# What drinking glass shape minimizes glass/liquid contact? **URL:** \ **Category:** Factual Questions\ **Created:** [May 22, 2007, 5:05am UTC](https://boards.straightdope.com/t/what-drinking-glass-shape-minimizes-glass-liquid-contact/405037 "2007-05-22T05:05:20Z")\ **Posts on this page:** 12\ **Page:** 3
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [May 25, 2007, 5:20pm UTC](https://boards.straightdope.com/t/what-drinking-glass-shape-minimizes-glass-liquid-contact/405037/41 "2007-05-25T17:20:52Z")
[QUOTE=KP] **I would very much appreciate someone double checking my figures.** Lunchtime is over. [/QUOTE] 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. > [@KP](#): > > 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. **Chronos’s** argument should be sufficient. ---
**Author:** ![KP](https://avatars.discourse-cdn.com/v4/letter/k/a9adbd/32.png) [@KP](https://boards.straightdope.com/u/KP)\ **Post date:** [May 25, 2007, 5:30pm UTC](https://boards.straightdope.com/t/what-drinking-glass-shape-minimizes-glass-liquid-contact/405037/42 "2007-05-25T17:30:14Z")
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!)_ ---
**Author:** ![Pasta](https://avatars.discourse-cdn.com/v4/letter/p/ecccb3/32.png) [@Pasta](https://boards.straightdope.com/u/Pasta)\ **Post date:** [May 25, 2007, 5:30pm UTC](https://boards.straightdope.com/t/what-drinking-glass-shape-minimizes-glass-liquid-contact/405037/43 "2007-05-25T17:30:26Z")
[QUOTE=KP] **I would very much appreciate someone double checking my figures.** Lunchtime is over. [/QUOTE] (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 And, 0.2505 \< 0.2606. ---
**Author:** ![KP](https://avatars.discourse-cdn.com/v4/letter/k/a9adbd/32.png) [@KP](https://boards.straightdope.com/u/KP)\ **Post date:** [May 25, 2007, 5:34pm UTC](https://boards.straightdope.com/t/what-drinking-glass-shape-minimizes-glass-liquid-contact/405037/44 "2007-05-25T17:34:39Z")
Your points are well taken, and correct. 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. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [May 25, 2007, 8:38pm UTC](https://boards.straightdope.com/t/what-drinking-glass-shape-minimizes-glass-liquid-contact/405037/45 "2007-05-25T20:38:14Z")
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. ---
**Author:** ![KneadToKnow](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/kneadtoknow/32/3999_2.png) [@KneadToKnow](https://boards.straightdope.com/u/KneadToKnow)\ **Post date:** [May 26, 2007, 11:17am UTC](https://boards.straightdope.com/t/what-drinking-glass-shape-minimizes-glass-liquid-contact/405037/46 "2007-05-26T11:17:57Z")
[QUOTE=Chronos] But of course [/QUOTE] Do you do the Fritz Feld mouth pop when you say that? ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [May 26, 2007, 4:34pm UTC](https://boards.straightdope.com/t/what-drinking-glass-shape-minimizes-glass-liquid-contact/405037/47 "2007-05-26T16:34:19Z")
Probably not, since I’m not sure who Fritz Feld is. ---
**Author:** ![KP](https://avatars.discourse-cdn.com/v4/letter/k/a9adbd/32.png) [@KP](https://boards.straightdope.com/u/KP)\ **Post date:** [May 26, 2007, 6:51pm UTC](https://boards.straightdope.com/t/what-drinking-glass-shape-minimizes-glass-liquid-contact/405037/48 "2007-05-26T18:51:23Z")
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: **Unit volume Sphere:** Volume = 4πr[sup]2[/sup] =1 r= cube root of (2\* 3/(4p)) = 0.7815926418 Surface = 4π/3 r[sup]3[/sup] V/S ratio = **0.2605308806** **Unit Volume Cone:** Surface: πhr [[grade school derivation](http://mathcentral.uregina.ca/QQ/database/QQ.09.96/millward1.html)] Volume: πhr[sup]2[/sup]/3 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] = (3/π)[sup]0.2[/sup] = .9908188867 h[sub]opt[/sub] = (3/π)[sup]0.2[/sup] = .9908188867 r[sub]opt[/sub] = kh = ((3/π)[sup]0.2[/sup])[sup]2[/sup] = (3/π)[sup]0.4[/sup] = .9817220663 S = πhr = πk[sup]3[/sup] = 3.0558547093 V/S = **0.3272406888** **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.** ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [May 26, 2007, 10:28pm UTC](https://boards.straightdope.com/t/what-drinking-glass-shape-minimizes-glass-liquid-contact/405037/49 "2007-05-26T22:28:31Z")
> [@](#): > > Volume = 4πr[sup]2[/sup] =1 > r= cube root of (2\* 3/(4p)) = 0.7815926418 > Surface = 4π/3 r[sup]3[/sup] 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]). ---
**Author:** ![KneadToKnow](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/kneadtoknow/32/3999_2.png) [@KneadToKnow](https://boards.straightdope.com/u/KneadToKnow)\ **Post date:** [May 28, 2007, 3:48pm UTC](https://boards.straightdope.com/t/what-drinking-glass-shape-minimizes-glass-liquid-contact/405037/50 "2007-05-28T15:48:38Z")
[Fritz](http://www.imdb.com/name/nm0270915/bio) [Feld](http://en.wikipedia.org/wiki/Fritz_Feld) ---
**Author:** ![Monty](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/monty/32/47_2.png) [@Monty](https://boards.straightdope.com/u/Monty)\ **Post date:** [May 28, 2007, 5:49pm UTC](https://boards.straightdope.com/t/what-drinking-glass-shape-minimizes-glass-liquid-contact/405037/51 "2007-05-28T17:49:21Z")
[QUOTE=Chronos] 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. [/QUOTE] Isn’t y=1/x also of infinite volume (assuming you’re talking about the area between the two parabolas)? ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [May 28, 2007, 6:15pm UTC](https://boards.straightdope.com/t/what-drinking-glass-shape-minimizes-glass-liquid-contact/405037/52 "2007-05-28T18:15:48Z")
Quoth **KneadToKnow** : > [@](#): > > [Fritz](http://www.imdb.com/name/nm0270915/bio) [Feld](http://en.wikipedia.org/wiki/Fritz_Feld) 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. [Previous page](https://boards.straightdope.com/t/what-drinking-glass-shape-minimizes-glass-liquid-contact/405037.md?page=2)