First of all, a quick check indicates that you miscalculated the “efficiency” of your hemisphere (your calculation is r/A rather than V/A).
Second, and more fundamentally, your definition of “efficiency” is not unitless–it has, in fact, units of length. There’s nothing intrinsically wrong with this (although it is a bit misleading, as “efficiency” normally indicates a unitless quantity). However, what it does mean is that, using your definition, larger volumes of fluid will have a larger “efficiency,” even when enclosed by the same shape.
This is easy to see by looking at just the simple sphere, with volume of 4π/3 r[sup]3[/sup] and area of 4πr[sup]2[/sup]. In this case, your “efficiency” is V/A = r/3, so a basketball filled with water has a larger “efficiency” than a ping-pong ball.
So… it’s not suprising that your larger-volume spherical section has a greater “efficiency” than the smaller-volume section. But the “efficiency” will be greater still if you reform it into a hemisphere of larger radius.
When I said “closure” was forbidden, I mean that “closing the sphere” is forbidden. In the topological sense of a closed surface, a Klein bottle (for example) obviously still works as a drinking glass.
Question for Chronos: as I understand it, 3D physical models of a Klein bottle are not Klein bottles, correct? Is there a proper term for these common objects?
(Man, I can’t keep coming back here every 15 minutes – oops, disproven again!)
(On preview, it looks like zut pointed out some of the following.)
V(1 unit)/A for the hemisphere should be 0.2606. You mistakenly used the radius where you meant to use volume.
The volume of a spherical cap has a “3” multiplying the r[sup]2[/sup][sub]base[/sub] term. This gives V[sub]goblet[/sub]=1.4926.
You didn’t scale the answer to a unit volume goblet. V/A will scale linearly with the radius, so (1.4926)/(5.221) must be scaled by 1/(1.4926)^(1/3), leaving: V(1 unit)/A for the goblet = 0.2502
I’ll have to recalculate it properly as Volume/Surface , renormalizing the goblet to a unit volume for the goblet. That would make the two figures comparable.
I can certainly prove that it’s not optimal to use more of the sphere than a hemisphere. Suppose our glass extends above the equator of a sphere. We could cut off the portion of the glass above the equator, and replace it with a cylinder of the same height. This would leave the surface area of the glass unchanged, but would increase the volume, and thus increase the efficiency. It would also increase the size of the open surface, but we don’t care about that.
As for the Klein question, I think that the 3-d things one sees are properly an embedding of the Klein bottle, but IANAM, and might have the terminology off.
Fiddling with my model of the spherical goblet, I found some minor errors (e.g. I had h as “r-sin(θ)”, but it should be “r(1-sin(θ)”. It now looks like you right about the hemisphere being the optimal filling of a spherical goblet.
However, my reasoning kept leading back to the inverted cone (which I wrongly thought we’d ruled out). Another “sanity check” calculation (on a spreadsheet this time) shows that it can exceed the Volume/Surface ratio of the hemisphere. If I may beg your indulgence once again:
To better explore various “shapes” of cone, state r in terms of h (r=kh) – then
for the sake of an example, take the case k=1 and solve for h=r and v=1
V: πh (kh)[sup]2[/sup]/3 = πk[sup]2[/sup]h[sup]3[/sup]/3 = 1
h = r = cube root (3/k²π) = cube root (3/(1²)π) = cube root (3/π) = 0.9847450218
Surface = πhr = 3.0464738927
V/S ratio = 1/3.0464738927 = 0.3232409193
which is substantially greater than the V/S (0.2605308806) calculated for the unit volume hemisphere
To find the optimum k:
V: πhr²/3
S: πhr
V = Sr/3 = Skh/3 = 1
S=3/kh
kh is maximized (and S minimized) when k=h
V = πk[sup]2[/sup]h[sup]3[/sup]/3 = 1
V = πk[sup]5[/sup]/3 = 1
k[sub]opt[/sub] corresponds to cone angle of 0.7807864690 rad (= 44.7357693773 deg). Does anyone know any special geometric significance for that value? It definitely isn’t 45 degrees.
First of all, you have volume and surface swapped there, and you’re doing a sphere, not a hemisphere. But you seem to have the correct value for r and for V/A.
Secondly and more importantly, though, your cone formulae are inconsistent. In the volume formula, you’re using h for the perpendicular height (as is customary), but in your surface formula, you’re using h for the height along the slant (sqrt(h[sup]2[/sup] + r[sup]2[/sup]).
Oh. Nope, not particularly familiar with him, and don’t do mouth-pops.
Monty, I’m not sure exactly what you’re asking. y=1/x doesn’t have volume, being two-dimensional. The area under the curve is infinite, which is key to showing that the surface area of the trumpet is infinite. The trumpet, which is the surface of revolution of y=1/x, does have volume, but it’s finite, since it’s proportional to Int(1/x^2, x=1…infinity), which is finite.