## Tree Physics 1: capillary action, the height of trees, and the optimal placement of branches

One of the fundamental problems trees have to solve is how to transport water up to the leaves. The ability to solve this problem poses a constraint on the maximum height of trees. The standard freshman physics textbook answer to the question, “What’s the mechanism that lifts the water?”, is capillary action (see Halliday and Resnick for example). Capillary action is the term for the ability of a narrow tube (a capillary) to draw liquid into itself via the adhesive force between the liquid and the tube walls. In plants, the relevant tubes are known as Xylem and the fluid is water. In trees, capillary action alone is not the only mechanism to transport water. First, without evaporation, capillary action would only transport water until mechanical equilibrium is reached and the flow stops. There’s also osmotic pressure at the roots drawing the water in from the soil. There are probably also osmotic effects throughout the tree itself.

Anyway, ignoring all that, let’s see where the freshman physics treatment gets us–what’s the maximum possible height of a tree if capillary action is the limiting factor? To solve this problem, we’d need to think about branching and the fact that pores in the leaves constitute the open ends of the “capillaries”. In the absense of any knowledge about trees, this is a hard problem. Let’s then ignore all the complications and just start with a single capillary tube. We’ll put branching back in to some extent a bit later.

So, what’s the height of a column of water in a single tube with a circular cross-section with radius $R$? The height of the tube results from the balance of two forces: the force of gravity pulling down on the column of water and the adhesive force between the walls of the tube and the water column pulling up. The force of gravity is easy, it’s simply the weight of a column of water of height $H$:

$F_{g,1}=\rho g \pi R^2 H$

where $\rho$ is the density of water and $g$ is the acceleration due to gravity (and the “1” indicates that this result is for a single tube). The adhesion force is due to the surface tension at the top of the column of water pulling against the walls of the tube. The surface tension, $\gamma$, is the force per unit length at the interface of two materials. If the surface of the water makes an angle $\theta$ with respect to the normal to the wall, the upward component of the adhesive force is:

$F_{ad,1}=\gamma 2 \pi R \cos{\theta}$

Since $\theta$ is somewhere between 0 and 90 degrees, let’s ignore the $\cos{\theta}$. Equating the two forces, we find the height of the water column is:

$H=\frac{2 \gamma}{\rho g R}\approx \frac{0.14 cm}{R}$ (1)

where the 0.14 cm comes from plugging in lookupable values. The question now remaining is, “what’s $R$?” According to Emmanual Mapfumo at the Department of Horticulture, Viticulture and Oenology at the Waite Agricultural Research Institute in Australia, the radius of xylem in vitus vinifera is $12 \mu m$ or so. While that’s not a tree, it’s a good enough estimate for me for now. So, the equilibrium height of a column of water with the dimensions of xylem is about $h\approx 10 m$, or, much too short to be the maximum height of a tree.

Okay, so a tree is not composed of a set of parallel tubes of constant diameter from top to bottom. If we bring back in some of the reality of trees, does capillary action postdict something observably true?

Branching should change the story because the height as calculated above is determined by the surface-to-volume ratio of the column of water. If that ratio goes up, the height goes up too. How can we crank the ratio up? If the xylem in upper branches are narrower than those lower down and there are enough of them (is this true? probably), that’d do the trick.

Let’s now model the tree as a very simplified system–a “tree” of height $H$, composed of a single bottom tube of height $h_1$ and radius $R$ that branches once into $N$ upper tubes of height $h_2=H-h_1$, each with radius $r$. To further simplify things, let’s assume the total radius of the assembly stays constant throughout the entire tree. This constrains $r$ to satisfy:

$N \pi r^2 = \eta \pi R^2$

where $\eta$ is the packing fraction. As an example, for a 2D hexagonal close packed arrangement, $\eta= \frac{\pi}{2\sqrt{3}} \approx 0.9$.* This model is consistent with the assumption that the mass per unit height of a tree is constant throughout. This assumption is not generally believed to be true (see next post), but anyway.

