How much water in the barrel?
 

So my Granpa has this water container buried on the property of his farm. The container is a perfect cylinder, flipped on its side (the curved surface creates top and bottom, the flat surfaces form two sides). One day we sat around trying to devise a way to measure the amount of water still within the cylinder. The only access we have to the contents of the cylinder is a small hole on the top curved surface. We figured that if we dropped a measuring stick down through the hole we could measure the depth of the water (or close to it), kinda like a dip stick. So here's what we know: the depth of the water(x), the length of the cylinder(L), the diameter of the cylinder(d). Although neither one of us is a math genius, we were able to develop two equations to figure out the volume of water in the cylinder, one equation if the cylinder was less than half full, one equation if the cylinder was more than half full. I know there must be an easier way to do this (with one equation describing all possible volumes) but we couldn't figure it out. Any of you math players wanna give it a shot?

For some reason, the thought of density-related integrals is coming to mind. I've had some baaad experience with those, which is probably why I'm blocking it out. Perhaps you can give us the two equations you'd come up with?

Ok, well I got one mother of an equation:

You take
Depth = what was measured on your dip stick
r = radius of barrel
L = length of barrel

Volume = L*r^2*(Cos^-1( (r-Depth)/r ) - (1/2)Sin(2*Cos^-1( (r-Depth)/r )))

Just to make it more readable
z = (r-Depth)/r

Volume = L*r^2*( Cos^-1(z) - (1/2)Sin(2Cos^-1( z) ))

Anyone care to confirm or deny this?

No, CSflim, I actually don't care to confirm or deny that...

synic213, I'd be very surprised if your "two formulas" didn't also work for each other's cases. What, pray tell, are these two formulas?

Also, CSflim, arcos(x) is less confusing than cos^-1(x), you know what I mean? You don't have the luxury of writing this down nor printing it up in LaTeX...

Furthermore, if you want to do some vague sanity checking for your formula, try it out for some extreme cases, like if L = 1 and Depth = 0, r, and 2r...

hmm, for z in radians, r=radius, x=depth:
z=arcos((r-x)/r)

A=occupied area of the circular end:

A=2r*z-(r-x)*sqrt(2*r*x-x^2)

where sqrt means square root

and so if V=volume, and L is length of cylinder

V=L*A

EDIT:oops, typo

hmm, it would nice if browsers could render LaTeX

Okay, I will now agree with CSflim's formula, as convoluted as it is. I have an alternate formula (using different choices of identities) and will display both his and mine for comparison (not that mine's any simpler)...

If x is the depth, r the radius and L the length, then the volume V is:

V = L*r² * ( arccos((r-x)/r) - sin( 2*arccos((r-x)/r) )/2 )
V = L*r * ( r*arccos((r-x)/r) - 2(r-x)*sin( arccos((r-x)/r) ) )

or, you can simplify it in a two step formula:

z = arccos((r-x)/r)
V = L*r² * ( z - sin( 2*z )/2 )
V = L*r * ( r*z - 2(r-x)*sin(z) )

Ah, isn't arcos nice...


So, what were those two formulas, again?

ahh, i see where i went wrong, i did a circumference instead of an area, oops.

hmm, for z in radians, r=radius, x=depth:
z=arcos((r-x)/r)

A=occupied area of the circular end:

A=r^2*z-(r-x)*sqrt(2*r*x-x^2)

where sqrt means square root

and so if V=volume, and L is length of cylinder

V=L*A.

My square root comes from expanding the sine of the arcosine.

Re: How much water in the barrel?
 

Personally, I would measure the current depth of the water then start pumping more water into the barrel (measuring how much is added) until the barrel is either twice the starting depth (or full if it's more than half full) and then extrapolate the starting volume from the ending volume. But that's just me.

Wow guys! Thanks for the long and glorious equations! I will have to try them out to see if any of them hold water (pun intended). The thinking behind my equations went something like this: If the water level is less than half the barrel, you can determine the chord length of the water level, and from that draw a triangle from the center of the "circle" to the chord tips, in essence giving you a pie piece. Subtract the area of the triangle from the entire area of the pie piece (calculated by computing the ration of angles between piece of pie to whole pie) and you're left with the cross sectional area of the water. Multiply by the length of the cylinder and you have your volume. This equation didn't work when the barrel was more than half full because the water was now on the opposite side of the chord, if that makes sense. For this scenario, we had a set value for the volume of water for a half full barrel, and then added in the amount of water sitting on top (using a method similar but not identical to equation 1). I'll try to rederive the equations if I can, but they were all done using simple geometry and trig. I'm sure there's away to describe the whole circle at once, using calc, which is probably what you guys did. Thanks again.

