# Why is aerodynamic resistance defined inversely?

When I think of wind resistance, I always think of the friction co-efficient, in $\mathrm{ms^{-1}}$, because that's always the way that it was presented in highschool and undergraduate physics.

But in land surface modelling, aerodynamic resistance (basically, how much the roughness of the surface slows air movement down), is defined inversely. For example, this presentation states on page 18 that:

• $r_a$ = $f($wind speed, atmospheric stability, surface roughness$)$
• $r_a$ decreases with increasing wind speed
• $r_a$ increases with increasing stability
• $r_a$ decreases with increasing surface roughness

and then gives a bunch of examples in $\mathrm{sm^{-1}}$:

Surface type   ra (s.m^-1)
Ocean          200
Grass          70
Crops          20-50
Forest         5-10


To me, it seems really counter intuitive ("resistance is higher over grass than over a forest"?). Is the some benefit to defining resistances this way, or is this just some annoying historical mistake?

• @naughth101 can you write the resistivity equations explicitly? because others may want to use it and the presentation link is not available. – Gemechu Fanta Garuma Mar 9 '16 at 19:50
• @GFG: I no longer have access to the presentation either, but the exact form of the equations is not relevant to the question. Also, relevant equations will be in any relavant textbook, and some are also in Deditos' answer below. – naught101 Mar 10 '16 at 1:27

It's partly historical, partly point-of-view, but it's not a mistake.

The friction coefficient emphasises the effect of the surface on a property of the boundary layer, i.e., greater surface friction slows the near-surface wind more. Aerodynamic resistance emphasises the effect of the boundary layer on surface-atmosphere exchange, i.e., greater mixing yields a lower resistance to evaporation. Note that it's evaporation not advection that's being resisted. You could turn the confusion around and argue that friction coefficient is a counter intuitive term: if there's more friction, why does evaporation increase? This would be making the same category error.

Historically (IIRC) the electrical circuit metaphor was introduced by Penman and Schofield (1951) in an early modification of the Penman evapotranspiration equation to account for how vegetation limits the transpiration rate. The original Penman equation is often quoted using aerodynamic conductance (or friction velocity or friction coefficient), i.e., a quantity using the the $m s^{-1}$ units that you're used to. This is a natural choice for micrometeorologists trying to understand the bulk mass transfer properties of the turbulent boundary layer because it can be calculated from the boundary layer quantities that they measure.

Once you start introducing other surface effects on evaporation it's not really appropriate to extend the analogy of a turbulent fluid friction velocity down from the boundary layer to vegetation stomata, roots and the soil. So we use the electrical metaphor instead, in which case it becomes slightly simpler to describe the system using resistances rather than conductances. In the first instance it was just a case of calculating the total system resistance by adding a stomatal resistance to the aerodynamic resistance:

$E = \frac{\rho}{r_s+r_a}\left(q_{sat} - q_a\right) = \frac{\rho}{r_t}\left(q_{sat} - q_a\right)$

You can express this as conductances, but it's not as neat:

$E = \frac{g_a g_s \rho}{g_a+g_s}\left(q_{sat} - q_a\right)$

You can see in the later slides of the presentation you linked to that these resistance networks have got progressively more complex, so nowadays this neatness point might be moot.

Penman and Schofield (1951) Some physical aspects of assimilation and transpiration, Symp. Soc. Exper. Biol., 5, 115-129.

• Having dabbled in electronics, that was exactly the answer I needed. Perfect, thanks! – naught101 Sep 9 '14 at 0:34

This is a good question, and the answer is, aerodynamic resistance is not defined inversely. It is rather, defined in a context that is often misinterpreted.

In your question, you state that aerodynamic resistance is basically how much the roughness of the surface slows air movement down. This statement is not correct, and it seems to stem from the misinterpretation of context.

A monograph by De Groot (1963) also shows that molecular transfer processes have the general form analogous to the electric circuit shown by @Deditos' answer:

$$Flux = \dfrac{Force}{Resistance}$$

This is also true for air-sea and other general fluid-fluid interfaces. Flux is the transfer of a quantity (say, momentum, enthalpy, mass, etc.) through the interface and the associated boundary layers (say, air, water, canopy, soil, etc.). In the electric circuit analogy, force has the character of potential gradient and resistance has that of the inverse conductivity. The important point to be made here is that resistance is not resistance of interface to the aerodynamic flow - what we would intuitively imagine as friction or stress. It is in fact, resistance of the interface to the forcing. In case of momentum, this means that given equal forcing, higher resistance yields lower flux. Thus, lower resistance translates to rougher surface. This is why forest has lower resistance values than grass or open ocean.

Example: Given equal forcing, rougher surface results in higher stress compared to smoother surface. It can be said that the rougher surface is "more permitting", or "less resistant" of momentum flux.

In the electric circuit analogy, Flux, Force and Resistance are symbolic, conceptual entities. Force is not necessarily in $\rm N$, and may be a temperature or humidity gradient like it is given on slide 16 in the presentation you linked. Resistance may thus take different formulations.

Note that nowadays, in both modeling and theory, we often use exchange coefficients to characterize momentum $(C_{D})$ and enthalpy $(C_H$, $C_E)$ fluxes through the interface, which act as conductivity and not resistance. For example, in case of momentum:

$$\boldsymbol{\tau} = \rho C_{D}|\mathbf{U}|\mathbf{U}$$

where $\boldsymbol{\tau}$ $(\mathrm{N/m^{2}})$ is vertical flux of horizontal momentum (wind stress), $\rho$ $(\mathrm{kg/m^{3}})$ is air density and $\mathbf{U}$ $(\mathrm{m/s})$ is wind vector at some reference height above the surface. $C_{D}$ (non-dimensional) has different values depending on the surface properties.

In that particular presentation that you linked in your question, it is not clear to me why resistance $r_a$ has units of $\mathrm{s\ m^{-1}}$. For sensible and latent heat flux $\mathrm{ (W / m^2) }$ formulations on slide 16, the units don't quite work out, but it is possible that the equations shown were more illustrative than exact. Because bulk flux formulae are most often based on theoretical, empirical and dimensional grounds, $r_{a}$ can be defined in various dimensions (units) depending on the bulk flux formulation.

Reference:

De Groot, S. R. Thermodynamics of Irreversible Processes. North Holland Publishing Co., 1963.

• Thanks for the answer. Not entirely sure I follow. In particular, the line "resistance is not resistance of interface to the aerodynamic flow ... It is in fact, resistance of the flux (tranfer) to the forcing" is pretty opaque to me. One point: the units make perfect sense: $F=m\cdot a = kg \cdot m\cdot s^{-2}$, so $flux = \frac{kg \cdot m\cdot s^{-2}}{m\cdot s^{-1}} = kg \cdot s^{-1}$. And mass can just be replaced with energy, I suppose. – naught101 Sep 9 '14 at 23:43
• @naught101 OK, I edited my answer in an attempt to make it more clear. Speaking of units not working out, I meant the units for resistance in the heat fluxes on slide 16. Nevertheless, the flux units you wrote are not correct, you are missing $m^{-2}$ :). The electric circuit analogy is conceptual rather than literal, see my edited answer for explanation. – milancurcic Sep 10 '14 at 2:40