Monday, March 25, 2013

 

Logic and proportion: scaling surface and volume


A recent article on aging liquor "reminded" the reader that larger barrels present relatively more surface to the contained volume (which is generally not true); a blog which republished an excerpt including that "reminder" let it pass without objection, then added an update to inform readers that the another reader had told them it was wrong.  It is hard to believe that the contrary of something so important could so easily be passed off as true.  Let's just hope it doesn't become a politico-religious question and end up debated on opinion pages! In any case, the point here is not to name and shame those without a keen grasp of the obvious but to provide clues to help others avoid the mistake.

As a first thought, consider why people (or penguins) huddle together to keep warm.  The total volume of people does not change, but the volume of the huddle presents less exposed surface, hence less heat loss. If we consider only two people of about the same size, the volume of their cuddle (or spoon) is twice their individual volume but their exposed surface is reduced. However, their combined girth will have increased while their height will not, so one might object that the huddle is not the same shape.  This will be formalized below for circular cylinders, an approximation for barrels; it will be shown that the surface may increase faster than volume for barrels of greater volume but only with changes to the height:diameter proportion.


Considering how to explain the principle rigorously in clear, simple, everyday terms, it is apparent that it is quite easy to do for regular solids with parametric mathematical descriptions but less so in general.  For instance, the volume and surface of a sphere depend only on the radius, which is the distance from the center to any point on the surface (by definition of a sphere). So, one need only consider how the surface increases and the volume increases as the radius increases. Although the constants (pi, 3, 4) don't change and so don't matter for the scaling effect, the formulas of interest are:

 (I don't use them often, so I checked in a book to be sure).  To simplify the mental calculation, let's assess the case of an eight-fold (that's 2*2*2 times) volume change of a first sphere to a bigger one; that volume increase is achieved by replacing R by (2*R) so its surface will be (2*R)*(2*R)*pi*4, which is 4 times the surface of the first one. 4 is less than 8, not greater.



Similarly, one can easily reason with cubes, but this is where the difficulty with rigorous expression in clear, simple everyday terms becomes apparent. How to state what "aligned stacking" of cubes means?  The idea is that for the experiment we'll want to stack cubes and place them next to each other, but in such a way that their faces in contact are totally in contact (edge-on-edge all around), not rotated leaving parts of the touching faces exposed. Hopefully, that says enough.


Clearly, if one takes eight identical cubes and stacks them two by two on two by two to make a cube, it will be 2*h * 2*h *2*h, where h is the height, width and depth of each of the eight component cubes. The eight little cubes have six faces each for a surface of 6*(h*h), which is a total of 8*6*h*h.  However, after stacking, the eight-fold larger cube composed of them has 6 sides each (2*h)*(2*h) for a total of 6*(2*h)*(2*h)= 4*6*h*h.  The volume of the composed cube is 8 times greater but its surface is only 4 times greater.

A comment on this is in order, because cubes, like spheres, are special cases: they each minimize the surface required to contain a volume.  Spheres do so when there is no constraint on the shape of sides, cubes when sides must be planar. [Note to self: obvious, but can I prove these assertions?]  The comment is this: it may be possible to increase volume and increase surface even faster but shape must change.  Shape of spheres can't change, so this comment does not apply to them, but it could to elliptic spheres (more football shaped).

To merge rectangular volumes less optimally (more pessimally) for instance, taking two of the (2*h)*(2*h)*(2*h) composed cubes, one might decompose them and compose a single layer (4*h)*(4*h)*(1*h). This new arrangement will have double the initial volume, since it is composed of two blocks. However, its surface will be
Thus, the shape change to a large, flat square preserved the surface-to-volume ratio (by doubling the surface as  well as the volume). Had we arranged the blocks in a single layer (2*h)*(8*h)*(1*h) or a row (or tower) (16*h)*(1*h)*(1*h), the surface could have more than doubled. The 2 by 8 arrangement exposes 8+2+8+2 = 20 square sides instead of 16, and the row arrangement exposes 16+1+16+1 = 34 square sides!  This suggests the sort of changes to barrel shape that would be necessary to increase surface for greater volume: taller barrels.

Finally, consider another nice, regular solid, and one more barrel-shaped:  the circular cylinder.  For such a cylinder, of height h and diameter 2*R:
Considering the implications for the relative surface and volumes:
 If R increases, the value of this ratio will decrease unless h decreases enough to compensate (or more).

To continue with the eight-fold volume change example, we need to consider at least two cases: one in which the shape, as expressed by the h:R proportion, is maintained; one in which h is adjusted to increase surface at least as fast as volume.

  1. Constant shape, by which we mean h=a*R for some constant a. If a barrel is twice as tall, its diameter must double, and vice versa.
    • volume =  (pi * R * R) * (a * R)
    • surface = (2 * pi * R) * (a * R) + 2 * (pi * R * R)
    • the surface/volume ratio becomes 2/a*R +2/R = (2+2*a)/(a*R)
    • if R doubles, that decreases (by half) to : (2+2*a) / (a*(2*R))
  2. Adjusting height/width to preserve surface-to-volume is more complicated. Let's call the initial radius and height R0 and h0 respectively, and express the values sought, R1 and h1 in terms of the eightfold volume and surface:
    • volume = pi *(R1 * R1)+R1*h1=8*R0+8*R0*h0
    • surface = R1 * R1*h1 = 8 * R0* R0*h0
    Sadly, this is turns out to be less tractable than I'd hoped, and I haven't found a closed-form giving (height, radius) pair solutions for given volume and surface constraints. Intuitively, increasing only the height 8-fold increases the surface almost 8-fold--it increases the surface of the side 8-fold-- but it does not modify the top and bottom areas, so the total is not quite eight times more surface. Similarly, increasing the section (top, bottom area) 8-fold to increase the volume 8-fold increases the side surface by 2sqrt(2), so not quite 8 total either. But I see no reason one couldn't increase the section a little more and reduce the height a smidgen in the second case, or decrease the section a little and increase the height more than 8-fold in the first case, arriving at 8-fold solutions both ways.
In any case, it should be pretty clear now that scaling up volumes scales up the skin surface less unless shape is changed (quite a lot).
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# posted by Maurice Lanselle @ 5:03 PM 0 comments

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