Raindrops are wet!

[thorfinn]

That's right, raindrops are wet. But how wet, that's the question! At various times and places, the question, "is it better to run or walk when it's raining?" has been raised... and I've never had a satisfactory answer from discussion. So, whilst in Sydney visiting the rels, the question came up again, and this time, I decided to take a physics approach to the problem.


First, we simplify the problem to something we can understand, then we frob the parameters until it looks more realistic.

So, we start by assuming a cuboidal human with a waterproof bottom side. This gives us five sides that can "get wet" to worry about... the top, the front, the back, and the left and right sides.

Next, we assume completely vertical rainfall at a constant rate. This means that if our cuboidal human is standing perfectly still, the only side that gets wet (ignoring drippage - let's also assume our initial cuboidal human is a perfect water-absorber) is the top side, and the rate is determined by how long our human stays in the rain.

Okay. Now lets play with the constraints. If our cuboidal human (OCH) moves horizontally towards the front, then OCH still picks up water on OCH's topside, at the same rate as if OCH were standing still. But what about OCH's sides? Now, if the rainfall is perfectly vertical, then OCH's front-side will be moving "into" raindrops as they fall down, and no other sides will be affected.

Since we're assuming constant speed perfectly vertical rainfall, then I think that this can be abstracted as a non-moving mist of droplets (the density of the mist being determined by the rate of downfall - faster rainfall, thicker mist)... What does that mean? That means that over the same distance, the amount of water arriving on OCH's front-side is identical, whether OCH moves fast or slow!

Whoa. Okay. But there's a third factor to consider... OCH is also warm. This means that water collected on OCH will evaporate at a certain rate... So, if OCH moves slowly, and the rainfall is light (ie, the mist is low density), water collected on OCH's front-side will evaporate as fast as it collects... and that's true of OCH's top-side as well!

Now, lets say that OCH's sides all have a certain amount of water they can collect before they become "soaked"... So, if it is raining lightly, and OCH runs fast, OCH's frontside collects a lot of water, and becomes soaked. At which point, OCH's wet on the inside, and starts to scream "I'm melting, I'm melting!". If OCH goes slowly, then the rate of evaporation vs the rate of water collecting on OCH's front is quite high, so OCH's never wet on the inside...

If, however, it's raining heavily, then OCH's better off running, because the amount of rainfall on OCH's top-side is quite high, and OCH's front-side is probably doomed to become wet no matter what.

Hrm. Okay. Now, how about constant non-vertical rainfall? What does that do to the problem? Well... if the rainfall is going in the same direction that OCH is, then OCH's in luck. OCH should move forward at the same rate (or possibly slightly faster than) the horizontal component of the rainfall's motion... How much faster depends on the rate of downfall vs the rate of collecting rain on OCH's front-side by outrunning the rain - minimising the top-side wetness by moving faster vs increasing the front-side wetness. Now, OCH's actually usually quite a bit smaller on top-side than on front-side, so that means that if there's horizontal rainfall from behind OCH, then OCH should outrun the rain, but not by very much.

What if the rainfall has a sideways component? Then that's the same as the topside... OCH can't do anything about that, but it becomes a factor in how fast to run, because the more sideways the rainfall, the faster OCH gets wet on the side, so the faster OCH should run to minimise the duration of wetting, but again, factoring in the fact that if OCH runs faster, OCH gets wetter on his front-side.

If the rain is coming from in front of OCH, then it's all bad. OCH's frontside is doomed to get soaked... Or is it? There's a solution to this! Lean forward, to minimise OCH's profile, so that OCH's topside is the one that's exposed to the rainfall... which then reduces it to the earlier problem, of topside rainfall rate vs how much to outrun the rain... That solution applies to sideways vectoring rainfall too, but OCH usually has much trouble moving forward whilst leaning sideways, so that's not very useful.

So... what's it all mean? Actually, the conclusions are quite useful! If it's not raining hard (and not blowing hard in any particular direction), then OCH's better off walking slowly, possibly with OCH's arm over OCH's head to improve the amount of of topside absorption before soaking, because the evaporation component is big enough relative to the rate of getting wet that OCH's probably going to stay mostly dry. If it is raining hard (or is blowing hard sideways), then OCH's better off running, to minimise the time spent in the hard rain, because the evaporation component is too small in comparison to the rate of wetting... and the only other slight side issue, is if OCH's very lucky, and the rainfall is coming from behind at slow enough for OCH to keep up, then OCH should try to match (and slightly overshoot) the horizontal speed of the rainfall.


Of course, OCH's probably better off carrying an umbrella and/or wearing a raincoat... but that's not as geeky as working out the problem. *grin* Oh, and just to note, the above theory matches my experiences with various rainfall.

Comments

[madigrrl.livejournal.com]

Reading this in lynx, as is my wont, I was
struck by one thing above anything else; the
word 'OCH' was constantly repeated in blue all
over the screen, as though the whole thing was
written by some touretic Scotsman. Which, given
the actual content of the post seems oddly
appropriate.

[thorfinn]

*chuckle* Guess lynx is doing <b> tags in blue for you. The post would be kinda amusing in a mad scottish accent...

[drwally.livejournal.com]

I'm your only friend, I'm not your only friend but I'm a little glowing friend, but really I'm not actually your friend but I am...

