In my story, the heroes are trying to prevent a global flood and I need to know how long would it take for constant rainfall to flood an Earth-like planet so that the only things left are the tips of mountains.

The rainfall would be heavy and fall constantly and the source is magical so no need to know where all that extra water is coming from.

  • $\begingroup$ You could make the rainfall from a large ring system and not magical at all or a powerful wizard could have drawn an icy moon too close to your planet where it broke apart into a ring system. - Just a thought. Saturn's rings rain onto Saturn. space.com/20595-saturn-rings-rain-water.html $\endgroup$
    – userLTK
    Sep 19, 2018 at 11:04
  • $\begingroup$ rainfall is very vague, some information an rate, quantity, and extent are necessary. $\endgroup$
    – John
    Oct 22, 2018 at 2:24

2 Answers 2


In a raw approximation, take the intensity of the rainfall, expressed in $mm/hour$.

That tells you the volume of rain falling in an hour over a given surface, thus the height of the water layer in mm over that area.

Now take the height of the mountains you want to be out of the flood, and do some simple math: $time = height /rain \ intensity $.

Let's say you want to cover all lower than 6000 meters with a rainfall of 30 mm/hour. It would take $t = 6000000/30 = 200000 \ hours = 22 \ years$.

The highest recorded rainfall is about 300 mm in 1 hour, and at that rate it would take something more than 2 years to do the job.

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    $\begingroup$ With rains that strong, the mountaintops might meet the rising water halfway... $\endgroup$
    – bukwyrm
    Sep 19, 2018 at 8:44
  • $\begingroup$ I appreciate the simplicity of this answer, but seeing how it is off by probably around 30%, I believe it should at least be pointed out in the comments that more accurate approximations are just as easy to make. $\endgroup$ Sep 19, 2018 at 13:31
  • $\begingroup$ Very Good, its also worth noting that if the water was appearing magically then the mass of the earth would be increasing, perhaps not by a cosmically impressive scale but it would still happen which would increase Gravity on the world which could potentially reach the point where the clouds would get lower and lower until it was more like dense fog then clouds $\endgroup$ Sep 19, 2018 at 13:46

28.1 years

So just as an interesting aside (and not sure how accurate of an answer you're looking for), I chose to go with volume measurements just to see how far off from L.Dutch's answer we'd get. I believe that my answer is more accurate, but still ignores some things like the mass which doesn't need to be displaced by water above sea level as well as volumes below sea level not already filled with water.

So the volume we need to fill is the volume of the radius of a sphere at 6km higher than earth's surface (interestingly enough, right about the height of most rain clouds... go figure). From this, we subtract out the volume of the earth. This leaves us with a necessary volume of 3.77E9 km3.

Next we want to convert this to liters, so using that every 1km3 of water is 1E12 liters, we can see that we have a whopping 3.77E21 L of volume to fill.

Here is why I posted a second answer mainly... the metric system can be a pain for some things, but for the overwhelming majority of calculations, it is just the tops. 1mm in depth per sqm of water is exactly 1L of water. If the surface area of earth is 510.1e6 km2, this gives us 5.101E14 sqm.

Using the above, this gives us a rate (at 1mm per hour) of 5.101E14L/hr. Doing a little bit more math (dividing volume needed by fill rate), and changing to a rate of 30mm per hour to keep consistent, we actually wind up with ~28.1 years as the fill rate here.

The difference (which actually surprised me, I was guessing around 10% tops) comes from the fact that the total surface area (and more importantly, volume which is why it is drastically longer than the above answer) needing to be 'filled' will change as you get higher off of the earths surface. Filling in those gaps of square meter plates perpendicular to earth's surface will take an additional 6 years.

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    $\begingroup$ (4 * 3,1415 * ((6000) ^ 2))-(4 * 3,1415 * ((6006) ^ 2)) should give the difference between the surfaces of a 6000km vs a 6006km radius sphere - it comes to about 1 million square kilometers.. compared to the ~450 million km² of a 6000km radius sphere, that would not add a quarter filling time. ----- At such huge radii, and such comparatively small delta r, surface of sphere or surround of circle can be approximated as staying the same approximated as $\endgroup$
    – bukwyrm
    Sep 19, 2018 at 12:27
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    $\begingroup$ Also, please try to stay consistent with notation within your answer. Sometimes it is XXEXX km3, then XXeXX sqm. And i would recomment using MathJax if you want to note calculations. much more readable (It's basically like Latex Math notation). One a side note: "the metric system can be a pain for some things", what things would that be? I have yet to see anything the imperial system does to benefit calculations over SI units. $\endgroup$ Sep 19, 2018 at 12:54
  • $\begingroup$ @bukwyrm You lost me, why are you speaking about differences in surface areas? We want to only use the surface area of earth because this is where the rate of rain number is calculated. $\endgroup$ Sep 19, 2018 at 14:26
  • $\begingroup$ @ArtificialSoul Fixed the E/e notations. And I don't want to start a holy war on imperial vs metric, but having lived a considerable amount of time in countries using both, I find myself having to use either really weird units (who uses decimeters?!) when referring to things in everyday real life or some strange numbers. As I said, in the science world, it's wonderful and imperial is just plain awful. $\endgroup$ Sep 19, 2018 at 14:32
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    $\begingroup$ @ColonelPanic I am in Germany. So the only argument for why you dislike the metric system is that it does not roll off the tongue as easily? Well, there is a preference for everything. But okay, let's leave it at that. A matter of preference. I wouldn't want to use two different systems at once, though. And there is no way in hell I am going to calculate anything using the imperial system. $\endgroup$ Sep 19, 2018 at 14:59

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