# PurpleAir Implementation of the Alternative (ALT) Method

## Introduction

In their article, “Calibration of low-cost PurpleAir outdoor monitors using an improved method of calculating PM2.5”, Lance Wallace (an independent researcher), Jianzhao Bi of the University of Washington, Wayne R. Ott of Stanford University, and both Jeremy Sarnat and Yang Liu of the Gangarosa Department of Environmental Health, Rollins School of Public Health, Emory University, describe their transparent and reproducible alternative method (ALT) of calculating \text{PM}_{2.5} from the particle numbers in three size categories. This was used in place of the estimates provided by Plantower, the manufacturer of the laser counters used in our PurpleAir sensors. We can view their scientific, peer-reviewed paper here: https://doi.org/10.1016/j.atmosenv.2021.118432

In this article, I will guide you through the researchers general description, to PurpleAir’s specific implementation. All the while I’ll make the transition through rigorous mathematical notation.

## General Outline of Alternative (ALT) Method

The approach adopted by the researchers here is identical to that recommended by manufacturers of optical particle monitors capable of collecting simultaneous data on multiple particle size categories; for example, the TSI Model 3330.

1. Calculate the particle count per dL of air, N_i, of particles in each of the size categories 0.3–0.5 \mu \text m, 0.5–1.0 \mu \text m, 1.0–2.5 \mu \text m, 2.5–5.0 \mu \text m and 5.0–10 \mu \text m. Since the Plantower output provides the total number of particles sized by at least 0.3 \mu \text m, at least 0.5 \mu \text m, etc., we must subtract the total number at least 0.5 \mu \text m in size from the total number at least 0.3 \mu \text m in size to arrive at the number of particles in the 0.3–0.5 \mu \text m size category. Similar subtractions will result in five size categories under 10 \mu \text {m:} 0.3–0.5 \mu \text m, 0.5–1.0 \mu \text m, 1.0–2.5 \mu \text m, 2.5–5.0 \mu \text m, 5.0–10 \mu \text m.

2. Choose a mean diameter, D_i, for each of the five size categories up to 10 \mu \text m. As an example, the mean particle diameter in the 0.3–0.5 \mu \text m size category must be between 0.3 and 0.5 \mu \text m, so we can approximate it by the arithmetic mean of the size boundaries:
\frac{0.3 \mu \text m + 0.5 \mu \text m}{2} = 0.4 \mu \text m
or by the geometric mean of the size boundaries:
\sqrt{(0.3)(0.5)} = \sqrt{\left(3\over 10\right)\left(5\over 10\right)} = \sqrt{3\over 10} \cdot \sqrt{5\over 10} = \frac{\sqrt{3}}{\sqrt{10}} \cdot \frac{\sqrt{5}}{\sqrt{10}} = \frac{\sqrt{3} \cdot \sqrt{5}}{10} = \frac{\sqrt{15}}{10} \approx 0.387 \mu \text m
What I suspect to be an industry standard, several manufacturers use the geometric mean, so the researchers, and we at PurpleAir, use it as well.

3. Calculate the total particle volume per dL of air, V_i, in each size category. Note that this is not strictly volume. This is given by multiplying the number of particles per dL of air, N_i, by the mean volume per particle:

V = N \frac{πD^3}{6}
1. Multiply the particle volume per dL of air, V_i, by a reference density, \rho_\text{ref}, to arrive at an estimate of the mass concentration M_i in each size category. Note that this does not involve strictly mass. The researchers chose \rho_\text{ref} to be the density of water, a choice that is also made by the Model 3330 manufacturers.
M = \rho_\text{ref} V
1. Add the appropriate size categories to estimate the desired particulate mass concentration among \text{PM}_1, \text{PM}_{2.5} and \text{PM}_{10}. \text{PM}_1 is the sum of the mass concentrations in the size categories up to 1 \mu \text m: 0.3–0.5 \mu \text m and 0.5–1.0 \mu \text m. \text{PM}_{2.5} is the sum of the mass concentrations in the size categories up to 2.5 \mu \text m: 0.3–0.5 \mu \text m, 0.5–1.0 \mu \text m, and 1.0–2.5 \mu \text m. \text{PM}_{10} is the sum of the mass concentrations of all these size categories: 0.3–0.5 \mu \text m, 0.5–1.0 \mu \text m, 1.0–2.5 \mu \text m, and 2.5–10 \mu \text m.

