M1.1
(a) Compare the appearance of the trajectories for the case with the curvature effect terms
and case without the curvature terms. Why is the trajectories not closed circle?
Input parameters:
 latitude = 60 degree
 zonal velocity = 0 m/s
 meridional velocity = 40 m/s
 runtime = 5 day
Circles are not closed in both cases. A displacement after each loop is to the right of the
initial position and the loop drifts eastward if the curvature effect term is neglected. This
direction is the opposite direction if the curvature effect is included. The radius of the
loop is V/f. Coriolis parameter (f) changes with latitude (the higher the latitude is, the
smaller coriolis parameter is and the greater the radius is). The radius of the top half of
the loop is larger than the one of the bottom half of the loop and thus the position of the
particle after one loop is displaced toward east. For the case where the curvature effect is
included, let’s divide the loop into quadrants (NW, NE, SE, SW). The table below shows
signs of each term in coriolis equations. In NW and SW, the arc is stretched out in zonal
direction. In NE and SE, the arc is stretched out in meridional direction. This causes the
displacement after one loop to drift westward.
quadrant
NW
NE
SE
SW
u
+
+
-
v
+
+
uv
+
+
-
-u2
-
Du/Dt
Coriolis
term
Dv/Dt
Curvature Coriolis
term
term
Curvature
term
+
+
+
+
-
-
+
+
(b) What is the difference if meridional velocity changed from 40 m/s to 80 m/s? By
varying the runtime, see if you can determine the period of the loop in each case and
compare this to the time given in Eq. (1.16) for 60 ° N
Input parameters
 latitude = 60 degree
 zonal velocity = 0 m/s
 meridional velocity = 80 m/s
 runtime =5 day
The radius of circle is given by V/f. As the velocity becomes larger, the radius of loop
becomes greater as seen in simulations. The frequency of the circle does not change, is
function of latitude, which is indicated by   sin.
Based on   sin, the time required to make a full loop is expected to be 49748 sec
(0.5758 day).
Input parameters:
 latitude = 60 degree
 zonal velocity = 0 m/s
 meridional velocity = 80 m/s
 runtime = 0.6 day
Temporal change in position (time around the end of one loop)
With the curvature effect
time,s
lat
long
Without the curvature effect
time, s
lat
long
42813.45
43320.78
43828.11
44345.06
44862.01
45378.95
45895.9
46410.73
46925.56
47440.39
47955.23
48327.63
48700.03
49072.43
49444.82
49730.4
50015.98
50301.56
50587.13
50900.35
51213.57
51526.78
51840
55.24
55.49
55.74
56.02
56.31
56.61
56.93
57.25
57.59
57.93
58.28
58.54
58.8
59.06
59.33
59.53
59.74
59.94
60.15
60.37
60.6
60.82
61.04
2.05
1.57
1.11
0.67
0.25
-0.14
-0.5
-0.83
-1.13
-1.39
-1.61
-1.75
-1.87
-1.97
-2.04
-2.08
-2.1
-2.12
-2.11
-2.09
-2.05
-1.99
-1.92
43356.54
43771.12
44185.71
44619.96
45054.2
45488.45
45922.7
46363.54
46804.38
47245.22
47686.05
48077.45
48468.84
48860.23
49251.62
49512.65
49773.69
50034.73
50295.76
50556.8
50817.83
51078.87
51339.9
55.74
55.95
56.17
56.41
56.66
56.92
57.19
57.47
57.76
58.06
58.36
58.63
58.91
59.18
59.46
59.65
59.84
60.03
60.21
60.4
60.59
60.78
60.96
7.64
7.27
6.9
6.54
6.21
5.89
5.6
5.33
5.08
4.87
4.68
4.54
4.42
4.33
4.26
4.24
4.22
4.21
4.22
4.24
4.28
4.32
4.38
Input parameters:
 latitude = 60 degree
 zonal velocity = 0 m/s
 meridional velocity = 40 m/s
 runtime = 0.6 day
Temporal change in position (time around the end of one loop)
With the curvature effect
time,s
lat
long
42223.29
57.62
1.78
42645.05
57.72
1.56
43089.29
