# Physics puzzle about the concentration of fog droplets in the air

• Hak
Hak
Homework Statement
Physics puzzle: Visibility on the road is 100 m. Assuming that the diameter of a fog droplet is 1 micron, estimate the concentration of fog droplets in the air.
Relevant Equations
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To estimate the concentration of fog droplets in the air, I thought of the mass of water in a given volume of air: I searched this value on the net, and I found that it is typically around ##0.05 g/m^3##.
The mass of a single fog droplet can be calculated from its diameter and density. Assuming that the fog droplet is spherical and has a diameter of ##1## micron (##10^{-6} m##), its volume is

$$V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi(0.5 \times 10^{-6})^3 \approx 5.24 \times 10^{-19}m^3$$

The density of water is about ##1000 kg/m^3##, so the mass of a single fog droplet is

$$m=\rho V=1000 \times 5.24 \times 10^{-19}m^3 \approx 5.24 \times 10^{-16} Kg$$

Now, we can estimate the concentration of fog droplets in the air by dividing the mass of water in a given volume of air by the mass of a single droplet. This gives us

$$n= \frac{0.05 \times 10^3}{5.24 \times10^{-16}} \approx 9.54 \times 10^{16} droplets/m^3$$

(Edit) This would mean that there are about 95 trillion fog droplets in every cubic meter of air...

This process does not seem convincing to me. In my opinion there is something deeper that I cannot decipher. Where am I going wrong? Could you give me some hints?

"Visibility on the road is 100 m."
Looks like your calculation has no relation to this critical piece of information.

Hill said:
"Visibility on the road is 100 m."
Looks like your calculation has no relation to this critical piece of information.
You are right, which is why my reasoning seems highly fallacious. I don't know how to use this data and I can't figure out how to put it into a development. Do you have any suggestions?

A spherical fog droplet of radius $a$ blocks an area $\pi a^2$ from view (this being the cross-section perpendicular to a striaght line between the dorplet and the observer). If the volume concentration of dorpelts is $C$, then the amount of fog droplets in a hemispherical shell of radius $r$ and thickness $\delta r$ is $2\pi Cr^2\,\delta r$, and the total area they block from view is (at most) $2C\pi^2a^2 r^2\,\delta r$. We find $C$ from the requirement that the droplets within a hemisphere of radius 100m entirely block the surface area of that hemipshere from view.

Chestermiller and nasu
pasmith said:
A spherical fog droplet of radius $a$ blocks an area $\pi a^2$ from view (this being the cross-section perpendicular to a striaght line between the dorplet and the observer). If the volume concentration of dorpelts is $C$, then the amount of fog droplets in a hemispherical shell of radius $r$ and thickness $\delta r$ is $2\pi Cr^2\,\delta r$, and the total area they block from view is (at most) $2C\pi^2a^2 r^2\,\delta r$. We find $C$ from the requirement that the droplets within a hemisphere of radius 100m entirely block the surface area of that hemipshere from view.
Thank you very much.
pasmith said:
A spherical fog droplet of radius $a$ blocks an area $\pi a^2$ from view (this being the cross-section perpendicular to a striaght line between the dorplet and the observer).
This part is clear to me.

pasmith said:
If the volume concentration of dorpelts is $C$, then the amount of fog droplets in a hemispherical shell of radius $r$ and thickness $\delta r$ is $2\pi Cr^2\,\delta r$, and the total area they block from view is (at most) $2C\pi^2a^2 r^2\,\delta r$.
I can't understand this part. How do you arrive at these symbolic relations that you have enunciated? What considerations did you start from? Thank you.

pasmith said:
We find $C$ from the requirement that the droplets within a hemisphere of radius 100m entirely block the surface area of that hemipshere from view.
Although I did not understand the previous part, I start from your formulas and try to develop an argument.
The requirement that the droplets within a hemisphere of radius ##R = 100 m## entirely block the surface area of that hemisphere from view is equivalent to saying that the total area blocked by the droplets is equal to the surface area of the hemisphere, which is ##2\pi R^2##.
Therefore, we can write an equation as follows:
$$\int_{0}^{R} 2 C \pi^2a^2r^2dr = 2 \pi R^2 \Rightarrow \ \pi Ca^2\int_{0}^{R} r^2dr = R^2 \Rightarrow C \int_{0}^{R} r^2dr = \frac{R^2}{\pi a^2} \Rightarrow \ C \left[ \frac{r^3}{3} \right]_0^{R} = \frac{R^2}{\pi a^2} \Rightarrow \ C \frac{R^3}{3} = \frac{R^2}{\pi a^2}\Rightarrow \ C\frac{R}{3} = \frac{1}{\pi a^2} \Rightarrow \ C = \frac{3}{\pi R a^2}$$.
Substituting numerically:
$$C = \frac{3}{\pi \times 100 \times 10^{-12}} \approx 9.5 \times 10^{9} droplets/m^3$$.

