Interesting question: pouring water out of a glass

[NickThissen]

Here's an interesting question I came across on another forum.

Suppose you have a cylindrical glass with a height h and diameter 2r (r = radius), filled with water to a height of H < h.

At what angle would you have to hold the glass before it pours out due to gravity?


Let's ignore any effects such as surface tension, then (I think!) the question is basically a geometrical question:


I'm not sure yet if the fact that the glass is cylindrical is important, but it might be.


I have tried coming up with a solution for this but failed... I was never good at geometrical stuff lol. The main problem I had was that the height of the water H does not come up in any equation I found...

I am starting to fear an analysis of the height of the waterlevel (when glass is tilted), which will probably involve the volume of the water, will be required. The volume is of course pi r²H, but if we can calculate the volume again when the glass is tilted under some angle theta, we might be able to find a relation between theta and the height of the waterlevel... Unfortunately, such a volume calculation appears to be extremely complicated at first sight (remember that the glass is cylindrical, not square!)...

I thought this was an interesting question, and other people at the other forum seem to think there is no easy relation. I think it should be possible however to find a relation, however complicated

Thanks for any input!

[jemidiah]

Find a depth function D(x,y) which gives the distance through the cylinder you would travel if you shot through it at the indicated point, looking side-on as in your picture. Integrate over the function in 2 dimensions on the area where water should be for a given theta piecewise--the limits of integration wouldn't be bad in Cartesian. This integral gives the volume of water as a function of theta. Since volume and height are equivalent information, invert to find theta as a function of height.

The trick is to find the depth function D for a given theta. I suppose it's not too difficult after all. Find one for theta = pi/2 (straight up and down), and rotate that function in the plane through a rotational transformation about the lower right pivot. The straight up and down version would be straightforward--it would just be a function giving the length of parallel secants cutting through a circle, which can be derived without much trouble geometrically.


None of the individual pieces I've listed seems overly complicated, but putting it all together would be a headache, and would result in some awful piecewise solution with trig functions flying around everywhere.

I thought of two other solution methods. The first is to brute force the volume through an integral in cylindrical coordinates, but that requires very complicated boundary equations. The second is to use dimensional analysis, but the angle is unitless and could show up in an arbitrarily complicated way. The above seems about as simple as you'll get, though of course there may be some unforeseen and magical shortcut .

[NickThissen]

Well, the guys at the other forum think they came up with a solution:


In the second image, the blue line is the waterlevel, had the glass not been tilted, while the red line is the current waterlevel. They seem to think that the red line will always cut the blue line exactly in half, so you end up with one right triangle, and:
tan theta = r / (h - H)


I'm not so sure about this answer though... I don't know how to justify the assumption that the blue line is cut in two equal pieces by the red line... That just doesn't seem true all the time, does it?

[wy125]

hi. I worked out the problem without that assumption and got the same answer so i guess that assumption is correct at least for the case when the cup is just about to spill.

[NickThissen]

Can you show me how? I'm interested to see

[jemidiah]

Conservation of volume should justify that assumption. If you rotate the red line up in your second picture, you'll add volume to the bottom triangular section and remove it from the top section, which would increase the volume of the tilted water compared to the untilted, which violates conservation of volume. Note that the volume in the two triangular sections is equal because my depth function is symmetric about a vertical line through the middle, which the triangles happen to also be symmetric around.

Clever solution! However it seems to break down at low water heights. My depth function approach would work down there, though maybe there's another nice trick.

[wy125]

Okay, take a look at my attachment. It's a scan of the illustration I used (i know it's crude but I hope it's clear enough)

Definitions:

h - the height of the cup

R - the cup radius

theta - the angle of cup rotation

L0 - the intial level of the water when cup is flat

L - the water height when rotated at an angle theta


V - the volume of water in the cup


A - Area of cross section of cup (pi * R^2)


(I'll explain s and d from the illustration later)

Okay so as Jemidiah points out the volume of water in the up remains constant.

This is given by:

(1) V = A L0 = constant


Now we need to figure out the water level for an arbitrary rotation theta.

To do this we can compute the volume of water again. The way I did this was to say okay there is a nice rectangular section (really cylindrical of course) and a triangular section (really a cylindrical section that is cut diagonally in half). The volume of the rectangular section is easy, that's just A(s - d). The volume of the triangular section is clearly 1/2 the volume of the complete cylindrical section, which means that is 1/2 Ad. So the volume of water is:

(2) V = A (s - d) + 1/2 A d

Setting (1) and (2) equal gives:


A L0 = A(s-d) + 1/2 A d

dividing by A and simplifying gives

L0 = s - d/2

The water level, L, is related to s by

s = L / sin(theta)

And d and R are related by:

d = 2R/tan(theta)

Substitution gives:

L0 = L/sin(theta) - 2R/tan(theta) * 1/2

L0 = L/sin(theta) - R/tan(theta)


solving for L gives:

(3) L = sin(theta)[L0 + R/tan(theta)]

this gives the water level given theta.

