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

2. 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 operations  Units conversions  Linear relationships

3. 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).

4. Subscripts and superscripts provide information about specific variables and define mathematical operations. k 1 and k 2 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

5. 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 x a x b = x a / x b = (x a ) b = x a+b x a-b x ab

6. Question 1.2a Simplify and where possible evaluate the following expressions - i) 3 2 x 3 4 ii) (4 2 ) 2+2 iii) g i. g k iv) D 1.5. D 2

7. 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

8. In exponential form... M E = 597 x 10 22 kg M M = 7.35 x 10 22 kg The mass of the moon (M M ) can also be written as 0.0735 x 10 24 kg 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

9. 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 m 3 The mass of the earth’s crust is 2.8 x 10 22 kg The volume of the earth’s crust is 1 x 10 19 m 3 =mass/volume = 2.8 x 10 3 kg/m 3

10. Differences in the acceleration of gravity on the earth’s surface (and elsewhere) are often reported in milligals. 1 milligal =10 -5 meters/second 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/s 2 ) 1 milligal = 10 -5 m/s 2 hence 1 m/s 2 in terms of milligals is found by multiplying both sides of the above equation by 10 5 to yield 10 5 milligals= 1m/s 2 - thus g=(9.8m/s 2 ) x 10 5 milligals/(m/s 2 ) =9.8 x 10 5 milligals

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

12. 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.

13. 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.

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/s 2. How many meters per second squared is that?

16. 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 A e is the age of the earth (our  t) But, what is the age of the earth in days?

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

18. The North Atlantic Ocean is getting wider at the average rate v s of 4 x 10 -2 m/y and has width w of approximately 5 x 10 6 meters. 1. Write an expression giving the age, A, of the North Atlantic in terms of v s and w. 2. Evaluate your expression to answer the question - When did the North Atlantic begin to form? Plate spreading

20. How thick was it originally? Over what length of time was it deposited?

23. 510,000 years 2.1 to 2.7 million years ago 0.5 to 2.1 million years ago

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

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

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

28. 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. 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=age, z=depth where

30. 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 geologic significance of A 0 - 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. sedimentation but instead would be the remains of sediments deposited at an earlier time A 0.

33. 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. tt xx 0 age at 0 depth: just one possibility

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

35. Consider the significance of A 0 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.

36. 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

37. You should be able to show that A 0 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.

38. Our new equation looks like this -

39. … 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. Is this a good model?

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

41. 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.