# 10.7: Are all the terms in these equations equally important? Let's use scale analysis.

Our equations of motion have several terms in them, and the question is, “Which ones are the largest and, thus, the most important?” The answer is, “It depends on the situation.” You can follow the following steps to determine which terms to keep and which you can safely ignore when you are trying to calculate the force balance for a specific atmospheric phenomenon. This process is called scale analysis and it can be applied to any conservation equation you would like to simplify.

1. Decide on the phenomenon of interest (e.g., cyclone, front, hurricane, tornado).
2. Determine the characteristic (i.e., typical) lengths and times over which the phenomenon occurs.
3. Determine the range of fluctuations of equation variables in space and time during the phenomenon.
4. Approximate derivatives (i.e., $$\left.\frac{\partial p}{\partial x} \sim \frac{\Delta p}{\Delta x}\right)$$
5. Compare the magnitudes of the terms in the equation.
6. Keep only the relatively large terms (say, the top two orders of magnitude) and neglect the smaller terms.

### The Coriolis Parameter

Define the Coriolis parameter as $$f \equiv 2 \Omega \sin \phi .$$ At $$45^{\circ} \mathrm{N}$$

$$f=(2)\left(7.27 \times 10^{-5} \mathrm{s}^{-1}\right) \sin 45^{\circ} \sim 10^{-4} \mathrm{s}^{-1}$$ We will use this parameter in the scale analysis in the next section and then throughout the rest of the lesson.

Example: Scale analysis of the averaged x-momentum equation for mid-latitude synoptic-scale flow in the free troposphere.

1. Phenomenon: mid-latitude synoptic-scale flow in the free troposphere.

2. $$L \sim 1000 \mathrm{km}=10^{6} \mathrm{m} ; H \sim 10 \mathrm{km}=10^{4} \mathrm{m} ;\left(\text { in boundary layer only, } C_{d} \sim 10^{-2} \text { and } h \sim 1000 \mathrm{m}\right) ; t \sim ?$$

3.

$$U \sim 10 \mathrm{m} \mathrm{s}^{-1} ; \Delta p \sim 10 \mathrm{hPa}=10^{3} \mathrm{Pa}$$ (in the horizontal)

$$t \sim L / u=\left(10^{6} \mathrm{m}\right) /\left(10 \mathrm{m} \mathrm{s}^{-1}\right)=10^{5} \mathrm{s} \sim 1$$ day

$$f_{o} \sim 2 \Omega \sin \phi_{o} \sim 2 \Omega \cos \phi_{o} \sim 10^{-4} \mathrm{s}^{-1}$$

$$W \sim H / t=10^{-1} \mathrm{m} \mathrm{s}^{-1} ; a \sim 10^{7} \mathrm{m} ; \nu \sim 10^{-5} \mathrm{m}^{2} \mathrm{s}^{-1} ; \rho \sim 1 \mathrm{kg} \mathrm{m}^{-3}$$

4 and 5.

 term magnitude (variables) magnitude (m s-2) $$\frac{D u}{D t}=$$ $$\frac{U}{t}$$ $$\frac{10}{10^{5}}=10^{-4}$$ $$2 \Omega v \sin \phi$$ $$f_{o} U$$ $$10^{-4} 10=10^{-3}$$ $$-2 \Omega w \cos \phi$$ $$f_{o} W$$ $$10^{-4} 10^{-1}=10^{-5}$$ $$\frac{u w}{a}$$ $$\frac{U W}{a}$$ $$\frac{(10)\left(10^{-1}\right)}{10^{7}}=10^{-7}$$ $$\frac{u v \tan \phi}{a}$$ $$\frac{U^{2}}{a}$$ $$\frac{10^{2}}{10^{7}}=10^{-5}$$ $$-\frac{1}{\rho} \frac{\partial p}{\partial x}$$ $$\frac{\Delta p}{\rho L}$$ $$\frac{10^{3}}{(1)\left(10^{6}\right)}=10^{-3}$$ $$-\frac{C_{d}}{h}|\vec{V}| \vec{V}$$ $$\frac{C_{d} U^{2}}{h}$$ $$\frac{10^{-2} 10^{2}}{10^{3}}=10^{-3}$$ $$\nu \nabla^{2} u$$ $$\frac{\nu U}{H^{2}}$$ $$\frac{\left(10^{-5}\right)(10)}{10^{8}}=10^{-12}$$

6. Using just the most important (i.e., biggest) terms and knowing that the turbulent drag term goes to zero above the atmospheric boundary layer, we can write a simplified x-momentum balance as:

$0=-\frac{1}{\rho} \frac{\partial p}{\partial x}+f v$

Using the same scale analysis with the y-momentum equation, we can write a simplified y-momentum balance as:

$0=-\frac{1}{\rho} \frac{\partial p}{\partial y}-f u$

Only two terms remain in both equations. One term is the pressure gradient force and the other is the Coriolis force. Since $$\frac{D u}{D t}$$

is much smaller than either the pressure gradient force or the Coriolis force, these two forces must be about in balance. We call this balance the Geostrophic Balance. It is very important for understanding atmospheric dynamics and we will talk about its consequences in more detail later.

The molecular friction term is the smallest of all the terms for the case of large-scale flow in the atmosphere. This term is almost always very small for most meteorological phenomena, which is why we had eliminated it from the averaged momentum conservation equation earlier.

We ignored the acceleration term, Du/Dt, because it is an order of magnitude smaller than the other two terms. We often must keep all terms that are within an order of magnitude of each other because our approximations may bias our results one way or the other. For instance, if we say that the velocity is 10 m s–1, the spatial scale is 100 km, and the pressure change is 10 hPa when more accurate numbers are more like 20 m s–1, 50 km, and 5 hPa, then we would be off almost an order of magnitude in our value for the centrifugal force, but get the same order of magnitude for the pressure gradient force. So, terms that are two orders of magnitude smaller than the rest you can easily neglect, but think carefully about terms that are only one order of magnitude different. For example, for very intense low-pressure systems, $$\frac{D u}{D t}$$ must be considered because it can become about as large as the pressure gradient force and the Coriolis force.

When scale analysis is applied to the z-momentum equation for mid-latitude synoptic-scale flow, the result is the simplified z-momentum balance:

$0=-\frac{1}{\rho} \frac{\partial p}{\partial z}-g$

If you rearrange Equation [10.26], you will get the hydrostatic equilibrium equation, Equation [2.18]. Previously we derived it using a balance of forces on a slab of air, but here it comes naturally out of the z-momentum equation.

The following video (4:19) provides further explanation on how to complete the above example:

Scale Analysis video