**4. The proposed terminology for infectious droplets' transmission mode classification**

The terminology suggested in [18] is adopted in this chapter, and it is the following:


**Figure 1** is an elaboration of the results calculated by [15]. The ordinate shows the vertical distance traveled by a droplet with an initial diameter of 20 microns (solid line) and by one of 40 microns (dashed line) falling from a height of 2 m. On the abscissa, the horizontal distance is shown. An air temperature equal to 20°C and an emission droplet velocity equal to 10 m/s are the boundary conditions. As shown, the time required for the complete evaporation (extreme points of the trajectories symbolized with a star), or the distance traveled, also depends on air relative humidity.

#### **Figure 1.**

*Vertical and horizontal distance traveled by droplets with an initial diameter equal to 20 microns (continuous line) and 40 microns (dashed line) for three air relative humidity (RH) of 30%, 50%, and 70%. Air temperature and droplets' emission velocity equal to 20°C and 10 m/s (cough case) are considered boundary conditions. Figure elaborated from the results of [15].*

Therefore, it is crucial to separate the two different transmission routes, since the evaporation of the infected droplet marks the boundary between two different modes of transport of the virions, i.e., through liquid or a solid carrier.

#### **5. Methodology to compare droplet and airborne transmission modes**

The data reported by [9], which identified and reported the link between ONP viral load and cell culture probability, were interpolated through the Matlab "curve fitting tool" by an interpolating equation of the fourth degree:

$$P(\lambda) = a\lambda^4 + \beta\lambda^3 + \gamma\lambda^2 + \delta\lambda + \tau \tag{12}$$

where α, β, γ, δ, and τ are coefficients calculated equal to �0.3314, 9.34, �91.98, 385.2, and � 586.4, respectively. Based on these data, an *R*<sup>2</sup> equal to 0.9994 was calculated.

To compare airborne and droplet transmission modes, another relevant and open research question RQ4 is: What happens to viruses after complete droplet evaporation and if they retain their full potential for infection? The droplets produced from body secretions such as sneezing and coughing are not constituted of pure water and have a significant amount of residue or dissolved substances, including virions. While pure droplets evaporate completely, the real droplet evaporates to form a solid droplet residue or droplet nucleus. The final size of the droplet nucleus, once the droplet has evaporated to its crystallization diameter, will depend on the amount of the material dissolved [19]. Those residues potentially give a means for the virus to be further transported, provided that it survives the drying process. There is evidence supporting that viruses coated by a lipid membrane tend to retain their infectivity longer at low relative humidity [20]. However, the opposite is true in relevant counterexamples as discussed by [21]. There are many literature examples, in which virions'spreading after droplet evaporation is modeled by considering the virions i) entrapped in the droplet nuclei and ii) preserving their infectivity [22, 23]. Nevertheless, since there is no empirical evidence that i) virions are entrapped in the droplet nucleus, ii) multiple droplet nuclei can be generated by one droplet, and iii) dissolved substances (including virions) can be expelled by the droplet during evaporation, in this chapter, a different approach is suggested to compare droplet and airborne transmission modes. The idea is to compute the number of total droplets that an infected subject should emit to have the same number of airborne virions that touch the host surface in Brownian motion as those that impact the same surface being carried on a droplet of diameter *DG* characterized by a medium concentration λ0. Virions must traverse the distance to the target cell to infect it [24]. The same authors, to support the hypothesis of diffusion-limited infection, declared that it is not the amount of virus in a medium overlay in a culture well that determines the infectivity but the viral load.

Although several complex models exist to investigate the results of the impact of a droplet against a surface [25], it is assumed, for simplicity, that the impact area on which the virions are deposited is equal to the cross-sectional area of the droplet. Therefore, in the proposed model, the impact area is expressed as

$$S = \frac{\pi}{4} D\_G^{-2}.\tag{13}$$

The number of virions deposited by the droplet *N*virions,droplet [RNA copies] on the surface *S* is calculated as

$$N\_{\text{virions,droplet}} = \frac{\pi}{6} D\_{G,0}{}^6 \times \lambda\_0. \tag{14}$$

Since the average SARS-CoV-2 virion diameter is 100 nm, the virions motion follows the Brownian mechanism once released from the droplet in the surrounding air. For the comparison, it is assumed that virions are dispersed in the volume of inhaled air: the underlying hypothesis is that all the virions can infect but are released in the air as single particles after droplet drying. Virions that can come into contact with the surface *S* are contained in a control volume *Vc* [m3 ], defined as

$$V\_c = \mathbb{S} \times d\mathbf{z} \tag{15}$$

where *dz* is the average diameter of the virion in [m]. The airborne viral load λairborne [RNA copies/mL] required to deposit the same number of virions as those contained by a droplet of diameter *DG* can be found as

$$
\lambda\_{\text{airborne}} = \frac{N\_{\text{virions,droplet}}}{V\_c}.\tag{16}
$$

It is assumed that the droplet's diameter distribution remains the same over time. Therefore, the time required to achieve the airborne viral load λairborne is calculated as

$$t = \frac{\lambda\_{\text{airborne}} \times V}{\sum\_{j=1}^{M} N\_{\text{droplet},j,\mu} \times \left(V\_j || \times \lambda\_0 \right) \times \mathbf{0.6}}.\tag{17}$$

where 0.6 is the correction factor used to convert the values in [droplets/(person � minute)] in the case of speech. The volume *V*, conservatively, can be assumed to equal the volume that a person emits during a conversation, which is 11.7 L/min, or 11,700 mL/min [26].
