• Member Statistics

    16,680
    Total Members
    7,904
    Most Online
    sereinwriter
    Newest Member
    sereinwriter
    Joined

Archived

This topic is now archived and is closed to further replies.

H2Otown_WX

Subgeostrophic/Supergeostrophic

Recommended Posts

I was mulling through T.N. Carlson's "Mid-Latitude Weather Systems" and came across a concept that confused me a bit. He was referring to the idea of quantifying vorticity advection via the method of solenoids and stated that it would tend to overestimate vorticity advection by the total wind in troughs because the wind is subgeostrophic and tend to underestimate it in ridges because the total wind in ridges is supergeostrophic. This seemed counter-intuitive to me as I would think wind is a trough would be supergeostrophic and vice versa. Does it have something to do with the direction of the velocity vector?

Share this post


Link to post
Share on other sites

There are a couple ways to explain this, but I think the easiest is through the ageostrophic wind. (You may already know this, so sorry if I am simply repeating it).

When we make the geostrophic assumption, we assume that the wind is not accelerating. This is clearly not true in the base of troughs or peaks of ridges since the wind is changing direction there.

The ageostrophic wind is given by

Vag = k/f X dV/dt

where the X is a cross product. Also, we know that

V = Vg + Vag

At the bottom of a trough, the acceleration vector, dV/dt, is toward the north in the positive v direction, since the wind is changing from NWly to NEly (in an idealized situation). When you take the cross product, the ageostrophic wind is pointing westward in the -u direction, and the geostrophic wind is in the positive u direction. This implies that the actual wind is less than the geostrophic wind since the ageostrophic component is acting in the opposite direction. The same applies to the tops of ridges, but you find that the ageostrophic wind points in the same direction as the geostrophic wind, making the actual wind > geostrophic wind.

Another way to examine this is by incorporating the centrifugal force at the base of a trough/top of a ridge because of the curvature. With the PGF given in one direction, (for a trough in this example) the Coriolis force and the Centrifugal force will point in the opposite direction. Because of the addition of the centrifugal force vector, for the flow to be "balanced", the Coriolis vector must shorten, which implies slower winds speeds than geostrophic flow, which would exclude the PGF). If one considers the gradient wind to be the actual wind, then the ratio of the geostrophic wind to the actual wind can be given by

Vg/V = 1+ V/fR

where f is the Coriolis parameter and R is positive for cyclonic flow. In this way, it is clear that since f is positive in the northern hemisphere and R is positive for troughs, the actual wind must be greater than the geostrophic wind. The opposite applies for the peaks of ridges.

grad3.gif

Hopefully this made an ounce of sense and was at least slightly related to your question! :lol:

Share this post


Link to post
Share on other sites

Wait I'm a little confused, at the end it seems like you agreed with what I thought. R is positive for troughs, the actual wind must be greater than the geostrophic wind.

Share this post


Link to post
Share on other sites

Wait I'm a little confused, at the end it seems like you agreed with what I thought. R is positive for troughs, the actual wind must be greater than the geostrophic wind.

The ratio is

Vg/V = 1 + V/fR

In the N.H., f is positive, and R is positive for troughs, so Vg/V > 1, meaning that the geostrophic wind is greater than the actual wind, making the actual wind subgeostrophic.

Share this post


Link to post
Share on other sites

The ratio is

Vg/V = 1 + V/fR

In the N.H., f is positive, and R is positive for troughs, so Vg/V > 1, meaning that the geostrophic wind is greater than the actual wind, making the actual wind subgeostrophic.

Ohhh okay..well that makes sense then. Thanks I like how that equation kind of just sums it up that is nice.

Share this post


Link to post
Share on other sites

  • Recently Browsing   0 members

    No registered users viewing this page.