### Geostrophic Balance

This balance occurs often in the atmospheric flow that is a straight line (R=±∞) well above Earth’s surface, so that friction does not matter. The Rossby number, R_{o}, is much less than 1.

Let’s think about how this balance might occur. Assume that the parcel is initially at rest ($\overrightarrow{V}=0$ ) in the Northern Hemisphere (see figure below). The Coriolis force is thus zero and the parcel begins to move from high pressure toward low pressure. However, as it accelerates and attains a velocity perpendicular to the pressure gradient, the Coriolis forces begin to grow perpendicular and to the right of the velocity vector and the PGF. The resulting acceleration is now the vector sum of the PGF and the Coriolis forces and turns the velocity to the right of the PGF. As the velocity continues to increase, the Coriolis forces increase but stay always perpendicular and to the right of the velocity vector while the PGF always stays perpendicular to the pressure gradient. Eventually, the PGF and Coriolis forces become equal and opposite and the air parcel will move parallel to the horizontal pressure gradient. This condition is called **geostrophic balance**. We have simplified somewhat the approach to geostrophic equilibrium because, in reality, air parcels would overshoot and undergo inertial oscillations (discussed below) and because the pressure field would evolve in response to the motion.

$$-f{V}_{}-\frac{\partial \Phi}{\partial n}=0\text{geostrophicwindbalance}$$

Let the Coriolis term be designated as $Co=-f{V}_{}$ and the pressure gradient term as $P=-\frac{\partial \Phi}{\partial n}$ . Then the force balances are shown in the figure below.

Note that the Coriolis force is always to the right of the velocity vector in the Northern Hemisphere. It is always to the left of the velocity vector in the Southern Hemisphere. When the pressure gradient force and Coriolis force are in balance, the PGF is to the left of the velocity vector and the Coriolis force is to the right in the Northern Hemisphere. Watch the video below (1:10) for further explanation:

Click here for a transcript of Into Geostrophic Balance Video

### Inertial Balance

In this case, the pressure gradient force is minimal and the centrifugal and Coriolis forces are in balance.

$$-\frac{V{}^{2}}{R}-fV=0\text{inertialbalance}$$

Let the centrifugal force be designated by $Ce=-\frac{V{}^{2}}{R}$ .

We can manipulate Equation [10.37] to find the radius of the circle:

$$R=-\frac{V}{f}$$

For* f* ~ 10^{-4} s^{-1} and *V *~ 10 m s^{-1}, *R* = - 100 km. Inertial balance is not a major balance in the atmosphere because there is almost always a significant pressure gradient, but it can be important in oceans.

### Cyclostrophic Balance

The balance in this case is between the pressure gradient force and the centrifugal force.

$$-\frac{V{}^{2}}{R}\text{\hspace{0.17em}}-\frac{\partial \Phi}{\partial n}=0$$

In this case, the scale of the motion is so small that Coriolis acceleration is not important. The Rossby number, R_{o}= centrifugal acceleration / Coriolis >> 1.

Examples are tornadoes, dust devils, water spouts, and other small circulations. These can be either cyclonic or anticyclonic and, in fact, a few percent of tornadoes are anticyclonic.

### Gradient Balance

In gradient balance, the pressure gradient force, Coriolis force, and horizontal centrifugal force are all important. This balance occurs as wind in a pressure gradient field goes around a curve.** **There are many examples of this type of flow on any weather map - any synoptic-scale pressure gradient for which the isobars curve is an example of gradient flow.

$$-\frac{V{}^{2}}{R}-fV-\frac{\partial \Phi}{\partial n}=0\text{}$$

To solve this equation for velocity, we can use the quadratic equation:

$$\begin{array}{l}a{x}^{2}+bx+c=0\text{\hspace{1em}}\to \text{\hspace{1em}}x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}\\ \frac{{V}_{}{}^{2}}{R}+f{V}_{}+\frac{\partial \Phi}{\partial n}=0\text{\hspace{1em}}\to \text{\hspace{1em}}{V}_{}=\frac{-f\pm \sqrt{{f}^{2}-4\left(\frac{1}{R}\right)\frac{\partial \Phi}{\partial n}}}{\frac{2}{R}}\end{array}$$

$$V=-\frac{f\text{\hspace{0.05em}}R}{2}\pm \sqrt{\frac{{f}^{2}{R}^{2}}{4}-R\frac{\partial \Phi}{\partial n}}$$

$\frac{{f}^{2}{R}^{2}}{4}$is always positive, so there are eight possibilities because R can be either positive or negative, $\frac{\partial \Phi}{\partial n}$can be positive or negative, and we have the ± sign in between the two terms on the right hand side of Equation [10.39].

Northern Hemisphere | R>0 |
R<0 |
---|---|---|

$\frac{\partial \Phi}{\partial n}$ >0 | no physical solutions | positive root; physical |

$\frac{\partial \Phi}{\partial n}$ <0 | positive root; physical | both roots; physical |

You are looking for whether positive or negative values of R and $\frac{\partial \Phi}{\partial n}$give non-negative and real values for V because only non-negative and real V is physically possible.

For R > 0 and $\frac{\partial \Phi}{\partial n}$> 0, V is always negative, so there are no physical solutions.

For R > 0 and $\frac{\partial \Phi}{\partial n}$< 0, only the plus sign gives a positive *V* and thus a physical solution.

For R < 0 and $\frac{\partial \Phi}{\partial n}$> 0, only the plus sign gives a positive *V* and thus a physical solution.

For R < 0 and $\frac{\partial \Phi}{\partial n}$< 0, both roots give positive *V* and thus physical solutions.

So there are four physical solutions. However, there is one more constraint. This additional constraint is that the absolute angular momentum about the axis of rotation at the latitude of the air parcel should be positive in the Northern Hemisphere (and negative in the Southern Hemisphere). Without proof, we state that only two of the four physically possible cases meet this criterion of positive absolute angular momentum in the Northern Hemisphere. They are:

- Regular low: R > 0 and $\frac{\partial \Phi}{\partial n}$< 0 and ${V}_{}=-\frac{f\text{\hspace{0.05em}}R}{2}+\sqrt{\frac{{f}^{2}{R}^{2}}{4}-R\frac{\partial \Phi}{\partial n}}$
- Regular high: R < 0 and $\frac{\partial \Phi}{\partial n}$< 0 and ${V}_{}=-\frac{f\text{\hspace{0.05em}}R}{2}-\sqrt{\frac{{f}^{2}{R}^{2}}{4}-R\frac{\partial \Phi}{\partial n}}$

These two cases are depicted in the second and third panels of the figure below.

The video below ( 3:22) explains these four force balances in more detail: