# Inches vs Inches^3

Peace to all. After gathering and calculating the measurements by volume (of a box), will the results be in regular inches or will they be in inches^3?

For example, if the dimensions of the item are as follows: L = 4.13in, W = 5.50in, H = 7.50in, will the result of the volume be 170.36in or 170.36in^3

When is inches^3 appropriate compared to inches?....Does the same rule apply to all measurements in this scenario (such as CM, MM, etc)?

• Cubic inches are a unit of volume, square inches are a unit of area and inches are a unit of length. Same goes for all units of length. Sep 6 '21 at 22:05
• It should be $\mathrm{in}^3$. When we say that the volume of an objectis $170 \, \mathrm{in}^3$, you can think of this as meaning that we can fit 170 cubes with a side length of $1$ inch into the box. (Technically, this might only possible if the cubes can be stretched and squished, though.)
– Joe
Sep 6 '21 at 22:09
• A cuboid with volume $8\textrm{ in}^3$ might have dimensions $(1\textrm{ in}\times1\textrm{ in}\times8\textrm{ in})$ or $(1\textrm{ in}\times2\textrm{ in}\times4\textrm{ in})$ or $(2\textrm{ in}\times2\textrm{ in}\times2\textrm{ in})$ or etc. Notice that both the numerical value and the unit of each side multiplicatively contribute to (get multiplied to obtain) the volume quantity $8\textrm{ in}^3.$ Sep 6 '21 at 22:29

You can sort of treat the units as quantities. The volume is $$4.13\text{ in}\times 5.50\text{ in}\times 7.50\text{ in}=(4.13\times 5.50\times 7.50)(\text{in})^3$$. Hence, the unit of volume is $$\text{inches}^3$$. In general, you can apply this strategy of treating units as a variable/quantity in order to find the units for another property/measurement calculated from a formula. It can also be helpful when figuring out unit conversions.

• Are you suggesting $3$ inches $\times 5$ feet equals 15 inchesfeet? This only works when all units are identical, else one needs to convert units to a common unit. Sep 27 '21 at 22:44
• @amWhy Sure, it doesn't sound conventional or meaningful, but it still works out mathematically. You can convert inchesfeet into inches^2 or any other more meaningful area unit using standard unit conversions. Sep 28 '21 at 2:16

A volume is always measured in length3. In your case, length is inches.

Outside of physics, people may not be quite so careful, or knowledgable. If you buy a chainsaw, you might be told it has a 1.6" engine. In context, it obviously means cubic inches. You're smart enough to figure that out. The chainsaw salesman may neither appreciate nor understand a correction.

• Sometimes the diagonal length is used instead, so be careful! Sep 7 '21 at 7:20
• @Trebor When has the diagonal length of a chainsaw ever been used to specify it? And which diagonal? Sep 7 '21 at 7:59
• I don't mean chainsaws specifically. For instance, the limit of luggage size in my city's public transport uses the diagonal length. It serves as an inaccurate but practical measure of volume in some sense. Sep 7 '21 at 8:12
• @Trebor Outside of physics, people may not be quite so careful, or knowledgable, or be worrying about volume per se. If you want to transport luggage, you might be limited by the diagonal length. In context, it is obviously vaguely related to volume, and to length, and to width, and to height, and to mass, and to difficulty of stowing. You're smart enough to figure that out. The luggage inspector may neither appreciate nor understand a lecture. Sep 7 '21 at 8:38

Indeed the measurement of the volume of the box will be in inches cubed, i.e. $$\text{in}^3$$.

The same goes for other units: If the side length of the box is in centimeters ($$\text{cm}$$), then the volume of the box will have units $$\text{cm}^3$$.

If $$X$$ is the number of cubic units $$\space (u)\space$$ it is always appropriate to write

$$X\space u^3\space$$ as opposed to $$X^3 u.\quad$$ The distinction is important where, for example, $$X>1\implies X^2 mi >Xmi^2.\quad$$ Oddly enough, casual language portrays it the other way: "$$X\space$$ square miles is less than $$X\space$$ miles squared," where the latter refers to a square that is $$X$$ miles on a side.

In your example each $$1D$$ measurement is expressed as a unit to the first power but the final expression shows the unit cubed. $$4,13\space in\space \times\space 5.5\space in\space \times\space 7.5\space in = 170.3625\space in^3$$

• You cannot compare $X^2\text{mi}$ to $X\text{ mi}^2$. Perhaps you were thinking $(X\text{ mi})^2\gt X\text{ mi}^2$.
– robjohn
Sep 7 '21 at 9:16