# How big are clouds? [closed]

How big are clouds? When I look up into the sky I have no frame of reference, so I don't know if they are 200 feet or 2 miles across. When I am in a plane looking out at a cloud, I try to use the wing as reference but I still don't have a good reference point, because the clouds are just a large white mass.

I realize that "cloud" is a very loose term, so interpret how you wish.

• This is a question of "measurement," but what you trying to measure, as you say, is not clearly defined, thus there is no value. In fact, clouds can not even be "counted." – user 147593 Sep 18 '15 at 4:37
• Pick a cloud. Measure its angular extent. Make a reasonable assumption about the altitude e.g. 2 miles. – CuriousOne Sep 18 '15 at 5:02
• @skillpatrol: There is nothing wrong with estimating the order of magnitude of things. So what if he is off by a factor of two or three? – CuriousOne Sep 18 '15 at 5:45
• Can't believe no-one has linked the appropriate xkcd cartoon yet... xkcd.com/941 – Oscar Bravo Sep 18 '15 at 12:03
• I'm voting to close this question as off-topic because the guy should go look up the answer. – John Duffield Sep 26 '15 at 14:58

How big are clouds? When I look up into the sky I have no frame of reference, so I don't know if they are 200 feet or 2 miles across.

An old astronomy based system, that I learned as a kid in the scouts, is to use your hands and arms as a rough guide measure, specifically by making your hand into a fist, then putting your arm as forward outwards as far as you can stretch.

Move your arms so that one fist is lined up with the horizon, and then put the other fist on top. Then holding your second arm steady move the first fist and put it on top of the second fist. Continue doing this until the top of the last fist is directly overhead.

Unless you have very small hands, that is assuming you are a normal sized adult, it might take eight fists to get from horizontal to 90 degrees, so each of your fists is about 11 degrees high. Turn your fists by 90 degrees as you stretch out your arms to measure horizontal angles of the clouds.

How accurate this idea actually is, I don't know, there are a lot of variables involved , such as the length of your arms, the size of your fist and the ability to define the edge of the cloud clearly.

Steve corrected me on this below, proportions are important re: arms and hands.

From Cloud Guide, which writes about the same technique, for a well defined cumulus cloud:

The key formula is: Mid Cloud Hor.Angle(deg)=C* Mid Cloud Altitude(mi)/Cloud Dist. or Cloud Dist.(mi) = C* (Mid Cloud Altitude(mi))/(Mid Cloud Hor.Angle(deg)). Here C is 57.3 degrees or 180 degrees divided by Pi (radians in 180 degrees). To get the cloud’s vertical or horizontal dimensions, the formula to use is:

The cloud dimension (either horizontal or vertical) = (Dimension angle)*(Cloud Dist)/(57.3 Deg.) So as soon as you know the cloud distance (in miles), you can find out both the vertical height and horizontal width of the cloud.

Take a well defined, clear edged cumulus cloud which happens to be the same dimensions both horizontally and vertically and whose middle is two fists above the horizon (2 x 11 deg.= 22 deg.), then the cloud distance = 57.3 deg.* 0.5 miles/(22 deg.) = 1.3 miles away. If this same cloud is one vertical fist by 1 horizontal fist across, then the cloud’s dimensions are both cloud dimensions, height and width = 11 deg. * 1.3 miles/ 57.3 deg.= .25 miles.

In this case, the cumulus cloud would be around 1.3 miles away from you and 0.25 miles both high and in width.

Because it's a cloud question, I just have to include a picture of a typical cumulus cloud, this might show how difficult it is to get clearly defined edges as well.

This method may give a rough width estimate for nearby cumulus clouds, but it gets a bit iffy for spread out and/or high altitude clouds. However, it passed the time for us, as kids with something to do, waiting for the skies to clear.

Wikipedia Cumulus Clouds does not say how wide cumulus clouds typically are, so I doubt you can get an accurate width, although a height estimate might be easier.

• "Unless you have very small hands" -- one advantage of the technique is that people with small hands usually also have short arms in proportion, and it all more or less cancels out such that a person's fist at arms length has about the same angular size whoever you are. It's only when you have small or large hands for your arm length that it goes wrong. – Steve Jessop Sep 18 '15 at 12:29
• @SteveJessop good point, at the time I tried we ALL had small hands and arms, as kids. – user81619 Sep 18 '15 at 12:31

I don't know if they are 200 feet or 2 miles across

Clouds are fractal. Cloud particles can be a few dozens of micrometres, and big tropical cyclones can be thousands of kms across. That's a range of more than 10 orders of magnitude! That's why clouds are a pain to represent in models — it is simply impossible to have a physics-based model of a cloud with a domain large enough to represent a full cloud system.

Realistically, it is not possible to give a lower limit for the size of a cloud. When you "see your breath" on a cold day outdoors, that is fundamentally no different from the fog you see above a lake, which again is fundamentally no different from a large system. It's all liquid and solid particles floating in the air.

So, to answer your question: a cloud can be as large as thousands of kms across. There is no practically applicable lower size limit.

• Excellent answer. The concept of size loses meaning. – user 147593 Sep 18 '15 at 14:19

The nature of a cloud makes answering this question a bit difficult. Clouds can have various sizes ranging from the size of a football field to that of a city, with thicknesses ranging from from a dozen to several hundreds of feet. The difficulty arises when considering what is 'a' cloud..

If, for example, the picture shows a single cloud, it's radius would be hundreds of miles. Average cumulonimbus clouds are around 1000 feet thick their dimension are in the order of several hundreds or thousands of feet. So the answer really depends on what cloud you're looking at, and how you define the boundary of a cloud.

