Geol 351 - Geomath

1. Geol 351 - Geomath Some basic review tom.h.wilson tom.wilson@mail.wvu.edu Department of Geology and Geography West Virginia University Morgantown, WV

2. 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

3. 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

4. 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.

5. 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

6. 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

7. 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 is73500000000000000000000 kg Express the mass of the earth in terms of the lunar mass.

8. 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 =81.2 lunar masses

9. Write the following numbers in exponential notation (powers of 10)? The mass of the earth’s crust is 28000000000000000000000kgThe volume of the earth’s crust is 10000000000000000000 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 10 19 m3 =mass/volume = 2.8 x 103 kg/m3

10. 2 What is g (9.8m/s ) in milligals? 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 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/s2hence 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

11. A headache, but very critical See http://spacemath.gsfc.nasa.gov/weekly/6Page53.pdf

12. 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

13. 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

14. Itokawa – a little asteroid Stereo pair – try it out

15. 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?

16. 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? 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

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

18. 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?

19. Try your hand at these questions.

20. Age versus depth relationship (s) How thick was it originally? Over what length of time was it deposited?

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

22. Troughs Layered Deposits of?

23. 510,000 years 0.5 to 2.1 million years ago 2.1 to 2.7 million years ago Variations induced by Astronomical Cycles?

24. Milankovitch Cycles

25. Astronomical forcing of global climate: Milankovitch Cycles http://www.sciencecourseware.org/eec/GlobalWarming/Tutorials/Milankovitch/ Take the quiz

26. http://www.nasa.gov/mission_pages/MRO/multimedia/phillips-20080515.htmlhttp://www.nasa.gov/mission_pages/MRO/multimedia/phillips-20080515.html http://www.sciencedaily.com/releases/2008/04/080420114718.htm

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

28. 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. What does this equation assume about the burial process? Is it a good assumption?

29. 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 Age = 1500 years 1m 2m 5.3m Age = 3000 years Age = 7950 years For k = 3000years/m Age = 3000 years Age = 6000 years Age = 15900 years

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

31. 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 -

32. 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 sedimentationbut instead would be the remains of sediments deposited at an earlier time A0. What is Ao?

33. 0 age at 0 depth: just one possibility The slope of this line is t/x =1500years/meter, what is the intercept? t x The intercept is the line’s point of intersection along the y (or Age) axis at depth =0.

34. 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?

35. 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 If the lake is currently 15 meters deep, how long will it take to fill up?

36. 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

37. 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

38. Our new equation looks like this -

39. 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.

40. Return to the group problems(open Day1GroupPbs.xlsx)

41. 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.