If you want to convert for example $2\,\mathrm{\frac{m}{s}}$ to $\mathrm{\frac{km}{h}}$ you have just to multiply it by $3.6$ and get $2\,\mathrm{\frac{m}{s}} = 7.2\,\mathrm{\frac{km}{h}}$.

Personally I memorize this by knowing the factor $3.6$ (makes sense since the hour has $3600$ seconds and $1\, \mathrm{km}$ has $1000$ meters) and that the value before the $\mathrm{\frac{km}{h}}$ has to be higher.

Is there a good way how can students memorize that the $\mathrm{\frac{km}{h}}$ has to be higher? I often see that the divide by 3.6 instead of multiplying or vice versa.

  • Is there some simple intuitve reason, you can easily memorize why the $\mathrm{\frac{km}{h}}$ has to be higher?
  • Is there any real life situation you can intuitively see that the $\mathrm{\frac{km}{h}}$ has to be higher?

I am not looking for a verbal memnomic but more for a conceptual one. I am also not looking for a formal derivation, which is simple and most of my studends can do it, if they have to (but which clearly is too long just for resonable memorization).

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    $\begingroup$ I teach them (university physics) to multiply by one. As in $1=3.281 ft/m$ or $1 = 5280 ft/mile$ and so forth. For real world, you might gain some conceptual grip by comparing both to mph which is more familiar due to our conventions on US roads ( I assume non-metric country for this comment) $\endgroup$ Commented Jun 18, 2014 at 15:08
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    $\begingroup$ @JamesS.Cook Julia's profile says "location: Germany". $\endgroup$
    – dtldarek
    Commented Jun 18, 2014 at 15:43
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    $\begingroup$ I wouldn't really advise anyone to memorise this. Because in similar situations (convert feet/second to miles/hour or yard/second to miles/hour) it may or may not work. They should remember that one hour is 3,600 seconds and therefore one meter per second is 3,600 meters per hour or 3.6 km/h. And 1 foot per hour is 3,600 feet per hour, 1 yard per hour is 3,600 yard per hour, and nobody can remember the conversion constants anyway; knowing how to do the calculation gives you a chance to get the right result. $\endgroup$
    – gnasher729
    Commented Jun 18, 2014 at 17:45
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    $\begingroup$ Do you often find yourself needing to convert m/s to km/hr? The process is important to understand, but there are hundreds of conversions (if not thousands) and one will remember those they use most often. $\endgroup$ Commented Jun 18, 2014 at 21:03
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    $\begingroup$ I'm not one of the down-voters, but I assume it is because (I feel) this is tangentially related to math education as written. $\endgroup$ Commented Jun 19, 2014 at 1:46

7 Answers 7


Better don't learn such conversion factors, learn how to derive such on the fly. This helps also in the case you want to change inches per second into feet per hour, or cubic feet per hour into cubic centimeters per second.

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    $\begingroup$ Or furlongs per fortnight. $\endgroup$
    – Linear
    Commented Jun 19, 2014 at 10:26
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    $\begingroup$ To amplify on vonbrand's answer, a typical way to do it without memorization is this: $$10\ \frac{\text{km}}{\text{hr}}\times\frac{1000\ \text{m}}{1\ \text{km}}\times\frac{1\ \text{hr}}{3600\ \text{s}}=0.3\ \text{m}/\text{s}$$ $\endgroup$
    – user507
    Commented Jun 19, 2014 at 13:17
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    $\begingroup$ @BenCrowell - your comment has the making of a full answer. I was taught this method as dimensional analysis (whether the term is correct, I don't know, the teacher called it that) and it is a great way to avoid say, dividing inches (instead of multiplying) by 2.54 to get mm. I use this method as a 'lower risk' way to make long conversions. $\endgroup$ Commented Jun 19, 2014 at 14:19
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    $\begingroup$ @JoeTaxpayer technically dimensional analysis is something else (I think), but people do often use the term to mean this sort of unit conversion. Anyway, this seems relevant: physics.stackexchange.com/q/8133/124 $\endgroup$
    – David Z
    Commented Jun 21, 2014 at 1:07
  • $\begingroup$ This link likes my use of the phrase en.wikipedia.org/wiki/Dimensional_analysis and a google search on for images shows the same series of equalities used to change units. What else did you think it meant? $\endgroup$ Commented Jun 21, 2014 at 1:33

To answer the question

Is there any real life situation where you can intuitively see that the km/h number is higher?

and seeing that you are based in Germany:

The ICE goes up to 300 km/h, but nobody suggests that it comes anywhere near the speed of sound.

This, of course, presupposes that one knows the rough speed of high speed trains in km/h, and also the rough speed of sound in m/s (whose value at sea level around 20 centigrade is about 340).

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    $\begingroup$ The advantage of this example is the ease of localisation: in France you have the TGV, in Japan the Shinkansen, in Britain you can cite the HS2 and get yourself in political hot water... :-) $\endgroup$ Commented Jun 19, 2014 at 13:51

It might help to make vivid what this conversion means. 10 km/h is a reasonable running speed. A soccer field is 105 m long. In one second, can you run a few meters, or can you run a third of a soccer field? (American football fans might have a better sense for this, as the number of yards traveled by a runner is key to the game, but I assume that soccer is more relevant in Germany.)

Similarly, 100 km/h is a slow highway driver. Does a driver travel 1/4 of a soccer field, or 3 and a half soccer fields, in a second?

For both of these computations, although I would guess the right way, I could imagine someone guessing wrong. But once someone knows the right answer, I imagine they would remember it, as running/driving down a playing field is more memorable than the numbers $3.6$ and $0.27$.

