page_title: Surface-area-to-volume ratio
section_title: SA:V for balls and N-balls
hierarchical_section_title: Surface-area-to-volume ratio / SA:V for balls and N-balls
caption_attribution_description: English: Plot of the surface-are:volume ratio (SA:V) for a 3-dimensional ball, showing the exponential decline as theradius of theball increases.
context_page_description: The surface-area-to-volume ratio, also called the surface-to-volume ratio and variously denoted sa/vol or SA:V, is the amount of surface area per unit volume of an object or collection of objects.
  In chemical reactions involving a solid material, the surface area to volume ratio is an important factor for the reactivity, that is, the rate at which the chemical reaction will proceed.
  For a given volume, the object with the smallest surface area is a ball, a consequence of the isoperimetric inequality in 3 dimensions. By contrast, objects with tiny spikes will have very large surface area for a given volume.
context_section_description: A ball is a three-dimensional object, being the filled-in version of a sphere ("sphere" properly refers only to the surface and a sphere thus has no volume). Balls exist in any dimension and are generically called n-balls, where n is the number of dimensions.
  For an ordinary three-dimensional ball, the SA:V can be calculated using the standard equations for the surface and volume, which are, respectively,  and . For the unit case in which r = 1 the SA:V is thus 3. The SA:V decreases exponentially as the radius increases - if the radius is doubled the SA:V halves (see figure).
  The same reasoning can be generalized to n-balls using the general equations for volume and surface area, which are:
  volume = ; surface area = 
  So the ratio reduces to . Thus, the same linear relationship between area and volume holds for any number of dimensions (see figure): doubling the radius always halves the ratio.
caption: Plot of the surface-area:volume ratio (SA:V) for a 3-dimensional ball, showing the exponential decline as the radius of the ball increases.
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