Problem Set 2a Precipitation 1. The shape of the drainage basin can be approximated by a polygon, whose vertices are located at the following coordinates: (5, 5), (-5, 5), (-5,-5), (0.-10), and (5,-5). The rainfall amounts of a storm were recorded by a number of rain gages situated within and nearby the basin as follows: All coordinates are expressed in kilometers. Determine the average rainfall on the basin by (a) arithmetic-mean method, (b) the Thiessen method, and (c) the isohyetal method. Hints: For the Thiessen method, begin by drawing the polygon around gage 9, then draw the polygons around gages 2,3,5,and 7; for the isohyetal method, draw isohyets with the maximum rainfall on a ridge running southwest to northeast through (-3,3). Solution: a.) Arithmetic Mean Method Σ(ππππππππ ππππππππ) π 63 + 59 + 41 + 39 + 105 + 98 + 60 + 41 + 81 π= 9 587 π= 9 π= π = 65.22ππ b.) Thiessen Method π΄πππ ππ π΅ππ ππ = 107.5 π΄πππ ππ πΌππππ’πππππ = 16,23,13,16,0,35,14,12,4 π (63 ∗ 16) + (59 ∗ 23) + (41 ∗ 13) + (39 ∗ 16) + (105 ∗ 0) + (98 ∗ 35) + (60 ∗ 14) + (41 ∗ 12) + (81 ∗ 4) = 107.5 π = 80.08ππ c.) Isohyetal Method 16(63 + 59) 23(59 + 41) 13(41 + 39) 16(39 + 105) 35(105 + 98) 14(98 + 60) 12(60 + 41) 4(41 + 81) ( )+( )+( )+( )+( )+( )+( )+( ) 2 2 2 2 2 2 2 2 π= 16 + 23 + 13 + 16 + 35 + 14 + 12 + 4 π= 9306.5 133 π = 69.97ππ Area of Basin 6 4 X-Axis 2 0 -6 -4 -2 -2 0 2 4 6 -4 -6 -8 -10 -12 Y-Axis Coordinates within and near the Basin 6 4 X-Axis 2 0 -12 -10 -8 -6 -4 -2 -2 0 2 4 -4 -6 -8 -10 -12 Y-Axis Area of basin= Area of basin= (ππ΄ ππ΅ −ππ΄ ππ΅ )+(ππ΅ ππΆ −ππ΅ ππΆ )+(ππΆ ππ· −ππΆ ππ· )β―β―β―β―(ππ ππ −ππ ππ ) 2 ⌊(25−−25)+(25−−25)+(50−−5)+(−5−−15)+(25−−25)⌋ 2 Area of influences are= 16,23,13,16,0,35,14,12,4 = 107.5 ππ2 6 8