I’m ssuming that we have a fixed volume of liquid we want to store, so we can’t just make the glass really big. The solution is a hemispherical glass, with the liquid filled to the brim. A simple way to prove this:
Take your glass, of whatever shape, and fill it up. Now take another identical full glass, and turn it over upside down, with the rim against the first glass. This doubles the volume, and also doubles the glass area, so the ratio of the two remains the same. But it also turns the problem into a completely-enclosed surface, with no “free” area. We already know that the solution to that problem is a sphere, so the problem reduces to finding a shape that, when doubled, forms a sphere. But of course, that’s a hemisphere.
What I don’t understand is that there should be another variable defining the “droopiness” which would be the length of the rope, which should always be greater than 2α. Perhaps the rope length is unitary. This makes sense, as as α gets smaller, the surface area gets smaller too.
As stated earlier, the catenary is defined as follows:
Now, all I have to do is integrate the volume of the cup, divide it by the surface area and find the global maximum.
I’m not at all confident that any of this is correct, but if I ever get the energy to figure it out, I’ll report back. I’m hoping that some mathematician will drop in tell me the damn answer.
I think you guys are right. I was kind of caught up in the calculation of the volume to surface area ratio of a rotated catenary as a calculus problem.
What’s the name of that shape that has a finite volume but infinite surface area? I remember this from calculus 20 years ago, but can’t remember the name. Anyway, I’ve decided that shape would be the worst possible shape for minimizing glass/liquid contact.
However, I’ve thought of another wrinkle. A hemisphere is only the most efficient shape when filled to the brim. I suppose there isn’t a definitive answer for “most efficient shape no matter how much liquid you put in”, since many shapes would vary in efficiency depending on how full they are. Even a hemisphere approximates a flat plane when it is really large relative to the amount of liquid.
No, it was shaped like a trumpet but with an infinitely long stem that asymtotically approached zero width. The volume equation gave a finite value, but the surface area gave an infinite value.
There are myriad such shapes. One candidate for your trumpet shape would be a simple surface of revolution of y=1/x, but you could also have infinite disks, or irregular shapes comparable to either of those, or combinations of them, or almost any fractal, with all sorts of weird shapes.
I don’t know jack about calculus, but I do know a lot about how to drink. Traditionally, the “correct” volume for a wine glass is to fill it only to the widest part of the globe, creating a virtual hemisphere. You could even think of the bottom hemisphere part as the true vessel, and the unused glass parts on top as a tool for holding the glass.
If you are interested in the math behind this problem, google for “Queen Dido’s problem”, which is related to a number of areas of mathematics, such as the calculus of variations and soap bubbles.
NO. The hemishpere holds the wine. The stem is the tool for holding the glass. The quasi-cylindrical part above the hemisphere is a carefully tapered duct to concentrate & direct the aroma to your nose. Never touch the bowl while handling wine.
I think Chronos is right but am not confident I see it clearly enough.
Wouldn’t a soap bubble on the surface of a bath take on the shape we want? That is, wouldn’t it minimize its surface area while constraining its interior volume?
Another way to look at Chronos’ explanation is this:
We know that the minimal surface area for containing a given volume is a sphere. So, why not make it a sphere?
Well, we also know that as you get “above” a hemisphere, you’re losing volume to surface area because the sphere is “closing in on itself”. So, you know you want to stop it before then.
For the same reason, you can conclude that you don’t want to stop “expanding” your sphere from the bottom until you get to a hemisphere.
At least that’s how I see it. Maybe it helps someone.
Any glass shape you choose can be turned into a closed volume by joining the open ends of two of them. That closed volume will have exactly the same surface/volume ratio as the individual glasses. 2x the volume and 2x the surface area.
If the shape you choose is non-hemispherical, then the closed volume will be non-spherical, and will therefore have a higher S/V ratio than the spherical version. The individual glasses will therefore have a higher S/V ratio than the hemispherical version.
