Question spurred by this thread. I feel like it may be an inverted cone (like a martini glass), but 1) my calculus is way too rusty to prove it, and 2) that doesn’t tell us what the angle should be.
Note: the answer is not a sphere (well, I’m pretty sure. feel free to prove me wrong). That’s the way to minimize surface area of a closed container. The question here is how to minimize the area where glass touches liquid in an open-topped container.
Is the container standing vertical? Do you mean to find the min S.A. of the inside of the container for a given volume of liquid?
I’m 99% certain it’s a hemisphere, I’ll do the math in a bit and be back.
I was 99% certain also, but for some reason I’m thinking I have to use the calculus of variations to get an answer and not first semester calculus methods 
Ok, for a unit volume I have a hemisphere giving a SA of 3.8 square units. The best cone will have a SA of 4.2 square units with a radius of 0.88 units which is an angle of ~55 degrees.
edit Used equations for a sphere and not hemisphere originally.
Bigger glass --> less contact liquid with glass.
(Assuming the shape stays the same)
A sphere is the most efficient shape for a closed surface. Assuming we have gravity, the opening should be at the top of the sphere and we want to maximize the volume to surface area ratio of the cone formed by connecting the center of the sphere to the edges of the hole in the sphere. My guess is that this can be done by making the opening equal to an arc of 90 degrees in length, with the glass being an arc of 270 degrees in length. The resulting surface is more efficient than a sphere for surface to volume ratio, and not surprisingly, looks something like the top part of a wine glass. The volume occupied by the cone formed by the edges of the hole and the center of the sphere is what makes this shape more efficient than a sphere.
A hemi-sphere can’t be the best shape because adding a cylinder on top of the hemisphere is going to give better efficiency. Taking infinitesimal slices of the hemisphere, there is a lower glass/liquid ratio at the top than at the bottom so simply extruding the top layer upwards is going to increase the total efficiency.
The difference between this problem and a normal area/volume maximization is that you get the top of the glass “free”. So your shape should be acting to take advantage of that as much as possible and get a large top.
Then coming at this from a purely intuitive basis, wouldn’t the most efficient solution be a conical container with the angle approaching 180 degrees (ie flat)?
How is that going to maximize volume?
No.
Half mile diameter glass, and you have a puddle, with the maximum amount of contact consistent with surface tension, and the shape plays no part.
For a given large diameter, flat, indented, or domed all give different values.
The statement is simply wrong.
Spherical is the geometric minimum surface/unit volume shape, but you have to decide what portion of the sphere is the optimum for your volume.
Tris
Wouldn’t that make the volume approach zero?
I’ve been struggling on this for an hour, and my calculus is damn rusty (31 years). This is a shot out of left field, but wouldn’t a cylindrical catenary be optimal? A catenary (the shape taken on by a rope strung between two points) is defined by
y = r* cosh(x/r)
where r is the radius of the opening.
I don’t think so. If your argument was correct, extending the height of the cylinder to infinity would maximize the efficiency. This would effectively eliminate the hemisphere, resulting in a cylinder. A sphere of equivalent volume would be more efficient.
It seems pretty clear to me that using a sort of test-tube shape with a hemispherical bottom will reesult in less glass-to-liquid contact than a spherical surface, since you’ve got a lot of liquid open to the air on top instead of touching the spherical walls (“free”, as someone said above). The volume of a sphere is 4/3(pi) r^3, and its surface area in contact is 4 (pi) r^2. The equivalent volume in a hemisphere + cylinder will have the hemisphere filled, but with a height of h = (2/3)r in the cylinder. Thec surface area of liquid touching glass will be 2 (pi) r^2 + (4/3) (pi) r^2, or (2/3) (pi)r^2 less than the sphere.
I think The Librarian nailed it. A big glass is always better.
Whatever the best shape is, don’t forget surface tension. That gives you the ability to slightly overfill the glass to form a shallow dome above the top.
By the way, by cylindrical catenary I meant rotated about the y axis. A catenary rotated about the x axis is the minimum surface area for a solid with 2 open circles at either end. It’s the shape that a bubble will take on when attached across two parallel concentric rings. The catenary looks kind of like a parabola. The glass would look somewhat like a whiskey sour glass, only wider.
It isn’t that a bigger glass is always better, it’s just that you don’t want a glass that is too big. A gigantic spherical glass is equivalent to a flat surface. And a flat surface is the worst possible shape. Actually, a flat plane covered by another flat plane is the worst possible. Assuming we’re ignoring capillary action here.
Actually, I just realized that’s not true. You could put little divisions in that sandwich of two planes and get some sort of star shape. So, given capillary action, the worst possible shape would be a series of glass tubes that suck the liquid up through capillary action. Millions of tiny pipettes around a reservoir. The size of the pipettes would depend on the exact physics of the liquid/glass interface.
Back to “best”. I don’t see how you can beat a hemisphere. Why exactly does introducing a cylinder on top of the hemisphere improve things?
My calculations say that the surface area of spherical glass holding a unit volume is (36 pi)[sup]1/3[/sup], about 4.836. For a hemispherical glass, it’s (18 pi)[sup]1/3[/sup], about 3.838 (as reported by Snarky_Kong).
For a cylinder mounted on top of a hemisphere I find the optimal height of the cylindrical part to be zero. The volume help by the glass is V = 2/3 pi R[sup]3[/sup] + pi H R[sup]2[/sup]. The surface area is S = 2 pi R[sup]2[/sup] + 2 pi H R. Solving for V = 1 and substituting H into the surface area yields S = 2/3 pi R[sup]2[/sup] + 2/R. The surface area is minimized when R = (2/3 pi)[sup]-1/3[/sup], which means that H = 0.
Darryl Lict, what is the surface area of a glass holding a unit volume, given your shape?
:smack:
Yeah, that’s what I get for trying to intuit anything before lunch.