A Biomechanical Model of Branch Shape in Plants
Catherine Jirasek, Przemyslaw Prusinkiewicz
Department of Computer Science, University of Calgary, Calgary, AB T2N 1N4, Canada
ABSTRACT
In this paper the effects of gravity and tropism are incorporated
into a model of plant branch growth, yielding a model of the curving of a plant axis in a vertical plane. The axis is assumed to consist of discrete segments connected together by rotational springs.
Initially, tangential loading on an axis is considered; a recursive
torque formula that specifies a segment’s displacement due to
external forces acting on the segment is developed. Radial growth
of an axis is considered as well; material properties of an axis
change as the discrete segments become thicker in girth. The
model is extended to branching structures, using an empirical relationship between branch width and branching angles.
Keywords: branch shape, physically-based modeling, plant biomechanics, tropisms.
1 INTRODUCTION
The shape of branches has a profound effect on the geometry of
plants. The positioning of branches directly affects the appearance
of plants, which is of primary relevance to the application of plant
models to image synthesis. The orientation of organs supported by
branches also affects biologically important plant characteristics,
such as the distribution of light in a canopy.
A plant axis is a sequence of segments (internodes) within a branch
[9], and may support higher-order (lateral) axes (see Figure 1). The
initial orientation of an axis (as it emerges from a bud) is determined by the phyllotactic pattern inherent in a particular plant species. Over time, this orientation may be modified by mechanical or
physiological factors. The reorientation of plant axes is usually
gradual, but the resulting change in shape of branches and positions of supported axes can be significant.
In the purely mechanical case an axis bends due to external forces.
For example, the twigs of a weeping birch or a weeping willow are
flexible and will hang down due to the force of gravity.
An axis may also bend as a result of a physiological change in the
axis structure. Specifically, a curvature change due to differential
growth occurs when one side of the axis grows faster or slower
than the other side [11]. When the opposite sides have unequal
lengths, the axis curves towards the shorter side.
Differential growth of an axis may result from internal or external
causes. If the differential growth of an axis is determined by an
internal process inherent in the plant, the movement is termed nastic [3]. If the curving is determined by an external stimulus that
regulates axis growth, the movement is termed tropic [13]; tropic
movements result in axes being positioned such that they are best
adapted to the conditions of their environment. Different organs
may assume quite different positions as a result of external stimuli
[13]. For example, a branch or stem may grow towards a light
source, whereas leaves may assume orientations perpendicular to
the direction of light.
The present work seeks to simulate the development of axes
affected by gravity and tropisms. Main interest lies in modeling the
shape of branch limbs in a biologically justified and aesthetically
pleasing fashion. The proposed model is limited to two-dimensions (reorientation of an axis in a vertical plane).
The paper is organized as follows. Previous work that is directly
related to the present research is discussed in Section 2. The basic
two-dimensional physically-based model is introduced in Sections
3.1-3.4, where the tangential loading of an axis made of segments
of equal length is considered. The model is generalized in Section
3.5 to axes that consist of segments of unequal length; local parameters of this model - the mass of individual segments and the elasticity of joints between them - are derived as a discretization of an
underlying continuous rod model. Tropic influences and growth of
segments in width are introduced in Section 4. In Section 5 the
model is extended to include branching using Murray’s (1927)
empirical relationship between branch widths and branching
angles. A model of gradual reorientation of branches is introduced
to reconcile the initial direction of axes determined by the branching angle, and the final orientation dictated by gravity and tropisms. Future work is presented in Section 6.
2 RELATED WORK
Ideas that have been presented in previously published papers have
been incorporated into the present work.
Fournier et al. (1994) take a biomechanical approach to describe
stem growth and orientation. They develop a mechanical model for
stresses and strains in the cross-section of an axis that is tangentially loaded as it grows in both a longitudinal and radial direction.
That is, Fournier et al. assume the cross-section of a stem grows
radially and longitudinally while it is loaded by external forces. In
particular, they consider a gravitationally strained form of a radially and longitudinally growing branch. As a stem grows older, a
axis 1
internode
Negative
Orthogravitropism
is
Positive
Orthogravitropism
2
ax
Negative
Plagiogravitropism
Positive
Plagiogravitropism
pipe
ch
bran
Diagravitropism
Figure 1: An axis that is attached
to a supporting axis.
Figure 2: An individual organ can pass through
different gravitropic phases.
given part of the stem will tend to lean downward due to it’s own
weight. The resulting shape of the stem is an “S” due to the effect
of age, load, and the initial orientation that the stem seeks to grow
in. The concept of a longitudinally and radially growing axis that is
subject to tangential loads, age, and tropisms was used in the formulation of the biomechanical model of axis growth that is introduced in this paper (see Sections 3.1-3.4).
