# Notation for rule of thumb, without breaking dimensional homogeneity?

I'd like to know how to write rules of thumb in a concise way, without breaking dimensional homogeneity.

For example, if a runner has an average speed of ~10 km / h, an approximation of the covered distance would be

$$\mathrm{distance} \approx \mathrm{duration} * 10 \frac{\mathrm{km}}{\mathrm{h}}$$

Is there any shorter way to write it? The goal would be to make it clear that you can simply multiply the number of hours by 10, and you'd get the number of kilometers.

$$\mathrm{km} = 10 * \mathrm{h}$$

is concise, but it's also obviously wrong because it breaks dimensional homogeneity.

There was a question on bicycle.stackexchange ("How to convert calories to watts on Strava rides?"), and one of the answers was Calories(kcal) = Watts * Hours * 4. This rule of thumb doesn't break homogeneity, but it still looks weird because one kcal is 1.163Wh, and not 4Wh. What would be a better way to write it?

• I'm not really sure I understand your question: Your first equation is exact except for the difference between distance and average distance, and does not depend on units. Your other equations use "km" etc. as shorthand for "distance in km", which you could write as $d/\mathrm{km}$ etc. And the factor 4 includes an efficiency factor, so it's not just a unit conversion. Is your question what would be a better way to write the last equation? Commented Mar 21, 2022 at 12:10
• @Toffomat: 1) Indeed. $\approx$ is not just just for "approximately 10km/h", but for assuming that the average speed is constant. 2) $d/km$ might be what I'm looking for. Thanks. Or is there another way to write "distance in km"? I suppose $\mathrm{km}$ will always mean the unit and not the distance in this unit. 3) Yes, I'd like to offer the answerer a better alternative than Calories(kcal) = Watts * Hours * 4, because, as you said, there's an implicit, hidden efficiency factor. Commented Mar 21, 2022 at 12:44

If your aim is to improve the equation Calories(kcal) = Watts * Hours * 4, there are a few things to consider:

• First of all, you want to take the input quantities (time $$t$$ , average power delivered to the bicyle $$P$$) and compute the energy $$E$$ consumed by the metabolism. (You could add subscripts to distinguish "pedal power" and "food energy", but for the purpose of this answer, this is not necessary.) Clearly, this will also involve the conversion efficiency $$\eta$$ (from food to pedal). Here, the energy is $$E=P t /\eta\,.\tag{+}$$ This equation is fine as it is, and independent of any units.
• Next, you have specific units in mind (W, h, kcal; $$\eta$$ has units of 1 or percent), so you'll like to have an easy way to compute the energy in your head while biking. The simplest way (I think) is to make Eq. ($$+$$) dimensionless (divide everything by the desired overall unit and do some algebra): $$\tag{\times} \frac{E}{\text{kcal}}=\frac{P t /\eta}{\text{kcal}} = \frac{P}{\text{W}}\,\frac{t}{\text{h}}\,\frac{1}{\eta}\underbrace{\frac{\text{W} \cdot \text{h}}{\text{kcal}}}_\lambda = \frac{P}{\text{W}}\,\frac{t}{\text{h}}\,\frac{1}{\eta}\,\frac{\text{W} \cdot 3600 \text{s}}{4184\text{J}} = \frac{P}{\text{W}}\,\frac{t}{\text{h}}\,\frac{1}{\eta}\cdot 0.8604 \,.$$ Note that the factor $$\lambda$$ is dimensionless from the start (this is a cross-check for such calculations). This equation is still exact (except for rounding the numerical factor), and still valid for all units, but it is most convenient to use for the ones we put in.
• Finally, you assume an average value for the efficiency of $$\eta=21.5\%=0.215$$. $$\tag{\ast} \frac{E}{\text{kcal}} = \frac{P}{\text{W}}\,\frac{t}{\text{h}}\,\frac{1}{0.215}\cdot 0.8604 = \frac{P}{\text{W}}\,\frac{t}{\text{h}}\cdot 4.002 \,.$$

So you can express the equation in any of the three ways and it will be exact (except for rounding) and general (except that the last form assumes a specific efficiency).

• Excellent, thank you very much. The last one will do. And since we're talking about a rule of thumb, we can assume an efficiency of 0.2151, and get a nice round 4 instead of 4.002. Commented Mar 21, 2022 at 13:31
• @EricDuminil Well, 4.002 is roude to three digit, but given the level of uncertainty on the other input quantities, 4 is fine anyway Commented Mar 21, 2022 at 13:33

My preference for such things is $$\left[\frac{\mathrm{distance}}{1\ \mathrm {km}}\right] = 10\left[\frac{\mathrm{duration}}{1\ \mathrm{hr}}\right]$$

As another example, the electron plasma frequency is given by $$\omega = \sqrt{ne^2/\epsilon_0 m}$$. Since all but one of the quantities on the right-hand side are constants, this can be written as a very straightforward rule of thumb:

$$\left[\frac{\omega}{1\ \mathrm{Hz}}\right] = 5.64 \times 10^{4} \left[\frac{n}{1\ \mathrm{cm}^{-3}}\right]^{1/2}$$

• Since the goal is to be as concise as possible : are the $1$s needed in front of the units? Do the square brackets have any special meaning? Commented Mar 22, 2022 at 18:49
• @EricDuminil well, the goal isn’t necessarily to be as concise as possible. In this case, it’s pretty unambiguous if you leave them out, but if e.g. one of the units was “meter”, then the unit symbol could be easily confused with a mass of some kind in the absence of the $1$. I’d call that a stylistic choice. Similarly, I think the brackets look better than the alternative, maybe because I’d pronounce the plasma frequency expression “the plasma frequency in Hz is equal to …” Commented Mar 23, 2022 at 1:50
• It makes sense, thanks. What is the alternative you're talking about? No brackets, or round brackets? This notation is indeed not the most concise, but if it was used regularly, I suppose it would become very natural and readable. Commented Mar 23, 2022 at 8:01

Your example is $$s\approx vt,\,v=10\text{km/h}$$.