Tom.h.wilson [email protected] Department of Geology and Geography West Virginia University...
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Transcript of Tom.h.wilson [email protected] Department of Geology and Geography West Virginia University...
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Num
ber
20 40 60 80 100 120 140 160
Grade
Written Exam
Average = 75
Average = 75
With a 30 point curve.
In-class test results and discussionProblem 1: Given the velocity determine the distance from the ridge.
Change in depth per million year time interval
0 20 40 60 80 100
Time (millions of years)
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-300
-250
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d
ep
th
(m
eters)
Ocean Floor Subsidence
-4000-3500
-3000-2500-2000
-1500-1000
-5000
Dep
th
(m
eters)
0 20 40 60 80 100Time (millions of years)
Change in depth per million year time interval
0 20 40 60 80 100
Time (millions of years)
-400
-350
-300
-250
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-100
-50
0
d
ep
th
(m
eters)
Ocean Floor Subsidence
-4000-3500
-3000-2500-2000
-1500-1000
-5000
Dep
th
(m
eters)
0 20 40 60 80 100Time (millions of years)
Velocity (spreading rate) = 3 cm/year, Age = ~31My
Distance to ridge = age.velocity=
930km
3rM Given > Evaluate
logM
Problem 2:
Refer to pages 55-57 for basic discussions of mathematical manipulations of logs and exponentials.
We’ve already worked with three bases - 2, 10 and e. Whatever the base, the logging operation is the same.
10.get to toraised bemust 5 power that theis what asks 10log5
? 10log5
How do we find these powers?
5log
10log 10log
10
105
431.1699.0
1 10log5
105 431.1 thus
Lecture 2 slides 31-33
31
In general, base
numberbase
10
10
log
)(log number) some(log
or
b
ab
10
10
log
)(log alog
Try the following on your own
?3log
)7(log 7log
10
103
8log8
21log7
7log4 32
subscript no with log, asten often writ is log above noted As 10
log10 is referred to as the common logarithm
ln. asten often writ is log e
2.079 ln8 8log e
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.
etc. ,logor og 25l33
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.
z
ez
0)(
Problem 4:
In problem 4 you had to solve for rather than z.
z
z ]ln[)](ln[ 0z
ez
0)(
zz ]ln[)](ln[ 0
zz ]ln[)](ln[ 0
zz )](ln[]ln[ 0
zz
)(
ln 0
)(0lnz
z
Take ln to get >
Then do algebra to solve for .
cz 0ln)ln( 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
8 kilometer crustal root
1 km topo relief
Crust = 2.67 g/cm3
moho
Mantle Lithosphere = 3.3 g/cm3
Mountain
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.
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
8 kilometer crustal root
1 km topo relief
Crust = 2.67 g/cm3
moho
Mantle Lithosphere = 3.3 g/cm3
Mountain
A B
ba MM
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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Num
ber
Histogram for Computer Exam Grades
30 40 50 60 70 80 90 100
Grade
Average Grade = 78
Average = 78%
Standard Deviation = 15.6
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.
0 5 10 15 20 25
r (meters)
10
11
12
13
14
H
Problem 1: Cone of depression
r = 300 meters
r = 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.
Ocean Floor Subsidence
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Time (millions of years)
-4000-3500
-3000-2500-2000
-1500-1000
-500
Dep
th (
met
ers)
0
Change in depth per million year time interval
0 20 40 60 80 100
Time (millions of years)
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0
d
epth
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eter
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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.
Ocean Floor Subsidence
0 20 40 60 80 100
Time (millions of years)
-4000-3500
-3000-2500-2000
-1500-1000
-500
Dep
th (
met
ers)
0
Change in depth per million year time interval
0 20 40 60 80 100
Time (millions of years)
-400
-350
-300
-250
-200
-150
-100
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0
d
epth
(m
eter
s)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.
Ocean Floor Subsidence
0 20 40 60 80 100
Time (millions of years)
-4000-3500
-3000-2500-2000
-1500-1000
-500
Dep
th (
met
ers)
0
Change in depth per million year time interval
0 20 40 60 80 100
Time (millions of years)
-400
-350
-300
-250
-200
-150
-100
-50
0
d
epth
(m
eter
s)
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.