2. Calculate a conversion factor (cf), k, for the air being monitored by some comparison to gravimetric studies of the same air, or to research-grade monitors that have themselves been calibrated using gravimetric measurements. The researchers have suggested the conversion factor of k = 3, referred by us at PurpleAir as “ALT cf=3”.

## An Implementation of the Alternative (ALT) Method

The ALT mass concentration is given by:

\begin{align*} \rho_\text{ALT} &:= k \cdot \sum \left(\rho_\text{ref.} V \right) \\ &= k \cdot \sum \left(\rho_\text{ref.} N \frac{\pi D^3}{6 }\right) \\ &= \frac{k \rho_\text{ref.} \pi}{6} \cdot \sum \left(N D^3 \right) \\ \end{align*}

where the conversion factor k=3, the reference density is that of water, \rho_\text{ref.} = 1 \text { g/cm}^3, the particle count per dL of air in the i^\text{th} size range is N_i, and the mean diameter of a sphere in the i^\text{th} size range is D_i.

### Unit Analysis

Before we continue to articulate the ALT mass concentration, let’s first understand the units that \rho_\text{ALT} is expressed in. Recall the following variables and their associated units:

\rho_\text{ref.}\!\!: \frac {\text{g}}{\text{cm}^3}

N\!\!: \frac{1}{\text{dL}}

D\!\!: \mu \text m

So since that:

\begin{align} \rho_\text{ALT} &= \frac{k \cancelto{\left( \frac {\text{g}}{\text{cm}^3} \right)}{\rho_\text{ref.}} \pi}{6} \cdot \sum \left( \cancelto{\left( \frac{1}{\text{dL}} \right)}{N} \cancelto{( \mu \text m)^3}{D^3} \right) \implies \left( \frac {\text{g}}{\text{cm}^3} \right) \left( \frac{1}{\text{dL}} \right) ( \mu \text m)^3 \\ &= \frac {\text{g}}{\cancel{\text{cm}^3}} \cdot \textcolor{#a4a}{\left( \frac{\cancel{\text{cm}}}{10^{-2} \text m} \right) ^3} \cdot \frac{1}{\cancel{\text{dL}}} \cdot \textcolor{#a4a}{\frac{\cancel{\text{dL}}}{10^{-1}\text{ L}}} \cdot \cancel{\mu \text m^3} \cdot \textcolor{#a4a}{\left( \frac{10^{-6} \text m}{\cancel{\mu \text m}} \right)^3} \\ &= \frac{\text g}{10^{-6} \cancel{\text m ^ 3}} \cdot \frac{1}{10^{-1} \cancel{\text L}} \cdot \textcolor{#a4a}{\frac{\cancel{\text L}}{10^{-3} \text m ^ 3}} \cdot 10^{-18} \cancel{\text m ^3} \\ &= \frac{10^{-18 - (-6) - (-1) - (-3)} \text g}{\text m ^3} \\ &= \frac{10^{-18 + 6 + 1 + 3} \text g}{\text m ^3} \\ &= \frac{10^{-8} \text g}{\text m ^3} \\ &= \frac{10^{-2} \cdot 10^{-6} \text g}{\text m ^3} \\ &= \frac{10^{-2} \mu \text g}{\text m ^3} \\ \end{align}

which is a hundredth of a microgram per cubic meter; not the form we want the units in. To instead express \rho_\text{ALT} in the units \frac{\mu \text g}{\text m ^3}, we’ll make note of the following unit fraction:

= \cancel{\left( \frac{10^{-2} \mu \text g}{\text m ^3} \right)} \cdot \textcolor{#a4a}{\frac{\frac{1}{100} \cdot \frac{\mu \text g}{\text m ^3}}{\cancel{\frac{10^{-2} \mu \text g}{\text m ^3}}}}

That is to say, dividing \frac{10^{-2} \cdot \mu \text g}{\text m ^3} by 100 to convert it to \frac{\mu \text g}{\text m ^3} is analogous to dividing inches by 12 to convert it to feet.

