3.12 The Coriolis Force
49
where n is time level and Δt is time step. The trajectory of our flui parcel can be
predicted with:
X
n+1
= X
n
+ Δt · U
n+1
and Y
n+1
= Y
n
+ Δt · V
n+1
(3.41)
Again, predictions from the momentum equations are inserted into the latter
equations as to yield an update of the locations. I decided to tackle this problem
entirely with SciLab without writing a FORTRAN simulation code.
3.12.8 Analytical Solution
Equations (3.37) and (3.38) can be combined to yield:
d
2
X
dt 2 = −Ω
2
X and
d
2
Y
dt 2 = −Ω
2
Y
(3.42)
The solution of these equations that satisfie initial conditions in terms of location
and velocity are given by:
X (t) = X o cos(Ωt) +
U o
Ω
sin(Ωt)
(3.43)
Y (t) = Y o cos(Ωt) +
V o
Ω
sin(Ωt)
(3.44)
This solution describes the trajectory of a parcel along an elliptical path. In the
absence of an initial disturbance (u = 0 and v = 0), and using (3.39), the latter
equations turn into:
X (t) = X o cos(Ωt) − Y o sin(Ωt)
Y (t) = Y o cos(Ωt) + X o sin(Ωt)
which is the trajectory along a circle of radius
X 2
o + Y 2
o , as expected.
3.12.9 The Coriolis Force
We can now reveal the Coriolis force by translating the trajectory seen in the f xed
frame of reference (see Fig. 3.16), described by (3.43) and (3.44), into coordinates
of the rotating frame of reference. The corresponding transformation reads:
x = X cos(Ωt) + Y sin(Ωt)
(3.45)
y = Y cos(Ωt) − X sin(Ωt).
(3.46)
Figure 3.18 shows the resultant fl w path as seen by an observer in the rotating frame
of reference. Interestingly, the flui parcel follows a circular path and completes the