Now we need the condition of the cup just before spilling. This is the case when the vertical projection of the cup height equals the water level, L. The horizontal projection of the cup height is:

h' = h sin(theta)

h' is the horizontal projection.

Setting this equal to L gives:

h sin (theta) = sin(theta)[L0 + R/tan(theta)]

solving for tan(theta) gives

tan(theta) = R/(h - L0)

And as Jemidiah points out is no good at low water levels.

[NickThissen]

I see, I should have simply tried the volume calculation, as I thought it would be much more involved than this... But of course, it's just a cylinder + half a cylinder...

You are also making it much more complicated than required. If you simply set s = h at the beginning, then you start at the condition that water is at the edge, and you arrive at the same answer:
(using H as h0, the initial water level)

V = pi R^2 H = pi R^2 (h - d) + 1/2 pi R^2 d
h - d/2 = H (*)

Since d = 2r/tan theta, we have
h - r/tan theta = H

or, solving for tan theta:
tan theta = r / (h - H)



(*) Actually, I think you were already done here. Doesn't this already tell us that the assumption we talked about is true?

But good job!


However, this method too breaks down when the initial water height is below a certain level (half h??), since then d > h, and the volume calculation is no longer correct.

I think we may need some calculus to calculate the volume of water in that instance, because you can no longer form 'easy' regions. It's more like a puddle at the bottom of the tilted glass, quite a complicated shape.

[wy125]

Yes you're right about setting s=h. Yeah it seems to be the case that the assumption is true. the step where we say that the volume of the cylindrical section with 1/2d is the same as the triangular section is making the same assumption actually so I guess I lied a bit about solving it without that assumption.

[NickThissen]

No, I don't think so actually. I think your method proves that the assumption is right.

The volume of the upper 'triangle' is always going to be half the volume of a cylinder with the same height, right?. How is this the same as our assumption?

[wy125]

Well why did they say that the line always goes through the midpoint? Visually? If so then I agree with you it's not the same assumption.

[NickThissen]

Yes, it was just an assumption out of thin air basically. But you proved it with the volume calculation I think!

[jemidiah]

The assumption isn't as "out of thin air" as you might at first think. Get a glass of water, fill it most of the way, and put your finger at the midpoint of the water level line. Tilt the water and (if the glass has certain properties; a cylinder works) you'll see the water level line rotates about your finger. From there the midpoint idea arises naturally, followed by the simple solution given.

I haven't given up on a decently simple solution for the other case. It would be some modification of the original formula, naturally; perhaps a quadratic in tan(theta)?

[Shaggy Hiker]

It seems to me that the assumption will only work for cases where the red line doesn't intersect the bottom of the glass.

[wy125]

It seems to me that the assumption will only work for cases where the red line doesn't intersect the bottom of the glass.

You're right. That solution breaks down when the water level isn't sufficiently high.

[NickThissen]

The general method (calculating the volume of water and equating it to the same value in non-tilted state) should still work though. The only difference is that the volume calculation is no longer easy...

[krtxmrtz]

You can even try a brute force approach and calculate the water volume of the tilted cylinder by integrating the limiting surfaces (I've taken time to do it) though there's no point in doing it this way if it can be more easily.
At low water levels the solution is different since the limiting surfaces change.

[jemidiah]

Using basically the approach I talked about above gave the following integral for the volume of water at low levels:

V = Integral from x=r-h*tan(theta) to x=r of the integral from y=0 to y=(x-r)/tan(theta)+h of 2*sqrt(r^2-x^2) dy dx

I don't trust myself to do the algebra and I don't have a computer algebra system handy, but there we are. I doubt this will simplify to anything terribly nice but of course you never know. If anyone's able to carry out the integrals we could at least see if the answer is simple. If it is, there's likely a physical trick to derive it easily. If it's not, there likely isn't.


This result was obtained by integrating over my depth function on the green shaded region. I found the equation of the line to be (x-r)/tan(theta)+h = y, with the line passing through the points (r-h*tan(theta), 0) and (r, h), which is where the limits of integration come from.

[NickThissen]

The integrals itself aren't really difficult. It's basically three integrals, one over y and one over x, but the one over y is easy since there is no y in the equation (right??). Of the other two, one can be done by substituting u = x^2, and the other is with a sine substitution.

The trouble is the limits of course... It's really hard to not make a mistake manually lol... So I let maple do it, and it spit out this for the volume:

(No, that's not a vector, note the minus sign at the end of the first line lol, it was too long to display in one line)

Unfortunately, we also need the initial height h0 before we can solve for theta... Since V = pi r^2 h0 should be solved for theta.

[jemidiah]

Yup, there's no y in the depth function (that is, the integrand). Does Maple have a way of simplifying algebraic expressions? We'd have solve for theta in the resulting volume equation, but if it's been as simplified as we can do, the inverted form wouldn't be any better--which pretty much precludes the possibility of there being a single simple physical trick to find the result.