Niklas (1992), who also takes a biomechanical approach to
describe plant behavior, considers the effect of geometry and material moduli of an axis on the mechanical performance of plants. He
considers plants as a structure whose shape, size, and elasticity
contribute to its mechanical behavior. Niklas uses the theory of
elasticity to determine the spatial distribution of forces within
structures, which in turn affects the shape and movement of plant
organs. Formulas that relate the material modulus to the shape of
an organ were used in the derivation of the formulas that describe
the rigidity of an axis that undergoes longitudinal and radial
growth (see Section 3.5 and Section 4.3).
Digby and Firn (1995) completed a biological study on plants that
were affected by gravitropism. They found that in certain plants an
individual organ can pass through different gravitropic phases that
are conventionally referred to as negative orthogravitropism, negative plagiogravitropism, diagravitropism, and positive plagiogravitropism (see Figure 2). This is due to the tendency of an organ to
grow in a specific direction. That is, recently formed internodes of
an organ tend to show a negative gravitropic response due to the
organ’s tendency to grow upward, whereas older internodes show a
positive gravitropic response due to the organ’s weight. The resulting shape of the organ is an “S” due to the effect of age, load, and
initial orientation that the organ seeks to grow in. Tropic influences
were included in our model of axis growth (see Section 4.1).
Shinozaki et al. (1964) introduced a model, commonly referred to
as the pipe model, that quantitatively analyzes plant form. Shinozaki et al. found that the sum of the cross-sectional areas of
stems and branches that are found at a certain horizontal level is
proportional to the amount of leaves existing at and above that
horizon. That is, the cross-sectional area of a branch is proportional to the number of leaves supported by it [5]. This led to a new
interpretation of plant form where a branch is taken to be made up
of several identical pipes of constant thickness or cross-sectional
area. Each pipe runs from a leaf, through all of the successive
branches, down to the trunk base. The pipe acts as both the vascular passage and as the mechanical support of the branching structure. Each branch is made up of a bundle of pipes, and the cross-
Figure 3: The pipe model.
section of the branch is proportional to the number of pipes that
run through the branch (see Figure 3). The pipe model was used as
the underlying concept for the simulation of branching structures
(see Section 5).
Holton (1994) used the pipe model to simulate branching structures, but he extended his model to include structures that are influenced by gravitropic and phototropic effects. Holton represents
tropisms by appropriately oriented vectors that are added into the
path of each branch; this in turn will orient the branch in the same
direction as the tropism vector. Prusinkiewicz and Lindenmayer
(1990) give a detailed description of how this can be done. The
bending of branches can be simulated by slightly rotating each
branch segment in the direction of the tropism vector. The angle of
orientation α is calculated using the formula α = e H × T
where e is the elasticity modulus of the limb, H is the branch segment, and T is the tropism vector (see Figure 4). The formula can
be interpreted in a physical sense. T can be rendered as a force
applied to the end of vector H . If H can rotate around it’s starting
point, then H × T specifies the torque applied to segment H . The
method for simulating tropisms was refined and incorporated into
the simulation of branching structures (see Section 5.2).
3 A TWO-DIMENSIONAL PHYSICALLYBASED MODEL OF AXIS SHAPE
3.1 Basic Concept of the Model
A plant axis is assumed to consist of discrete segments connected
by rotational springs. The proposed model uses mechanical factors
to orient individual segments of these axes. Each segment is
assigned a length, width, and mass, and each spring is assigned a
unique spring constant. The axis as a whole is subject to gravity
and, possibly, other predefined external forces.
3.2 Deformation of a Rod in Two-Dimensions: A Recursive Torque Formula
In this section we consider the simplest case of an axis consisting
of a sequence of equal-length segments with identical masses. A
recursive formula is derived that will indicate the torque about
each joint, given that the axis is subject to external forces.
mi
yi−1
mi−1
i−1
spring
T
i−2
rod segment
yi−2
mi−2
H’
i−3
α
yi−3
mi−3
H
xi−1
xi−2
(a)
Figure 4: Due to a tropism T , segment H is oriented around it’s starting point by an angle α .
xi−3
(b)
Figure 5: (a) A rod consists of discrete segments that are connected by rotational springs. (b) The graphical interpretation of a rod consisting of several
discrete segments.
