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I have calculated the mass of the storm cloud for Hurricane Sandy (year 2012, in the United States of America). The dimensions of Hurricane Sandy of 1800 km in diameter and 13 km in height. Considering the density of the storm cloud multiplied by its volume. However the rain that I drop is much smaller than the rain that I would expect to spill over the territory hit by the storm. The mass, which I get for Hurricane Sandy (2012 U.S.A.) is 3.25x10^4 kg/m^2 . I would appreciate the help that would allow me an approach to solve my problem.

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  • $\begingroup$ Do you mean any rain, or just rain that reaches the surface? $\endgroup$ – gerrit Jun 27 '17 at 13:24
  • $\begingroup$ @Carlos. I am considering amount of rain that causes damage, flood, etc. I could think that about 4.33 percent of the total mass (tons) of the hurricane are raindrops that can precipitate from the clouds. The rest dissipates and does not turn into rain. However I can not justify the above. Just my guesswork. $\endgroup$ – Carlos Jun 28 '17 at 18:17
  • $\begingroup$ If you want to add information to the question, you can do so by clicking on the edit link. $\endgroup$ – gerrit Jun 28 '17 at 18:18
  • $\begingroup$ The information in this question is also pertinent. The answer really may well be "all of it and then some"... because there is a continuous pump of moisture into a cloud. $\endgroup$ – JeopardyTempest Oct 3 '17 at 8:06
  • $\begingroup$ Be wary, a storm cloud isn't a floating full volume of water, but instead a bunch of those tiny drops with tons of air in between. So you can't just calculate it as the volume of a cloud * the density of water. Instead you need to know how much moisture is in those clouds (which is discussed in the previous comment) $\endgroup$ – JeopardyTempest Oct 3 '17 at 8:16

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