Exercise 3: Averaging methods for vector time series.

Page 4

To summarize the situation up to this point: We have two ways to describe the observed data, either through speed and direction, or through the east and north components.

Let us now assume that we want to determine the mean wind over the two day observation period. We can calculate the mean in two ways:

The obvious question is: Do both methods give the same result, and if not, which method gives the correct result?

To answer the first question we look again at our model data from the Trade Wind zone and from the region of the Westerlies. Here are the data from the figures of the previous pages in tabular form. Verify that you understand the data - the direction of both wind systems, why the east component of the two wind systems is of different sign, why the north component is relatively small in both wind systems, and which system displays the larger variability in speed. Go back a page or two if you cannot think of reasons for these observations.

time Trade Wind Westerlies
(hours) speed
(m s-1)		 direction
(°)		 east comp.
(m s-1)		 north comp.
(m s-1)		 speed
(m s-1)		 direction
(°)		 east comp.
(m s-1)		 north comp.
(m s-1)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24		 7.79
7.92
8.05
7.94
7.47
7.63
7.58
7.34
6.63
7.06
7.15
7.45
7.48
7.76
8.20
8.07
7.87
7.48
7.46
7.31
7.04
7.03
7.23
7.60
6.98		 89.0
86.0
85.5
95.0
113.5
115.0
88.0
65.5
57.5
71.5
83.0
95.5
71.0
86.0
85.5
95.0
113.5
115.0
88.0
65.5
57.5
71.5
83.0
95.5
71.0		 7.79
7.90
8.03
7.91
6.85
6.92
7.58
6.68
5.59
6.70
7.10
7.42
7.08
7.74
8.17
8.03
7.22
6.78
7.45
6.65
5.93
6.67
7.18
7.56
6.60		 0.13
0.55
0.63
-0.69
-2.98
-3.23
0.26
3.05
3.56
2.24
0.87
-0.71
2.44
0.54
0.64
-0.70
-3.14
-3.16
0.26
3.03
3.78
2.23
0.88
-0.73
2.27		 10.11
5.39
1.08
7.76
5.35
4.89
0.98
11.44
13.01
14.25
6.81
1.87
3.92
6.11
13.94
8.23
9.66
9.99
13.24
6.14
11.23
6.80
7.47
10.08
7.00		 268.500
259.500
253.125
283.250
289.875
274.750
248.750
252.125
278.625
274.375
265.500
268.375
251.250
259.500
253.125
283.205
289.875
274.750
248.750
252.125
278.625
274.375
265.500
268.375
251.250		 -10.11
-5.30
-1.04
-7.55
-5.03
-4.87
-0.91
-10.89
-12.87
-14.21
-6.79
-1.87
-3.70
-6.00
-13.34
-8.01
-9.09
-9.96
-12.34
-5.84
-11.11
-6.78
-7.45
-10.08
-6.63		 -0.26
-0.98
-0.31
1.78
1.82
0.40
-0.35
-3.51
1.95
1.08
-0.53
-0.05
-1.26
-1.11
-4.04
1.89
3.28
0.83
-4.80
-1.88
1.68
0.52
-0.59
-0.29
-2.25
mean 7.50
7.20*		 85.72 7.18 0.48 7.87
7.68*		 266.7 -7.67 -0.28
(hours) speed
(m s-1)		 direction
(°)		 east comp.
(m s-1)		 north comp.
(m s-1)		 speed
(m s-1)		 direction
(°)		 east comp.
(m s-1)		 north comp.
(m s-1)
time Trade Wind Westerlies

*: This is the value for the mean wind speed if it is calculated from the mean east and north components.

Do the mean values for each column correspond to the estimates of your notes?

We can now answer the first question with a definite no, the two methods (calculating the mean wind speed from the speed, and calculating it from the mean of the components) do not give the same result.

The obvious next question is: Which result is correct? Select one answer only from the following options:

Mean wind speed has to be calculated from the mean of the components, because the components do not change as dramatically as the wind speed during a storm.
Mean wind speed has to be calculated from the observed wind speed, because if you want to determine mean wind speed, you start from the observations.
Mean wind speed has to be calculated from the components, because wind is a vector quantity, and vector means are derived from component means.
Mean wind speed has to be calculated from the observed wind speed, because the important quantity is speed; the direction can be ignored in this calculation.

back to contents © 2000 M. Tomczak
contact address:
home page: http://www.es.flinders.edu.au/~mattom
web address of Exercises in Oceanography: http://www.es.flinders.edu.au/~mattom/IntExerc
This page last updated 13 May 2000