Can a glass of water be filled to its edge?

[russ_watters]

Dave’s diagram first diagram capturs my thought. There is water “over” the rim, when the rim is exactly flush with the ocean’s surface.

https://www.physicsforums.com/attachments/1716258259557-png.345644/

That diagram does not show elevations. Draw the rest of the Earth, with lines tracing the elevations.

 

[davidjoe]

I can’t do a better job than Dave’s, Russ. That pan’s rim would define a “flat spot” on a sphere, unless water does assume the contour of the earth, which I thought we were all in agreement on.

And if we raised that pan up out of the ocean, it’s hard not to envision water running over the rim.

 

[hutchphd]

I thought we were all in agreement on.

I, too, assumed that this was stipulated by all. Now I am lost in the arguement. Can we just agree that there are local methods to differentiate a radial field from an absolutely uniform one? What else is important here (in fact what is the question??). The OP never indicated the original motivation for the question.......

 

[russ_watters]

I can’t do a better job than Dave’s, Russ.

I told you what it's missing and you didn't even try to add it. This does not help convince me you are serious.
[Edit]
Heck, you can also tell me the numbers: what is the elevation at each edge and the center? After you lift the pan 1m, what is the new elevation of each side and the center?

That pan’s rim would define a “flat spot” on a sphere, unless water does assume the contour of the earth, which I thought we were all in agreement on.

The Earth's surface is 70% water. Obviously the contour of the ocean is the contour of the earth.

 

[davidjoe]

I told you what it's missing and you didn't even try to add it. This does not help convince me you are serious.
[Edit]
Heck, you can also tell me the numbers: what is the elevation at each edge and the center? After you lift the pan 1m, what is the new elevation of each side and the center?


The Earth's surface is 70% water. Obviously the contour of the ocean is the contour of the earth.

Usually I think pictures are worth a 1,000 words. But here I think words might do a better job. It’s agreeable, right, that because the earth is a sphere, that even if we used a small wire cheese cutter to cut a “slice” off of the surface of the ocean, it would be incredibly thin but it would be “dome shaped”, an invariable trait of spheres.

The “pan” simply replicates this on a larger scale. The water over the center of a pan will be deeper as shown in Dave’s drawing, just like the “tallest” part of that “dome” will be the “middle”.

 

[hutchphd]

The thickness of such a slice is known as the sagitta (a term familiar to mirror grinders. @davidjoe you have not said what motivates the question.......

 

[russ_watters]

Usually I think pictures are worth a 1,000 words.

Then draw the picture I describe.

But here I think words might do a better job.

Clearly not since the conclusion you have reached obviously contradicts reality.

The water over the center of a pan will be deeper as shown in Dave’s drawing, just like the “tallest” part of that “dome” will be the “middle”.

Water flow has nothing to do with depth, and no, "deepest" and "tallest" are not the same thing.

You seem to be actively avoiding the correct answer. Last chance. Post the diagram I asked for or directly answer the questions that followed or I will lock the thread.

 

[davidjoe]

The thickness of such a slice is known as the sagitta (a term familiar to mirror grinders. @davidjoe you have not said what motivates the question.......

Fair question, admittedly totally inapplicable knowledge to me as a greying attorney, but I’m curious what the true contour of the surface of still liquids really would be, and why, ignoring factors unique to any specific liquid, like viscosity or surface tension.

I had the thought that it might actually be convex in a flat pan, which means you could never fill up a glass totally, because the liquid would run over the edge, first.

 

[DaveC426913]

If the pan were to be elevated, still full of water, such that only the center point of its bottom touched the surface, would the profile of the water in the pan change, possibly being inverted from convex, following the earth’s contour, to concave, for the reason that gravity pulling the water downward more forcibly in the middle of the pan, displaces water not being pulled down as strongly toward the pan’s wall, because that part of the pan is further from the source of gravity, and experiences weaker gravity.

Well, now we're talking about a very different scenario: the height difference between the centre of the pan and its edges is so great that there is a measurable gravitational gradient.

I think Russ is correct: diagrams with approximate measurements (we're talking scores of miles now, instead of feet) will mitigate a lot of the communication confusion we've been having.

 

[DaveC426913]

I had the thought that it might actually be convex in a flat pan, which means you could never fill up a glass totally, because the liquid would run over the edge, first.

You are talking about measurements in a setup that simultaneously spans

• the millimetres of a liquid's meniscus in a wine glass, to
• a baking pan, miles wide enough that the curvature of the Earth comes into play.

You may not be able to get away with loose descriptions and approximations much longer...

 

[davidjoe]

You are talking about measurements in a setup that simultaneously spans

• the millimetres of a liquid's meniscus in a wine glass, to
• a baking pan, miles wide enough that the curvature of the Earth comes into play.

You may not be able to get away with loose descriptions and approximations much longer...

Your first diagram, with ships, the pan with a uniform rim and the curvature of the earth is 100% what I would have drawn, except I might have extended the rim to be flush with with the surface.

I can’t pass a test of generating a diagram on my iPhone, and that’s why I was glad that you did. If the conversation must end because I can’t draw diagrams on my phone, even though it would be the same diagram that you did, then I suppose it is my loss for not being more technically adept, and if it gets locked, I still appreciate the input.

 

[davidjoe]

I told you what it's missing and you didn't even try to add it. This does not help convince me you are serious.
[Edit]
Heck, you can also tell me the numbers: what is the elevation at each edge and the center? After you lift the pan 1m, what is the new elevation of each side and the center?


The Earth's surface is 70% water. Obviously the contour of the ocean is the contour of the earth.

I can answer the question under your edit. If we had a pan that was 250 miles wide, for simplicity, then 100 of these pans would touch each other, along the equator.

Each pan would represent 1/100 of 360 degrees, or 3.6 degrees. There is a trig function to determine the height of a line that extends 250 miles at 3.6 degrees, then cut it in half, because half of the dome is going back down.

It’s also doable by right triangle computation.

 

[russ_watters]

If the conversation must end because I can’t draw diagrams on my phone, even though it would be the same diagram that you did, then I suppose it is my loss for not being more technically adept, and if it gets locked, I still appreciate the input.

Here you're not just saying you can't draw diagrams, you're saying that even if you could you wouldn't draw the diagram I asked for(edit: we cross posted and the diagram you described in your last post is also wrong/not what i asked for). Nor are you answering the questions I asked. You aren't trying to move towards the answer, you're actively avoiding it. That's why the thread is now locked.

For the record, the answers to the questions I asked are:

-The elevation of the water at the edges and center of the pan are the same: zero meters above sea level (or 6378km from the center of Earth).

-If you raise the elevation of the pan by 1m, the elevations at sides and edges are now 1m.

So the water at the center of the pan is not "higher" than at the sides.

Caveat on this answer that it only works for relatively small pans of a few km since beyond that the sides aren't rising directly vertically. But you couldn't have a flat bottom with a much larger "pan" anyway (like an ocean basin).

There's another caveat/catch as the altitude gets large, but I'm skipping that.

[edit] Thread will remain locked for the reasons described above.