1) The document discusses isostasy and how it relates to balancing forces in geological basins and mountain belts. It provides examples of calculating equilibrium forces and depths of compensation. 2) One example shows that thinning the crust and mantle lithosphere by half would form a basin around 1.3 km deep if filled with water or 2.7 km deep if filled with sediment. 3) Experiments are described to find rifting parameters that could produce the crustal structure observed in the Upper Delfin basin, including initial thicknesses and sediment densities.

1. Isostasy in Geology and
Basin Analysis
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 -
Solid
rs
1 2
h2
h1
Fluid (rf)
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.
Solid
rs
1 2
h2
h1
Fluid (rf)
depth of equal
compensation

6. Solid
rs
1 2
h2
h1
Fluid (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)

7. Because m=rxv (density x volume),
convert to m=rh, and: rfh1=rsh2
Solid
rs
1 2
h2
h1
Fluid (rf)
depth of equal
compensation
This equation
correctly
describes
equilibrium
isostatic
balance in the
diagram.

8. Onward to geology –
Courtesy Sue Cashman

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:
rc=2800 kg/m3
rm=3400 kg/m3
ra=3300 kg/m3
rc
elevation=3km
rm
hc=35km
hm=?
asthenosphere (rc)
Z=?

10. 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=

11. 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

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

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?
rC=
2800
rm=
3400
ra= 3300
water
crust
mantle
lithosphere
rs=
2200
sediment
crust
mantle
lithosphere
depth of equal
compensation
Let rw= 1000
kg/m3
Let rs= 2200 kg/m3
ho=
1km
hc
hm
hs=?
rC=
2800
rm=
3400
1 2

14. rC=
2800
rm=
3400
ra= 3300
water
crust
mantle
lithosphere
rs=
2200
sediment
crust
mantle
lithosphere
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

15. rC=
2800
rm=
3400
ra= 3300
water
crust
mantle
lithosphere
rs=
2200
sediment
crust
mantle
lithosphere
depth of equal
compensation
ho=
1km
hc
hm
(hs-ho)
hs=?
rC=
2800
rm=
3400
rwho + rchc + rmhm + ra(hs-ho) = rshs + rchc + rmhm
rwho + ra(hs-ho) = rshs
rwho + rahs – raho = rshs
hs(ra- rs) = raho – rwho
hs = ho(ra- rw)
(ra- rs)

16. rC=
2800
rm=
3400
ra= 3300
water
crust
mantle
lithosphere
rs=
2200
sediment
crust
mantle
lithosphere
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)
(3.3-2.2)
= 2.1 km

17. rC=
2800
rm=
3400
ra= 3300
water
crust
mantle
lithosphere
rs=
2200
sediment
crust
mantle
lithosphere
depth of equal
compensation
ho=
1km
hc
hm
(hs-ho)
hs=?
rC=
2800
rm=
3400
How much sediment would it take to fill a water-
filled basin up to sea level?
“rule of thumb”: the thickness of sediment needed to fill
a basin is ~ 2.1 times the depth of water that the
sediment replaces

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.
(Kluesner, 2011)

19. rC=
2800
rm=
3400
ra= 3300
water
crust
mantle
lithosphere
rs=
2200
sediment
crust
mantle
lithosphere
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)
(3.3-2.2)
= 6.3 km

20. Calculate how much sediment would be needed to
fill the Alarcon Basin up to sea level:
Maximum water depth in the Alarcon Basin = 3km.
Our calculation shows that ~6.3 km of sediment would
be needed to fill the basin up to sea level.

21. crust
mantle
litho-
sphere
crust
mantle
litho-
sphere
EXAMPLE 3 How does crustal thinning affect
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

22. crust
mantle
litho-
sphere
crust
mantle
litho-
sphere
1 2
hc1=
30km hc2
hm1=
90km
hm2
ha
Note: ha = hc1 + hm1 - hc2 - hm2 - Z
Z
rc(hc1) + rm(hm1) = rs(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

23. crust
mantle
litho-
sphere
crust
mantle
litho-
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)

24. crust
mantle
litho-
sphere
crust
mantle
litho-
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)
=
=
3.00
2.29
= 1.31 km
for water

25. If crust and mantle lithosphere are thinned to half
of original thickness, how deep a basin would
form in response to thinning?
Our calculation shows that the newly-formed basin
would be ~1.3 km deep if it was filled by water.

26. crust
mantle
litho-
sphere
crust
mantle
litho-
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, how deep will it be? -
B: Fill with sediment
(rs = 2.2 g/cm2)
Z=
60ra-45rm-15rc
(ra – rs)

27. crust
mantle
litho-
sphere
crust
mantle
litho-
sphere
1 2
hc1=
30km hc2
hm1=
90km
hm2
ha
Z
Z(ra-rw) = 60ra-45rm-15rc
Basin is formed:
A: fill with water
B: fill with sediment
B: Fill with sediment
(rs = 2.2 g/cm2)
Z=
60ra-45rm-15rc
(ra – rs)
60(3.3)- 45(3.4)-15(2.8)
(3.3-2.3)
=
=
3.00
1.0
= 2.7 km
for sediment

28. If crust and mantle lithosphere are thinned to half
of original thickness, how deep a basin would
form in response to thinning?
Our calculation shows that the newly-formed basin
would be ~2.7 km deep if it was filled by sediment.