We can now calculate the forces for this model and see what the results might tell us about trees. The upward force—the total due to the adhesive force in each of the N upper tubes—is:

$F_{ad,N}=N \gamma 2 \pi r=\gamma 2 \pi R \sqrt{\eta N}=\sqrt{\eta N} F_{ad,1}$ (2)

We see that the upward force scales like the square-root of the number of tubes in the upper branches and is independent of the height. The gravitational force is due to the mass of all the water in the branched tree and is given by:

$F_{g,N}=\rho g \left(h_1 \pi R^2 + N h_2 \pi r^2\right) = F_{g,1} \left(1-\left(1-\eta\right)\frac{h_2}{H}\right)$ (3)

We see that the downward force is sensitive to the overall height of the tree and the location of branching, but it is not sensitive to the number of upper tubes.

Actually making a numerical prediction about the maximum height would be futile here as trees don’t branch only once and I’d have to assume a few more numbers I don’t know, but we can use this analysis to make a qualitative prediction about branching.

Combining the results of Equations (2) and (3) makes a prediction for the structure of a tree if the tree’s structure tries to both maximize the height (as may be the case in dense forrests, where maximizing the height may also maximize the amount of light getting to the leaves) and maximize the amount of water transported simultaneously. The upward force on the water grows like $\sqrt{N}$ regardless of the height of branching as long as branching takes place. The volume of water taken up into the tree is maximized when the branching happens high up the tree–when $h_2\rightarrow 0$. Thus, to maximize both the height of the tree given that the tree is also trying to maximize its water uptake, the tree should have lots of branches very high up and none very low. Does that prediction plan out? Take a look at the Sequoia and see for yourself.**

*Here’s how to calculate the packing fraction for circles in a 2D hexagonal close packed (HCP) arrangement. In english, the HCP structure is the “honeycomb nesting”, in which a row of touching circles sits in the interstices of the adjacent rows (consider just the “A” sheet in this picture). The packing fraction, $\eta$, is the ratio of the area of the circles to the total area covered. Consider a lattice with M rows, each composed of N disks of radius r. The combined area of the disks in the lattice is $M N \pi r^2$. The total area covered is the area of the rectangle covered by the M rows of N disks, which is the length of the rows times M times the height per row. The height per row is the same as the distance between the centers of the rows. The centers of 3 adjacent disks mark the vertices of an equilateral triangle with side length 2r. The height of that triangle and thus the distance between rows is $\sqrt{3}r$. So, the area of the covered rectangle is $(2 N r)(M \sqrt{3} r)$. The packing fraction is the ratio of those areas, $\eta= \frac{\pi}{2\sqrt{3}}$.

There’s probably lots of interesting things about packing of objects. I remember hearing about M&M packing in my undergrad condensed matter course, and that’s pretty fun. The big thing there is oblate spheroids randomly pack a lot tighter than spheres do.

**Do I really believe this is the mechanism behind the branch distribution of the sequoia? I don’t know. It’d help if I actually knew something about trees. It’s at least consistent with this paper that claims that tree branch ratios are all about getting light to the leaves efficiently, where efficiently is defined in the paper..

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### 22 Responses to “Tree Physics 1: capillary action, the height of trees, and the optimal placement of branches”

1. AlexM Says:

Keep up the good work!

2. alice Says:

Hello, I am a mechanical engineering graduate and i found you article very interesting. Although, first time round i did not follow all of you workings i will most certainly be reading it again. Im particularly interested in trees and their capacitity to shift water using simple fluidic principles. keep up the good work and thanks again for taking the time to put this up.

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4. Itay Says:

Hi, I think your assumption that the relevant size is the radius of the xylem is incorrect. See the following article by Harvard Prof. M. Holbrook about this rather intricate problem:

Physics Today, January 2008, page 76

The real problem seems to be not how come trees are as tall as they are, but rather how come they are not even taller (the answer seems to do more with what happens to water when you exert an immense pull on it)

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16. Dantam Rao Says:

plugging R=12.5e-4 cm into equation (1) yields h ~ 1m. But your blog says maximum height h~10 m. Which is the correct?

17. Marc Says:

In equation 1 if you do 14e-2 / 12e-6 you get more like 116.7 meters. I think you missed a zero during the conversion and could have saved yourself a bit of time.

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19. Joe Vreeland Says:

OK. So what’s the ‘normal-physical’ height of a tree in the real world?

Imagine that the math is correct, then: 116.7 Meters = 382.87401633 Feet, which is pretty close to the tree called “Hyperion” in Sequoia National Forest. He is 115.7 Meters/379.7 Feet.

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