tom.h.wilson

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Geology 161 - Geomath
Post-test wrap-up
tom.h.wilson wilson_at_geo.wvu.edu
Department of Geology and Geography West Virginia
University Morgantown, WV
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Average 75 With a 30 point curve.
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In-class test results and discussion
Problem 1 Given the velocity determine the
distance from the ridge.
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Velocity (spreading rate) 3 cm/year, Age
31My Distance to ridge age.velocity 930km
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Problem 2
Given gt
Evaluate logM
Refer to pages 55-57 for basic discussions of
mathematical manipulations of logs and
exponentials.
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Problem 2
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Problem 3
Evaluate the log25. Show work on attached paper.
See page 39
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Lecture 2 slides 31-33
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Lecture 2 slides 31-33
Weve already worked with three bases - 2, 10 and
e. Whatever the base, the logging operation is
the same.
How do we find these powers?
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In general,
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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.
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Geologic Application?
Sediment grain size classification using the phi
scale -
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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.
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See pages 56 and 57
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In problem 4 you had to solve for ? rather than z.
Take ln to get gt
Then do algebra to solve for ?.
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See slides 8 and 9
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Can you evaluate the natural log of
Slide 8
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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 well 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
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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?
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See slides 14 through 19
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Back to isostacy- The ideas weve 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
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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
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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
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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
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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
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Take a few moments and verify that
Slide 19
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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 -
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Average 78 Standard Deviation ?15.6
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Computer exam -
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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.
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Basic PsiPlot Setup-
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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.
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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.
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Using Excell -
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Its always a good idea to make one computation
by hand!
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Prolem 2 -PsiPlot
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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.
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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.
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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.
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Problem 2 - EXCEL
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For Next Time
Finish Reading Chapter 7