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Rain measuring simple unit with a measuring jar fitted with a funnel collect rain pouring. Here I presume the rain falling vertically down with zero speed. However, usually the wind is blowing. If it blows with higher speed, then naturally rain pours down not vertically, but at some angle. With more speed of wind, slanting of rain increases. In that case will this not affect the collected amount in measuring the rain? Or is there any correction one has to apply considering the wind speed and angle made by the rain with vertical line?

  • Imagine you are collecting in a bucket. If the rain falls vertically you will collect the maximum amount. If the rain falls horizontally, the walls of the bucket will block the water and you will collect none. In between, you will collect some fraction of the maximum amount. – pentane May 05 '15 at 18:11
  • @pentane - I am sorry but I think you are wrong. The limit of purely horizontal velocity is only possible if the rain has no vertical velocity; the assumption here is that the vertical component of velocity is unaffected by the wind. See the answer that Soham wrote. – Floris May 05 '15 at 18:28
  • @Floris how about instead of tilting the rain, you keep the rain completely vertical and tilt the bucket so that it is not completely upright. Is the amount of rain that enters the bucket not reduced as the effective orifice area of the bucket is reduced? – pentane May 05 '15 at 20:07
  • @pentane - It is not the same thing, as the velocity of the rain perpendicular to the surface changes in the case you describe. The key here is that the same amount of rain has to his the same area of ground per unit time, whether there is wind or not. If you take a piece of paper and hold it (almost) vertical, it stays dry... but if you put it on the ground it will get equally wet whether there is wind or not. – Floris May 05 '15 at 20:13
  • @Floris Hit the ground yes, enter the bucket no. OP wants to know "does wind affect rain collection?", the answer is "yes it does", see the ~1000 or so papers on wind-induced undercatch – pentane May 05 '15 at 20:32
  • @pentane - you are right. To the extent that a horizontal wind velocity can result in a change in the vertical velocity of the rain, wind will affect rain collection efficiency. This happens when the shape of the container is such that it "bends" the wind, and is therefore more important when the precipitation meter is installed at some distance above the ground. People do use wind shields around high precision weather stations to minimize this effect. I don't think that is what OP is after but I could be wrong. If wind velocity is "purely horizontal", there is no effect. – Floris May 06 '15 at 01:25
  • @Floris "This happens when the shape of the container is such that it "bends" the wind" Actually this happens in absence of any container above the ground as well, because the terminal velocity of rain drops are constant, any increase in horizontal component will result in a decrease of vertical component – Azad May 06 '15 at 05:53
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    Sorry I was wrong. Rain drops horizontal velocity is the same as air so there's no horizontal drag and vertical component will be equal to terminal velocity regardless of angle – Azad May 06 '15 at 06:06
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    @Azad right - so as long as the wind velocity is constant there is no effect. When it is variable (what? Wind speed changing? Never!) the effect you describe will impact the vertical velocity (because of quadratic drag). – Floris May 06 '15 at 11:45
  • @Floris now that I think about it even wind speed change with height has an effect. Because if wind speed is higher in upper layers (which is usually) then the drop experience horizontal drag as it gets to a lower layer. – Azad May 06 '15 at 12:21
  • @Azad - you are right that the wind speed changes with height, and that this induces vertical drag; nn the other hand - the extent to which this is the same everywhere (not just at the meter) it just reflects the actual precipitation (thus it is a source of drag, but not a source of error). Only where the variations around the meter are different does this drag result in a measurement error. – Floris May 06 '15 at 13:24
  • @Floris If I understand you correctly you are saying that it's second derivative (and higher) that matters but I think even the first derivative alone can have an effect because if you have no wind at all (unlikely I know) you wouldn't have this source of extra drag so it's a source of error relative to no wind case. – Azad May 06 '15 at 13:38
  • It is only a source of error (as opposed to a source of drag) if the effect is different for the rain collector than for the soil. I am saying that if all the rain gets the same drag, it will fall at the same rate everywhere - on the soil and on the collector. Thus, no error in the measurement just because of a velocity gradient, if that is the same everywhere. And "error or no error" is what we are trying to establish. – Floris May 06 '15 at 13:44
  • The rain gauge may not collect same amount of water in winds , imagine at lower angle at 0° or 15° , it will collect no water or very little of what is raining. Correct me if I am wrong. – MUQEET SOHEB Jun 21 '22 at 06:37

2 Answers2

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Two parts to this answer. First - the idealized situation where wind adds a purely horizontal component to the rainfall; next, what actually happens around a pluviometer when there is wind.

