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.