Isostasy & basin analysis powerpoint

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Isostasy in Geology andBasin AnalysisThis 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 -

Solidrs12h2h1Fluid (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.

Solidrs12h2h1Fluid (rf)depth of equal compensation

Solidrs12h2h1Fluid (rf)depth of equal compensationTo 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

Solidrs12h2h1Fluid (rf)depth of equal compensationThis equation correctly describes equilibrium isostatic balance in the diagram.

Onward to geology Courtesy Sue Cashman

EXAMPLE 1Estimate thickness of lithosphere:In this example, weve 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/m3rm=3400 kg/m3ra=3300 kg/m3

rcelevation=3kmrmhc=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) - rmhcrc(hc+e) - rmhc(ra-rm)


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) - rmhcrc(hc+e) - rmhc(ra-rm)2800(35+3) 3400(35)(3300-3400) -12,600 -100


hm=?

asthenosphere (rc)

Z=?

Z===Z = 126 km

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

Courtesy Scott Bennett

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=2800rm=3400

ra= 3300 water

crust

mantlelithospherers= 2200sediment

crust

mantlelithosphere

depth of equal compensation

Let rw= 1000 kg/m3Let rs= 2200 kg/m3 ho= 1km hc

hmhs=?rC=2800rm=340012

rC=2800rm=3400

ra= 3300water

crust

mantlelithospherers= 2200

sediment

crust

mantlelithosphere


ho= 1km hc

hmhs=?rC=2800rm=3400Remember, 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. 12

rC=2800rm=3400

ra= 3300water

crust


crust

mantlelithosphere


ho= 1km hc

hm

(hs-ho)

hs=?rC=2800rm=3400rwho + 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=2800rm=3400

ra= 3300water

crust


crust

mantlelithosphere


ho= 1km hc

hm

(hs-ho)

hs=?rC=2800rm=3400hs = ho(ra- rw) (ra- rs) = 1.0 (3.3-1.0)(3.3-2.2)= 2.1 km

rC=2800rm=3400

ra= 3300water

crust


crust

mantlelithosphere


ho= 1km hc

hm

(hs-ho)

hs=?rC=2800rm=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

EXAMPLE 2BSediment filling Alarcon Basin exampleDetermine 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)

Image from page 166, Jared W. Kluesner, 2011, Marine Geophysical Study of Cyclic Sedimentation and Shallow Sill Intrusion in the Floor of the Central Gulf of California, Ph.d. thesis, UNIVERSITY OF CALIFORNIA, SAN DIEGO, 231 pages.

(Used with permission from J. Kluesner, 2015)

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rC=2800rm=3400

ra= 3300water

crust


crust

mantlelithosphere


ho= 1km hc

hm

(hs-ho)

hs=?rC=2800rm=3400hs = ho(ra- rw) (ra- rs) = 3 (3.3-1.0)(3.3-2.2)= 6.3 km

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.

crustmantlelitho-spherecrustmantlelitho-sphereEXAMPLE 3 How does crustal thinning affect the depth of sedimentary basins?12hc1=30kmhc2hm1=90kmhm2haNewly-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 - ZZ

crustmantlelitho-spherecrustmantlelitho-sphere12hc1=30kmhc2hm1=90kmhm2haNote: ha = hc1 + hm1 - hc2 - hm2 - Z

Zrc(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- ZraZ(ra-rw) = 60ra-45rm-15rc

crustmantlelitho-spherecrustmantlelitho-sphere12hc1=30kmhc2hm1=90kmhm2haZZ(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)

crustmantlelitho-spherecrustmantlelitho-sphere12hc1=30kmhc2hm1=90kmhm2haZZ(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

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.

crustmantlelitho-spherecrustmantlelitho-sphere12hc1=30kmhc2hm1=90kmhm2haZZ(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)

crustmantlelitho-spherecrustmantlelitho-sphere12hc1=30kmhc2hm1=90kmhm2haZZ(ra-rw) = 60ra-45rm-15rc Basin is formed:A: fill with waterB: 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)==3.001.0= 2.7 kmfor sediment

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.

(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.

Martin-Barajas, A., Gonzalez-Escobar, M., Fletcher, J.M., Pacheco, M., Oskin, M., and Dorsey, R., 2013, Thick deltaic sedimentation and detachment faulting delay the onset of continental rupture in the Northern Gulf of California: Analysis of seismic reflection profiles, Tectonics, 32, 1-18, doi: 10.1002/tect.20063.29

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, A., Gonzalez-Escobar, M., Fletcher, J.M., Pacheco, M., Oskin, M., and Dorsey, R., 2013, Thick deltaic sedimentation and detachment faulting delay the onset of continental rupture in the Northern Gulf of California: Analysis of seismic reflection profiles, Tectonics, 32, 1-18, doi: 10.1002/tect.20063.30

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

Arturo Martn-Barajas, Mario Gonzlez-Escobar, John M. Fletcher, Martn Pacheco, Michael Oskin, and Rebecca Dorsey, 2013, Thick deltaic sedimentation and detachment faulting delay the onset of continental rupture in the Northern Gulf Of California: Analysis of seismic reflection profiles: Tectonics, v. 32, pp 1-18, doi:10.1002/tect.20063.

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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


MartnBarajas, Arturo, et al. "Thick deltaic sedimentation and detachment faulting delay the onset of continental rupture in the Northern Gulf of California: Analysis of seismic reflection profiles." Tectonics 32.5 (2013): 1294-1311.32

crustmantlelitho-sphere12hc1=__kmhc2=0hm1=90kmhm2=0haDelfin basin

Solve for Z. Then see how well your answer matches sediment thick-ness Z on the cross-sectionNote: ha = hc1 + hm1 - hc2 - hm2 - ZZ(=hs)Experiment #1Use 90 km for a starting thickness of mantle lithosphere, and hc1 and hc2 values measured from the cross-section.






Solve for Z. Then see how well your answer matches sediment thick-ness Z on the cross-sectionNote: ha = hc1 + hm1 - hc2 - hm2 - ZZ(=hs)Experiment #4Use same thickness values you used in experiment #1, but change rs to 2400. rs varies with total sediment thickness.

Experiment #1rc(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)



Solution: Z = 11.4 km (about right, matches cross-section)



Solution: Z = 14.1 km (too large, does not match cross-section)



Solution: Z = 9.4 km (about right)

Delfin BasinLocation map, Northern Gulf of CaliforniaNote: These solutions assume sufficient sediment supply to keep the basin filled. Thats a good assumption here because of the very high sediment input from the Colorado River. (Martin-Barajas et al., 2013)

Arturo Martn-Barajas, Mario Gonzlez-Escobar, John M. Fletcher, Martn Pacheco, Michael Oskin, and Rebecca Dorsey, 2013, Thick deltaic sedimentation and detachment faulting delay the onset of continental rupture in the Northern Gulf Of California: Analysis of seismic reflection profiles: Tectonics, v. 32, pp 1-18, doi:10.1002/tect.20063.

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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 BasinGoogle Earth

Isostasy & basin analysis powerpoint

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