We assume that each segment i embodies a mass m i at it’s end.
applied to mass m i , then the torque about joint i due to the forces
Let x i be the horizontal distance between masses m i and m i + 1
Fi – 1
(see Figure 5). The torque about joint i due to a vertical force
applied to the end of segment i – 1 is equal to
Fi – 1
vert
τ̂ i
vert
= Fi – 1
vert
xi – 1 .
vert
τ̃ i
vert
= Fi – 1
vert
τi
, Fi – 2
hor
vert
xi – 1 + F i – 2
vert
( xi – 1 + xi – 2 ) .
, ... , F 0
hor
is:
hor
hor
+ F 0 ( yi – 1 + yi – 2 + … + y0 ) .
When both the horizontal and vertical forces are included, the total
torque about joint i becomes:
external
= F i – 1 yi – 1 + F i – 2 ( yi – 1 + yi – 2 ) + …
hor
hor
+ F 0 ( yi – 1 + yi – 2 + … + y0 )
hor
(2)
xi – 1 + F i – 2
( xi – 1 + xi – 2 ) + …
+ Fi – 1
vert
By extending this reasoning to all masses m i – 1 , m i – 2 , ... , m 0 ,
+ F0
( xi – 2 + xi – 3 + … + x0 ) .
we obtain the complete formula for the torque about joint i due to
a vertical force:
This yields the recursive formula:
τi
= Fi – 1
vert
+ F0
vert
vert
xi – 1 + F i – 2
vert
τi
( xi – 1 + xi – 2 ) + …
( xi – 1 + xi – 2 + … + x0 ) .
external
(3)
Furthermore, by considering the torque about joint i – 1 ,
τi – 1
vert
= Fi – 2
+ F0
vert
vert
xi – 2 + F i – 3
vert
= xi – 1 ( F i – 1
vert
vert
to τ i – 1
+ Fi – 2
vert
vert
(4)
:
+ … + F0
hor
= yi – 1 ( F i – 1
hor
+ Fi – 2
+ xi – 1 ( F i – 1
vert
+ τi – 1
.
external
hor
+ Fi – 2
(7)
+ … + F0 )
vert
hor
+ … + F0
vert
)
(8)
τi
external
= x i – 1 ( m i – 1 + m i – 2 + … + m 0 )g + τ i – 1
external
, (9)
where g is the acceleration of gravity.
vert
) + τi – 1
vert
. (5)
A similar reasoning holds true for horizontal forces that may be
applied to the system. If y i is the vertical distance between masses
m i and m i + 1 , and F i
vert
vert
In particular, the torque about joint i in the presence of gravity and
no other forces is
( xi – 2 + xi – 3 ) + …
( xi – 2 + xi – 3 + … + x0 ) ,
we can relate recursively τ i
τi
vert
(6)
hor
τi
acting on mass i – 2 :
hor
= F i – 1 yi – 1 + F i – 2 ( yi – 1 + yi – 2 ) + …
(1)
The torque about joint i by a rod consisting of two segments is the
sum of the torque due to force F i – 1
acting on mass i – 1 and
the torque due to force F i – 2
hor
is the horizontal force component
3.3 Introducing the Torque of the Spring
and the Frictional Damping Torque
To keep a spring stretched beyond it’s rest state, a force must be
applied to each end of the spring. As the force increases in magnitude, the elongation of the spring will increase. If the elongation is
not too great, the force F that is being applied to the spring is
directly proportional to the displacement x of the spring beyond
it’s rest state:
F = kx ,
(10)
ignored since they do not affect the equilibrium state of the segments (in equilibrium, there is no acceleration, which, in turn,
means that there are no pseudo forces acting on the segments).
Knowing the torque about joint i , the angular acceleration of segment i – 1 can be found using the equation
where k is the spring constant that is a measure of the material
properties of the spring. Equation (10) is known as Hooke’s Law
[14].
In the equilibrium, the spring at joint i that connects segments i
and i – 1 exerts a torque in the negative direction to the torque
τi
in Equation (8). The torque exerted by the spring at joint i
external
can be found using the rotational analogue of Hooke’s law
τi
spring
= – κθ i ,
(11)
τi
total
α i = ---------- ,
Ii
where I i is the moment of inertia of the rod about joint i .