Part 1: ideal situation

Imagine the density of rain drops per unit volume is $N$, and the vertical velocity is $v$. The rate at which drops arrive on a unit area of ground or rain gage will be

$$R = N \cdot v$$

This is illustrated by this diagram: all these rain drops will make it into the box in unit time:

enter image description here

Now we add a horizontal component (due to wind) to the rain velocity. Let us call this additional component $v_h$. As a result, the rain will be falling at an angle $\theta$ to the vertical, and it will have a new velocity $v_t = \sqrt{v^2 + v_h^2}$. The angle $\theta$ is given by

$$\theta = \cos^{-1}\frac{v}{v_t}$$

or

$$v_t = \frac{v}{\cos\theta}$$

enter image description here

In this diagram you see the same number of rain drops as before; they start in different places, but they all arrive in the collector in the next unit of time.

But there is another way of looking at this - we can rotate the entire picture (and move a few drops - I colored them green; but there is once again the same number of drops as before):

enter image description here

Since this rain is arriving at the collecting aperture of the pluviometer at an angle, the apparent area is smaller - by $\cos\theta$. But the velocity $v_t$ is greater - by $\cos\theta$. These two terms exactly cancel out - and for this ideal case, the rate at which rain is collected is independent of the speed of the (horizontal) wind. The only thing this ignores is the situation where the size of the drops becomes significant compared to the area: the chance of hitting an edge becomes nonzero, and how such a drop breaks up (and what fraction ends up in the container) is hard to predict. In particular, the behavior may be different on the leading edge compared to the trailing edge. But that really belongs in

Part 2: real world

So - why doesn't it work out like this in real life (as pointed out by @pentane, and supported by numerous references)?. The problem is that there is no such thing as "perfectly horizontal wind" - especially not in the vicinity of objects. Wind tends to tumble and turn; in short, it has vertical components which will change the local rate at which rain falls. Looking for example at figure 6.11 from [these lecture notes](http://www.jma.go.jp/jma/jma-eng/jma-center/ric/material/1_Lecture_Notes/CP6-Precipitation.pdf), we see the following cartoon:

enter image description here

This shows that the direction of the rain drops is a function of the position relative to the gage, because the wind direction changes around the gage. As a result, some rain is pushed away, leading to a collection deficit. But quoting from the same source:

Wind exerts a significant influence on the observation of precipitation with snow and rain gauges, and there is no way to avoid its effects. However, accurate collection of precipitation in a rain gauge is possible when the wind around the receptacle is horizontal and its speed is equal to that at ground level or when no vortices develop near the gauge. A windshield is effective in reducing the influence of wind.

If you have a typical rectangular collector (a large tin can, for example) the air flow might look something like this:

enter image description here

(image adapted from this website)

Typically, the air draft will be up, leading to an underestimate of the collection efficiency. There are companies dedicated to solving this problem for you - see for example NovaLynx Alter-Type Wind Screens which shows the following:

enter image description here

and states on their website:

The 260-952 Rain Gauge Wind Screen minimizes the formation of strong updrafts that can distort the trajectories of precipitation particles falling toward a gauge. The screen also generates turbulent air motions over the gauge orifice to break up streamlines and thus improve the catch. Use of a wind screen is recommended with all precipitation gauges located in windy areas.

After reading @Azad's comment above, another factor sprang to mind. As long as horizontal wind velocity is constant, the rain drops will continue to fall at their average terminal velocity. However, if the wind speed changes (either because that is what wind does, or systematically because if accelerates around an obstacle) then the rain will experience a greater vertical drag (because of the quadratic nature of drag, a "cross wind" increases forward drag). And that in turn would slow the rate at which the rain falls and make the reading low.

In short - the answer is "it depends". There are various mechanisms that make rain collection dependent on the wind, but there is no "easy formula" as it depends on the details of the geometry and the magnitude as well as variation (temporal fluctuations), of the wind speed

Floris
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Change in direction won't impact the amount collected per unit area. Amount collected per unit area can be affected only if there is some lensing/dispersion. Since, mass of water that falls, and the area on which it falls (any increase in area in direction of wind is compensated by corresponding decrease in opposite direction) are both constant w.r.t. direction and speed of wind, therefore mass/area is also constant.

Soham
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    Correct. It might be even better with a diagram, but even without, the argument that all the water that fell before still has to fall, and still covers the same total area (the size of the cloud it fell from) means that it must have the same rate per unit area, regardless of horizontal velocity. Very nice first answer - may there be many more!! – Floris May 05 '15 at 18:25