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