# Physical representation of volume to surface area

I was looking at this XKCD what-if question (the gas mileage part), and started to wonder about the concept of unit cancellation. If we have a shape and try to figure out the ratio between the volume and the surface area, the result is a length. For example, a sphere of radius 10cm has the volume of $\approx 4118 cm^3$ and an area of $\approx 1256 cm^2$. Therefore, the volume : surface area is $\approx 3.3 cm$.

My question is: what is the physical representation of length in this ratio?

• By physical representation do you mean some sort of a physical description/interpretation of what sort of qualitative features this ratio indicates about the object at hand? – joshphysics May 20 '13 at 18:12

For the case of a sphere the ratio you found is:

$$\frac{V}{S} = \frac{ \frac{4}{3} \pi R^3}{4 \pi R^2} = \frac{R}{3}$$

We can actually pass off the volume as being the integral of the surface area here. That's passable when you check the calculus.

One approach is then to ask "what is a function divided by its derivative". This is really similar to the area to perimeter ratio of a circle.

$$\frac{A}{P} = \frac{ \pi R^2}{ 2 \pi R} = \frac{R}{2}$$

Of course you see the "2" because of the value of the exponent, which comes from there being two dimensions, just like the sphere. So now we have explained part of the answer, which is that the linear dimension is divided by the number of dimensions. This is still unsatisfactory because we have no clear sense of how we should define this particular "characteristic length".

One attempt at resolution of this problem would be to test the idea for a square-cube system.

$$\frac{V}{S} = \frac{ R^3}{ 6 R^2} = \frac{R}{6}$$

$$\frac{A}{P} = \frac{ R^2}{4 R} = \frac{R}{4}$$

You can see that it still follows our required rule, but the "characteristic length" is now half of the length of a side. Of course we want to make a statement general to all shapes. This is still muddled by the definition of "characteristic length". So let's avoid it by making a statement about the ratio of "insideness" to "outsideness" for any class of shapes, moving from one dimension to another.

$$\left( \frac{I}{O} \right)_{n+1} = \frac{n}{n+1} \left( \frac{I}{O} \right)_{n}$$

This begs the definition of the "characteristic length", which I'll call $l$.

$$l \equiv n \left( \frac{I}{O} \right)_{n}$$

Unfortunately I can't claim to have invented something new. This is the idea behind Hydraulic diameter. The only difference is a factor of 2. A 4D being with a pipe of constant 3D cross-section would use your formula to calculate hydraulic radius. Wikipedia also includes the same observation I just made:

For a fully filled duct or pipe whose cross section is a regular polygon, the hydraulic diameter is equivalent to the diameter of a circle inscribed within the wetted perimeter.

I've shown that this is also true thinking of a cube versus sphere. So provided we correct your $3.3 cm$ by multiplying by the number of dimensions, you've obtained a sort of generalized radius. Other, more exotic, shapes won't be so simple to explain. If you had a sphere with a bumpy surface and counted the area you had to paint, this would shrink the shape's hydraulic radius.

One way we could justify this concept is referring to fluid dynamics. The hydraulic diameter is used because pushing fluid through a "bumpy" pipe is like pumping it through a smaller smooth pipe. So the number is sort of a proxy for viscous resistance. Well, that can be one use.

let's consider some simple examples: a sphere, a cube, and a rectangular parallelepiped. Let's denote the radio of the volume to the surface area of a given object by $\ell$, then we have \begin{align} \ell(\mathrm{sphere}) &= \frac{\frac{4}{3}\pi R^3}{4\pi R^2} = \frac{1}{3} R = \frac{1}{6}D \\ \ell(\mathrm{cube}) &= \frac{L^3}{6L^2} = \frac{1}{6}L \\ \ell(\mathrm{parallelepiped}) &= \frac{LWH}{2(LW + LH + WH)} \end{align} where $R$ is the radius of the sphere, $D$ is the diameter of the sphere, $L$ is the side length of the cube, and $L,W,H$ are the length, width, and height of the rectangular parallelepiped. Notice that in the case of the cube and the sphere, we get a length that roughly tells us about the lateral dimensions of the object in any given direction, what one might be inclined to call the "characteristic length." On the other hand, consider a the parallelepiped with $W=H=\epsilon$ where $\epsilon$ is small. In this case we can ignore terms of order $\epsilon^2$ relative to order $\epsilon$ terms, and we get $$\ell = \frac{L\epsilon^2}{2(L\epsilon + L\epsilon + \epsilon^2)}\sim \frac{1}{4}\epsilon$$ and we see that $\ell$ becomes very small. So in this case, for a very long, narrow parallelepiped, the ratio gives a good idea of the lateral dimensions in two of the dimensions, but not the third. In general, if you have an object that is approximately spherically symmetrical (and not pathological in other ways) then the ratio gives a good idea of how big all of the dimensions of the object are, but if this symmetry is absent, then the concept of characteristic length defined by this ratio somewhat breaks down.

The physical representation depends on the geometry of the system. In the case of a sphere, then we have the simple result $$\frac{V}{A} = \frac{(4\pi/3)R^3}{4\pi R^2} = \frac{R}{3}.$$ That is, the ratio is one-third of the radius.

Now spheres are special in that they maximize this ratio. For example suppose you had a cube of side length $s$. Then $$\frac{V}{A} = \frac{s^3}{6s^2} = \frac{s}{6}.$$ Of course, to make sure we're comparing apples to apples, we should relate $s$ to $R$ in some meaningful way, say by equating volumes. If $(4\pi/3)R^3 = s^3$, then $s = (4\pi/3)^{1/3} R$, and so the volume-to-surface area ratio for a cube is $$\frac{V}{A} = \frac{1}{6} \left(\frac{4\pi}{3}\right)^{1/3} R \approx 0.27 R.$$

Many things in nature assume spherical shapes because this minimizes potential energy associated with surface tension, subject to the constraint that all your "stuff" has a fixed volume.

By the way, the ratio of surface area to length becomes manifestly important when studying capacitance in CGS units. The capacitance of a sphere of radius $R$ is $A/(4\pi R) = R$ (yes, centimeters are the unit of capacitance in CGS), and the capacitance of a plane-parallel conductor of area $A$ and separation $d$ is (neglecting edge effects) $A/(4\pi d)$.

If you divide a volume by an area you get a length (as you have found), this length is physically just the length of a cylinder, using the XKCD example (you could use any n-sided prism) where the circle face has an area equal to the surface area (of your original shape).

you can see this image that demonstrates it: NOTE: The scale between the sphere and the cylinder is not correct, the length would be a lot smaller than presented here.