)
ho=
1km
hc
hm
(hs-ho)
=
water
C=
2800
m=
3400
a=
3300
1.0 (3.3-1.0)
(3.3-2.3)
hs=?
crust
s=
2300
C=
2800
mantle
lithosphere
m=
3400
= 2.3 km
sediment
crust
mantle
lithosphere
depth of equal
compensation

17. “rule of thumb”: the thickness of sediment needed to fill
a basin is ~ 2.3 times the depth of water that the
sediment replaces
ho=
1km
hc
hm
(hs-ho)
water
C=
2800
m=
3400
a=
3300
hs=?
crust
mantle
lithosphere
s=
2300
C=
2800
m=
3400
sediment
crust
mantle
lithosphere
depth of equal
compensation

18. 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.

19. ho( a- w)
hs =
a- s)
ho=
1km
hc
hm
(hs-ho)
=
water
C=
2800
m=
3400
a=
3300
3 (3.3-1.0)
(3.3-2.3)
hs=?
crust
mantle
lithosphere
s=
2300
C=
2800
m=
3400
= 6.9 km
sediment
crust
mantle
lithosphere
depth of equal
compensation

20. EXAMPLE 3 How does crustal thinning effect
the depth of sedimentary basins?
1
hc1=
30km
2
Z
crust
mantle
hm1=
90km lithosphere
hc2
crust
hm2
mantle
lithosphere
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

21. c(hc1)
+ m(hm1) = w(Z) + c(hc2) + m(hm2) + a(ha)
30 c + 90 m = w(Z) + 15 c + 45 m + a(120-15-45-Z)
30 c + 90 m = Z w + 15 c + 45 m + 60 a- Z a
Z( a- w) = 60 a-45 m-15 c
1
hc1=
30km
2
Z
crust
mantle
hm1=
90km lithosphere
hc2
crust
hm2
mantle
lithosphere
ha
Note: ha = hc1 + hm1 - hc2 - hm2 - Z

22. Z( a-
w)
= 60 a-45
m-15 c
If the new basin is filled by water,
what is its depth (Z)? 1
hc1=
30km
2
Z
crust
mantle
hm1=
90km lithosphere
hc2
crust
hm2
mantle
lithosphere
ha
A: Fill with water
( w = 1.01 g/cm2)
60 a-45 m-15
Z=
( a – w)
c

23. Z( a-
w)
= 60 a-45
m-15 c
If the new basin is filled by water 1
hc1=
30km
2
Z
crust
mantle
hm1=
90km lithosphere
hc2
crust
hm2
mantle
lithosphere
ha
A: Fill with water
( w = 1.01 g/cm2)
60 a-45 m-15
Z=
( a – w)
c
60(3.3)- 45(3.4)-15(2.8)
=
(3.3-1.01)
3.00
=
2.29
= 1.31 km
for water

24. Z( a-
w)
= 60 a-45
m-15 c
B: Fill with sediment
If the new basin is filled by sediment, ( = 2.3 g/cm2)
w
what is its depth? 60 a-45 m-15 c
1
2
Z=
( a – w)
Z
hc1=
crust
hc2
30km
crust
hm1= mantle
litho90km
sphere
hm2
ha
mantle
lithosphere

25. Z( a-
w)
= 60 a-45
m-15 c
Basin is formed:
A: fill with water
B: fill with sediment
1
Z
hc1=
crust
hc2
30km
mantle
hm1=
90km lithosphere
hm2
ha
B: Fill with sediment
( s = 2.3 g/cm2)
2
crust
mantle
lithosphere
60 a-45 m-15
Z=
( a – s)
c
60(3.3)- 45(3.4)-15(2.8)
=
(3.3-2.3)
3.00
=
1.0
= 3.0 km
for sediment

26. EXAMPLE 3B Use Upper Delfin Basin
sediment fill and crustal thickness to estimate
amount of thinning of mantle lithosphere.
1
hc1
hm1
crust
mantle
lithosphere
2
hc2 =
Sediment
thickness
crust
mantle
hm2= lithosphere
Mantle lithosphere
removed

27. Location map, Northern Gulf
of California
Delfin Basin
Tiburon Basin

28. Interpreted seismic line, Delfin Basin and Tiburon
Basin
Low-angle detachment
fault
intrusions
Thickness of sediments and crust are interpreted
from seismic lines

29. Minimum thicknesses, Delfin Basin
Sediments
4 km
Metasediments
4 km
Intrusions
0.4 km
Other
1 km
from Dorsey, 2010, Table 1

30. EXAMPLE 3B Use Upper Delfin Basin
sediment fill and crustal thickness to estimate
amount of thinning of mantle lithosphere.
1
hc1=
35km
crust
hs=4km
hc2=
~10km?
mantle hm2=
hm1=
litho- ?km
?km
sphere
2
Sediment
thickness
crust
mantle
lithosphere
Mantle lithosphere
removed

33. Notes, 9-20-13
Example 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]