Transcript What are the intercepts? - West Virginia University

Geol 351 - Geomath
Some basic review
tom.h.wilson
[email protected]
Department of Geology and Geography
West Virginia University
Morgantown, WV
Going back over the basics
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
operations
 Units conversions
 Linear relationships
Tom Wilson, Department of Geology and Geography
Let’s take another look at those example
problems
I’ve linked an Excel file on the
class page that contains some
analysis of these problems
(Group Problems).
Tom Wilson, Department of Geology and Geography
Subscripts and superscripts
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.
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
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 helps simplify the
computation
In exponential form ...
ME = 597 x 1022kg
MM = 7.35 x 1022kg
The mass of the moon (MM) can also
be written as 0.0735 x 1024kg
Hence, the mass of the earth expressed as
an equivalent number of lunar masses is
M E(M )
ME
597 x1022
597



MM
7.35 x1022
7.35
=81.2 lunar masses
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 - thus
g=(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
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.
Tom Wilson, Department of Geology and Geography
Columbus thought he’d made it to Asia
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.
Tom Wilson, Department of Geology and Geography
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
M
 6 x105 kg
day
t
Ae  4.5 x109 years
M/ t is the rate of mass gain
Ae is the age of the earth (our  t)
AE  365 days
year
But, what is the age of
the earth in days?
 1.6425x10 12 days
x 4.5 x 109 years
1) What is the total mass gained?
2) Express the mass-gain as a fraction of
the earth’s present day mass
i.e. AE (M / t )
M E  5.95 x10 24 kg
9.855 x1017
7
M

1
.
66
x
10
ME
E
24
5.95 x10
Plate spreading
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?
a 5  5a 5 log a a log 5 | none of these 
a b  c   a b  a c | a b  a c | ca b | (b  c ) log a 
f


f g
g
a
|a
| none of these 
g   ( f  g ) log a | a
a


a 0   0 |1| a | none of these 
f
( a b ) c   a b  a c | a b  c | a bc | none of these 
How thick
was it
originally?
Over what
length of time
was it
deposited?
510,000 years
0.5 to 2.1 million
years ago
2.1 to 2.7 million
years ago
Astronomical forcing of global climate:
Milankovitch Cycles
http://www.sciencecourseware.org/eec/GlobalWarming/Tutorials/Milankovitch/
Take the quiz
http://www.nasa.gov/mission_pages/MRO/multimedia/phillips-20080515.html
http://www.sciencedaily.com/releases/2008/04/080420114718.htm
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
Age  k x depth
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?
symbolic notation  a  kz
where
a=age, z=depth
Example - if k = 1500 years/m calculate sediment age at
depths of 1m, 2m and 5.3m. Repeat for k =3000 years/m
1m
2m
5.3m
Age = 1500 years
Age = 3000 years
Age = 7950 years
For k = 3000years/m
Age = 3000 years
Age = 6000 years
Age = 15900 years
A familiar equation
Age  k x Depth
a straight line.
The general equation of a
straight line is y  mx
b
In this equation -
y  mx  b
which term is the slope and which is the
intercept?
In this equation Age  k x Depth
- which term is the slope and which is
the intercept?
k is the slope of the line
the intercept must be zero
A more generalized representation of the
age/depth relationship should include an
intercept term A  kD  A 0
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.
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
sedimentation but instead
would be the remains of
sediments deposited at an
earlier time A0.
50000
AGE (years)
40000
A  kD  A 0
30000
20000
10000
0
0
20
40
60
Depth (meters)
80
100
0 age at 0 depth: just one possibility
A  kD  A 0
AGE (years)
150000
The slope of this line is
t/x =1500years/meter,
what is the intercept?
100000
t
50000
x
0
-50000
-20
0
20
40
60
depth (meters)
80
100
The intercept is the line’s
point of intersection
along the y (or Age) axis
at depth =0.
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.
70000
AGE (years)
60000
50000
40000
30000
20000
What are the
intercepts?
10000
0
-10000
0
20
40
60
80
Depth (meters)
Are all these curves realistic?
100
Consider the case for sediments
actively deposited in a lake.
A  kD  A 0
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?
A  kD  A 0
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.
What is the intercept?
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.
1x10
Age (years)
5x10
5
present day depth
at age = 0.
4
0
Age =?
-5x10
-1x10
-15,000
4
5
-50
0
50
Depth (meters)
100
Our new equation looks like this -
A  1000 D - 15,000
1x10
Age (years)
5x10
-5x10
-1x10
5
4
0
4
5
-50
0
50
Depth (meters)
100
A  kD  A 0
Is this a
good
model?
… 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.
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 it
to solve some problems related to the
material covered in Chapters 1 and 2.