tom.h.wilson
wilson@geo.wvu.edu
Department of Geology and Geography
West Virginia University
Morgantown, WV
Written Exam
5
Average = 75
Number
4
3
2
1
0
20
40
60
80
100
120
140
160
Grade
Average = 75
With a 30 point curve.
In-class test results and discussion
Problem 1: Given the velocity determine the
distance from the ridge.
Depth (meters)
Ocean Floor Subsidence
0
-500
-1000
-1500
-2000
-2500
-3000
-3500
-4000
0
20
40
60
80
Time (millions of years)
100
Depth (meters)
Ocean Floor Subsidence
0
-500
-1000
-1500
-2000
-2500
-3000
-3500
-4000
0
20
40
60
80
100
Time (millions of years)
Velocity (spreading rate) = 3 cm/year,
Change in depth per million year time interval
Age = ~31My
0
depth (meters)
Distance-100to ridge = age.velocity=
930km
-50
-150
-200
-250
Problem 2:
Given >
M  r 3
Evaluate logM
Refer to pages
55-57 for
basic
discussions of
mathematical
manipulations
of logs and
exponentials.
Problem 2:
M  r 3
log M  log r
3
log M  log   log r 3
log M  log   3 log r
Problem 3:
Evaluate the log25. Show
work on attached paper.
See page 39
Lecture 2 slides 31-33
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.
log 5 10 asks what is the power that 5 must be raised to to get 10.
log 5 10  ?
How do we find these powers?
log 5 10 
log10 10
log10 5
1
log 5 10 
 1.431
0.699
thus 51.431  10
31
In general,
log10 (number)
log base (some number) 
log10 base
or
log b a 
log10 (a)
log10 b
Try the following on your own
log 3 7 
log10 (7)
?
log10 3
log 8 8
log 7 21
log 4 7
32
As noted above log 10 is often writ ten as log, with no subscript
log10 is referred to as the common logarithm
log e is often writ ten as ln.
thus
log e 8  ln8  2.079
loge or ln is referred to as the natural
logarithm. All other bases are usually
specified by a subscript on the log, e.g.
log 5 or log 2 , etc.
33
Sediment grain size classification
using the phi scale    log 2 ( d )
Problem 4:
Porosity as a function of depth is
z
expressed as ,
 ( z )  0 e 
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.
See pages
56 and 57
In problem 4 you had to solve for 
rather than z.
z
z
ln[ ( z )]  ln[0 ] 
 ( z )  0e  Take ln to get >

 ln[ ( z)]   ln[0 ]   z
 ln[ ( z)]  ln[0 ]   z
 ln[0 ]  ln[ ( z)]  z
 0 
  z
  ( z) 
 ln 
Then do algebra to
solve for .

z
 0 

ln
  ( z) 
See slides 8 and 9
Can you evaluate the natural log of
  0 e-cz


ln    ln 0 e-cz  ?
Slide 8
ln( )  ln 0   cz
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
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?
Mountain
1 km topo relief
 Crust = 2.67 g/cm3
moho
 Mantle Lithosphere = 3.3 g/cm3
8 kilometer
crustal root
See slides 14 through 19
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
 cl   m d   c L   m D
cl   m d  c e  l  r    m D
Take a few moments and verify that

c
r  
  m  c

e

Slide 19
A
B
Mountain
1 km topo relief
 Crust = 2.67 g/cm3
moho
 Mantle Lithosphere = 3.3 g/cm3
8 kilometer
crustal root
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
Ma  Mb
Will work in class -
Histogram for Computer Exam Grades
7
Average Grade = 78
6
Number
5
4
3
2
1
0
30
40
50
60
70
80
90
100
Grade
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.
Basic PsiPlot Setup-
Problem 1: Cone of depression
10
r = 300 meters
11
H
12
r = 2000 meters
13
14
0
5
10
15
r (meters)
20
25
Summary of results1. 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.
Using Excell -
It’s always a good idea to
make one computation by hand!
Prolem 2 -PsiPlot
Depth (meters)
Ocean Floor Subsidence
0
-500
-1000
-1500
-2000
-2500
-3000
-3500
-4000
0
20
40
60
80
100
Time (millions of years)
Change in depth per million year time interval
 depth (meters)
0
-50
-100
-150
-200
-250
-300
-350
-400
0
20
40
60
80
Time (millions of years)
100
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.
Depth (meters)
Ocean Floor Subsidence
0
-500
-1000
-1500
-2000
-2500
-3000
-3500
-4000
0
20
40
60
80
100
Time (millions of years)
Change in depth per million year time interval
 depth (meters)
0
-50
-100
-150
-200
-250
-300
-350
-400
0
20
40
60
80
Time (millions of years)
100
Summary of the resultsOcean 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.
Depth (meters)
Ocean Floor Subsidence
0
-500
-1000
-1500
-2000
-2500
-3000
-3500
-4000
0
20
40
60
80
100
Time (millions of years)
Change in depth per million year time interval
 depth (meters)
0
-50
-100
-150
-200
-250
-300
-350
-400
0
20
40
60
80
Time (millions of years)
100
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.
Problem 2 - EXCEL