Although the rod is not a rigid body, we approximate it’s moment
of inertia as if it were a rigid body; this approximation will not
affect the validity of the static solution that will be obtained. Thus,
the moment of inertia about joint i is calculated using
i–1
where κ is the spring constant and θ i is the angle between the
I =
spring’s rest angle and the spring’s current angle. That is, θ i is the
displacement angle of the spring with respect to it’s rest state.
damping
= – bω i ,
(12)
where b is the constant that specifies the strength of the damping
force and ω i is the angular velocity of segment i – 1 with respect
τi
damping
spring
external
The angular acceleration of segment i – 1 is related to it’s position
θ i (the angle of rotation of segment i – 1 with respect to segment
i ) and angular velocity ω i by equations:
as defined
in Section 3.2, the total torque about joint i is:
τi
total
= τi
external
+ τi
spring
= yi – 1 ( F i – 1
hor
+ xi – 1 ( F i – 1
+ τi – 1
external
+ τi
vert
hor
+ Fi – 2
– κθ i – bω i .
(17)
dω
---------i = α i .
dt
(18)
hor
+ … + F0
vert
tuting Equation (14) for τ i
total
, we obtain a system of differential
equations from which we can find the equilibrium state of the rod
as a solution to the initial value problem. We find this solution
numerically using (for simplicity) Euler’s method [8]:
+ … + F0 )
vert
dθ i
-------- = ω i ,
dt
By expressing the acceleration α i using Equation (15), and substi(13)
damping
+ Fi – 2
(16)
(see Figure 6).
, the damping torque
, and the torque due to external forces τ i
,
m j to the center of rotation. The sum is taken from j = 0 to i – 1
to segment i .
Taking into account the spring torque τ i
∑ m j r 2j
j=0
where m j is the mass of a particle j and r j is the distance from
The damping torque caused by the medium in which the rod bends
acts in the opposite direction to the angular velocity of a segment.
We assume that the damping torque about joint i is
τi
(15)
)
θ kj + 1 = θ kj + ω kj ∆t ,
(14)
3.4 Finding the Final Rotational Angle of a
Segment
Using dynamics, the angle that segment i – 1 makes with respect
to segment i can be found as the angle at which the rod is in static
equilibrium. Consider the dynamic system of segment i and segment i – 1 . The frame of reference attached to segment i is noninertial since the segment may be accelerating due to a force that is
being applied to it. If the motion of segment i – 1 is considered
with respect to that frame of reference, then pseudo forces will be
acting on segment i – 1 . For simplicity, these pseudo forces are
ω kj + 1 = ω kj + α kj ∆t
(19)
for j = 1, 2, …, i ,
(20)
where k is the number of iterations. The initial angle θ i0 of segment i – 1 about joint i is set to a predefined elevation angle. It is
assumed that the initial values θ 0j for j = 1, 2, …, i – 1 are set to
0
with respect to θ i0 , and the initial velocity ω 0j
for
j = 1, 2, …, i are set to 0 .
Each axis will be subject to forces and will assume a predefined
spring constant. The simulation of the growth of an axis begins
with one segment. For k iterations, Euler’s method is used to find
mi
segment i
ri−1
mi−1
ri−2
segment i−1
r
θ est s
ta
te
mi−2
r0
m0
Figure 7: Segment i – 1 is rotated by angle θ with respect
to it’s rest state.
Figure 6: The moment of inertia is calculated using the
distance r j from mass m j to the axis of rotation.
the change in position θ kj of segment j due to the forces acting on
the axis. When the equilibrium state has been obtained, a new segment that assumes an orientation that is parallel to the previous
segment is added to the end of the axis (see Figure 7). The change
in position of each segment is calculated until a state of equilibrium is reached, and the process repeats for any number of segments. Figure 8 illustrates the growth of an axis that is initially
oriented upward at 50° with respect to the horizontal plane and
that is subject to gravity.
where V is the volume of a segment, A is the area of the crosssection, l is the length of a segment, m is the mass of a segment,
and ρ is the density of a segment. Substituting Equation (21) into
Equation (22) yields
m = ρAl ,
(23)
which specifies that the mass of a segment is proportional to the
length of the segment:
m∼l .
3.5 Altering the Lengths of Internodes
(24)
Therefore, as the length of a segment is increased, the mass of the
segment will increase proportionally as well.
3.5.1 Length and Mass of an Internode
The lengths of the segments that constitute the rod do not need to
be the same. Intuitively, it is apparent that as the length of a segment increases, the mass of the segment should increase proportionally as well. To derive this relationship begin with the
equations
(21)
m = ρV ,
(22)
Our model of a branch axis can be viewed as a discretization of a
(uniform) continuous rod. Consequently, the rigidity of the spring
that connects segment i and i – 1 depends on the length of segment i – 1 . This relationship is derived as follows.
in
iti
al
or
ien
ta
tio
n
V = Al and
3.5.2 Length and Spring Constant of an Internode
0
50
(a)
(b)
(c)
(d)
Figure 8: The simulation of the growth of an axis that is initially oriented upward at 50° with respect to the horizontal plane and that is
subject to gravity.