Re: Re: How much water in the barrel?
 

And I'm not quite sure how this would work so maybe you could explain it better. Putting twice the water depth into the barrel doesn't equate to twice the water volume.

No, I suspect none of us used calculus. You see, in order to have used calculus, I would have had to find the integral of sqrt(r^2 - (r - x)^2), which is not an easy task!

No, we all used the same goemetric approach, as I suspected, and that's why I pressed you for your actual formulas (and why I'm still surprised you haven't posted them). You see, it would actually be a bit of a trick to find a formula that didn't work for both cases. Ours works because, if the barrel is more than half full, we would be subtracting a negative triangle and, thus, we would be properly adding it.

So, again, I would be rather suprised if your formula didn't work for both cases...

Remember, a sideways cylinder creates rectangular cross sections, so it's actually easy to integrate. X is a constant, length of the cylinder, y is the variable cord length of the circle parallel to the ground and you just integrate over Z the height of the circle (or water level). Bingo. Take Defninite integral. Or plug in X for the upper bound, take the antiderivative and then plug it into a Second Fundamental theorm equasion (F(b)-F(a), F being the antiderivative of your equasion for area f)and you have an equasion for the volume.

(Didn't do it, maybe the formula for the circle is your tough sqrt. Remember, sqrt is just x^1/a so deriving is easy.

The solution is now trivial.

Enjoy.

...so trivial that you proceed to not state the antiderivative...

Oh, and sqrt(x) = x^(1/2). It's a square root and not an a'th root, right?

The fact of the matter is that almost everyone here knows enough calculus to find the area given the antiderivative. The hard part is finding that antiderivative. I can tell that you're new to calculus because the rather simple looking sqrt(r^2 - (r - x)^2) is actually quite hard to find an antiderivative for.

For example, here's the antiderivative of a very similar expression:

∫ sqrt(r^2 - x^2) dx = sqrt(r^2 - x^2)/2 + r^2*arcsin( x/r )/2 + C

I got this by looking it up in an integral table (there happens to be a physics text book in the office). How would you "trivially" do it, tehhappyboy--integration by parts?

Re: Re: Re: How much water in the barrel?
 

Ahh, you're right. I was just trying to save time. I would fill the tank and measure what it took to fill it compared to the depth that was in it first.

volume to fill/increased depth = original volume/original depth

solve for original volume and add to the volume added to fill.

There are two problems with your approach, onetime2.

First, it will only work if the volume you are filling has a constant horizontal cross-sectional area. For instance, it will work with a cylinder on its flat side or a rectangular prism. In fact, it will work with any polygonal prism with its flat side down. However, it will not work for shapes like a sphere, or a cone.
In this instance, we are talking about a cylinder with the curve "side" down and, in this instance, your idea simply will not work. Think about it...

Secondly, your method is quite involved and will disturb the very system you're trying to observe. You would find the volume of the water by filling the container with water. So, if you had that specific amount of water in the barrel for a reason, you've just that up by filling the barrel. Besides, filling a tank with a measured amount of fluid is a lot of work. Isn't it simpler to just figure out the volume as a function of fluid depth?

Yes, if I can find my textbook, problems like this are in there. It was like 8 chapters ago or something. All I can remember is that it definitely didn't involve integrals and such.

Ok, somebody here has to have a TI-89 floating around...

synic,

Since you seem to have a real world problem that needs an answer (not just a philosophical discussion).....

L=Length of the tank, R=Radius of the tank, H=Height of the water

V = L [ r^2 cos^-1((R-H)/R) - ((R-H) * sqrt(2RH - H^2)]

Note that when h=0, V=0 and when H=2R, V=pi*R^2*L as expected.

Q.E.D.

I have never understood how this happens. DDDDave, were you not reading this thread? The only reason why this thread went on a philisophical digression is because the question was already answered--and several ways, too! For instance, does this look familiar?
Secondly, QED is used to signify the end of a proof. What was your proof, exactly? That you couldn't be bothered to read the thread you're replying to?

Oops. He didn't.

As one of my math professors said: if you have to take the the sin of arccos (or any arc function aside from sin) you are convoluting the equation.

sin(arccos x) is just nasty!
sin(arcsin x) is however fundamental.

