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Geology 161 - Geomath. Post-test wrap-up. tom.h.wilson wilson@geo.wvu.edu. Department of Geology and Geography West Virginia University Morgantown, WV. Average = 75 With a 30 point curve. In-class test results and discussion.
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Geology 161 - Geomath Post-test wrap-up tom.h.wilson wilson@geo.wvu.edu Department of Geology and Geography West Virginia University Morgantown, WV
Average = 75 With a 30 point curve.
In-class test results and discussion Problem 1: Given the velocity determine the distance from the ridge.
Velocity (spreading rate) = 3 cm/year, Age = ~31My Distance to ridge = age.velocity= 930km
Problem 2: Given > Evaluate logM Refer to pages 55-57 for basic discussions of mathematical manipulations of logs and exponentials.
Problem 3: Evaluate the log25. Show work on attached paper. See page 39
Lecture 2 slides 31-33 We’ve already worked with three bases - 2, 10 and e.Whatever the base, the logging operation is the same. How do we find these powers? 31
or Try the following on your own In general, 32
log10 is referred to as the common logarithm thus loge or ln is referred to as the natural logarithm. All other bases are usually specified by a subscript on the log, e.g. 33
Geologic Application? Sediment grain size classification using the phi scale -
Problem 4: Porosity as a function of depth is expressed as , where (z) is the porosity at depth z, is a constant term, 0 is the porosity at z = 0, e is the natural base and z is the depth and has units of kilometers. Given that 0 is 0.6 and that (100meters) =0.57, a) determine and b) determine the porosity at a depth of 1 kilometer. Show your work on the attached paper.
In problem 4 you had to solve for rather than z. Take ln to get > Then do algebra to solve for .
is a straight line. Power law relationships end up being straight lines when the log of the relationships is taken. In our next computer lab we’ll determine the coefficients c (or ) and ln(0) that define the straight line relationship above between ln() and z. We will also estimate power law and general polynomial interrelationships using PsiPlot. Slide 9
Mountain 1 km topo relief Crust = 2.67 g/cm3 moho 8 kilometer crustal root Mantle Lithosphere = 3.3 g/cm3 5. The figure on the next page shows a highly simplified mountain range having 1 km of relief above the surrounding area. A crustal root extends 8km into the mantle lithosphere. Assume the crust has a density of 2.67 g/cm3 and that the mantle lithosphere has a density of 3.3 g/cm3. Determine whether the kilometer topographic relief is compensated. If it is not, how deep should the crustal root be to compensate the mountain?
Back to isostacy- The ideas we’ve been playing around with must have occurred to Airy. You can see the analogy between ice and water in his conceptualization of mountain highlands being compensated by deep mountain roots shown below. The Airy Hypothesis Slide 14
In the diagram below left we have an equilibrium condition. In the diagram below right, we have upset this equilibrium. How deep must the mountain root be to stabilize a mountain with elevation e? Slide 15
In the diagram below we refer to the compensation depth. This depth is the depth above which the combined weight of a column of mantle and crust of unit horizontal cross section is constant. Regardless of where you are, the total mass of material overlying the compensation depth will be constant. Slide 16
If the weight of material above a reference depth is not constant then the crust is not in equilibrium or crustal roots will have to extend below that depth to compensate for the mass excess. The relationship that must hold for the combined weight of crust and mantle above the compensation depth allows us to solve for r (see below) ... Slide 17
Again we have simplified the equation by assuming that the horizontal cross section of these vertical columns has equal area in all cases, hence the areas cancel out and the mass equivalence relationship reduces to the product of the density and thickness (l, d, L or D). Slide 18
Take a few moments and verify that Slide 19
Mountain 1 km topo relief Crust = 2.67 g/cm3 moho 8 kilometer crustal root Mantle Lithosphere = 3.3 g/cm3 A B The mass between the surface and the compensation depth at A must be equal to that at point be from the surface to the compensation depth. The basic equation to solve is Will work in class -
Average = 78% Standard Deviation = 15.6
Computer exam -
Statement of the problem- Calculate the depth to water table as a function of distance from the well for cones of depression having radii of 300 meters and 2000 meters.
Summary of results- 1. The computations reveal that the cone of depression rises sharply in the vicinity of the well. A rise of nearly 1.5 meters occurs within a distance of 5 meters from the well center. From 5 to 25 meter distances from the well, the water table rises much more gradually by only about 1/2 meter.
Summary of results (cont.)- 2. The water level drop associated with the 2km radius cone of depression is greater than that associated with the 300 meter radius cone of depression by about 0.75 meters. Otherwise the two curves are very similar in shape.
Statement of the problem - Compute the subsidence of the seafloor relative to the ridge crest for a 100 million year time span. Also compute the change in depth per million year time intervals during that 100 million year time frame.
Summary of the results- Ocean crust drops rapidly from the ridge crest by about 1000 meters during the first 10 million years after its formation. Thereafter, it drops more gradually at the rate of about 1000 meters per 40 million years.
The plot of change in depth per million year time interval provides a nice illustration of the variations in subsidence rate. Initially the subsidence rate exceeds 350 meters per million years and drops quickly - within about 10 million years to subsidence rates of 50 - 20 meters per million years.
For Next Time Finish Reading Chapter 7
