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When isobaric surfaces surround the Northern Hemisphere high in the south and low
in the north, the motion follows the trajectory in Fig6. 4.
Now I would like to calculate how much kinetic energy an air parcel obtains when evolving
from its motionless state �� to state �� where the speed reaches maximum. The force a unit volume of air parcel undergoes and the distance it covers constitute work. The energy of this work is converted into kinetic energy. Of the two forces the Coriolis force is perpendicular to the direction of the motion and does not contribute to the work. ( refer to Fig6. 5 )
Therefore, kinetic energy obtained by a unit volume of air parcel during this motion
is provided only by the pressure gradient force. The pressure gradient force is always perpendicular to contours of height. So the total amount of the work during this motion is obtained by integrating from �� to �� ( in Fig.4 ) the inner product of the pressure gradient force and the line segment along the path of the air parcel.
The amount of work given is
∫ −ρ・g・∂h／∂n・ds =ρ・g( height � �height ��) where ρ is the density of air, h is the height of isobaric surface and n and s denote the unit vector directed to the steepest slope of isobaric surface and the unit vector directed to the path of the parcel, respectively.
This equation means that the energy produced during the parcel’s transferring from ��
to �� is equal to the lost amount of potential energy. This amount of work becomes kinetic energy, which is 1／2・ρ・v2
That means an air parcel in ageostrophic motion obtains kinetic energy by being compressed
by the pressure gradient force, and this kinetic energy is equal to the potential energy which is lost while moving down the isobaric surface. In other words, an air parcel obtains
the velocity by moving on isobaric surface and lowering its height.
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6. Ageostrophic wind (1)
Here, we have to learn about ageostrophic motion.
the air parcel has the same density as the surrounding air at that height. How does this
Suppose you put an air parcel quietly on a surface which has the pressure gradient and
air parcel behave? To make things easier, I assume that the surrounding air never changes
its pressure gradient and the air parcel never mingles with the surrounding air.
At first the Coriolis force doesn’t act on it, because the speed of the air parcel is zero.
The forces which act on the air parcel are the pressure gradient and Coriolis forces.
Only the pressure gradient force acts on it. So the air parcel starts to sink perpendicular
to contours of height.
∂V／∂t＝―f・V−g・∇h
The following equation of motion can be obtained:
whereV is the air parcel’s velocity after t time and f, g and ∇h denote the Coriolis
parameter, the acceleration of gravity and the gradient of the geopotential height,
respectively.
Geostrophic wind at this moment Vg gives
∂Vg／∂t＝0＝−f・Vg−g・∇h
Subtracting the latter from the former leads to
∂（V−Vg）／∂t＝−f・（V−Vg）
This equation reduces to
∂A／∂t＝−f・A where a vector A keeps on rotating with a frequency f. Its period T is 2・π／f＝2・π／2・ω・sinφ≒12／sinφ where φ means latitude and ω denotes the rotating angular velocity of the earth, which is approximately equal to 2・π／24 hours. So at 30゜N the rotation has a 24-hour period. Fig6. 1 shows what the above equations mean.
1. �� : An air parcel is at rest on the pressure field. 2. �� : First, it starts to move toward low height perpendicular to height contours in response to the pressure gradient force, but as soon as motion develops, the Coriolis force also acts on it. So the parcel moves acceleratingly with both forces acting on it. The pressure gradient force remains unchanged, whereas the Coriolis force acts deflecting the parcel’s motion toward the right in proportion to its speed. So descending along contours of isobaric height, the parcel gradually accelerates parallel to contours of height. 3. �� : Eventually the parcel’s motion becomes parallel to height contours and has the same direction as geostrophic wind’s. Converting potential energy into kinetic energy, the parcel moves downward along the slope of the pressure surface and its speed becomes twice as high as geostrophic wind’s. The Coriolis force also becomes twice as strong as the one needed for geostrophic balance and acts on the parcel the way it makes
the parcel move upward perpendicular to contours of height, just as strongly as the
force acting on the initial motionless air parcel, but reversely. 4. �� : The parcel’s direction is gradually deflected to the right and the parcel
decelerates upward across contours of isobaric height.
5. �� : Getting to its original height, the parcel is again at rest for a moment. Then the parcel repeats the motions from �� to ��.
Fig. 2 shows one period of ageostrophic motion and the forces acting on it. There contours
of isobaric height are not drawn in straight lines but in curved lines.
Here the differental vector means ageostrophic component in Chapter 5.
to learn more about ageostrophic motion |
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