I am working with several different datasets with different units. I want to convert the distances to mm for comparison. The units I'm dealing with are $\mathrm{kg/m^2/s}$, $\mathrm{mm/day}$, and $\mathrm{mm/(3hr)}$. I'm getting mixed up on how to do it and make sure everything is converted correctly for each.

For this analysis I'm doing, I'm looking at the median of the yearly maximum rainfall over a 25 year time period, if that might have an impact on converting.

  • $\begingroup$ The first step's going to be to figure out what your units actually are. I mean, if you're actually trying to convert $\frac{\mathrm{mm}}{\mathrm{day}}$ to $\mathrm{mm}$, you're probably going to be out of luck as that doesn't really make sense. However, it's more likely that you actually want to convert $\frac{\mathrm{mm}}{\mathrm{day}}$ into, say, $\frac{\mathrm{mm}}{\mathrm{year}}$, and that you're merely omitting the $\frac{}{\mathrm{year}}\text{-qualifier}$ for brevity. So to start with, is that the case? Or do you mean $\frac{\mathrm{mm}}{25\,\mathrm{year}},$ etc.? $\endgroup$
    – Nat
    Jul 5, 2018 at 20:38
  • $\begingroup$ Thank you. Yeah that's fine it's all in the comments. Haha. I think you're right though. I think I'm thinking of it the wrong way because I was told it needs to be in mm, but they actually meant mm/yr or mm/25 yr. Which does make the thought process a lot easier. $\endgroup$ Jul 5, 2018 at 21:04
  • $\begingroup$ And actually going back to my code I'm using to do this, converting to mm/yr gets values that are way to high, so they must have meant mm/day. Gotta love when bad communication is what causes your problem... $\endgroup$ Jul 5, 2018 at 21:22
  • $\begingroup$ Hey Alex, thanks for your concern about whether this is on topic. As it turns out, unit manipulation is one of the topics that's not quite straight physics but which the community has decided is fine here, so this particular question is on topic. Welcome to the site! I'm going to edit out your first paragraph to allow future readers to get right to the question. $\endgroup$
    – David Z
    Jul 5, 2018 at 21:49

4 Answers 4


You can convert it using the density of water, which is roughly $1000~\mathrm{kg}/\mathrm{m}^3$. If you multiply the $\mathrm{mm}/\mathrm{day}$ with this density, you get the right conversion.

So $1~\mathrm{kg}/\mathrm{m}^2/\mathrm{day} \equiv 0.001~\mathrm{m}/\mathrm{day} = 1~\mathrm{mm}/\mathrm{day} = 0.125~\mathrm{mm}/(3\mathrm{h})$.
Or $1~\mathrm{mm}/(3\mathrm{h})=8~\mathrm{mm}/\mathrm{day} \equiv8~\mathrm{kg}/\mathrm{m}^2/\mathrm{day}$

  • $\begingroup$ Thank you! It seems this is exactly what I need to do and a big part of my confusion came from me being told I needed to convert to mm when they actually meant mm/day. $\endgroup$ Jul 5, 2018 at 21:23

Here is the thing. I am not totally sure about the formulas because I don't know what they represent, so I will that with you.

But I am going to do is, give you better understanding of how conversion works so you can solve this and many other similar problems.

Lets say you are trying to find the speed of wind with $\frac{m}{s}$ - where m is meter and s is second. You are given a speed 30$\frac{km}{h}$. The way conversion works is you replace the current unit with (KM in our case) with its equivalent with new unit(m in our case). Since 1km = 1000m and 1hour = 60 mins = 60 * 60 seconds our new formula will be like this:

$$30\frac{km}{h} = 30*\frac{1000}{60*60} = 30 * \frac{1000}{3600} = 30* \frac{5}{18} = 8.3 \frac{m}{s}$$

Now all you need to is apply the same logic to your case. If you want to change a day to hour, replace d with 24h. If you are given $2\frac{inch}{Day}$ to get $\frac{mm}{h}$ replace inch with equivalent mm which is 25.4(according to google) and 1 day = 24Hour. So formula will like like this $$2\frac{inch}{Day} = 2*\frac{25.4}{24} \approx 2.1$$

  • $\begingroup$ Written another way it's: kg m^-2 s^-1 (a mass flux per unit area per unit time). It's a unit used for precipitation in certain models. $\endgroup$ Jul 5, 2018 at 20:36
  • $\begingroup$ Specifically, it is kilograms per square meter per second of rainfall. $\endgroup$
    – ohwilleke
    Jul 5, 2018 at 20:54

I thought I would add my own solution for anyone else and just show what I ultimately ended up doing. Part of the issue was I was instructed to convert to mm when they meant mm/unit of time I ended up converting all my units to mm/hr.

(kg/m^2/s) * 3600 = mm/hr

$\frac{1kg}{m^2*s}$ * $\frac{m^3}{1000 kg}$ = $\frac{0.001m}{s}$

$\frac{0.001m}{s}$ * $\frac{1000mm}{1m}$ = $\frac{1mm}{s}$

$\frac{1mm}{s}$ * $\frac{60s}{min}$ = $\frac{60mm}{min}$

$\frac{60mm}{min}$ * $\frac{60min}{hr}$ = $\frac{3600mm}{hr}$

(mm/3hr)/3 = mm/hr

$\frac{mm}{3hr}$ = $\frac{mm}{hr}$ * $\frac{1}{3}$

(mm/day)/24 = mm/hr

$\frac{mm}{day}$ * $\frac{day}{24hr}$ = $\frac{mm}{24hr}$

$\frac{mm}{24hr}$ = $\frac{mm}{hr}$ * $\frac{1}{24}$

Hopefully that's a simple way to understand things for anyone that's interested.


To get from kg to a distance you will need to have a density of presumably rain or snow. So if you get 1 kg/m^2, and rain has (look up the exact value) a density of 1000 kg/m^2, then 1 kg/m^2 is 1 kg / (1000 kg/m^3) = .001 m of rain. So 1 kg/m^2/s is 0.001 m/s.


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