I’m still thinking (not very effectively) about surface tension. I may be on the wrong track.
You can get a bit of convexity on the open surface, so for a given volume you could use a slightly smaller hemisphere. Wouldn’t this bonus vary with the size of the opening, meaning that as you increased the size of the glass/ liquid* you’d lose a bit of the volume-holding surface convexity? Wouldn’t that mean that at some small volume of liquid there would be a benefit to having just that little bit less of the sphere to keep the opening a bit smaller?
*[sub] for a given liquid, material, temperature, pressure etc[/sub]
With all due respect to Chronos, who knows much more math than I, I’m not sure I agree. In some cases it is beneficial to “cancel out” part of the equation, but in other cases, this eliminates an essential premise, resulting in a simpler, but different problem, as I believe it does here.
In crude verbal terms, I could say this:
Using Chrono’s argument of reduction to two hemispheres (= 1 sphere) then moving the dividing plane on the spherical to create a spherical cap and a spherical goblet decreases the “efficiency” of BOTH parts, relative to a hemisphere, but since the total glass area and total volume remain constant, this is not possible. If one becomes less efficient, the other must become correspondingly more efficient to compensate, and achieve the same totals.
However, though I think it might be on the right track, I could poke holes in that exact wording myself, so I ran an easy test case as a reality check. If I were on the right track, it would be a disproof by example.
r = (h[sup]2[/sup]+r[sub]Base[/sub][sup]2[/sup])/(2h)
A = 2πrh
V = hπ/6 (r[sub]Base[/sub][sup]2[/sup]+h[sup]2[/sup]
We have already calculated r for a 2-unit sphere (= two 1-unit hemispheres) as ~0.7816
so let’s make the cut a little “above” the equator of the sphere at r[sub]Base[/sub] =0.75
The above cap equations need h, which we calculate via θ, the angle “above the equator”
θ = arccos(r[sub]Base[/sub]/r) = arccos(.95957) = 16.348°
h = r-sin(θ) = 0.7816 - sin (16.348°) = **0.5001 **
A = 2πrh = 2π (.7816)(.5001) =2.456
V = h(π/6)(r[sub]Base[/sub][sup]2[/sup]+h[sup]2[/sup]
= 0.3073
efficiency = 0.3073/2.456 = 0.1251
We obtain the volume and surface of the “goblet” by subtracting the cap from the whole sphere
A= 7.677 - 2.456 = 5.221
v= 2 - 0.3073 = 1.7927
efficiency = V/A = 0.3434
Unless I’ve blown the math(very possible, calculating on the back of a napkin using online tables)a spherical goblet subtending 212.7° is more efficient than a hemisphere. There is no special significance to this angle, it was just a post-facto result of choosing a ‘round’ value of r[sub]Base[/sub] that was slightly lower than the r of the unit hemisphere, for easy comparison. In fact, I originally meant to guestimate a value corresponding to θ < 10°, but I “missed” – so much for my guesstimation! (θ < 1° would have been better, but not a “round number”)
I would very much appreciate someone double checking my figures. Lunchtime is over.
BTW, a way of envisioning how this could be true is this:
A sphere is the most efficient closed body, but some part of it (the “wider” parts) are more efficient than others. We are required to accept two less efficient regions to achieve closure at “top” and “bottom”. If I add more of the “efficient” segments to the hemisphere, the average efficiency of the entire glass should increase. In fact, as some people intuited, adding a cylindrical extension to the hemisphere would increase the efficiency – but only to a certain length. Arguments taking it to infinity are reductio ad absurdum.
In the problem posed we are NOT required to achieve closure (Indeed, closure is forbidden for a usable glass, which must be filled/emptied) so not only are we not obligated to accept the inefficient “top” of the sphere, but we get a “free surface” bonus for not using it!. (we still need the spherical “bottom” to hold the liquid against gravity)
Even if we require a spherical glass, no one has offered any argument that filling it halfway is optimal, and it can be (has been?) shown that it isn’t.