Correct me if I'm wrong here but a "cylinder with the curved side down" has a curved side up as well. If it's split in the middle aren't both sides mirror images of each other? Additionally, wouldn't a sphere split in the middle have each half as mirror images?

The purpose of most tanks is to hold fluid obviously. If they didn't need this water for some reason then why have the tank?Obviously if they can't or don't want to add water to it then they shouldn't. But if their goal is simply to figure out how much water the tank holds AND they plan on filling it anyway then it's a perfectly acceptable approach. Rather than criticize others for their answers, how about accepting that there are other paths that can be taken besides the formulaic approach?

You seem to be taking this question far too personally.

This is simply not the issue. You want to use ratios to determine the original volume of water but you can only use ratios if the change in water level is proportional to the change in water volume. If you were to think about the geometry of the barrel, you'd see that this is not the case. Thus, your method is insufficient...

If you knew the volume of the container and the amount of water used to fill the tank, you can simply take the difference and that would be the volume of water oringally in the tank.

However, and again, not only is it a difficult thing to fill a tank with a measured amount of water but you are also disturbing the system. I mean, what's the point of knowing how much water was in the tank if you're going to fill it to find out?
There's much to be said for determining the state of a system without changing it.


The only time I have taken this thread personally is when it became apparent that DDDDave didn't read the thread and that has very little to do with the question and more to do with a peeve of mine.

Instead of defending yourself from criticism, perhaps you should try learning from them?

First I'll address your bashing of DDDDave. There are plenty of threads around where people didn't read all the responses before responding themselves. There's nothing wrong with it. Although, I'm sure you must read every response to every thread before you respond since it would be ridiculous to criticize someone else if you are guilty on occasion as well, right?

As far as your criticism of my responses, I could care less if you criticize them and I did learn from the criticism and that's the point of a discussion board. Post something, if it doesn't make sense then there's a good chance someone else can help you work your way through it. Of course most try to do it in a constructive rather than arrogant and inflammatory way, but I guess everyone is entitled to their own style.

The original poster never added the criteria that the original condition could not be disturbed. That's simply a requirement you feel compelled to abide by.

I see your point about my method and will modify it to:

Let's say the water depth at the start is 4". Fill the tank to 4" from the top, measuring the amount of fluid added (which can easily be done if you know the rate of flow of the pump you're using and the time spent pumping). Then measure the volume of water needed to fill the remaining 4" of tank. Double it and add it to the volume you added to get the tank to 4" from the top.

And to answer an earlier question of yours:

As you pointed out, the method of calculating the volume was already outlined earlier in the thread before my first response. I was simply throwing out a mechanical method of determining it as an alternative. If you don't like it that's fine but you are not the arbiter of what's allowable.

Wanna know how _I_ would measure the volume of fluid remaining? (ask a physicist)

option 1:

I would find the total mass and then the mass of the container itself. From the difference, we would know the mass of the contents. Next, determine the temperature of the contents. Thirdly, determine the specific gravity of the contents. Lastly, using the relation between density and volume, determine the volume using the specific gravity (density) and mass.

:D

Not the least bit practical in this case.

option two:

1) Find the total volume of the container geometrically
2) fill the container until it is full, noting how much fluid it requires (using the same compound as the contents of the container).
3) subtract the volume required to fill the container from the geometric volume
4) account for a 0.05% difference due to material displacement (contianer)
5) have a beer and be satisfied that I know how much volume _was_ in the container

option 3:

1) insert a hose into the container
2) drain the contents into a number of vessels, adding the volumes as required


You see, simply put ... I am far too lazy to determine the volume of the contents using calculus. I am sure that the shape is very unruly and would be a pain in the ass to integrate - this said, I doubt very highly that the vessel is actually a perfect cylinder and is in fact a streched cylinder (more like a rectangle with rounded corners). Meh!

As an engineer by schooling, my first thought was this question has been around since the first underground tank was installed and there is a straightforward formula written down somewhere.

I googled 'volume cylindrical segment' and came up with what I posted earlier. I made no claim that this was my original thinking and derivation. I have no desire to debate the higher math involved. There was an excellent little graph included with the discussion that I have posted here.

http://www.chevelles.com/showroom/Da.../c4img2072.gif
If you highlight the graph the numbers show up a lot better. Sorry.


It's nice to be important,
but it's more important to be nice.