Where does the error, if any, lie? Thank you for any helpfulness.

Hak said:
Where does the error, if any, lie?
If you know it to be wrong, do you know what the answer is supposed to be?
You have confused diameter with radius.

@pasmith's argument gives a lower bound on the least R for which objects at that distance could be completely obscured. In practice, many droplets will be wholly or partly behind nearer droplets, increasing the visibility. But then, statistically , your view will never be completely obscured , so you would have to decide what fraction of the field of view needs to be obscured to constitute an effective limit of visibility.

haruspex said:
If you know it to be wrong, do you know what the answer is supposed to be?
You have confused diameter with radius.
@pasmith's argument gives a lower bound on the least R for which objects at that distance could be completely obscured. In practice, many droplets will be wholly or partly behind nearer droplets, increasing the visibility. But then, statistically , your view will never be completely obscured , so you have to decide what fraction of the field of view needs to be obscured to constitute an effective limit of visibility.
Thanks.
haruspex said:
If you know it to be wrong, do you know what the answer is supposed to be?
You have confused diameter with radius.
I don't know the final result. You are right, I confused the radius with the diameter: there must be a coefficient ##0.5^2 =0.25## in the denominator of the final expression, if it is correct.
@pasmith's argument gives a lower bound on the least R for which objects at that distance could be completely obscured. In practice, many droplets will be wholly or partly behind nearer droplets, increasing the visibility. But then, statistically , your view will never be completely obscured , so you would have to decide what fraction of the field of view needs to be obscured to constitute an effective limit of visibility.
I understand the reasoning and agree with it. How to choose a particular fraction, though? How to physically formalise this purely descriptive reasoning? If you would like to give me some advice, I would be grateful. Thank you very much.

Hak said:
Thanks.

I don't know the final result. You are right, I confused the radius with the diameter: there must be a coefficient ##0.5^2 =0.25## in the denominator of the final expression, if it is correct.

I understand the reasoning and agree with it. How to choose a particular fraction, though? How to physically formalise this purely descriptive reasoning? If you would like to give me some advice, I would be grateful. Thank you very much.
If we take the limit of visibility as being the distance at which only fraction ##\alpha## of the field is visible then @pasmith's formula is equivalent to ##\alpha=e^{-3}##. That might be reasonable.

haruspex said:
If we take the limit of visibility as being the distance at which only fraction ##\alpha## of the field is visible then @pasmith's formula is equivalent to ##\alpha=e^{-3}##. That might be reasonable.
Sorry, how did you achieve this? I'm a bit confused: where does the term ##\alpha## come from? Thank you.

Hak said:
Sorry, how did you achieve this? I'm a bit confused: where does the term ##\alpha## come from? Thank you.
In a shell thickness r at distance dr there are ##C4\pi r^2dr## droplets, as @pasmith wrote, except I'm using spheres, not hemispheres.
##dr## being small, we can take these as non-overlapping to the viewer, so they present an area ##C4\pi^2 a^2 r^2dr##. As a fraction of the shell area, that is ##C\pi a^2dr##.
If ##V(r)## is the visible fraction of a shell at distance r then we have##V(r+dr)=V(r)(1-C\pi a^2dr)##, whence ##dV=-C\pi a^2drV##. Solving the ODE, ##\ln V=-C\pi a^2R##.
If we choose the limit of visibility as ##V=e^{-3}## then ##R=\frac 3{C\pi a^2}##.

pasmith
haruspex said:
In a shell thickness r at distance dr there are ##C4\pi r^2dr## droplets, as @pasmith wrote, except I'm using spheres, not hemispheres.
##dr## being small, we can take these as non-overlapping to the viewer, so they present an area ##C4\pi^2 a^2 r^2dr##. As a fraction of the shell area, that is ##C\pi a^2dr##.
If ##V(r)## is the visible fraction of a shell at distance r then we have##V(r+dr)=V(r)(1-C\pi a^2dr)##, whence ##dV=-C\pi a^2drV##. Solving the ODE, ##\ln V=-C\pi a^2R##.
If we choose the limit of visibility as ##V=e^{-3}## then ##R=\frac 3{C\pi a^2}##.
I understood almost everything. Your process assumes that my previous reasoning and calculations (the ones after @pasmith's post) are right. It's correct? I'm right? I understood the steps that led to that result of ##V##, but now it's not clear to me what to do: you understood that only a particular fraction needs to be considered, but how to include this in the calculations? I don't know how to connect ##\alpha = e^{-3}## with the value of ##C## that I found (and you confirmed). Thanks for any clarification.

