y Manual in Physical Geology Activity 9.5 Relief and Gradient (Slope) Analysis N
ID: 288818 • Letter: Y
Question
y Manual in Physical Geology Activity 9.5 Relief and Gradient (Slope) Analysis Name: On a vinit to Yourmite, you and wome friends find a nice place along the Merced River for a prmik, Vevs hane a sopegrapic con tour map (Fig. A9.5.1) in which the indes csurs are lalbeled in feet. From bertter vantage point ny walking up to point B 6200 -6 Merced River Figure A9.S.1 A Using the labwked elevatis of the index cont, what is the coonr ineral (he cevalion ilence -f. between two adjacent contour) in Fig. AaS1- degandnapna aa Ha tuba O'ou have a har scale on the ma that ird rates adistance of linna har sale on the map in mani 30 (Hnt:Measare the disance from A to B What is the horisontal distance from A to B measused along a straight line in t using a metric ruler on the map, and then C armeasuremen,ofhow man, mm onthe map equalloon inthe tofindthe answer using proportiondthe to find the answer using proportions.) aat i. the average podmt (elevation change dinikd h, hirinontal diunte, beth r prewd ati ding from A to B? 262Explanation / Answer
The average gradient along the straight line A to B is = elevation change / horizontal distance
We see in the map that each contour interval is of 40 ft. And the map is 1:1000 ft in scale. Therefore the elevation change between A and B would be approximately 6600-6100 = 500 ft and the horizontal distance (according to the image provided in the question which might not be the actual linear distance in the actual map in the paper of the student) = 8 cm in linear scale = 1000/30.82 x 8 = 259.5717 ft. (as 30.89 cm = 1 ft)
therefore, gradient = 500/259.5717 = 1.92
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