<p>As I said in my comment, viruses aren't my area (concerning units). But the answer is actually unitless. The exact answer you seek is dependent on a couple of different variables:</p>
<ul>
<li>The speed of the exhalation (cough or breathing)+speed of wind</li>
<li>The atmospheric stability</li>
</ul>
<p>Of course, the viral load of the exhalation is also important, but since you're asking for a ratio, I won't list that.</p>
<p>The <a href="http://courses.washington.edu/cewa567/Plumes.PDF" rel="nofollow noreferrer">Gaussian Plume model</a> can be used to describe pollutants. The Gaussian Plume model is: <span class="math-container">$$\frac{C}{Q}=\frac{1}{2\pi U \sigma_y \sigma_z}\exp\left(-\frac{1}{2}\left[\frac{y}{\sigma_y}\right]^2-\frac{1}{2}\left[\frac{z-H}{\sigma_z}\right]^2\right)$$</span>,</p>
<p>where <span class="math-container">$C$</span> is the concentration, <span class="math-container">$Q$</span> is the emission rate, <span class="math-container">$U$</span> is the wind speed (or worst case-scenario is cough speed), <span class="math-container">$x$</span> is the downwind distance, <span class="math-container">$z$</span> is the distance from the ground, and <span class="math-container">$\sigma_x, \sigma_z$</span> are the horizontal and vertical diffusion constants.</p>
<p><span class="math-container">$\sigma_y, \sigma_z$</span> are functions of stability and wind speed. The actual formulas are... iffy. Theoretically they can be described as <span class="math-container">$$\sigma_{y,z}=\sqrt{\frac{2K_{y,z}x}{U}}$$</span>, where <span class="math-container">$K$</span> is the eddy viscosity. <a href="http://courses.washington.edu/cee490/DISPCOEF4WP.htm" rel="nofollow noreferrer">There are empirical formulas based on stability classifications too.</a></p>
<p>Lets assume that the receiver is directly downwind and at the same height (z=H and y=0). Therefore the equation collapses to: <span class="math-container">$$\frac{C}{Q}=\frac{1}{2\pi U \sigma_y \sigma_z}$$</span>. Since the distance downwind is expressed only by <span class="math-container">$\sigma_{y,z}$</span>, then to answer your question, we need to isolate <span class="math-container">$\sigma_{y,z}$</span>. <span class="math-container">$$\sigma_y \sigma_z=\frac{1000}{2\pi U}$$</span></p>
<p>Let's just say, for the sake of argument, consider two different scenarios:</p>
<ol>
<li>Empirical <span class="math-container">$\sigma_{y,z}$</span> (refer to <a href="http://courses.washington.edu/cee490/DISPCOEF4WP.htm" rel="nofollow noreferrer">this</a> for values of constants)</li>
</ol>
<p><span class="math-container">$$(ax^b)\times (cx^d+f)=acx^{b+d}+afx^b=\frac{1000}{2\pi U}$$</span></p>
<p>And this is complicated to solve. So I'll leave solving for <span class="math-container">$x$</span> as an exercise to the reader.</p>
<ol start="2">
<li>Constant <span class="math-container">$K_{y,z}$</span></li>
</ol>
<p><span class="math-container">$$\sqrt{\frac{2K_{z}x}{U}}\sqrt{\frac{2K_{y}x}{U}}=\frac{1000}{2\pi U}$$</span>
<span class="math-container">$$\frac{4K_{z}K_{y}x^2}{U^2}=\left(\frac{1000}{2\pi U}\right)^2=\frac{10^6}{4\pi^2U^2}$$</span>
<span class="math-container">$$x=\sqrt{\frac{10^6 K_{z}K_{y}}{16\pi^2}}$$</span>
<span class="math-container">$$x=\frac{10^3}{4\pi}\sqrt{K_{z}K_{y}}$$</span></p>
<p>If we assume that <span class="math-container">$K_z=K_y=K$</span>, then the equation simplifies further. <span class="math-container">$$x=\frac{K10^3}{4\pi}$$</span>
There are many ways on how to calculate <span class="math-container">$K$</span>. One simple formula for <span class="math-container">$K$</span> can be used if mixing length theory is invoked</p>
<p><span class="math-container">$$K=\frac{ku_*z}{\phi(\frac{z}{L})}$$</span>, where <span class="math-container">$k$</span> is the <a href="http://glossary.ametsoc.org/wiki/Von_k%C3%A1rm%C3%A1n%27s_constant" rel="nofollow noreferrer">von-Karman constant</a>, <span class="math-container">$u_*$</span> is the <a href="http://glossary.ametsoc.org/wiki/Friction_velocity" rel="nofollow noreferrer">friction velocity</a>, <span class="math-container">$\phi_M(\frac{z}{L})$</span> is the <a href="https://gibbs.science/efd/handouts/dyer_1974.pdf" rel="nofollow noreferrer">Businger-Dyer function</a> for momentum, and <span class="math-container">$L$</span> is the Monin-Obukhov length. If we assume neutral atmospheric stability,<span class="math-container">$\phi_M(\frac{z}{L})=1$</span> and the resultant equation can be written as:</p>
<p><span class="math-container">$$x=\frac{ku_*z10^3}{4\pi}$$</span></p>
<p>More assumptions! Let's assume z=1.5 m and <span class="math-container">$u_*=1.11$</span> m s<span class="math-container">$^{-1}$</span> <a href="https://en.wikipedia.org/wiki/Log_wind_profile" rel="nofollow noreferrer">(corresponds to a 5 kt wind speed at 10 m with a roughness length of 0.1 m under neutral stability)</a>. Plugging those numbers in gets</p>
<p><span class="math-container">$x=35$</span> m</p>
<p>Of course, there were a LOT of assumptions to get to this point. But this is what I have to offer.</p>