### Expand Notations

We now return to articulate the ALT mass concentration. We may now narrow our calculations to \text{PM}_{2.5}. Let’s introduce a size category index as follows.

0.3–0.5 \mu \text m\!\!: i=1

0.5–1.0 \mu \text m\!\!: i=2

1.0–2.5 \mu \text m\!\!: i=3

Continuing the ALT \text{PM}_{2.5} mass concentration:

\begin{align} \rho_\text{ALT} &= \frac{k \rho_\text{ref.} \pi}{6} \cdot \sum _{i=1}^{3} (N_i D_i^3) \\ &= \frac{k \rho_\text {ref.} \pi}{6} \left( N_1 D_1^3 + N_2 D_2^3 + N_3 D_3^3 \right) \end{align}

To define the particle count per dL of air, N_i, of particles in each of the size categories, the ALT mass concentration becomes:

\begin{align} \rho_{\text{ALT}} &= \frac{k \rho_\text{ref.} \pi}{6} \left[ (n_{0.3} - n_{0.5}) D_1^3 + (n_{0.5} - n_{1.0}) D_2^3 + (n_{1.0} - n_{2.5}) D_3^3 \right] \\ &= \frac{k \rho_\text{ref.} \pi}{6} \left( n_{0.3} D_1^3 - n_{0.5} D_1^3 + n_{0.5} D_2^3 - n_{1.0} D_2^3 + n_{1.0} D_3^3 - n_{2.5} D_3^3 \right) \\ &= \frac{k \rho_\text{ref.} \pi}{6} \left[ D_1^3 n_{0.3} + (D_2^3 - D_1^3) n_{0.5} + (D_3^3 - D_2^3) n_{1.0} - D_3^3 n_{2.5} \right] \end{align}

where the particle count per dL of air of particles sized by at least 0.3 \mu \text m is n_{0.3}, by at least 0.5 \mu \text m is n_{0.5}, by at least 1.0 \mu \text m is n_{1.0}, and by at least 2.5 \mu \text m is n_{2.5}.

### Choose Mean Diameter for Each Size Category

Let’s now define a geometric mean diameter, D_i, for each of the five size categories as follows:

0.3–0.5 \mu \text m\!\!: \text D_1 = \frac{\sqrt{15}}{10} as expressed previously.

0.5–1.0 \mu \text m\!\!: \text D_2 = \sqrt{(0.5)(1.0)}

\begin{align} &= \sqrt{\frac{1}{2}} \\ &= \frac{1}{\sqrt{2}} \cdot \textcolor{#a4a}{\frac{\sqrt{2}}{\sqrt{2}}} \\ \implies \text D_2 &= \frac{\sqrt{2}}{2} \end{align}

1.0–2.5 \mu \text m\!\!: \text D_3 = \sqrt{(1.0)(2.5)}

\begin{align} &= \sqrt{5 \over 2} \\ &= \frac{\sqrt 5}{\sqrt 2} \cdot \textcolor{#a4a}{\frac{\sqrt{2}}{\sqrt{2}}} \\ \implies \text D_3 &= \frac{\sqrt{10}}{2} \end{align}

