Imagine, when no wind is blowing, rain is falling on an ideal tree: It has a one-dimensional vertical rigid trunk with one-dimensional single rigid branches perpendicular to the trunk (see picture 1, the side view) with spherical leaves such that in the upper part of the tree four of them can be put to the trunk just touching each other (see picture 2, the upper view). Without rain, the leaves are in horizontal position but they can rotate downward around the point of attachment to the branches when it rains. Each leave has a mass $m_{leave}$, and a "spring constant" $k_{leave}$, which tries to pull it back in horizontal position when rotating.

Suppose the tree is formed as in picture 1 and has eight layers of branches. Each layer consists of four perpendicular branches (see picture two, the upper view). From the top, each layer of branches is rotated such that the leaves from the layer above touch the leaves that are in the same position (see picture 3). The leaves have friction with the rain (otherwise, no energy could be transmitted to the leaves).

When it's not raining, seen from above, you can't see through the leaves and see part of the ground the tree is standing on.

When it starts to rain the leaves will rotate into the equilibrium state, with as a consequence the leaves will cover each other less.

I think it's a very difficult problem, but if the rain [carrying a moment density of $\rho(\frac{kg}{(sec)(m)^2})$] increases, does it stay dry under the tree, and if so, where?

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  • $\begingroup$ As the leaves overlap and are all horizontal, wouldn't you have to know the total surface area within each hemisphere of the sphere, and the size of each leaf, to determine how much water would pool on the leaves before surface tension at the edges was broken and water fell to the next leaf, and so on, 'till water exits the sphere completely? Water exiting the top hemisphere would take as long to fall into the bottom hemisphere as that water would take to exit the bottom hemisphere. So you could determine the time through one hemisphere, and double that to find time of dryness at the trunk. $\endgroup$ – Ernie Feb 21 '17 at 1:09

Pardon me for the "smart aleck" answer, but the question as stated has some problems. I'll address the most serious problem, and "leave" the others out (pun intended).

The most serious problem is that the question, as stated, is modeling leaves as some kind of torsional spring and says nothing as to whether the torsional-leaf-spring is frictionless or not. There are two cases here to address: either the leaf-spring has friction or it does not.

a.) If the leaf-spring(s) have some friction, then some kind of frictional constant has to be given. Or at the very least, some kind of threshold has to be given that would indicate at what point a water droplet would roll off the leaf. Since no such parameter is given, the question is incomplete if there is any kind of friction or viscosity involved.

b.) If the leaf-spring(s) are frictionless, and there is no capillary attraction (fluid friction) between the raindrops and the leaf, then any amount of rainfall above zero will make its way to the bottom of the tree as soon as rainfall at any level begins. This leads to the vacuous answer of "the tree cannot possibly keep anything dry underneath it". Put another way, even the slightest dm of rainfall landing on any leaf for any amount of time would be a pencil-standing-on-its-point situation.

Moreover, the leaves have mass and area, but there is insufficient information given to calculate any sort of moment of inertia. How readily the leaves roll over and dump any accumulated rainfall would depend on this moment of inertia.

Sorry for such a cop-out, but the way I see it, there simply is not enough information given (disregarding how overly idealized the problem is in the first place).

  • $\begingroup$ @photon-You're right. I'll edit the question. $\endgroup$ – descheleschilder Feb 21 '17 at 5:11
  • $\begingroup$ @photo-I like the "leave" pun! $\endgroup$ – descheleschilder Feb 21 '17 at 9:51

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