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Geology 351 – Mathematics for Geologists Computer lab – Evaluating erosion as a function of relief Erosion The following problem is taken from Ahnert (1970) - Functional relationships between denudation, relief, and uplift in large mid-latitude drainage basin, American Journal of Science, v. 268, p. 244-263. In earlier presentations we considered the influence of isostacy on variations in crustal thickness needed to compensate topographic features. In one exercise we considered the interplay between erosion and isostacy. We found, for example, that removing two kilometers of material from the mountain did not reduce the mountain height by 2km, because there was rebound of the crustal root in response to erosion. In the present example, we visit the issue of erosion rates from a quantitative point of view. Erosion rates are considered to be proportional to relief, as expressed by the following equation - dh h dt (1 In this equation, h is the relief, dh/dt is the erosion rate (change of relief with change in time) and is an erosion constant that has units of inverse length. Although we haven't gotten into integration, but you have probably already dealt with the integration of functions like this in your introductory calculus class. Integration of this equation yields a result we've seen before. Rearrange the above to obtain dh dt h (2 and integrate - i.e. h h0 t dh dt h 0 (3 where h0 is the initial relief, h the final relief and t the interval of time over which the erosional process is followed. This integration yields ln( h) ln( h0 ) t h ln( ) t h0 (4 (5 Then take both sides of the above equation to be exponents of e (the natural base) yields h exp(t ) h0 (6 and thus h(t ) h0 exp( t ) (7 Note that the erosion process behaves similar to porosity variations as a function of depth resulting from compaction (page 56 and exam problem), the growth rates of reefs (problem 2.15), and the radioactive decay processes (problem 2.13). Data presented in the table below are considered to be representative of erosion rates typical of mid-latitude drainage basins. Note that denudation rate is dh/dt and relief is h. Referring to equation (1), note that dh/dt is our y-variable and h is our x-variable. Relief (meters) 199 406 804 897 1189 1384 1585 1871 2073 Denudation Rate (m/ma) 55 68 127 129 150 160 219 247 295 Problem: 1) Compute the erosion coefficient , and 2) determine how many millions of years are required to reduce a mountain whose initial elevation is 2000 meters to an elevation of only 100 meters. In part 2 compute h(t) for times ranging from 0 to 70 My at 1My intervals. Plot your result and note the location of the point in time where relief has dropped to 100 meters. Open Excel and enter the above data into the spreadsheet. Your spreadsheet will look something like that shown at right These data should follow the relationship defined in equation (1) i.e. dh h . This is dt just a straight line, and the intercept in this case is 0. The general form of the linear regression relationship has both intercept and slope, but we know that the equation we wish to solve for has an intercept of 0. It is in the form: y=ax. So plot up your data, highlight the line, left click and select the trendline option. Under the options tab (see right) set intercept to 0, display the equation and, to give us a feel for how well the data fit a line, display R2. Variations of Relief through Time Denudation Rate dh/dt (m/My) 350 300 250 200 150 100 50 0 0 500 1000 1500 2000 Relief h (m) In your presentation, show the calculated regression line in your graph and indicate what the value of is Part 1) is complete and for part 2) we need to generate computations of h over a 70 million year time period with initial elevation of 2000 meters. We can use that data to determine the time it takes to wear this 2000 meter mountain down to 100 meters or we could compute it directly by solving for t in equation 4). This is something you should be good at by now. Just as a reminder, do the following: Create a column of times extending from 0 to 70 My in increments of 1 million years. You could label the column t or time. Then in the adjacent column enter the equation =2000*exp(-0.____*c2) and enter. This assumes you put your times in column c. Then copy the formula into the cells down to the end-time of 70 million years. Plot your data and note the point on the curve where h =100 meters as shown below. Erosion through Time 2000 Relief (m) 1500 1000 500 0 0 10 20 30 40 50 60 70 Time (My) Finally, compute t for h = 100 meters. As noted above, we could simply solve equation 4) ln h ln h0 t for t, when h = 100. That yields ln(20)/ to obtain ______ My. Summary of results: Erosion rate is assumed to vary in direct proportion to relief. This leads to exponential reductions of relief with time. In the present example, the height of a 2000 meter mountain is reduced by 1900 meters in ________ million years. If we look forward in time an additional ________ million years, what is the additional drop in relief? It is interesting to note that while the erosion rate decreases rapidly with time, the percentage of the remaining material that is eroded in unit time remains the same. For example, you have just found that in _______ million years, the mountain relief decreased 95% (from 2000 meters to 100 meters). You will also find that in the succeeding period of the same duration, mountain relief will decrease an additional 95% (form 100 meters to 5 meters), and so on. The same proportion of the remaining relief (95%) is removed in consecutive time intervals you’ve determined in the foregoing analysis. This is a basic property of the exponential decay process whether it be accumulation rate, erosion rate, porosity variations with depth, radioactivity …etc. Notes Assignment: a. Provide a plot of relief versus denudation rate. Plot the trendline, trendline equation, and value of R2 in your graph window. Follow the procedures outlined above. b. From the trendline equation what is the erosion coefficient . c. Given the erosion coefficient , calculate the variations of h through time and graph your results. Hand in a labeled graph. d. Circle the point on the graph where the relief drops to 100 meters. In revision: additional questions to be posted