CC BY-SA 4.0
2020-07-15T21:43:42.683Z
Side note: In way 2, I just assumed that OP is requesting $\frac{C}{Q}=\frac{1}{1000}$ even though they aren't even the same unit (emission rate vs concentration). But it reduces the dimensionality of the problem. I couldn't get to that point in way 1. I also invoked the Log-wind profile in way 2.
BarocliniCplusplus
CC BY-SA 4.0
2020-07-15T21:47:33.440Z
That is very helpful. I would assume that the most turbulent flow would travel the least distance and the most laminar flow would travel the most distance. What I am trying to find out is a safe distribution of large crowds outdoors, erring on the safe side.
polcott
CC BY-SA 4.0
2020-07-16T03:42:24.727Z
Before I mark that as an accepted answer I want to know your best estimate of ± error variance. It looks like a really great answer. It turns out to be exactly the same as my 100 feet wild guess.
polcott
CC BY-SA 4.0
2020-07-16T15:20:26.457Z
Since it's an equation, and the estimated value of 35 m just used typical values, an error variance isn't really a thing. But if you plug in some more extreme values, I get $(0.6, 318)$ m. When I tried the empirical approach (instead of solving for x, I just graphed the solution) I found answers ranging from 7 m to 199 m, depending on the stability classification. Certainly those have errors too, but I am not entirely sure the uncertainty of the model itself.
BarocliniCplusplus
CC BY-SA 4.0
2020-07-16T15:48:42.983Z
That is great. So 100 feet away is probably reasonably safe. I am also estimating that with a constant >= 3 MPH crosswind (perpendicular to the alignment of two people) would probably be an effective force field of protection between them. Does this make sense?
polcott
CC BY-SA 4.0
2020-07-16T17:36:11.553Z
If there is a crosswind, then it isn't really downwind. If there is a crossword, then the distance is probably shorter, but not a "force field." You would still have molecular diffusion. I should also note: this answer does not take into consideration that the virus has a finite lifetime in the air. And I am not a virologist, but I question if there is a minimum safe concentration of a virus, like there are for various pollutants.
BarocliniCplusplus
CC BY-SA 4.0
2020-07-16T23:17:39.913Z
I was thinking that there is virtually no chance that any virus particles could make it from person "A" to person "B" standing 20 feet apart with a 5 MPH cross wind between them. Although there may be no minimum safe concentration of the virus probability of infection would probably decrease as concentration decreases.
polcott
CC BY-SA 4.0
2020-08-25T10:29:06.760Z
Delightful answer that sets out assumptions, reasoning, qualifications and uncertainties. Exemplary.
Anton
CC BY-SA 4.0
2020-07-15T21:39:01.810Z
2020-07-15T21:39:01.810Z
BarocliniCplusplus
<p><strong>How far downwind of a covid-19 emitter has a 1000-fold reduction in PPM?</strong></p>
<p>Assuming that covid-19 (0.12μm diameter) particles have the aerodynamic properties of a gas:</p>
<blockquote>
<p>Particles &lt; 20 μm behave same as gas – Low settling velocity
<a href="http://home.engineering.iastate.edu/%7Eleeuwen/CE%20524/Presentations/Dispersion_Handout.pdf" rel="nofollow noreferrer">http://home.engineering.iastate.edu/~leeuwen/CE%20524/Presentations/Dispersion_Handout.pdf</a></p>
</blockquote>
<p>How far would someone have to be downwind of a covid-19 particle emitter for an at least 1000-fold reduction in PPM?</p>