Hak said:
Your process assumes that my previous reasoning and calculations (the ones after @pasmith's post) are right. It's correct?
Not your calculations. As I wrote, you have confused radius with diameter when plugging in the numbers.
Hak said:
you understood that only a particular fraction needs to be considered, but how to include this in the calculations?
According to https://en.wikipedia.org/wiki/Visibility, an object is considered visible if the contrast is at least 2%, and visibility is defined in terms of perceiving black objects. In terms of this thread, that means at least 2% of the object is still directly visible between the droplets, i.e. ##\alpha=0.02##. That leads to ##R=\frac {\ln(50)}{C\pi a^2}##, which approximates to ##R=\frac 4{C\pi a^2}##. Use that to calculate C.

haruspex said:
Not your calculations. As I wrote, you have confused radius with diameter when plugging in the numbers.

Of course. By 'calculations' I meant the derivation of the symbolic expression, not numerical calculations. There I mistakenly substituted confusing radius and diameter. We agree.

haruspex said:
According to https://en.wikipedia.org/wiki/Visibility, an object is considered visible if the contrast is at least 2%, and visibility is defined in terms of perceiving black objects. In terms of this thread, that means at least 2% of the object is still directly visible between the droplets, i.e. ##\alpha=0.02##. That leads to ##R=\frac {\ln(50)}{C\pi a^2}##, which approximates to ##R=\frac 4{C\pi a^2}##. Use that to calculate C.

OK, but how can I arrive at a factor ##2##% if I don't look in Wikipedia? I don't think I can deduce it from something ideal or theoretical, it would have to be empirical estimates unknown to me.
Perhaps, in general, it is more correct to say that
$$C = - \frac{\ln V}{\pi a^2 R}$$?

There is a maximum value (right?) for ##V = e^{-3}##, so that $$C= \frac{3}{\pi a^2 R}$$? Is this correct to say?

For the estimate you make in this post, you would similarly have:
$$C = \frac{\ln 50}{\pi a^2 R} \approx \frac{4}{\pi a^2 R}$$, but I could never get there without Wikipedia.

What do you think? What is right and what is wrong? Thank you very much, as always.

Hak said:
There is a maximum value (right?) for ##V = e^{-3}##,
@pasmith's model supposes the object has to be completely hidden by droplets, which tends to underestimate C, but also that droplets do not hide other droplets, which tends to overestimate C.
The main reason I tried the more realistic model was to check that, overall, the simpler model was not wildly wrong. And it isn't, so use it.

haruspex said:
@pasmith's model supposes the object has to be completely hidden by droplets, which tends to underestimate C, but also that droplets do not hide other droplets, which tends to overestimate C.
The main reason I tried the more realistic model was to check that, overall, the simpler model was not wildly wrong. And it isn't, so use it.
What is the "simpler model"?

Hak said:
What is the "simpler model"?
post #5

haruspex said:
post #5
Is post #5 the one concerning @pasmith's model, i.e. the one implying a value of ##C## with the ##3## in the numerator that I calculated in post #6, or not?
But didn't you just say that it tends to be underestimated and overestimated, so it is wrong? Or have I misunderstood? Thanks for any clarification.

Hak said:
Is post #5 the one concerning @pasmith's model, i.e. the one implying a value of ##C## with the ##3## in the numerator that I calculated in post #6, or not?
But didn't you just say that it tends to be underestimated and overestimated, so it is wrong? Or have I misunderstood? Thanks for any clarification.
Yes, the simpler model is the one you used. There is a way in which it underestimates C and a way in which it overestimates it. Which of these dominates, and how serious the error is, were not clear to me, so I developed the more accurate model in post #13. But my model requires an extra piece of information, namely, at what fraction of obscuration do we consider an object no longer visible.

