What is the angle liquid will pour out of a glass?

[playboy]

Hi, this is not a homework question. Been out of school for way to long and can't remember my physics.

I am developing something and I need an equation.

What is the angle at which liquid will pour out of a glass? The input I think would only be the % full the glass is. I do not think the size or shape of the glass would matter. The output would be the angle at which the glass must be tilted to start the output of water.

Thoughts?

Thanks!

 

[tiny-tim]

Welcome to PF!

What is the angle at which liquid will pour out of a glass?

Hi playboy! Welcome to PF!

I think the angle depends on the shape of the edge of the glass …

the "escape" direction of the water will be whatever direction it was going in when it lost contact with the glass.

 

[playboy]

thanks. I am not interested in the escape direction, only at what angle the liquid will start to expel from the container. Not looking for something exact just a ballpark.

 

[glueball8]

Do you mean that it doesn't come out at very very small angle because of water tension. And what angle would overcome that water tension?

 

[playboy]

Not sure if that's what I mean. I am talking in simple terms.

There is a glass of water that has a radius of r. The glass has a height of h. Water is poured into the glass to a height of hw. You pick up the glass and tilt it to pour the water out. At what angle due to gravity does the water pour out? Can anyone derive a formula for that? Thanks!

 

[Andy Resnick]

I can't imagine there will be a simple formula that is reasonably accurate- there are too many variables, including the detailed geometry of the lip of the glass.

The only variable to solve for is the velocity of the fluid at the lip.

I'm sure if some simplifying assumptions are made- constant volume flux out, constant 'pour' angle, cylindrical glass with square lip, an experiment would reveal reasonably simple relationships.

 

[maverick_starstrider]

I'm afraid it would depend on the shape of the glass and the angle you're tilting it while pouring (in addition to how full it was and actually how pure it is, which determines surface tension (for example a juice with lots of sugar has a notably different surface tension then tap water))

 

[Nick89]

Aren't we over complicating things a lot here? It seems to me all the OP is asking is at what angle the water will reach the edge of the glass.

Maybe I'm missing something (is the fact that the glass is round important?) but it seems to me it is basically a geometrical problem:

I haven't found a solution yet so quickly though, so maybe I am missing something...

 

[cosmo123]

I think this one of the answers, but someone will have to check this. I assumed that you can treat the glass as a simple box.

http://img23.imageshack.us/img23/2062/glassp.png

If the glass is more than half full, then the surface will sort of rotate around the centre of the glass. From the right angled triangle that means θ=arctan(r/(h-x)).

Ignore the second diagram because i just realized it doesn't work for a circular glass.

 

[Nick89]

I suppose x in your diagram is the initial height of the liquid when not tilted?

Are you sure this height remains the same (in the center) when you tilt the glass? That simplifies things alot, but I'm not so sure if that is valid...

 

[cosmo123]

I suppose x in your diagram is the initial height of the liquid when not tilted?

Are you sure this height remains the same (in the center) when you tilt the glass? That simplifies things alot, but I'm not so sure if that is valid...

I just got out a glass to try it and it looks like its at least close to true. I probably should have double checked it was true before starting to work this out.

 

[Danger]

A fellow showed me something really cool when I was still beginning as a bartender. We used a 2-oz. shot glass to pour 1-oz. shots. Hold the glass at 45° and speed pour until the booze reaches the edge. It yields exactly 1 ounce, about 4 or 5 times faster than trying to fill a vertical glass precisely to the 1-oz. line or a 1-oz. glass to the brim. I never got anywhere near as fast as he was; I once watched him pour 43 rye/Cokes in one minute flat, and as a team we built 50 B52's in less than 2 minutes. Unfortunately, he was also the sloppiest bartender on the face of the planet and it was my job to clean up after him.

 

[cosmo123]

On second thought, I am sure its true. Imagine that the glass was sealed at the top, so no air could get in or out. You'd expect it to act in exactly the same as if it were open, until it were about to overflow, so the volume of air, like the volume of water, must stay the same. This means that for the section at the top, with the two triangle shaped bits, one with the top bit of the water and one with the air, must be symmetrical. This means that the bit down the middle must alwats meet the centre of the glass at half-way. If it were lower, or higher, then there would be a change in the volume of air, and so a change in the volume of water, which obviously can't happen.

Thats my reasoning at least

 

[TurtleMeister]

A = \arccos \left(\frac{c \left( \sin B \right)}{\sqrt{a^2 + c^2 - 2ac \left(\cos B \right)\right)}}

where:

A = the angle of the glass when the water reaches the spill point
B = the angle of the container wall on the spill side (when the glass is level)
c = the radius of the container at the water level in the direction of tilt (when glass is level)
a = the distance from water level to the top of the wall of the container on the spill side (when the glass is level).

I used the law of sines and the law of cosines to arrive at this. But anyone please verify that it is correct.

Edit:
I think this solution only works if the water level does not reach the bottom of the container when tilted to the spill point.

 

[Andy Resnick]

Aren't we over complicating things a lot here? It seems to me all the OP is asking is at what angle the water will reach the edge of the glass.

Maybe I'm missing something (is the fact that the glass is round important?) but it seems to me it is basically a geometrical problem:
<snip>

I haven't found a solution yet so quickly though, so maybe I am missing something...

I think this one of the answers, but someone will have to check this. I assumed that you can treat the glass as a simple box.
<snip>

If the glass is more than half full, then the surface will sort of rotate around the centre of the glass. From the right angled triangle that means θ=arctan(r/(h-x)).

Ignore the second diagram because i just realized it doesn't work for a circular glass.

I don't see why these are candidate solutions- the situation involves the flow of fluid, not a static equilibrium. Or maybe I misunderstood the OP...

 

[maverick_starstrider]

We have to different interpretation, I believe me and Andy thought the OP was asking if you start pouring liquid out of a jug what angle will the flow out of the jug make.

 

[Nick89]

The output would be the angle at which the glass must be tilted to start the output of water.

I am not interested in the escape direction, only at what angle the liquid will start to expel from the container.

Seems to me he is talking about the angle of the glass.

 

[Borek]

I am a latecomer to the thread, but it looks to me like playboy was asking the simplest thing - pure geometry, no physics.

 

[Nick89]

I can now confirm that cosmo's answer is correct, here is why:

Since the volume of water should be the same before and after tilting, we can simply calculate it in both conditions, and derive an expression for the angle from that.

Symbols used as in the picture. The initial water height when not tilted is H (not shown in image).

The volume when not tilted is of course just:
V = \pi r^2 H

When tilted at an angle, we can basically form two regions, one rectangle and one triangle (actually: cylinders). The 'rectangle's volume is just \pi r^2 (h-x), while the 'triangle's volume is just half the volume of a cylinder: \frac{1}{2} \pi r^2 x

Of course, this must be equal to the volume when not tilted, so we have:
V = \pi r^2 H = \pi r^2 (h-x) + \frac{1}{2} \pi r^2 x
H = h - x/2

Finally we know that x is equal to:
x = \frac{2r}{\tan \theta}

So:
H = h - \frac{r}{\tan \theta}

Or, solving for the angle:
\tan \theta = \frac{r}{h-H}Still, this method too breaks down when the initial height H is below a certain level (h/2??), because the volume calculation is then no longer correct (since you cannot form the 'rectangular' part anymore).