CC BY-SA 4.0
2020-07-15T19:41:26.630Z
One thing I want to note is a 'unit miscommunication': Viruses aren't measured in PPM (parts per million or molecules per million molecules). PPM is reserved for gases. Aerosols are usually expressed in units of mass per unit volume. Viruses, I can't exactly speak about.
BarocliniCplusplus
CC BY-SA 4.0
2020-07-15T19:50:58.137Z
Related: covid-19 from the perspective of fluid mechanics: https://doi.org/10.1017/jfm.2020.330
Jean-Marie Prival
CC BY-SA 4.0
2020-07-15T21:12:32.807Z
@BarocliniCplusplus That make sense. I forgot that the PPM refers to molecules.
polcott
<atmosphere><turbulence><aerosol>
How far downwind of a covid-19 emitter has a 1000-fold reduction in PPM?
polcott
# How far downwind of a covid-19 emitter has a 1000-fold reduction in PPM?

(Asked by: username_0 on 2020-07-15)

**How far downwind of a covid-19 emitter has a 1000-fold reduction in PPM?**

Assuming that covid-19 (0.12μm diameter) particles have the aerodynamic properties of a gas:

> Particles < 20 μm behave same as gas – Low settling velocity
> [http://home.engineering.iastate.edu/~leeuwen/CE%20524/Presentations/Dispersion\_Handout.pdf](http://home.engineering.iastate.edu/%7Eleeuwen/CE%20524/Presentations/Dispersion_Handout.pdf)

How far would someone have to be downwind of a covid-19 particle emitter for an at least 1000-fold reduction in PPM?

**Tags:** atmosphere, turbulence, aerosol

> **Comments:**
> **username_1**: One thing I want to note is a 'unit miscommunication': Viruses aren't measured in PPM (parts per million or molecules per million molecules). PPM is reserved for gases. Aerosols are usually expressed in units of mass per unit volume. Viruses, I can't exactly speak about.  
> **username_2**: Related: covid-19 from the perspective of fluid mechanics: https://doi.org/10.1017/jfm.2020.330  
> **username_0**: @username_1 That make sense. I forgot that the PPM refers to molecules.  

---

## 1 answers

### Answer by username_1 on 2020-07-15. Score: 3 | Accepted answer

As I said in my comment, viruses aren't my area (concerning units). But the answer is actually unitless. The exact answer you seek is dependent on a couple of different variables:

* The speed of the exhalation (cough or breathing)+speed of wind
* The atmospheric stability

Of course, the viral load of the exhalation is also important, but since you're asking for a ratio, I won't list that.

The [Gaussian Plume model](http://courses.washington.edu/cewa567/Plumes.PDF) can be used to describe pollutants. The Gaussian Plume model is: $$\frac{C}{Q}=\frac{1}{2\pi U \sigma\_y \sigma\_z}\exp\left(-\frac{1}{2}\left[\frac{y}{\sigma\_y}\right]^2-\frac{1}{2}\left[\frac{z-H}{\sigma\_z}\right]^2\right)$$,

where $C$ is the concentration, $Q$ is the emission rate, $U$ is the wind speed (or worst case-scenario is cough speed), $x$ is the downwind distance, $z$ is the distance from the ground, and $\sigma\_x, \sigma\_z$ are the horizontal and vertical diffusion constants.

$\sigma\_y, \sigma\_z$ are functions of stability and wind speed. The actual formulas are... iffy. Theoretically they can be described as $$\sigma\_{y,z}=\sqrt{\frac{2K\_{y,z}x}{U}}$$, where $K$ is the eddy viscosity. [There are empirical formulas based on stability classifications too.](http://courses.washington.edu/cee490/DISPCOEF4WP.htm)

Lets assume that the receiver is directly downwind and at the same height (z=H and y=0). Therefore the equation collapses to: $$\frac{C}{Q}=\frac{1}{2\pi U \sigma\_y \sigma\_z}$$. Since the distance downwind is expressed only by $\sigma\_{y,z}$, then to answer your question, we need to isolate $\sigma\_{y,z}$. $$\sigma\_y \sigma\_z=\frac{1000}{2\pi U}$$

Let's just say, for the sake of argument, consider two different scenarios:

1. Empirical $\sigma\_{y,z}$ (refer to [this](http://courses.washington.edu/cee490/DISPCOEF4WP.htm) for values of constants)

$$(ax^b)\times (cx^d+f)=acx^{b+d}+afx^b=\frac{1000}{2\pi U}$$

And this is complicated to solve. So I'll leave solving for $x$ as an exercise to the reader.

