# Re: Rain droplets falling on a table

This questions is almost exactly similiar to the the following question, with an extra condition :

Rain droplets falling on a table

Suppose you have a circular table of radius R. This table has been left outside, and it begins to rain at a constant rate of one droplet per second. The drops, which can be considered points as they fall, can only land in such a way such that they impact the surface of the table. Once they strike the table, they form a puddle of radius r, centered at their point of impact. What is the expected number of droplets it takes to cover the table in water?

Now, suppose every water droplet that falls on the table dries out after $(R/r)^2$ seconds. What will the graph of the propability $P(N)$ that the table will be covered with N droplets versus N look like? Can someone please help me with this?

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after the drop hits the table, should we assume the drop spreads out to cover the area $\pi r^2$ in addition to that which is already covered? For example, if the first drop is hit by the second drop then is the total area covered by those two drops $2\pi r^2$? Also, is the drying event discontinuous or should we assume some linear evaporation to zero taking $(R/r)^2$ seconds? I need to know the rules before I play. –  James S. Cook Sep 1 '12 at 4:47
" if the first drop is hit by the second drop then is the total area covered by those two drops $2πr^2$ ?", i that that is why we are talking about probability, the 2nd drop can fall in any way, the area covered need not always be $2πr^2$. The water droplets evaporate after $(R/r)^2$ seconds. Thats all. I don't think the mechanism will matter. Sorry for late reply. –  Swapnanil Saha Sep 1 '12 at 13:40
You're standing on a flat floor. Think about pouring a bucket of water on your feet. Pour another bucket on your feet. Surely the puddle is larger for two buckets than it would be for a single bucket. If we apply this idea to raindrops then it doesn't matter if the new drop hits where an old drop already hit. The new drop will just spread out to cover $\pi r^2$-additional area. The number of drops should be measured by area of water divided by $\pi r^2$. –  James S. Cook Sep 1 '12 at 16:00
I dont think this is a math question in its current form because it requires the knowledge of physical behavior of water. Specifically, we need to be given information on the area covered by two drops of water that fall within distance $r$ of each other. Physically, these drops will spread over $2\pi r^2$. However if one thinks of the covered areas as sets then the collectively covered area is the area of union of the two sets, which is less than $2\pi r^2$. –  Ankur Sep 2 '12 at 0:29