Do it, the effect is FAR FROM MARGINAL
“…a reduction from 2 metres to 1 would raise infection risk only marginally, from 1.3% to 2.6%.”
I don’t know about “marginally”. To me it looks like the odds of getting infected just doubled.
However, I don’t think this is the correct model for estimating the effect of distance on the infection risk.
First, if by “infection risk” we mean “probability of getting infected given various conditions”, then the model has to be logistic — conditional probabilities don’t add, log-odds do.
If p is your infection risk, then
is a linear function of the conditions, in this case the total exposure.
Exposure rate is proportional to the density of the viral droplets in the vicinity of your nose/mouth. The density itself depends on distance from the Covid19 carrier and has an explicit time dependence due to droplet coagulation, diffusion and falling out of the air due to gravity. Then total exposure is the integral along your trajectory X(t) of the exposure rate. Putting all that together, your exposure risk
One BIG flaw with the unstated model (in either the article or the study itself) is that it talks about infection risk “at” a certain distance, as if the amount of time you send there doesn’t matter or doesn’t need to be mentioned. However, I recall an early study stating that the infection risks were calculated for being in the presence of an infected person for “10 minutes”. I’ll just make a steady state assumption so time as a factor drops out.
- The viral droplets spread in a sphere or a solid angle. In either case, assuming no explicit time dependence, the density will reduce inversely with the square of the distance r from the carrier’s oral cavities.
- We are in a steady state situation where the rate at which viral droplets coagulate or fall to the ground or diffuse away is the same as the rate at which they are expelled by the carrier into the air. This would probably add another power to the dependence on distance, due to the finite velocity of propulsion and some linear decay rate of the viral particles.
- You are “helicoptered” to that distance from the carrier, directly to 1m or 2m, and that you spend the same amount of time at that distance before being helicoptered out. This relieves us of doing the integral. Again, if we were to take your finite velocity of approach into account, it would increase the reciprocal effect of distance since it would take you longer to get deeper in and add to the time you are in the sphere of influence.
With these assumptions, the model for the infection risk log-odds simplifies to
Where a is the logistic coefficient and x_0 is the background infection rate infinitely far away from any carriers.
Great, so we have a model. What data do we have? Unfortunately, we only two data points, so we can’t test our model, only fit it.
Fitting the above data yields
a = 0.947 and
x0 = -4.567, which corresponds to a background infection rate of 1.03%.
What does this mean? Is the increase in infection risk marginal? Well, no. The added risk above background at 2m is only (1.3% — 1.03% = 0.27%). The added risk above background at 1m is (2.6% — 1.03% = 1.57%). The increase in the added risk above background of moving from 2m to 1m is a factor of 6 higher!
Is that marginal? Doesn’t seem “marginal” to me.
Furthermore, if you look at how the model predicts growth in the added infection risk, as you go closer than 1m, say to handshake distance at 0.5m, the odds quadruple!
Do not underestimate the effect of social distancing, especially as you come closer than 2m.
But also, since the time-scale for the infection risk is ~10 minutes, if you are just walking past someone in the outdoors and are in the “sphere of influence” for only a few seconds, you are not increasing your exposure odds. So all that one-way trail stuff, wearing a mask while running or biking, walking off the sidewalk into the street etc. are BS recommendations by politicians to pretend they are doing something.
The further apart you are, the longer you can stay, by the square of the distance factor.