Atmospheric Science 4320 / 7320

1. Atmospheric Science 4320 / 7320 Anthony R. Lupo

2. Day one • The orders of magnitude of PGF, CO, and Ro for a synoptic scale disturbances: • Synoptic scale disturbance: 2000 - 6000 km (space) 1 - 7 days (time)

3. Day one • Typical values:

4. Day one • Mesoscale disturbances: 10 - 2000 km, 1h - 1d

5. Day one • Hurricane

6. Day one • Tornadoes: ][

7. Day one • The horizontal Equation of motion for inviscid flow in terms of the geostrophic departure vector.

8. Day one • The horizontal acceleration vector is always normal and to the right of the geostrophic departure vector • It’s magnitude is directly proportional to the magnitude of the departure vector.

9. Day one • And here:

10. Day one • Then consider a case in which there are negligible spatial variations in Vgeo (spatially uniform PF) such that along the trajectory, Vg = constant • then,

11. Day one • becomes: • The time rate of change of the departure is always 90 degrees to the right of the Vag, and it’s magnitude is dependent on the magnitude of V departure!

12. Day one • The departure vector is of constant magnitude but rotates with time to the right - clockwise - with an inertial period of

13. Day one • The Inertial Oscillation  Is a consequence of Unbalanced flow! Thus, it is an oscillation whose frequencey is tied to the Coriolis parameter. • This is a “gravity-type” wave that is emitted from an unbalanced flow situation. More appropriately, this is a “Kelvin” type wave or oscillation.

14. Day one • Interial oscillation influential in: • Intense convection (MCSs) • Unbalanced jet streaks associated with highly curved flow and rapidly developing cyclones

15. Day one/two • Balanced flow in the friction layer. • Model I assume:

16. Day two • Thus the horizontal equations of motion become • The equation of motion in natural coords.

17. Day two • Thus for balance we must have for the s-component, • the frictional force balances the downstream component of the PGF.

18. Day two • Then for the n-component; • the CF balances the normal component of the PGF in Cartesian coordinates

19. Day two • The equation (BL)

20. Day two • The diagram • W/out Friction With Friction

21. Day two • Does our assumption for FR above hold up? • Balanced friction layer flow omitting the assumption of Fr acting directly opposite the horizontal velocity.

22. Day two • The balance equation (Eq 1) 

23. Day two • Thus we must determine by observation and calculation the direction of the friction vector: • PGF can be observed • the coriolis force can be calculated

24. Day two • and Fr can be determined from equation 1 as a residual: • Diagrammatically, you’ll get a result similar to our other one, at least it will be “CEFGW”

25. Day two • Friction in terms of the ageostrophic velocity vector • but recall that by geostrophy,

26. Day two • thus, • or,

27. Day two • so in this context,

28. Day two • Note of caution: some folks adhere to this position, that Vageo is equal to the difference between the Vfric and Vgeo. • We have shown this mathematically to be the case. We also know that Vgeo is non-divergent in it’s most basic form. • Thus what we have stated above is that Vfric is COMPLETELY the divergent part of the wind in a balanced flow.

29. Day two • However, you can also show mathematically, by Helmholtz partitioning that Vgeo contains a rotational and divergent part, and Vageo contains a divergent and rotational part (Keyser et al., 1989, MWR; Loughe et al., 1995 MWR, May, or Lupo, 2002, MWR )

30. Day two • Example: • Where “chi” is the velocity potential.

31. Day two • Diagramatic Example of Vfr development • Start with a system in geostrophic balance.

32. Day two • Wind adjusts such that PGF is balanced by CO, and is steady state. • Introduce a Frictional decelerating force (Fric.) (opposite the wind)

33. Day two • Friction reduces Vh to a value less that Vgeo which then reduces the CF, to a value less than that of the PGF resulting in a net force in the normal direction across the isobars towards the lower pressure causing air to turn to it’s left.

34. Day two • Thus final balance can be achieved if Vfr adjusts so that:

35. Day two • Properties of balanced flow in the friction layer • A) Horizontal non accelerating flow • B) Vfr < Vgeo or subgeostrophic wind speeds • C) Vfr has a component across the isobars from high to low • D) cross isobaric component is proportional to the Frictional force

36. Day two • Friction layer convergence and divergence results in vertical motions. • This will have important implications for when we examine the Vorticity, Omega Equations, etc....

37. Day two • Low pressure (surface convergence) gives upward motion. However, with regard to a developing system where vertical motion is forced aloft, the friction acts against cyclone development.

38. Day two • High pressure (surface divergence) gives downward motion. However, with regard to a developing system where vertical motion is forced aloft, the friction acts against anticyclone development.

39. Day three • Gradient Flow • (Holton 65 - 68 and Hess p 180 ff) • Gradient flow is unbalanced frictionless flow along a curved path at a constant speed and parallel to the isobars (isoheights)

40. Day three • Recall we observe that Vh tends to be parallel to the isobars in straight and curved flow regions. • Assumptions: 1) 2) 3)

41. Day three • Normal component of centripetal acceleration only. Rc (Radius of curvature) is positive for cyclonic flow and negative for anticyclonic flow. • 4) We’ll also assume that Rc = Ri, or the curvature of the flow is the same as the curvature of the isobars. • 5) acceleration = PGF - CF, normal acceleration due to unbalanced flow between PGF and CO.

42. Day three • The gradient wind equation • The horizontal equation of motion reduces to only the normal (n) component equation. (No (s) component, since there is no coriolis component in that direction)

43. Day three • The expression: • so, In scalar form • Ri = radius of curvature of the isobars

44. Day three • Equation above in form of (see it?): • where Vgr:

45. Day three • Recall: We must solve for the gradient wind by either “completing the square”, i.e., add 0.5 the “b” term and squaring, then adding to both sides. Alternatively, we can use the quadratic formula:

46. Day three • Examine the possible solutions of the Gradient Wind (Vgr) equation. • In agreement with polar coordinate convention, counterclockwise (clockwise) flow is a positive (negative) value of Vgr.

47. Day three • Consider positive root: • As the pressure gradient goes to zero, Vgr goes to zero

48. Day three • Oh yeah? Show me!

49. Day three • But, if the pressure gradient force > 0, then Vgr is positive. (Observed Low, counterclockwise flow)

50. Day three/four • Diagram!