According to the link I posted, the threshold is around 98% obscured. That makes the numerator 4 instead of 3, so the simpler model was in the ballpark. I don't know how accurate your answer needs to be, or whether the form of submission allows you to discuss the assumptions.

haruspex said:
Yes, the simpler model is the one you used. There is a way in which it underestimates C and a way in which it overestimates it. Which of these dominates, and how serious the error is, were not clear to me, so I developed the more accurate model in post #13. But my model requires an extra piece of information, namely, at what fraction of obscuration do we consider an object no longer visible.

According to the link I posted, the threshold is around 98% obscured. That makes the numerator 4 instead of 3, so the simpler model was in the ballpark. I don't know how accurate your answer needs to be, or whether the form of submission allows you to discuss the assumptions.

I don't think we need a very high degree of accuracy for this problem. Maybe we have already found the solution... Thanks for everything.

Suppose you have a 1D flow of photons with an intensity of I photons/sec. m^2 at the source. If there are water droplets in the air, this flow is attenuated by absorption and scattering of the photons. Let A be the area of a column of air oriented perpendicular to the photon flow. The volume of the differential air column between z and z + dz in the direction from source to receiver is Adz, and the number of water droplets in this volume is NAdz, where N is the number density (concentration of water droplets). The area of water droplets in the column is ##\pi a^2 NAdz##. So the loss of intensity between z and z + dz in the column is ##I\pi a^2NAdz##. Therefore, we have $$(I_z-I_{z+dz})A=I_z AN\pi a^2dz$$or $$\frac{dI}{dz}=-IN\pi a^2$$or $$I=I_0\exp{(-N\pi a^2L)}$$If we say that the visibility L corresponds to the distance at which the intensity decreases by a factor of e, we have $$N=\frac{1}{\pi a^2 L}$$

I think the only value in such a question is not the answer you arrive at, but your thought process in arriving at it.

The only information we are given is the visible radius $R$ and the size of a droplet (measured by a length $a$). It seems reasonable that the fraction $A(r)$ of the surface of the sphere of radius $r$ which is obscured depends directly on the number of droplets within the visible radius, so we are looking for something proportional to $C$ times a volume. We have two lengths we can use to form a volume, and the relevant combination seems to be the cross sectional area of a droplet (proportional to $a^2$) and the radius $r$ of the sphere. This leads us to $$A(R) = kCa^2 R = k'C\pi a^2 R$$ where in the last we assume a spherical droplet. (Visiblity must also depend on the opacity of the droplets, but since we are given no information about that we ignore it and assume an entirely opaque droplet.) So $$C = \frac{K}{\pi a^2 R}$$ for some constant $K = A(R)/k'$. We don't have any information with which to determine $K$; myself and @haruspex have proposed models with $$A(R) = \frac{\int_0^R 4\pi^2 a^2 C r^2 \,dr}{4\pi R^2} = 1 \Rightarrow K = 3$$ and $$A(R) = 1 - e^{-C\pi a^2 R} = 0.98 \Rightarrow K = \ln(50) \approx 3.91$$ respectively, but a rough order of magnitude calculation with $K = 1$ and $$C = (2\pi)^{-1} \times 10^{10}\,\mathrm{droplets}\,\mathrm{m}^{-3} \approx 1.6 \times 10^{9}\,\mathrm{droplets}\,\mathrm{m}^{-3}$$ would probably suffice.

Chestermiller said:
Suppose you have a 1D flow of photons with an intensity of I photons/sec. m^2 at the source. If there are water droplets in the air, this flow is attenuated by absorption and scattering of the photons. Let A be the area of a column of air oriented perpendicular to the photon flow. The volume of the differential air column between z and z + dz in the direction from source to receiver is Adz, and the number of water droplets in this volume is NAdz, where N is the number density (concentration of water droplets). The area of water droplets in the column is ##\pi a^2 NAdz##. So the loss of intensity between z and z + dz in the column is ##I\pi a^2NAdz##. Therefore, we have $$(I_z-I_{z+dz})A=I_z AN\pi a^2dz$$or $$\frac{dI}{dz}=-IN\pi a^2$$or $$I=I_0\exp{(-N\pi a^2L)}$$If we say that the visibility L corresponds to the distance at which the intensity decreases by a factor of e, we have $$N=\frac{1}{\pi a^2 L}$$
Thank you very much, but why are the two solutions at odds? Does one exclude the other, or is there some deeper reason I have not grasped? Thank you.

Edit: However, I did not quite understand how you arrived at the final result of ##N##. Could you explain it to me? Thank you.

Last edited:
pasmith said:
I think the only value in such a question is not the answer you arrive at, but your thought process in arriving at it.