By the way: My dad is a lawyer. At one point, I was reading one of his law textbooks, and it recommended memorizing 60 mph = 90 ft/sec in order to catch witnesses who give impossible descriptions of car accidents. So there is someone in the world who finds this worth memorizing, although it feels like a pretty specialized case.

  • $\begingroup$ 60 miles per hour is exactly 88 feet per second. 90 feet per second is good enough for catching unreasonable oral testimony, though. $\endgroup$
    – Jasper
    Commented Jun 5, 2019 at 0:05

I have faced this teaching problem before, but I am not very experienced a teacher.
This is what I think right now.

Rather than try to get them to memorize both the factor (3,6) and the "side" (multiply from m/s to km/h) it might be better to have a standard conversion that they think about each time (till it becomes automatic)

"If I walk 1 meter each second, I walk 60 meters per minute.
There are 60 minutes in a hour. I walk 60*60=3600 meters in an hour.
We have about 3 kilometers. Actually, 3,6.
When I walk 1 meter per second, I walk 3,6 kilometers in an hour"

  1. It's important to be able to do every conversion of complex units knowing base unit conversions. (Competence)
  2. It's very practical in physics courses to have the conversion of km/h and m/s memorized.

That said, let's go to answer your question:


As a side effect of the above 1., students should already know, that the bigger the unit the smaller the number before when doing conversions. For km/h and m/s it's intuitively seeming to be the other way around: In km/h the number is larger than in m/s, although km/h contains the larger units than m/s. Do an example with them. From this moment on, the students will remember: for these units it's the other way around.


For the number: It will stick after a small number of computations. Explicitely do exercises with speeds of 1 m/s, 10 m/s, 100 m/s, 1000 m/s. The students will 3.6 so often, they will just remember it.

Side Remark

I guess, the above only works in countries using the metric system. Imperial units have so many different conversion numbers, that 3.6 doesn't seem any special and won't stick around. In the metric system, the only conversions are 10,100,1000 except for times. And as velocities relate to times, we have another association that helps remembering 3.6.

  • $\begingroup$ It's very practical in physics courses to have the conversion of km/h and m/s memorized. No way, this is a really bad idea. $\endgroup$
    – user507
    Commented Jul 19, 2019 at 22:23

You should calculate with magnitudes:

m/s = (1/1000)km/(1/3600)h = 3.6 km/h

My experience is that math teachers detest calculations with magnitudes. They left them to physicists. Their problem is that standard mathematical education does not explain what are magnitudes and how to calculate with them. In the paper linked below I define magnitudes and show that calculations with them are respectable mathematical calculations.


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    $\begingroup$ I am used to "magnitude" being just a number, while "units" being units of measurement, so your answer baffled me. After some googling I found that another school of thought is to consider "magnitude" as a number together with units of measure. Since it is the units that make a difference, I would use "units" instead of "magnitudes". I would also mention dimensional analysis. Also, your equation is missing magnitude, er, a numeric value on the left. You cannot compare units to numeric values :) No, I did not read your PDF. $\endgroup$
    – Rusty Core
    Commented Jul 18, 2019 at 16:49
  • $\begingroup$ I think it is standard that 3 cm is magnitude, 3 is number and cm is unit. Concerning "a numeric value on the left" it is standard to omit factor 1, i.e. 1a=a and 1m/s= m/s. $\endgroup$ Commented Jul 18, 2019 at 17:20
  • $\begingroup$ @RustyCore - is this a regional issue? I agree with your comment, and am curious if "magnitude" with meaning/use shared by Zvonimir is more common in Europe? $\endgroup$ Commented Jul 18, 2019 at 18:41
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    $\begingroup$ I think it is more math-physics issue. My mathematical colleagues from Europe often identify numbers and magnitudes but physicist never identify numbers and magnitudes. For them there is no magnitude 5, but only 5 cm or 5 km or 5 g/cm*3. Mathematics behind magnitudes is as follows $\endgroup$ Commented Jul 19, 2019 at 11:04
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    $\begingroup$ A magnitude is defined by an equivalence relation ≈ which is defined on objects of interest that are comparable (<) and additive (+). Furthermore, + must be in harmony with <, and both < and + must be in harmony with ≈. By Hoelder-Cartan theorem Archemedean systems of magnitudes ( (∀a, b)(∃n) a < n⋅b) with no minimal magnitude, are isomorphicaly and densely embeddable in R+ (if they have minimum they are isomorphic to Z+ ). If a system of magnitudes is continuous, i.e. does not have empty Dedekind cuts, then it is isomorphic to R+. Here lies the connection between numbers and magnitudes. $\endgroup$ Commented Jul 19, 2019 at 11:09

I think if you have a job/task where you need to do this conversion routinely (e.g. moving between two different data sets), than it makes sense to just have this small conversion factor memorized. Derive it once and then practice with it a lot and just have it in ready access memory as a known quantity (like sin of 30 degrees).

But in the likely case you are not doing very routine work emphasizing just these two units, I think it's actually better to just do it the "long way" each time. After all, there will be a blizzard of units (grams, moles, mm Hg, BTUs, horsepower, tons of refrigeration, electon-volts, etc. etc. that need conversion).

It is a normal and helpful muscle-building part of chemistry class to get practice with "railroad track" problems.


Look, for example, at one of the later, longer, examples in this article. Note also that one of the good things about this method is, because of the emphasis on unit cancellation, you are less likely to "use a ratio upside down" (something I have seen award-winning scientists, McKinsey/Goldman sharpies, and even dummies like myself do...it is a common error).

So, yeah...make em line up the whole railroad track of conversions. 1000m/km, 60 s/min, 60min/hr. And rattle them all down. Yes, the answer is surprisingly small/simple (3.6). But unless students are routinely using just that conversion, rather than having many to deal with, I think it's wrong move to memorize this specific conversion.


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