2. Constant $K\_{y,z}$

$$\sqrt{\frac{2K\_{z}x}{U}}\sqrt{\frac{2K\_{y}x}{U}}=\frac{1000}{2\pi U}$$
$$\frac{4K\_{z}K\_{y}x^2}{U^2}=\left(\frac{1000}{2\pi U}\right)^2=\frac{10^6}{4\pi^2U^2}$$
$$x=\sqrt{\frac{10^6 K\_{z}K\_{y}}{16\pi^2}}$$
$$x=\frac{10^3}{4\pi}\sqrt{K\_{z}K\_{y}}$$

If we assume that $K\_z=K\_y=K$, then the equation simplifies further. $$x=\frac{K10^3}{4\pi}$$
There are many ways on how to calculate $K$. One simple formula for $K$ can be used if mixing length theory is invoked

$$K=\frac{ku\_\*z}{\phi(\frac{z}{L})}$$, where $k$ is the [von-Karman constant](http://glossary.ametsoc.org/wiki/Von_k%C3%A1rm%C3%A1n%27s_constant), $u\_\*$ is the [friction velocity](http://glossary.ametsoc.org/wiki/Friction_velocity), $\phi\_M(\frac{z}{L})$ is the [Businger-Dyer function](https://gibbs.science/efd/handouts/dyer_1974.pdf) for momentum, and $L$ is the Monin-Obukhov length. If we assume neutral atmospheric stability,$\phi\_M(\frac{z}{L})=1$ and the resultant equation can be written as:

$$x=\frac{ku\_\*z10^3}{4\pi}$$

More assumptions! Let's assume z=1.5 m and $u\_\*=1.11$ m s$^{-1}$ [(corresponds to a 5 kt wind speed at 10 m with a roughness length of 0.1 m under neutral stability)](https://en.wikipedia.org/wiki/Log_wind_profile). Plugging those numbers in gets

$x=35$ m

Of course, there were a LOT of assumptions to get to this point. But this is what I have to offer.

> **Comments:**
> **username_1**: Side note: In way 2, I just assumed that OP is requesting $\frac{C}{Q}=\frac{1}{1000}$ even though they aren't even the same unit (emission rate vs concentration). But it reduces the dimensionality of the problem. I couldn't get to that point in way 1. I also invoked the Log-wind profile in way 2.  
> **username_0**: That is very helpful. I would assume that the most turbulent flow would travel the least distance and the most laminar flow would travel the most distance. What I am trying to find out is a safe distribution of large crowds outdoors, erring on the safe side.  
> **username_0**: Before I mark that as an accepted answer I want to know your best estimate of ± error variance. It looks like a really great answer. It turns out to be exactly the same as my 100 feet wild guess.  
> **username_1**: Since it's an equation, and the estimated value of 35 m just used typical values, an error variance isn't really a thing. But if you plug in some more extreme values, I get $(0.6, 318)$ m. When I tried the empirical approach (instead of solving for x, I just graphed the solution) I found answers ranging from 7 m to 199 m, depending on the stability classification. Certainly those have errors too, but I am not entirely sure the uncertainty of the model itself.  
> **username_0**: That is great. So 100 feet away is probably reasonably safe. I am also estimating that with a constant >= 3 MPH crosswind (perpendicular to the alignment of two people) would probably be an effective force field of protection between them. Does this make sense?  
> **username_1**: If there is a crosswind, then it isn't really downwind. If there is a crossword, then the distance is probably shorter, but not a "force field." You would still have molecular diffusion. I should also note: this answer does not take into consideration that the virus has a finite lifetime in the air. And I am not a virologist, but I question if there is a minimum safe concentration of a virus, like there are for various pollutants.  
> **username_0**: I was thinking that there is virtually no chance that any virus particles could make it from person "A" to person "B" standing 20 feet apart with a 5 MPH cross wind between them. Although there may be no minimum safe concentration of the virus probability of infection would probably decrease as concentration decreases.  
> **username_3**: Delightful answer that sets out assumptions, reasoning, qualifications and uncertainties. Exemplary.  

---