The only information we are given is the visible radius $R$ and the size of a droplet (measured by a length $a$). It seems reasonable that the fraction $A(r)$ of the surface of the sphere of radius $r$ which is obscured depends directly on the number of droplets within the visible radius, so we are looking for something proportional to $C$ times a volume. We have two lengths we can use to form a volume, and the relevant combination seems to be the cross sectional area of a droplet (proportional to $a^2$) and the radius $r$ of the sphere. This leads us to $$A(R) = kCa^2 R = k'C\pi a^2 R$$ where in the last we assume a spherical droplet. (Visiblity must also depend on the opacity of the droplets, but since we are given no information about that we ignore it and assume an entirely opaque droplet.) So $$C = \frac{K}{\pi a^2 R}$$ for some constant $K = A(R)/k'$. We don't have any information with which to determine $K$; myself and @haruspex have proposed models with $$A(R) = \frac{\int_0^R 4\pi^2 a^2 C r^2 \,dr}{4\pi R^2} = 1 \Rightarrow K = 3$$ and $$A(R) = 1 - e^{-C\pi a^2 R} = 0.98 \Rightarrow K = \ln(50) \approx 3.91$$ respectively, but a rough order of magnitude calculation with $K = 1$ and $$C = (2\pi)^{-1} \times 10^{10}\,\mathrm{droplets}\,\mathrm{m}^{-3} \approx 1.6 \times 10^{9}\,\mathrm{droplets}\,\mathrm{m}^{-3}$$ would probably suffice.
OK, thank you very much. As I understand it, it is not the end result that is important, but the different methods applied. However, consider the following result you wrote.
pasmith said:
but a rough order of magnitude calculation with $K = 1$ and $$C = (2\pi)^{-1} \times 10^{10}\,\mathrm{droplets}\,\mathrm{m}^{-3} \approx 1.6 \times 10^{9}\,\mathrm{droplets}\,\mathrm{m}^{-3}$$ would probably suffice.
Shouldn't it be ##(4 \pi)## instead of ##(2 \pi)##? The radius is ##\frac{1}{2} \times 10^{-6}##, and squared it gives ##\frac{1}{4} \times 10^{-12}##. There should be a ##4## instead of ##2##, no?

@pasmith @haruspex @Chestermiller I have just found what is supposed to be the solution to this problem. I attach it at the bottom of the page.
It considers the diameter instead of the radius, so it conforms to the solution with K=1 advanced by @Chestermiller and @pasmith. What do you think? Thank you.

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Hak said:
@pasmith @haruspex @Chestermiller I have just found what is supposed to be the solution to this problem. I attach it at the bottom of the page.
It considers the diameter instead of the radius, so it conforms to the solution with K=1 advanced by @Chestermiller and @pasmith. What do you think? Thank you.
It makes the untrue assumption that the visibility distance is the same as the mean free path. At the mean free path distance, the probability of any given photon making it all the way is ##e^{-1}##, so more than a third of the object would be visible.

haruspex said:
It makes the untrue assumption that the visibility distance is the same as the mean free path. At the mean free path distance, the probability of any given photon making it all the way is ##e^{-1}##, so more than a third of the object would be visible.
Thank you.

## What factors influence the concentration of fog droplets in the air?

The concentration of fog droplets in the air is influenced by several factors, including temperature, humidity, air pressure, and the presence of condensation nuclei. Temperature and humidity determine the air's capacity to hold moisture, while condensation nuclei provide surfaces for water vapor to condense into droplets.

## How is the concentration of fog droplets measured?

The concentration of fog droplets is typically measured using instruments such as a fog droplet spectrometer or a cloud condensation nuclei counter. These devices can count and size the droplets, allowing scientists to determine their concentration in a given volume of air.

## What is the typical size range of fog droplets?

Fog droplets typically range in size from about 1 to 50 micrometers in diameter. Most fog droplets are on the smaller end of this range, generally between 5 and 15 micrometers.

## How does the concentration of fog droplets affect visibility?

The concentration of fog droplets directly affects visibility. Higher concentrations of droplets scatter more light, reducing visibility. When the concentration of fog droplets is high enough, it can create dense fog conditions, significantly impairing visibility.

## Can the concentration of fog droplets be altered artificially?

Yes, the concentration of fog droplets can be altered artificially through techniques such as cloud seeding, which involves dispersing substances like silver iodide or salt into the air to encourage droplet formation. However, these methods are more commonly used for cloud modification rather than fog control.

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