### Finalize Our Implementation of the Alternative (ALT) Method

\begin{align*} \rho_{\text{ALT}} &= \frac{k \rho_\text{ref.} \pi}{6} \left[ \left(\frac{\sqrt{15}}{10} \right)^3 n_{0.3} + \left[\left(\frac{\sqrt{2}}{2} \right)^3 - \left(\frac{\sqrt{15}}{10} \right)^3 \right] n_{0.5} \right. \\ & \quad + \left. \left[\left(\frac{\sqrt{10}}{2} \right)^3 - \left(\frac{\sqrt{2}}{2} \right)^3\right] n_{1.0} - \left(\frac{\sqrt{10}}{2} \right)^3 n_{2.5} \right] \\ &= \frac{k \rho_\text{ref.} \pi}{6} \left[ \frac{\cancelto{3}{15} \sqrt{15}}{\cancelto{200}{1000}} n_{0.3} + \left(\frac{\cancel{2} \sqrt{2}}{\cancelto{4}{8}} - \frac{\cancelto{3}{15} \sqrt{15}}{\cancelto{200}{1000}} \right) n_{0.5} \right. \\ & \quad + \left. \left(\frac{\cancelto{5}{10} \sqrt{10}}{\cancelto{4}{8}} - \frac{\cancel{2} \sqrt{2}}{\cancelto{4}{8}} \right) n_{1.0} - \frac{\cancelto{5}{10} \sqrt{10}}{\cancelto{4}{8}} n_{2.5} \right] \\ &= \textcolor{#59c2d3}{\frac{1}{4}} \cdot \frac{k \rho_\text{ref.} \pi}{6} \left[ \frac{3 \sqrt{15}}{\cancelto{50}{200}} n_{0.3} + \left(\frac{ \sqrt{2}}{\cancel{4}} \cdot \textcolor{#a4a}{\frac{50}{50}} - \frac{3 \sqrt{15}}{\cancelto{50}{200}} \right) n_{0.5} \right. \\ & \quad + \left. \left(\frac{5 \sqrt{10}}{\cancel{4}} - \frac{\sqrt{2}}{\cancel{4}} \right) n_{1.0} - \frac{5 \sqrt{10}}{\cancel{4}} n_{2.5} \right] \\ &= \frac{k \rho_\text{ref.} \pi}{24} \left[ \frac{3 \sqrt{15}}{50} n_{0.3} + \left(\sqrt{2} - \frac{3 \sqrt{15}}{50} \right) n_{0.5} + \left(5 \sqrt{10} - \sqrt{2} \right) n_{1.0} - 5 \sqrt{10} \cdot n_{2.5} \right] \\ &= \frac{k \rho_\text{ref.} \pi}{24} \left( \frac{3 \sqrt{15}}{50} n_{0.3} + \frac{50 \sqrt{2} - 3 \sqrt{15}}{50} n_{0.5} + \left(5 \sqrt{10} - \sqrt{2} \right) n_{1.0} - 5 \sqrt{10} \cdot n_{2.5} \right) \end{align*}

## Conclusion

Finally, let k = 3, \rho_\text{ref.} = 1 \frac{\text g}{\text{cm}^3}, and convert the units of \rho_{\text{ALT}} from \frac{10^{-2}\mu \text g}{\text m^3} to \frac{\mu \text g}{\text m^3} by dividing by 100 so that:

\begin{align*} \rho_{\text{ALT}} &= \frac{\cancelto{3}{k} \cancelto{1}{\rho_\text{ref.}} \pi}{24} \cdot \textcolor{#59c2d3}{\frac{1}{100}} \cdot \left( \frac{3 \sqrt{15}}{50} n_{0.3} + \frac{50 \sqrt{2} - 3 \sqrt{15}}{50} n_{0.5} \right. \\ & \quad + \left. \left(5 \sqrt{10} - \sqrt{2} \right) n_{1.0} - 5 \sqrt{10} \cdot n_{2.5} \right) \\ &= \frac{\cancel{3} \pi}{\cancelto{8}{24}} \cdot \frac{1}{100} \cdot \left( \frac{3 \sqrt{15}}{50} n_{0.3} + \frac{50 \sqrt{2} - 3 \sqrt{15}}{50} n_{0.5} \right. \\ & \quad + \left. \left(5 \sqrt{10} - \sqrt{2} \right) n_{1.0} - 5 \sqrt{10} \cdot n_{2.5} \right) \end{align*}
\implies \boxed{\rho_{\text{ALT}} = \frac{\pi}{800} \left( \frac{3 \sqrt{15}}{50} n_{0.3} + \frac{50 \sqrt{2} - 3 \sqrt{15}}{50} n_{0.5} + \left(5 \sqrt{10} - \sqrt{2} \right) n_{1.0} - 5 \sqrt{10} \cdot n_{2.5} \right)}

whose units are now in \frac{\mu \text g}{\text m^3}, where the particle count per dL of air of particles sized by at least 0.3 \mu \text m is n_{0.3}, by at least 0.5 \mu \text m is n_{0.5}, by at least 1.0 \mu \text m is n_{1.0}, and by at least 2.5 \mu \text m is n_{2.5}.