29. (Martin-Barajas et al., 2013)
EXAMPLE 3B What rifting parameters can
produce the crustal structure observed in the
Upper Delfin basin?
Experiment with different sediment densities and
different crust and mantle lithosphere initial
thicknesses to find values that could produce the
thicknesses shown on this cross-section.

30. This cross-section spans the Delfin and Tiburon
basins. The core of the Baja California peninsula (NW
end of cross-section) is un-rifted crust, and its
thickness (35-40 km) is a reasonable approximation of
pre-rift crustal thickness.
(Martin-Barajas et al., 2013)

31. Delfin Basin
Location map, Northern Gulf
of California
(Martin-Barajas et al., 2013)

32. From the cross-section, measure and record:
hc1, initial crustal thickness (use crustal thickness
at the NW end of the cross-section as an estimate)
hc2, crustal thickness under Delfin basin after
extension (0!)
hs, thickness of sediments (and intrusions at base
of sediments) in Delfin basin
(Martin-Barajas et al., 2013)

33. crust
mantle
litho-
sphere
1 2
hc1=
__km
hc2=0
hm1=
90km
hm2=0
ha
Delfin basin
Solve for Z. Then
see how well your
answer matches
sediment thick-
ness Z on the
cross-section
Note: ha = hc1 + hm1 - hc2 - hm2 - Z
Z(=hs)
Experiment #1
Use 90 km for a starting thickness of mantle
lithosphere, and hc1 and hc2 values measured from
the cross-section.

34. crust
mantle
litho-
sphere
1 2
hc1=
__km
hc2=0
hm1=
50km
hm2=0
ha
Delfin basin
Solve for Z. Then
see how well your
answer matches
sediment thick-
ness Z on the
cross-section
Note: ha = hc1 + hm1 - hc2 - hm2 - Z
Z(=hs)
Experiment #2
Use 50 km for a starting thickness of mantle
lithosphere, and hc1 and hc2 values measured from
the cross-section.

35. crust
mantle
litho-
sphere
1 2
hc1=
__km
hc2=0
hm1=
20km
hm2=0
ha
Delfin basin
Solve for Z. Then
see how well your
answer matches
sediment thick-
ness Z on the
cross-section
Note: ha = hc1 + hm1 - hc2 - hm2 - Z
Z(=hs)
Experiment #3
Use 20 km for a starting thickness of mantle
lithosphere, and hc1 and hc2 values measured from
the cross-section.

36. crust
mantle
litho-
sphere
1 2
hc1=
__km
hc2=0
hm1=
20km
hm2=0
ha
Delfin basin
Solve for Z. Then
see how well your
answer matches
sediment thick-
ness Z on the
cross-section
Note: ha = hc1 + hm1 - hc2 - hm2 - Z
Z(=hs)
Experiment #4
Use same thickness values you used in experiment
#1, but change rs to 2400. rs varies with total
sediment thickness….

37. Experiment #1
rc(hc1) + rm(hm1) = rs(Z) + rc(hc2) + rm(hm2) + ra(ha)
and ha = hc1 + hm1 - hc2 - hm2 - Z
Solve for Z.
Plug in
thickness values: hc1=35km, hm1=90km, hc2=0, hm2=0
and density values: rc=2800 kg/m3, rm=3400 kg/m3,
ra=3300 kg/m3, rs=2200 kg/m3
Solution: Z = 7.7 km (too small, does not match cross-
section)

38. Experiment #2
rc(hc1) + rm(hm1) = rs(Z) + rc(hc2) + rm(hm2) + ra(ha)
and ha = hc1 + hm1 - hc2 - hm2 - Z
Solve for Z.
Plug in
thickness values: hc1=35km, hm1=50km, hc2=0, hm2=0
and density values: rc=2800 kg/m3, rm=3400 kg/m3,
ra=3300 kg/m3, rs=2200 kg/m3
Solution: Z = 11.4 km (about right, matches cross-section)

39. Experiment #3
rc(hc1) + rm(hm1) = rs(Z) + rc(hc2) + rm(hm2) + ra(ha)
and ha = hc1 + hm1 - hc2 - hm2 - Z
Solve for Z.
Plug in
thickness values: hc1=35km, hm1=20km, hc2=0, hm2=0
and density values: rc=2800 kg/m3, rm=3400 kg/m3,
ra=3300 kg/m3, rs=2200 kg/m3
Solution: Z = 14.1 km (too large, does not match cross-
section)

40. Experiment #4
rc(hc1) + rm(hm1) = rs(Z) + rc(hc2) + rm(hm2) + ra(ha)
and ha = hc1 + hm1 - hc2 - hm2 - Z
Solve for Z.
Plug in
thickness values: hc1=35km, hm1=90km, hc2=0, hm2=0
and density values: rc=2800 kg/m3, rm=3400 kg/m3,
ra=3300 kg/m3, rs=2400 kg/m3
Solution: Z = 9.4 km (about right)

41. Delfin Basin
Location map, Northern Gulf
of California
Note: These solutions assume
sufficient sediment supply to
keep the basin filled. That’s a
good assumption here
because of the very high
sediment input from the
Colorado River.
(Martin-Barajas et al., 2013)

42. Our assumption that there
is sufficient sediment
supply to keep the basin
filled would not be a good
assumption further south in
the Gulf of California where
basins are sediment
starved.
Delfin Basin
Google Earth