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This question is not so much asking about how to do something, but about whether it makes sense.

For instance, converting miles per hour to kilometres per hour makes sense, because both terms have the same unit 'dimension' or 'signature', which can be seen when both are converted to SI—in this case both are, $ x\ ms^{-1} $.

In the same way, it makes sense to convert more complicated and seemingly different terms, like 4µJ / min (microjoules per minute) to N kph (Newton kilometres-per-hour), because both are compatible:

$$ \begin{split} \text{4µJ/min} &= 0.0000\overline6 \text{ g m}^2 \text{ s}^{-3} \\ \text{nN kph} &= 0.0000002\overline7 \text{ g m}^2 \text{ s}^{-3} \\ \therefore\ \text{4µJ/min} &\equiv 240\ \text{nN kph} \end{split} $$

Ok. So far so good. Units can be converted if they have the same unit signature, but not otherwise (so 1 m/s is incompatible with kg/L or m/s²). However, it also makes intuitive sense that this statement is true:

$$ 5 \text{ m s}^{-1} \equiv 0.2 \text{ s m}^{-1}$$

"Travelling five meters for each second is the same as spending a fifth of a second for each meter, right?" Well, that makes perfect sense, but the units are different: $\text{m s}^{-1} \neq \text{s m}^{-1}$. However, $ \text{m s}^{-1} = (\text{s m}^{-1})^{-1}$, and $ 5 \text{ s m}^{-1} = (0.2 \text{ s m}^{-1})^{-1}$, so I guess that does makes sense, right?

But, following the same sort of logic, it shouldn't make sense that:

$$ 5 \text{ m} \equiv 25 \text{ m}^2$$

because $ \text{m} = ({\text{m}^2})^{1/2} $, and $ 5 \text{ m} = ({25\text{ m}^2})^{1/2} $ ? But clearly meters are not at all compatible with square meters, because it makes no sense for them to be physically comparable. Meters are one-dimensional, and square meters are two-dimensional, and that's that, right?


I am writing a program that takes two terms like '60 mph' and 'mm/s' and converts the first to the second, assuming the units are compatible. (It does this by first converting both to pure SI units.) But my question is, what exactly makes units compatible?

Immediately, I assumed it was if the unit signatures were equivalent, but that excludes seemingly valid conversions like 'x m/s' to 'y s/m'. Those types of conversions can be done if the difference in power of the units is applied to the coefficient (for example, −1 and 1/2 in the two previous examples, respectively).

Furthermore, would it make sense to convert $ \text{m/s}^2 $ to $ \text{m}^{1/2}/\text{s} $, or something similar? Should that be allowed as a valid conversion, like this?

$$ \begin{split} 9 \text{m/s}^2 &= (3\text{ m}^{1/2}/\text{s})^2 \\ \therefore 9 \text{m/s}^2 &\equiv 3\text{ m}^{1/2}/\text{s} \end{split} $$

I guess I'm struggling to see where physical context breaks down and nothing but abstract (meaningless) math is left… where is this 'line'? What counts as a meaningful conversion and what doesn't?

Thanks, and apologies for a very waffly question!

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    $\begingroup$ Related: Wolfram Alpha helpfully does conversions from things such as speed (km/h) to pace (min/km) even though they are different quantities. See for instance wolframalpha.com/input/?i=30+km%2Fh+in+min%2Fkm $\endgroup$ – Samuel Apr 20 '15 at 8:55
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    $\begingroup$ Since you are looking at this from the programming perspective, you might be interested in the Haskell units package, which "defines an embedded type system based on units-of-measure". $\endgroup$ – A.P. Apr 20 '15 at 9:01
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For an equation to make sense, or, in technical term, dimensionally compatible, each term in it has to have the same dimension, or, in your words, signature of units. So your equation $$5 \text{ m s}^{-1} \equiv 0.2 \text{ s m}^{-1}$$ does not make sense. Although you can say, "Travelling five meters for each second is the same as spending a fifth of a second for each meter." However, you cannot write it as the above equation. "Equivalent" statement does not always end up with "equation".

