rain gauge

[Norman Albers]

I have a box of glass plates, thick and maybe 3"x8" or so. I want a rain gauge so I am figuring the relation between rain infall and height at the bottom of a 'V' configuration described by two plates, with two others plating the flat sides. I have to figure how to mark the scale of height coming up from the bottom so it corresponds to rain. If I decide how high it will be, say, six inches, does it matter how wide the opening is?

[MrMongoose]

The wider it is, the lesser the miniscus. If its too narrow, capillary action could come into play. Also in a brief shower, there may not be a large enough sample of rain droplets for a good statistically sound test unless you use a reasonably large collection area.

 

Reducing size can keep costs down and save space.

 

I'd go for 4-5 inches between the tops of the two plates.

[Norman Albers]

Good practical answer, MrMongoose. If I was smart I would have seen there is only one relevant parameter without working it out.

[MrMongoose]

Partial volumetric flow rate? I have to admit, I only really understood the last sentence of the first post.

[Norman Albers]

How do you mark the scale? That's what I was wondering. It is a square-power relationship, yah? Does it involve the width of the opening??? I thought this was a good elementary logic problem.

[Mr Skeptic]

Multiply the height of the rainwater at the tube by (area of tube / area of collector).

 

Edit: I didn't realize the thing was a V shape. That is rather unusual for a rain collector. I'll hide the answer in case you actually meant this as a puzzle.

[hide]

Divide the volume of water by the area of the collector. The volume will be V = 1/2 h * w * L. Those are height, width, and length of the rainwater. L will be either 3'' or 8'' depending on how you arrange the plates. The ratio h/w will be equal to H/W, the ratio of height to width of the collector, and will be constant. The collector area will be A = W * L. Then you want V/A = 1/2 h * w / W = h^2 / 2H, which will be (volume of rainwater)/(area it fell) = (depth of rainwater). As you said, it will be square-power relationship, and it will involve the total height or width of the collector -- a larger collector will collect more rain.

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[Norman Albers]

This is not a tube, it is a 'V'. Such a design is cool because the first millimeters are "amplified"...what Mr. Skeptic just said is only differentially true.

[MrMongoose]

Well, whats the volume of a triangular prism? Why are the first few millimetres so important? Are you actually trying to make this or is this intended to be a "brainteaser"?

[Norman Albers]

Both. These are the kinds of things I figure out when I'm fried from tensor calculus in GR.

 

Ah, Mr.Skeptic is getting hot. Can you tell me, oh skeptic, what is the significance of the opening of the 'V' beyond what MrMongoose said? He was also quite warm. Everything I have said is germain and I am not playing games with metric conversion a la crashing Mars landers.

[MrMongoose]

Thinking in terms of the miniscus again, the v section will make the boundary component of surface tension be at a finite angle to the vertical, so the innaccuracy will be lower. Also, as the water height will be projected onto the angled side it will be amplified with a gain equal to the secant of the angle to increase the resolution.

 

In terms of the entry of a v, the only other thing I can think of is that rain approaching the collector obliquely may bounce out again around the edges if the angle of the plates is large. Also, having a reduced volume for the same given free surface area will increase evaporative effects on the meaurement.

 

There's just so much to say. If this is realy a puzzle rather than you trying to get us to write an essay for you, maybe you could point us in the right direction?

[Norman Albers]

Write the equation for "amount of rain entering the top of the V" relating to how far up it fills the V starting at the bottom. What terms turn out to be significant? MrMongoose, you are focussing on details. Back off and solve the basic geometry. Mr.Skeptic, if it looks like a bar fight is erupting, you should dramatically lift your veil.

[MrMongoose]

I just wrote this post, then realised I was using the instantaenous free surface area rather than the area of the opening, but rather than delete it I'll leave it here just in case you made the same mistake as I did when I got excited about my "result"

 

Well, after a given time, and excusing the very word equations,

 

Volume of water in bottom of v= volume of incoming rain per unit area x area at top.

 

Volume of water in bottom of v = height of water x area of opening x 2sin (angle of v to vertical).

 

So,

 

Height of water= volume of incoming water per unit area/2sin(angle of v to vertical)

 

So water height is proportional to the amount of water that has been added, rather than the amount of water squared.

[Norman Albers]

I disagree. The first little bit goes into a small (not wide) place. The last bit filling up the gauge is one-to-one, yah? Mr.Skeptic, feel free to lay it on us. Without spilling the answer I can say to MrMongoose that there is a square root. Like I said, being not smart, I wrote ratios to express [math]\frac1 2 bh[/math].

[MrMongoose]

Well in reality, water entering the gap at the top of the v plates will run down the sides of the plates until it sits in the pool in the bottom of the v.

 

Volumetric flow rate per unit area x area of opening = water volume = rate of increase of water volume x time spent collecting

 

water volume is proportional to height x area of free surface

 

And area of free surface is proportional to height, so if you shove everything together,

 

rate of increase of water volume (i.e. rain per second per area) is proportional to height squared

 

You could equally say height is proportional to the square root of the amount of rain per area per second, which I suspect is where your square root comes from.

 

Now wheres the puzzle?

[Norman Albers]

Mongoose, you've spent too much time in the smoker.

[MrMongoose]

Is smokiness proportional to genius?

[Norman Albers]

Jerkiness.

[MrMongoose]

Sorry for trying to help you fix your rain gauge...

[Norman Albers]

Horsefeathers.

[Norman Albers]

Where is Mr Skeptic? What is your solution?

[Norman Albers]

The relationship that made me stupid was that the angle of the 'V' does not matter; it drops out as things are proportional. There are realistic considerations such as offered by MrMongoose, but these are secondary.

 

Next time you're bored, solve the cone.

[MrMongoose]

Well I'll save time on the actual calculations, but the realtionship will be a cube root one.

[Norman Albers]

What about the ratio and scale stick?

[MrMongoose]

Well, height is equal to the cube root of (3 times the volume divided by pi cos theta)

 

Where theta is the angle of the sides from the vertical.

 

You'll have to excuse the lack of symbols.

[Norman Albers]

Yes, MrMongoose, but given a height, does the angle affect the scale stick? Do you understand what I said in post#21? (Incidentally, click on this to see how to compose it: [math]cos \theta [/math].) Actually, I don't think the cosine is useful. Express the tangent ratio, and then also express the amount of rain in the top opening. Here's how I solved the first example, the 'V': width at the top is D, total height is H, rainfall is R, width of the water in the gauge is b, and measured height is h. Everything is proportional to the depth of the 'V' plates so it drops out obviously in this case and we can just talk about areas. The filled bottom of the gauge has area of: [math] A=\frac 1 2 bh[/math] as per a triangle. The quantity b can be expressed in terms of a ratio: [math]b=\frac D H h[/math]. Then, equate this area with the rainfall in the top: [math] A=DR[/math], and you get: [math] h=\sqrt{2RH}[/math]. Do you see how D drops out?