### 10.7 Natural coordinates are better horizontal coordinates.

Understanding the results of a balance of forces can often be easier if we choose a horizontal coordinate system that is aligned naturally with the air flow, and not just set up in Cartesian coordinates *x* and *y* or spherical coordinates *λ* and *φ*. We can choose one direction—let’s call it *s* for streamline—so that it is aligned with the streamline (and is thus always parallel with the flow) and increases in the direction of the flow (i.e., downwind). The second direction—let’s call it *n* for normal—increases to the left of the flow. An interesting and noteworthy feature of natural coordinates is that the wind velocity is always positive because, by the definition of natural coordinates, the velocity vector is always pointed in the positive *s* direction.

For the horizontal momentum equation without friction:

$$\frac{D\overrightarrow{V}}{Dt}=-f\text{\hspace{0.17em}}\overrightarrow{k}\times \overrightarrow{V}-{\overrightarrow{\nabla}}_{p}\Phi $$

where $\overrightarrow{V}=\overrightarrow{i}u+\overrightarrow{j}v$ and ${\overrightarrow{\nabla}}_{p}$ is the horizontal gradient operator on a pressure surface, not a height surface.

Let’s look at each term in the equation and put it in natural coordinates.

The **acceleration of the air parcel**: $$\frac{D\overrightarrow{V}}{Dt}$$

There are two ways that $\overrightarrow{V}$ can change. It can change by changing its speed ($V=\left|\overrightarrow{V}\right|$), $\frac{DV}{Dt}$ , which occurs in the streamline direction *s,* or by changing its direction, which occurs in the normal direction, *n*. Recall that *n* points to the left of the velocity vector.

Remember your physics, which showed that the acceleration due to rotation (the centripetal acceleration) is the velocity squared divided by the radius of curvature of the rotation. In this case, the rotation is the rotation of the air parcel as it moves horizontally over the Earth (not the rotation of the Earth itself). The acceleration is directed perpendicular to the motion and is directed towards the center of curvature. *R* is the radius of curvature of the changing direction, and by convention, *R *> 0 if the curvature is in the counterclockwise direction, and *R *< 0 if the curvature is in the clockwise direction. Hence the sign always works out if the centripetal acceleration is *V*^{2}/*R*.

The **Coriolis force**:

$$-f\text{\hspace{0.17em}}\overrightarrow{k}\times \overrightarrow{V}$$

The *n* component equals $-\mathrm{fV}\text{\hspace{0.17em}}$ because when *f* > 0, the force is to the right of the motion (i.e., streamline) and is, thus, negative because positive *n* points to the left of the motion. When *f* < 0, the force would be to the left in the positive *n* direction.

The **pressure gradient force (PGF)**:

$$-{\overrightarrow{\nabla}}_{p}\Phi $$

The *n* component of $-{\overrightarrow{\nabla}}_{p}\Phi $ is just $-\frac{\partial \Phi}{\partial n}$ .

Putting these pieces all together results in the equation for the *n* component of the horizontal momentum equation:

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

This equation is called the **Gradient Wind Equation**. The form of the equation in *z* coordinates is sometimes useful:

$$\frac{{V}^{2}}{R}=-fV-\frac{1}{\rho}\frac{\partial p}{\partial n}$$

Keep in mind that *V* is always positive in Equations [10.34] because they are cast in natural coordinates.

Let’s rearrange Equation [10.34a] slightly by moving the centripetal acceleration over to the right hand side, where it represents a centrifugal force. Then, by considering which of the terms are the most important, we can apply this equation to different situations:

$$\begin{array}{l}-\frac{{V}^{2}}{R}-fV-\frac{\partial \Phi}{\partial n}=0\text{,gradientbalance}\hfill \\ \text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{0.17em}}-fV-\frac{\partial \Phi}{\partial n}=0\text{,geostrophicbalance}\hfill \\ -\frac{{V}^{2}}{R}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\frac{\partial \Phi}{\partial n}=0\text{,cyclostrophicbalance}\hfill \\ -\frac{{V}^{2}}{R}-fV\text{\hspace{1em}}\text{\hspace{1em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0\text{,inertialbalance}\hfill \end{array}$$

We can use the **Rossby number** to help determine which of the above balances dominates. The Rossby number is defined to be the magnitude of the acceleration divided by the magnitude of the Coriolis force. When applied to the gradient wind equation, the Rossby number is given by

${R}_{o}=\frac{\text{acceleration}}{\text{Coriolisforce}}=\frac{{V}^{2}/R}{fV}=\frac{V}{fR}$

If *R _{o} *<< 1, then the centrifugal force is much smaller than the Coriolis force and the Coriolis force must balance the PGF (geostrophic balance).

If *R _{o} *>> 1, then the Coriolis force is much smaller than the centrifugal force and the centrifugal force must balance the PGF (cyclostrophic balance).

If *R _{o} *~ 1, then the Coriolis force and the centrifugal force balance each other (inertial balance).

#### Quiz 10-2: Coordinates and scales.

- Find
**Practice Quiz 10-2**in Canvas. You may complete this practice quiz as many times as you want. It is not graded, but it allows you to check your level of preparedness before taking the graded quiz. - When you feel you are ready, take
**Quiz 10-2**. You will be allowed to take this quiz only**once**. Good luck!