To see whether each term has the same dimension, usually we look at the the quantity in dimensions $M,L,T$. Here $M$ represents mass, $L$ represents length, $T$ represents time. There are more, but these are the often-used ones.

For your above equation, the left hand side has dimension $LT^{-1}$ and the right hand side $TL^{-1}$. This is dimensionally incompatible.

This is the topic of dimension analysis.

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  • $\begingroup$ Wolfram Alpha will convert 5 m/s to .2 s/m, so why does that make sense, if they are dimensionally incompatible? (Pointed out by @Samuel) $\endgroup$ – Jollywatt Apr 20 '15 at 8:59
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    $\begingroup$ @Joseph: That's the same idea as your statement that they are equivalent. Notice that when they write the equations, they used $1/$ on the left hand side. That makes the equations dimensionally compatible. $\endgroup$ – KittyL Apr 20 '15 at 9:04
  • $\begingroup$ So does that mean I could put a √ on the left hand side and convert 9m² to 3m? They're dimensionally compatible, but that doesn't make sense, does it? $\endgroup$ – Jollywatt Apr 20 '15 at 9:12
  • $\begingroup$ @Joseph: Of course it makes sense. Think about area of a square with side length $3$m. It is exactly this case. $\sqrt{L^2}=L$. $\endgroup$ – KittyL Apr 20 '15 at 9:19
  • $\begingroup$ Oooh… of course! So 3m is equivalent to 9m², in the same sort of way 5m/s is equivalent to 0.2s/m… so my program should allow that type of conversion? Saying "convert $ 3 \text{m} $ to $ 3 \text{m}^2 $" should simply give "$ 9 \text{m}^2 $"—that's not a invalid conversion. Same goes for something like $ 4 \text{ m/s}^2 $ to $ 2\ \sqrt{\text{m}}/\text{s} $. $\endgroup$ – Jollywatt Apr 20 '15 at 9:46
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To put it plainly, two measure units are (dimensionally) compatible if they measure the same thing. This is the object of study of dimensional analysis. Quoting the Wikipedia page:

The SI standard recommends the usage of the following dimensions and corresponding symbols: mass $[M]$, length $[L]$, time $[T]$, electrical current $[I]$, absolute temperature $[\Theta]$, amount of substance $[N]$ and luminous intensity $[J]$.

In your question you are asking if $$ [LT^{-1}] = [L^{-1} T] $$ which is not the case. On the other hand, you could indeed get the same information from either saying that you are covering $5$ meters every second or that you are taking $0.2$ seconds to cover a meter, since you are measuring two reciprocal quantities.

A nice example of this is that in the US car fuel efficiency is usually measured in miles per gallon, while in Europe it is often measured in litres per $100$ kilometers. You can find a humoristic (though accurate) interpretation of this at the end of this blog post. (note that, again, these are reciprocal measures)

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  • $\begingroup$ Also, note that fractional exponents are conventionally not allowed for dimensions. This is not a problem, though, because taking appropriate powers you can always reduce to the case of integer exponents. $\endgroup$ – A.P. Apr 20 '15 at 9:08
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Very interesting question. This not-quite-answer is too long for a comment.

I would interpret your assertion that $5 m s^{-1}$ and $0.2 m^{-1}s$ are equivalent (you carefully and correctly don't call them equal) as meaning that they convey the same information about the underlying physical situation you are modeling. In that situation the variables "distance" and "time" are real valued functions of the physical state once you've chosen units for those quantities. Then since $$ \left( 5 m s^{-1} \right)^{-1} = 0.2 m^{-1}s $$ knowing the velocity is the same as knowing its inverse.

You can make a similar argument for your second seemingly wrong example. There the "physical system" is the set of all squares. If you know the numerical value of the side of a square (in, say, meters) then you know its area (in square meters).

You will certainly enjoy the answer to this question at https://what-if.xkcd.com/11/

If you went outside and lay down on your back with your mouth open, how long would you have to wait until a bird pooped in it?

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