• "100 feet thick their dimension are in the order of several hundreds or thousands of feet" huh? Cumulonimbus is many kilometers high! – Mithoron Sep 18 '15 at 16:50
• You are correct. Sorry for the error. I missed a '0'. Have updated answer. – Praneet Srivastava Sep 18 '15 at 17:01
• Still don't get it, it looks like you're talking about another king of clouds – Mithoron Sep 18 '15 at 17:06

We need two things to estimate the size of a cloud: 1) it's angular extent $\theta$ 2) its height h above the ground. Then, on a rough basis, linear extent r=h$\theta$

This could also be estimated using the size of the cloud's shadow and the angle of the sun in the sky. A sextant and a tape measure/other length measuring instrument can help. The linear extent of the cloud can be calculated using trigonometry and a concept of similar triangles.

An easier answer: Suppose we get 3cm of rain fall over an area of one square kilometer. The mass of water (which came from the cloud) is about 30,000 tonnes

• This gives us mass of the cloud, not it's volume. The densities of various types of cloud vary, and the variation is not small. Also this tells us nothing about clouds that do not precipitate. – Praneet Srivastava Sep 18 '15 at 8:58
• It provides a good indication of the order of magnitudes of the mass of water in clouds. I am also taking the author at his word: "...so interpret how you wish." – user56903 Sep 18 '15 at 10:07
• This is not an accurate model of physical reality at all. Except in the own reality of climate and NWP models, a cloud does not turn into rain until it's empty. Rather, advection and convection continue before and during the precipitation, all particles fall but only some are large enough so that falling speed exceeds uplift, some particles evaporate before reaching the ground, etc. – gerrit Sep 18 '15 at 13:01
• The mass of water in a cloud is one of the dimensions and the quantity is quite surprising to most people. – user56903 Sep 18 '15 at 14:58
• If Denali is a granite (2.74 g/cm3) hemisphere 18k ft high, then it has a mass of 10^17 tonnes. Sometimes I hear people say clouds are heavier than mountains, which doesn't seem to be typically true. – jejorda2 Sep 18 '15 at 18:32

What... on... earth???

Clouds come in various sizes and are fractal in size.

It would be very difficult to answer the question "what is the average cloud diameter in the above picture", and even harder if it were an animated video with clouds constantly growing, shrinking, splitting, merging, changing shape, disappearing and forming anew, with fuzzy outlines and some "clouds" of vapor forming that block almost no light so don't show up visually, and smoke and smog and volcano plumes that look much the same as clouds, and...

But none of this this makes this question "UNANSWERABLE" or "MEANINGLESS". Mathematics is capable of handling far more than just single numbers. We have the tools available to describe fractals, distributions, and more.

There are plenty of resources on cloud height and dimension distributions. Now, it's admittedly hard to find them on Google because any search for "cloud size distribution" will find you a near-infinite number of papers on particle or droplet size distributions within clouds.

But papers do exist, such as this one by Jianjun LIU et al: http://www.iapjournals.ac.cn/aas/article/2015/0256-1530-32-7-991.html

• An excellent answer. I would just point out that OP is, almost certainly, talking about "cumulous clouds". And in any event, the one and only technique you can use to judge cloud size (from a single position on the ground) is a standard candle. Say there was a competition for judging cloud sizes. It's this simple: the nature of the skill of doing that, would be, identifying different cloud types, and then, knowing the distributions such as given in the excellent answer above. – Fattie Sep 19 '15 at 14:25
• @JoeBlow True on all counts :) – Dewi Morgan Sep 21 '15 at 0:01

It couldn't be easier to answer this question.

It's utterly commonplace in English that words have different meanings. When you, OP, say "clouds" you are almost certainly talking about "cumulus clouds" ... cartoon clouds.

A key thing to know is the clouds you are seeing, the bottoms are typically

about one or two kilometers off the ground.

The tops are typically

3 or 4, up to about 10 kilometers off the ground.

Any number of simple references about cumulous clouds can be googled up to gain these facts.

You can read-up online about the visual characteristics of "cumulous clouds" so that when you see one you can be more certain it likely conforms to these approximate sizes. Just as you say, you'll now know whether it is merely 100 meters across and close, or conversely 100s of kilometers away.

As I say, a good thing to bear in mind is that the bottoms of cumulous are usually 1 to a few km off the ground - that will give you a sense of scale, at least give you an order of magnitude anyway.

Note that - of course, obviously - there are many types of clouds from small wispy ones to planet-sized clouds.

Regarding non-cumulous clouds - you know nothing from this post :) Read up on them separately. But it's very likely you meant, and from your description you likely meant, "cumulous clouds".

Note too that you make an excellent point, OP, that it is in fact impossible to judge the size of many clouds you see. You, me, and a scientific measuring team would be equally clueless. (The only way to do so would be to have multiple cameras far apart, and perhaps use fast aircraft.) It is, as you say, quite simply impossible to know how far away a cloud is - unless you use "standard candle" techniques such as identifying a cumulous, as mentioned here.

(Comments that "clouds can't be measured - because of Fractal!" are naive. You can't measure ANYTHING - say a coastline length - "because of Fractcal!" Of course, when you ask the length of a coastline, there's a reasonable fractal dimension that is meant, in context, and you can give a reasonable answer in context -- as with every issue of language.)

Note. If you want to know

# The apparent angular size

of a cloud (or a UFO, or of anything), that is trivial.

Always remember the full moon is about one half degree. It's that simple.