Blue Canary in the alley by the lightswitch, who watches over you?

Make a little birdhouse in your soul... not to put to fine a point on it, say I'm the only bee in your bonnet...

Make a little birdhouse in your soul!

[delwyn.livejournal.com]

you sir, have too much spare time ;)

[thorfinn]

Heh. :) I don't, really... I'm just too much of a geek for my own good.

ObGeek

[lederhosen.livejournal.com]

First: if it's raining, you can probably ignore evaporation - humidity is likely too high to allow it.

> Hrm. Okay. Now, how about constant non-vertical rainfall? What does that do to the problem? Well... if the rainfall is going in the same direction that OCH is, then OCH's in luck. OCH should move forward at the same rate (or possibly slightly faster than) the horizontal component of the rainfall's motion... How much faster depends on the rate of downfall vs the rate of collecting rain on OCH's front-side by outrunning the rain - minimising the top-side wetness by moving faster vs increasing the front-side wetness. Now, OCH's actually usually quite a bit smaller on top-side than on front-side, so that means that if there's
horizontal rainfall from behind OCH, then OCH should outrun the rain, but not by very much.

To quantify that: let the rain's velocity be characterised by the vector (Vx, Vy, Vz). Pick the axes so that the required journey is in the positive x-direction, z-axis is vertical and inverted, so Vz = positive vertical velocity, and the y-axis is at right angles to the other two & Vy is non-negative. (Trying to make the velocity components positive here to avoid mucking around with absolute values later.)

So our travel velocity will be (D/t, 0, 0), where t is the time taken (assuming we're moving in a straight line, which turns out to be the best option) and D is the total distance to be travelled.

After eliminating a couple of constant factors (namely: density of rain, and area of cube's face): the rate at which the rain hits the top face of our cubical person is simply Vz. Total rain hitting this face is then Vz*t. Similarly, rate at which it hits the side face is Vy, and total on the side face is Vy*t.

Rate at which it hits the front/back faces is |D/t - Vx|, with total here equalling |D - tVx| (abs value signs.)

So total wettage is t*(Vy + Vz) + |D - tVx|. If D > tVx, this translates to
t*(Vy + Vz - Vx) + D, otherwise t*(Vx + Vy + Vz) - D. (And if D = tVx, the two formulae are equal.)

Case (a): Vx is negative or zero. Then D > tVx, total wettage is t*(Vy + Vz - Vx) + D. Since Vy, Vz, and -Vx are all positive, wettage is minimised by minimising t.

Translation: if the rain's blowing against your path, run as fast as you can.

Case (b): Vx is positive, i.e. the rain is blowing with you. Divide into three subcases:

(b.i): 0 < Vx < Vy + Vz. (i.e. Vx is positive, but not as much as Vy + Vz combined.) Then Vy + Vz - Vx > 0. In either case total wettage is t * (positive) + constant, the wettage is continuous with respect to t, so the optimal solution is to minimise t (i.e. once again, run as fast as you can.)

(b.ii). Vx = Vy + Vz. Total wettage is then either constant D (if D > tVx) or t*2Vx - D (if D < tVx). In the latter case, optimum is to minimise t - which brings us back to the first case. This tells us that the solution is to walk *at least as fast* as the rain - beyond that, it makes no difference how fast, total wettage is the same.

(b.iii) Vx > Vy + Vz. If D > tVx, then wettage = t*(negative) + constant, so within this range wettage is minimised where t is maximized - i.e. at the end of the range, t = D/Vx.

OTOH, if D < tVx, wettage = t*(positive) + constant, so wettage is minimised where t is minimised - within this range, that means t = D/Vx. So the solution here is t=D/Vx. (i.e. you walk just fast enough to keep up with the rain.)

Summary: If Vx < Vy + Vz, run as fast as you can. If Vx > Vy + Vz, walk at speed Vx (just keeping up with it.) And if Vx = Vy + Vz, you can pick either of these or anything in between, it won't make any difference to the total amount of water that hits you.

Re: ObGeek

[thorfinn]

First: if it's raining, you can probably ignore evaporation - humidity is likely too high to allow it.

I'm not certain about that... Rainfall doesn't always mean 100% humidity on the ground. It only means 100% humidity where the rain is forming... which is usually quite a long way up. The difference between humidity up there and humidity on the ground can be quite significant (especially given the air temperature variation).

Ground level humidity during rainfall often is not at saturation, especially during light rainfall and for areas where the wind isn't blowing in over the coast or from some other large body of water.

Re: ObGeek

[lederhosen.livejournal.com]

Ah, I'm used to Sydney weather. No such thing as 'light' rainfall here ;-) When it raineth, it raineth hard enough to bring it pretty close to 100% humidity even if it wasn't already. (Also causes cooling, lowering the threshold for saturation and the rate of evaporation.)

May be some evaporation going on, but I'd be surprised if it was enough to make much of a difference here. Plus, factoring in evaporation rates would require consideration of wind velocity, which would be horrible :-)

Re: ObGeek

[thorfinn]

Indeed. But, don't forget too - OCH is warmer than air temperature. That means you get convection and evaporation, even at 100% humidity. It's not much, but it is there. :)