57.82
1.33
43533.53
57.93
1.12
43977.78
58.05
0.91
44422.02
58.17
0.72
44882.78
58.31
0.53
45343.53
58.44
0.36
45804.29
58.59
0.2
46265.05
58.74
0.05
46717.67
58.88
-0.08
47170.28
59.04
-0.19
47622.9
59.19
-0.29
48075.52
59.35
-0.37
48467.66
59.49
-0.43
48859.81
59.63
-0.47
Without the curvature effect
time, s
lat
long
42583.78
57.74
3.13
42964.11
57.83
2.93
43368.19
57.93
2.73
43772.26
58.03
2.54
44176.34
58.14
2.36
44580.41
58.25
2.18
45006.41
58.37
2.01
45432.41
58.5
1.85
45858.41
58.64
1.71
46284.41
58.77
1.57
46695.42
58.91
1.46
47106.42
59.05
1.36
47517.43
59.19
1.27
47928.44
59.33
1.19
48295.96
59.46
1.14
48663.48
59.59
1.09
49251.96
49644.11
49968.02
50291.93
50615.85
50939.76
51164.82
51389.88
51614.94
51840
59.77
59.91
60.02
60.14
60.26
60.37
60.45
60.53
60.61
60.69
-0.5
-0.51
-0.52
-0.51
-0.49
-0.47
-0.44
-0.42
-0.38
-0.34
49031.01
49398.53
49715.16
50031.8
50348.43
50665.07
50931.68
51198.29
51464.9
51731.52
59.72
59.85
59.97
60.08
60.19
60.31
60.4
60.5
60.59
60.68
1.06
1.04
1.03
1.04
1.05
1.07
1.09
1.12
1.16
1.2
When latitude comes back to 60 degree, one loop ends. The followings are period for
each case
Velocity, m/s With curvature effect
Without curvature effect
49968.02
sec
49715.16
40
50301.56
50034.73
80
There are slight differences between the two period (the equations and simulations), but
not significant.M1.2.
Compare the magnitudes of the lateral deflection for ballistic missiles fired eastward and
westward at 43 degree latitude. The missile is launched at a velocity of 1000m s-1 and
travel 1000 km.
Eastward launch
Input parameters:
 latitude = 43 degree
 zonal velocity = 1000 m/s
 meridional velocity = 0 m/s
 runtime = 0.0116 day (1000 sec)
Final latitude
Magnitude of deflection
Without curvature terms
42.554 degree
-0.446 degree
Westward launch
Input parameters:
 latitude = 43 degree
 zonal velocity = -1000 m/s
 meridional velocity = 0 m/s
 runtime = 0.0116 day (1000 sec)
With curvature terms
41.909 degree
-1.190 degree
Final latitude
Magnitude of deflection
Without curvature terms
43.448 degree
0.448 degree
With curvature terms
42.79 degree
- 0.21 degree
Without the curvature effect term, the deflection is to the right of the velocity as seen in
matlab simulation. With the curvature effect, this may not occur as follow. Based on the
2
equation 1.11b, the Curvature effect term ( u tan a) contributes to negative
acceleration in the meridional direction for zonal movement. The degree of this effect
depends on latitude, and magnitude of zonal velocity regardless of its direction. If zonal
velocity is positive (eastward movement), coriolis force is negative (southward) in the
meridional direction and so is the force due to the curvature effect. Therefore, with the
curvature effect, deflection becomes greater in the negative direction as seen in the
matlab result. If zonal velocity is negative (westward movement), coriolis force directs in
the positive (northward). However, the curvature effect still works in the negative
direction in the meridonal direction. If zonal velocity is large enough or movement occurs
at high latitude, the magnitude of curvature effect becomes greater than coriolis effect,
resulting in deflecting in the negative direction.
Neglecting the curvature terms cause less error at lower latitude and/or for movement
with low velocity. In the case of westward launch given parameters (latitude, velocity) in
this simulation, neglecting the curvature effect cause huge error. It cannot be neglected.