Gradient Formula In Earth Science

WORKED-OUT Profile MAP Problem

On the map, the blue represents streams, the chocolate-brown lines are contours. The tiptop of one bespeak is given as 537 anxiety higher up sea level.

THE PROBLEM ASKS ME DO THE Following:

1. Print out the map.
2. Determine towards which management the stream is flowing.
3. Label all contours.
4. Calculate the boilerplate slope forth the stream from the blood-red dot at A to the cerise dot at B.
5. Make up one's mind the range of possible elevations for the pinkish dots at C, D and East.

HERE ARE MY ANSWERS:

1. The map is shown to the correct. ii. STREAM FLOW DIRECTION. The contours that cross the stream grade '5's that generally point towards the northward, so the stream flows generally towards the south. (The 'rule of 5's' states that where a contour crosses a stream, the contour forms a '5' that points upstream.) 3. LABEL CONTOURS. The contour interval is 20'. Possible contours include 500', 520', 540', 560', 580', 600', 620', etc. The given bespeak, 537', therefore, must lie betwixt the 520' and 540' contours (contours a and b). Because the stream flows south, contour a, which lies downstream of contour b, must be the lower of the 2. Thus, a is 520'; b is 540'. All contours that cross the stream must be in progression, getting higher in the upstream management. Thus, c is 560'; and d is 580'. Betoken q has an elevation greater than 560'. It lies outside closed hachured profile e, so e must be lower than q; so e = 560'. Similarly, indicate q lies outside of unhachured contour f, so f must exist higher than q; so 'f' = 580'. Point r is on the loftier side of the 580' contour, so is higher than 580'. Information technology lies ouside contour g, so one thousand must be college than r, so m = 600'. Contour h is inside the 600' contour then is higher; h = 620'. Bespeak southward lies inside the 620' profile and then is higher than 620'. Signal southward lies outside of hachured contour i; therefore contour i is less than due south; so i = 620'.
4. Slope. Gradient = vertical difference in elevation / horizontal distance. Then, to calculate the average gradient along the stream from the cerise dot at B to the scarlet dot at A (or vice versa) two facts demand to be known: The difference in acme between B and A. The altitude forth the stream from B to A. Difference IN Meridian: Bespeak A is on the 580' profile; point B is on the 520' contour. The difference in top is 60 feet. Altitude Forth THE STREAM: On my printed out map, I have marked off the number of inches betwixt B and A. Information technology equals 4.75 inches. The scale of the map is 1:2400, which means that each inch on the map = 2400 inches on the ground, or 200 feet. Then, iv.75 inches on the map = 4.75 x 200 feet = 950 anxiety. So, the gradient = deviation in elevation/horizontal distance = 60 vertical feet/950 horizontal feet. The last step is to express gradient in 'vertical feet' per 'horizontal mile'. To do this, I first convert the 950 horizontal feet into miles. There are 5,280 feet in a mile, so 950 anxiety = 950/5,280 miles = 0.180 miles. My gradient at present is 60 vertical feet/0.180 horizontal miles. This is a very awkward expression. I want to limited the gradient in terms of vertical feet/one horizontal mile. This can be done by solving for 'Ten' in the post-obit formula: X/lx = ane /0.180 or Ten = (60*1)/0.180 Thus, X = 333 and the gradient is 333 feet/mile, which is the desired answer. Important: If your printout is a different size than mine, and so the distance you measure betwixt points A and B will exist dissimilar from the distance I get. As a upshot, the slope you lot get volition be different than the answer shown in a higher place. To bank check if your slope is correct, multiply your gradient by your distance divided by my distance. If the result y'all go is equal to my gradient, then your respond is correct.
5. POSSIBLE ELEVATIONS FOR POINTS C, D, AND E: Point C lies between the 540' and 560' contours. It must, therefore take an peak that is more than 540' and less than 560'. Point D lies inside a 580' contour. There are no contours nested within that 580' contour. Since D lies within a non-hachured contour, it must exist college than that contour. The elevation of signal D must, therefore, be higher than 580'. Because there are no contours nested inside the 580' contour, there is no indicate within the 580' profile that has an elevation equal to or greater than the sum of 580' plus the contour interval (580' + xx' = 600'). The elevation of D must, therefore, be less than 600'. Point East lies inside a hachured 620' contour. There are no contours nested within that 620' contour. Since E lies within a hachured contour, information technology must be lower than that profile. The elevation of point E must, therefore, be less than 620'. Because there are no contours nested inside the hachured 620' contour, there is no point within that contour that has an elevation equal to or less than 620' minus the contour interval (620' - 20' = 600'). The elevation of E must, therefore, be greater than 600'.

Gradient Formula In Earth Science,

Source: http://academic.brooklyn.cuny.edu/geology/leveson/core/linksa/map_sample_answer2.html

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