Post on 02-Jan-2016
description
tom.h.wilsontom.wilson@mail.wvu.edu
Department of Geology and GeographyWest Virginia University
Morgantown, WV
Geol 351 - Geomath
Some basic review
Going back over the basics
Tom Wilson, Department of Geology and Geography
A lot of the trouble most of us have with math really boils down to problems with basic operations – with the algebra and even the arithmetic
Today, we’ll work through some simple problems to review use of
Exponential notation and exponential math operationsUnits conversionsLinear relationships
Let’s take another look at those example problems
Tom Wilson, Department of Geology and Geography
I’ve linked an Excel file on the class page that contains some analysis of these problems (Group Problems).
Subscripts and superscripts provide information about specific variables and define mathematical operations.
k1 and k2 for example could be used to denote sedimentation constants for different areas or permeabilities of different rock specimens. See Waltham for additional examples of subscript notation.
Subscripts and superscripts
The geologist’s use of math often turns out to be a necessary endeavor. As time goes by you may find yourself scratching your head pondering once-mastered concepts that you suddenly find a need for.
This is often the fate of basic power rules.
Evaluate the following
xaxb =xa / xb = (xa)b =
xa+b
xa-b
xab
Question 1.2a Simplify and where possible evaluate the following expressions -
i) 32 x 34
ii) (42)2+2
iii) gi . gk
iv) D1.5. D2
Exponential notation is a useful way to represent really big numbers in a small space and also for making rapid computations involving large numbers - for example, the mass of the earth is
5970000000000000000000000 kg
the mass of the moon is 73500000000000000000000 kg
Express the mass of the earth in terms of the lunar mass.
Exponential notation
In exponential form ...ME = 597 x 1022kgMM = 7.35 x 1022kg
The mass of the moon (MM) can also be written as 0.0735 x 1024kg
22
( ) 22
597 10 597
7.35 10 7.35E
E MM
M xM
M x
Hence, the mass of the earth expressed as an equivalent number of lunar masses is
=81.2 lunar masses
Exponential notation helps simplify the computation
Write the following numbers in exponential notation (powers of 10)?
The mass of the earth’s crust is 28000000000000000000000kg The volume of the earth’s crust is 10000000000000000000 m3
The mass of the earth’s crust is 2.8 x 1022 kg The volume of the earth’s crust is 1 x 10 19 m3
=mass/volume = 2.8 x 103 kg/m3
Differences in the acceleration of gravity on the earth’s surface (and elsewhere) are often reported in milligals. 1 milligal =10-5 meters/second2.
This is basically a unit conversion problem - you are given a value in one system of units, and the relation of requested units (in this case milligals) to the given units (in this case meters/s2)
1 milligal = 10-5 m/s2 hence 1 m/s2 in terms of milligals is found by multiplying both sides of the above equation by
105
to yield 105 milligals= 1m/s2 - thusg=(9.8m/s2) x 105 milligals/(m/s2) =9.8 x 105 milligals
A headache, but very critical
See http://spacemath.gsfc.nasa.gov/weekly/6Page53.pdf
The Gimli Glider
Tom Wilson, Department of Geology and Geography
http://hawaii.hawaii.edu/math/Courses/Math100/Chapter1/Extra/CanFlt143.htm
They calculated the required fuel weight in pounds instead of kilograms and added only about a third of the required fuel.
Columbus thought he’d made it to Asia
Tom Wilson, Department of Geology and Geography
6. Even Columbus had conversion problems. He miscalculated the circumference of the earth when he used Roman miles instead of nautical miles, which is part of the reason he unexpectedly ended up in the Bahamas on October 12, 1492, and assumed he had hit Asia. Whoops.
The roman mile is about 4,851 feet versus the nautical mile which is 6,076 feet.
Itokawa – a little asteroid
Stereo pair – try it out
Itokawa
Retrieved 2000 particles 1500 of them from the asteroid
Acceleration due to gravity on Itokawa is about 6 x 10-6 mm/s2.
How many meters per second squared is that?
The earth gains mass every day due to collision with (mostly very small) meteoroids. Estimate the increase in the earth’s mass since its formation assuming that the rate of collision has been constant and that
yearsxA
daykgx
t
M
e9
5
105.4
106
M/ t is the rate of mass gainAe is the age of the earth (our t)
But, what is the age of the earth in days? days 1.6425x10
years10 5.4 365
12
9
xxyeardaysAE
1) What is the total mass gained?2) Express the mass-gain as a fraction of the earth’s present day mass
kgxM E241095.5
. . ( / )Ei e A M t
Total
Mass
Gained
AE is a
t
Fractional
Mass EE MxMx
x 724
17
1066.11095.5
10855.9
The North Atlantic Ocean is getting wider at the average rate vs of 4 x 10-2 m/y and has width w of approximately 5 x 106 meters.
