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) $\endgroup$ Jun 18 '14 at 15:08
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    $\begingroup$ @JamesS.Cook Julia's profile says "location: Germany". $\endgroup$
    – dtldarek
    Jun 18 '14 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
    Jun 18 '14 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. $\endgroup$ Jun 18 '14 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. $\endgroup$ Jun 19 '14 at 1:46

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. $\endgroup$ Jun 19 '14 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
    Jun 19 '14 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. $\endgroup$ Jun 19 '14 at 14:19
  • $\begingroup$ @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
    Jun 21 '14 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? $\endgroup$ Jun 21 '14 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... :-) $\endgroup$ Jun 19 '14 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
    Jun 5 '19 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
    Jul 19 '19 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.


  • $\begingroup$ Hi Zvonimir and welcome to the site. It's generally a bad idea to leave a link in the way you did; instead you should paraphrase or explain what is behind that link and then leave the link only if there is a much deeper discussion there. $\endgroup$
    – Chris Cunningham
    Jul 18 '19 at 6:30
  • $\begingroup$ I hope it is better now. $\endgroup$ Jul 18 '19 at 16:31
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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
    Jul 18 '19 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. $\endgroup$ Jul 18 '19 at 17:20
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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. $\endgroup$ Jul 19 '19 at 11:09

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