1. Geology 351 - Geomath Basic Review tom.h.wilson
Department of Geology and Geography
West Virginia University
Morgantown, WV

2. Course description Revised
GEOL 351. Geomathematics. 3 Hr. PR:GEOL 101 and (MATH 150 or MATH 155). Mathematical methods and applications in geology, geochemistry, geophysics, and environmental science. Review of integral calculus, differential equations, and non-linear systems. Use of computers as geological problem-solving tools.
Revised
GEOL 351. Geomathematics. 3 Hr. PR:GEOL 101 and (MATH 150 or MATH 155). Mathematical methods and applications in geology, geochemistry, geophysics, and environmental science. Review of derivatives and integrals. Use of computers as geological problem-solving tools.
What do you get out of it? Hopefully an increased comfort level with and increased appreciation of math applications in geology. You should be able to look forward to your next math class!

3. Warm-up Exercise: Part 16’ 8’

4. Warm-up Exercise: Part 1

5. 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.

6. This is often the fate of basic power rules.
Superscripts - powers
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
xa+b
xaxb =
xa / xb =
(xa)b =
xa-b
xab

7. Try your hand at these questions.

8. i) 32 x 34 ii) (42)2+2 iii) gi . gk iv) D1.5. D2
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

9. 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 kg
the mass of the moon is kg
Express the mass of the earth in terms of the lunar mass.

10. 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 x 1024kg
Hence, the mass of the earth expressed as an equivalent number of lunar masses is
=81.2 lunar masses

11. Write the following numbers in exponential notation (powers of 10)?
The mass of the earth’s crust is kg The volume of the earth’s crust is m3
What is the density of the earths crust?
The mass of the earth’s crust is 2.8 x 1022 kg
The volume of the earth’s crust is 1 x m3
=mass/volume = 2.8 x 103 kg/m3

12. What is g (9.8m/s ) in milligals? 2
Differences in the acceleration of gravity on the earth’s surface (and elsewhere) are often reported in milligals. 1 milligal =10-5 meters/second2. g
What is g (9.8m/s ) in milligals? 2
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

13. Itokawa – a little asteroid
Stereo pair – try it out
So very small gravity

14. 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?
Satellite Hayabusa
6x10-9 m/s2 or

15. Additional Questions from Waltham
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/ t is the rate of mass gain
Ae is the age of the earth (our  t)
But, what is the age of the earth in days?

16. 17 kg 1) 9.855 x 10 2) AE is a  t 1) What is the total mass gained?
2) Express the mass-gain as a fraction of the earth’s present day mass
1) x 10 17 kg
AE is a  t
Total Mass Gained 2)
Fractional Mass

17. 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?

18. Age versus depth relationship (s)
How thick was it originally?
Over what length of time was it deposited?
In this slide we’re looking at a coal about 4-5 feet thick. At the base of the coal lies an underclay. So you’re beginning to form swamp like conditions, but the pH is too high to preserve organics. The WT is probably right at the surface. At the time these swamps were formed, this area was in the subtropical locations 20 degrees N Lat.
The peat begins to accumulate as the pH decreases to 2-3. At this point, the organics are preserved. Not all of them but a large percentage. The low pH knocks out the microbes.
Now you’ll see within the coal, there are some non-coaly partings of ash. Basically as the pH increases, the organics are removed and you basically get what’s left in the fireplace when you burn your logs. This can happen by a change in climate or by local incursion of ground water which is generally neutral pH.
Now as we get up toward the top, you can see there’s a very sharp boundary. Most likely what you have here is distributary channel deposits. Streams probably swept across the area and also probably eroded some of the swampy material. This shale might grade into a sand some distance left or right.
Deposition rates of the peat are about 1-2 mm per year, but this 4-5 foot section has been compacted. The percent compaction for peats is about 90 %, do the original thickness was about 40 to 50 feet. SO we’re looking at an age difference from the top to base of the layer of (12,192 years at 1mm per year) as much as 24,000 to 30,000 years.

19. Consider another depositional environment
Any ideas where this is? ?
It's cold there

20. Troughs
Layered
Deposits of?

21. Variations induced by Astronomical Cycles? 510,000 years
0.5 to 2.1 million years ago
2.1 to 2.7 million years ago
Variations induced by
Astronomical Cycles?

22. Milankovitch Cycles

23. Astronomical forcing of global climate: Milankovitch Cycles
Take the quiz

24. http://www. nasa. gov/mission_pages/MRO/multimedia/phillips-20080515
Above images are from the Mars Reconnaisance Orbiter
Mars Climate Orbiter: "The 'root cause' of the loss of the spacecraft was the failed translation of English units into metric units in a segment of ground-based, navigation-related mission software

25. Mini Maunder & global cooling
Mini Maunder & global cooling? Energy output declines during periods of time with little sunspot activity.

26. Mathematics as a tool for solving geological problems
Chapter 1
Mathematics as a tool for solving geological problems
The example presented on page 3 illustrates a simple age-depth relationship for unlithified sediments
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.
So in the text Waltham presents us a simplified expression which may have pretty limited usefulness. It may be pretty good in one area but not in another.
What does this equation assume about the burial process? Is it a good assumption?

27. 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 = years

28. Common relationships between geological variables
Chapter 2 -
Common relationships between geological variables
A familiar equation
a straight line.
The general equation of a
straight line is

29. In this equation -
which term is the slope and which is the intercept?
In this equation -
which term is the slope and which is the intercept?
A more generalized representation of the age/depth relationship should include an intercept term -

30. slope Intercept What is Ao?
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.
slope
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
Intercept
sedimentation but instead would be the remains of sediments deposited at an earlier time A0.
What is Ao?

31. 0 age at 0 depth: just one possibilityx
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.

32. Are all these curves realistic?
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.
What are the intercepts?
Are all these curves realistic?

33. Consider the significance of A0 in the following context
Consider the case for sediments actively deposited in a lake.
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?
and
Start here and note that the whole problem depends on where we stick y=0. If y=0 is the surface of the water then the age of the sediments at the surface is equal to -15,000 years; that’s how long it takes the lake to fill up – to reach a depth of y=0.
If the lake is currently 15 meters deep, how long will it take to fill up?

34. The slope (k) does not change
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

35. present day depth at age = 0.
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.
-15,000
present day depth at age = 0.

36. Our new equation looks like this -

37. 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.

38. Return to the group problems

39. For Next Time
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.