1. Write an expression giving the age, A, of the North Atlantic in terms of vs and w.2. Evaluate your expression to answer the question - When did the North Atlantic begin to form?
Plate spreading
5
0
5 5log log5 | none of these
| | | ( ) log
( ) log | | | none of these
0 |1| | none of these
( ) | | | none of these
a
b c b c b c b
ff f ggg
b c b c b c bc
a a a
a a a a a ca b c a
a f g a a aa
a a
a a a a a
How thick was it originally?
Over what length of time was it deposited?
510,000 years
2.1 to 2.7 million years ago
0.5 to 2.1 million years ago
Astronomical forcing of global climate: Milankovitch Cycles
Take the quiz
http://www.sciencecourseware.org/eec/GlobalWarming/Tutorials/Milankovitch/
http://www.sciencedaily.com/releases/2008/04/080420114718.htm
http://www.nasa.gov/mission_pages/MRO/multimedia/phillips-20080515.html
Mini Maunder & global cooling?Energy output declines during periods of time with little
sunspot activity.
The example presented on page 3 illustrates a simple age-depth relationship for unlithified sediments
depthxkAge This equation is a quantitative statement of what we all have an intuitive understanding of - increased depth of burial translates into increased age of sediments. But as Waltham suggests - this is an approximation of reality.
What does this equation assume about the burial process? Is it a good assumption?
Example - if k = 1500 years/m calculate sediment age at depths of 1m, 2m and 5.3m. Repeat for k =3000 years/m
1m2m5.3m
Age = 1500 years
Age = 3000 yearsAge = 7950 years
For k = 3000years/m
Age = 3000 yearsAge = 6000 yearsAge = 15900 years
kzanotationsymbolic a=age, z=depthwhere
A familiar equation
Depth x kAge
a straight line.
bmxy The general equation of a straight line is
In this equation -
which term is the slope and which is the intercept?
bmxy
Depth x kAge In this equation - which term is the slope and which is
the intercept?
line theof slope theis k
zero bemust intercept the
A more generalized representation of the age/depth relationship should include an intercept term -
0AD kA
The geologic significance of A0 - the intercept - could be associated with the age of the upper surface after a period of erosion. Hence the exposed surface of the sediment deposit would not be the result of recent
The slope of the line is, in this case, an inverse rate. Our dependant variable is depth, which would have units of meters or feet, for example. The equation defines depth of burial in terms of age. K, the slope transforms a depth into a number of years and k must have units of years/depth.
0
10000
20000
30000
40000
50000
AG
E (
year
s)
0 20 40 60 80 100
Depth (meters)
0AD kA
sedimentation but instead would be the remains of sediments deposited at an earlier time A0.
The slope of this line is t/x =1500years/meter, what is the intercept?
The intercept is the line’s point of intersection along the y (or Age) axis at depth =0.
-50000
0
50000
100000
150000
AG
E (
year
s)
-20 0 20 40 60 80 100
depth (meters)
t
x
0AD kA
0 age at 0 depth: just one possibility
If only the relative ages of the sediments are known, then for a given value of k (inverse deposition rate) we would have a family of possible lines defining age versus depth.
-10000
0
10000
20000
30000
40000
AG
E (
year
s) 50000
60000
70000
0 20 40 60 80 100
Depth (meters)
Are all these curves realistic?
What are the intercepts?
0AD kA
Consider the significance of A0 in the following context
If k is 1000 years/meter, what is the velocity that the lake bed moves up toward the surface?
If the lake is currently 15 meters deep, how long will it take to fill up?
Consider the case for sediments actively deposited in a lake.
0AD kA
What is the intercept?
The slope (k) does not change. We still assume that the thickness of the sediments continues to increase at the rate of 1 meter in 1000 years.
Hint: A must be zero when D is 15 meters
You should be able to show that A0 is -15,000 years. That means it will take 15,000 years for the lake to fill up.
Age =?
-1x105
-5x104
0
5x104
1x105
Age
(ye
ars)
-50 0 50 100
Depth (meters)
-15,000
present day depth at age = 0.
Our new equation looks like this -
15,000-D1000A
-1x105
-5x104
0
5x104
1x105
Age
(ye
ars)
-50 0 50 100
Depth (meters)
… we would guess that the increased weight of the overburden would squeeze water from the formation and actually cause grains to be packed together more closely. Thus meter thick intervals would not correspond to the same interval of time. Meter-thick intervals at greater depths would correspond to greater intervals of time.
0AD kA
Is this a good model?
Return to the group problems(open Day1GroupPbs.xlsx)
Finish reading Chapters 1 and 2 (pages 1 through 38) of Waltham
After we finish some basic review, we’ll spend some time with Excel and use itto solve some problems related to the material covered in Chapters 1 and 2.