# Chemistry - Is the amount of cubic inches in a cubic foot exact up to infinite significant figures?

## Solution 1:

Dealing with imperial measures in science is problematic. The main reason for that is that you have (at least) two different definitions (fixed values) for a lot of them, one based on the laws of the United States and one based on the laws of the United Kingdom. While the differences are quite insignificant for everyday use (as in baking a cake), in the science world these small deviations can cause further inaccuracies. On Wikipedia you can find more, but for the argument, let's look at the values for the yard:

```
United States 0.914401829 m
United Kingdom 0.9143993 m
International 0.9144 m
```

In principle **the definition of cubic inch to cubic foot is exact**, while the conversion to the SI system may carry an error.

```
Unit Abbreviation Definition SI equivalent (rounded 4sf)
Cubic inch cu in or in³ 16.38 cm³
Cubic foot cu ft or ft³ 1728 cu in 0.02831 m³
Cubic yard cu yd or yd³ 27 cu ft 0.7646 m³
```

Most of the time this shouldn't affect your measurements more than the use of significant figures, which is used as stating an error implicitly. That means it is in itself inaccurate. In any scientific capacity you would state the error explicitly.

When a value is stated as $1.00$ it has three significant figures it actually means that the measurement falls within the interval $0.99$ to $1.01$, or $1.00\pm0.01$.

Let's have a quick example. You measure the volume of water in a beaker to be $0.35~\pu{in^3}$. The beaker doesn't specify how accurate the measurement is, so we simply assume $\pm0.01~\pu{in^3}$. You would state that value as $0.35\pm0.01~\pu{in^3}$, or $0.34-0.36~\pu{in^3}$, or simply $0.35~\pu{in^3}$ with implied error. On the other hand, when we know significant figures are used, we would interpret a value of $0.35~\pu{in^3}$ as $0.35\pm0.01~\pu{in^3}$.

**When you convert this to the SI system, your value doesn't become more or less accurate. You cannot use significant figures to express the error in one system in the same way you would do it in the other, because the implicit error is different.**

The exact conversion (of the international yard) is
$$\frac{0.9144~\pu{m}}{1~\pu{yd}}=\frac{0.9144~\pu{m}}{3~\pu{ft}}=\frac{0.9144~\pu{m}}{36~\pu{in}}\implies1~\pu{in^3}=\left(\frac{0.9144}{36}\right)^3~\pu{m^3}.$$

The actual trust of the value is $\pm0.01~\pu{in^3}=\pm0.16387064~\pu{cm^3}$. When you convert your value $0.35~\pu{in^3}=5.74~\pu{cm^3}$ (keeping the same number of significant figures since the conversion is exact), you would imply an error of only $0.01~\pu{cm^3}$, which is far less than the actual trust. Instead your measured value falls somewhere between $5.58-5.90~\pu{cm^3}$. So the correct way of implicitly stating the error would be $6~\pu{cm^3}$.

### TL;DR

The conversions are exact, the system of significant figures is not. In doubt, state error margins explicitly.

## Solution 2:

Yes, there are exactly $12 \times 12 \times 12$ = 1728 cubic inches in a cubic foot.

Also, an inch is exactly 2.54 centimeters.

A foot is exactly 0.3048 meters.

A yard is exactly 0.9144 meters.

See NIST Special Publication 1038 for more information.

## Solution 3:

**Within** a certain system of measures, conversion factors are typically exact.

In imperial units, this means that a foot is *always* twelve inches, a yard is *always* three feet and a mile is *always* 1760 yards. With the exact conversion, we can use multiplication to see that:

$$1~\mathrm{yd} = 36'' \pm 0''\\ 1~\mathrm{m} = 5280' \pm 0'\\ 1~\mathrm{m} = 63360'' \pm 0''$$

Likewise, the conversion from *cubic* imperial units — which are defined as $\text{imperial unit}\times\text{imperial unit}\times\text{imperial unit}$ — to other cubic imperial units is equally exact.

$$1~\mathrm{yd^3} = 1~\mathrm{yd}\times 1~\mathrm{yd}\times 1~\mathrm{yd} = 3' \times 3' \times 3' = 27~\mathrm{ft^3}$$

This is akin to the exactness of the conversion of metric units, except that the metric factors are always multiples of 10. But there are also semi-metric units, such as the German ‘metric pound’, by definition $500~\mathrm{g}$. Conversion from grams to German metric pounds is as exact as conversion from grams to kilograms.

This becomes different if you leave your system of measures and compare **different** systems. Except in a few very special cases, where two systems chose physically identical starting and/or ending points but with different step sizes,^{[1]} the conversion factors will be nonexact.

Any nonexact conversion factor will introduce an error of its own into the equation and how to deal with these is best shown in Martin’s answer.

Note that even if the transformation is exact at a starting and ending point, measured values will still have an inherent uncertainty in them, as Martin pointed out. The uncertainty is transformed across the conversion and does *not* change magnitude.

[1]: See for example temperature. A measured temperature of $15~\mathrm{^\circ C}$ can be transformed into an *exact* value in kelvin because the kelvin and celsius scales are identical except for an addition factor. Thankfully, also the error is identical.

$$(15.00 \pm 0.005)~\mathrm{^\circ C} = (288.16 \pm 0.005)~\mathrm{K}$$

Similarly, the Réamur temperature scale is defined such that the boiling point of water is $80~\mathrm{^\circ r}$ while the zero-point is identical to the celsius scale. Therefore, the step is different and $1~\mathrm{^\circ C} = 0.8~\mathrm{^\circ r}$. Upon converting, we get a similarly exact value but a different error.

$$(15.00 \pm 0.005)~\mathrm{^\circ C} = (12.00 \pm 0.004)~\mathrm{^\circ r}$$