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# Isostasy and basin analysis powerpoint

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### Isostasy and basin analysis powerpoint

1. 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. 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. 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. 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. 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. 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. 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. 8. Onward to geology –
9. 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. 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. 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. 12. EXAMPLE 2 What is the effect of filling a basin with sediment?
13. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 27. Location map, Northern Gulf of California Delfin Basin Tiburon Basin
28. 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. 29. Minimum thicknesses, Delfin Basin Sediments 4 km Metasediments 4 km Intrusions 0.4 km Other 1 km from Dorsey, 2010, Table 1
30. 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
31. 31. 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]