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3.12 The Coriolis Force 49where n is time level and Δt is time step. The trajectory of our flui parcel can bepredicted 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 latterequations as to yield an update of the locations. I decided to tackle this problementirely with SciLab without writing a FORTRAN simulation code.3.12.8 Analytical SolutionEquations (3.37) and (3.38) can be combined to yield:d 2 Xdt 2=−Ω2 X and d2 Ydt 2 =−Ω2 Y (3.42)The solution of these equations that satisfie initial conditions in terms of locationand velocity are given by:X(t) = X o cos(Ωt) + U osin(Ωt)Ω(3.43)Y (t) = Y o cos(Ωt) + V osin(Ωt)Ω(3.44)This solution describes the trajectory of a parcel along an elliptical path. In theabsence of an initial disturbance (u = 0 and v = 0), and using (3.39), the latterequations 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 √ Xo 2 + Y o 2 , as expected.3.12.9 The Coriolis ForceWe can now reveal the Coriolis force by translating the trajectory seen in the f xedframe of reference (see Fig. 3.16), described by (3.43) and (3.44), into coordinatesof 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 frameof reference. Interestingly, the flui parcel follows a circular path and completes the