θ
s
r
θ
(a)
Figure 9: The relationship between
arclength s , radius r , and angle θ .
(c)
Figure 10: The simulation of the growth of an axis that consists of longer intermediate
internodes.
The rotational analogue of Hooke’s Law states that
τ = κθ ,
(25)
where κ is the spring constant and θ is the angle between the
spring’s rest orientation and the spring’s current orientation. For a
continuous rod it is known that [7]
τ = EIK ,
(26)
where E is the elasticity modulus, I is the moment of area, and K
is curvature. In a circular arc of length s and radius r , we have
(see Figure 9)
s
θ = r
(27)
thus
1
θ
K = --- = --- .
r
s
(28)
By substituting Equation (28) into Equation (26) we obtain:
EI
τ =  ------ θ .
 s
(b)
(29)
From a comparison of Equation (25) and Equation (29), it follows
that the spring constant that approximates the elastic property of a
segment of length s is equal to
Figure 10 illustrates a simulation of the growth of an axis that has
shorter end internodes and longer intermediate internodes. The
curves in Figure 8 and Figure 10 are approximately the same
shape, which confirms that both models are valid discretizations of
the continuous rod.
4 TROPISMS AND RADIAL GROWTH
4.1 Adjusting the Initial Rest Angle of a
Segment
As shown by Digby and Firn (1995), the organs of many plant species have a tendency to grow in a specific direction. The effect of
tropisms on plant axis was used in the context of modeling by
Fournier et al. (1994). This preferred tendency of growth has also
been incorporated into our model by orienting new segments in the
direction in which the axis seeks to grow in. Over time, the orientation of the internodes are altered due to external forces acting on
the axis, such as the force of gravity. Figure 11 shows an example
of plant axes that are inclined to grow in an upward direction but
that are subject to the force of gravity.
Instead of initially attaching segment i – 1 at an orientation that is
parallel to segment i , segment i – 1 assumes a rest angle that is in
the direction that the axis seeks to grow in. The rest angle φ of
segment i – 1 is the angle between the desired orientation of the
axis and segment i (see Figure 12a).
(30)
The rest angle is calculated when a new segment is attached to the
end of an axis. The calculations made in Section 3 to find the angle
θ still hold, but the segment is rotated by – φ + θ (see Figure
12b).
Thus, the spring constant κ is inversely proportional to the length
s of the segment:
Figure 13 illustrates the growth of an axis that has a tendency to
grow upward at an angle of 45° with respect to the horizontal
plane and that is subject to gravity.
EI
κ = ------ .
s
1
κ ∼ --- .
s
(31)
(a)
(b)
Figure 11: Plant axes that seek to grow in an upward direction but that are subject to the force of gravity.
4.2 Radial Growth of a Segment
Not only do axes grow longitudinally, they also grow radially. We
capture the radial growth using the pipe model introduced by Shinozaki et al. (1964). Each segment is thought of as a collection of
concentric pipes, where all pipes have the same cross-sectional
area. As a new segment is attached to the end of an axis, each previous segment will grow in a radial direction by obtaining an additional pipe (see Figure 14). That is, the same cross-sectional area is
added to each segment when a new pipe is added to the axis. A
new segment that has just been attached to the end of the axis initially consists of one pipe.
The addition of pipes to an axis will affect three things: the rest
angle of each segment will be altered, the material properties of the
axis will change, and the mass of each segment will increase.
As an axis grows radially, the rest angle of each segment will be
altered due to the addition of new layers of material to the segment. Each new layer will have a distinct rest angle, which, in turn,
will affect the rest angle of the segment as a whole. When a new
pipe is added to an axis, the new segment i – 1 that is attached to
the end of the axis assumes a rest orientation that is defined by the
tropic angle (see Section 4.1). As more segments are added to the
axis, segment i – 1 will obtain new pipes. We assume that each
new pipe that is added to segment i – 1 will have a rest angle equal
to the current angle between segment i and segment i – 1 (see
Figure 15).
The radial growth of segment i – 1 will also result in the increase
of the rigidity of the spring at node i . That is, the node will
become less elastic.
To derive a relationship for the rest angle and spring constant of a
segment that consists of several pipes, begin with the rotational
analogue of Hooke’s Law; the torque exerted by an individual
spring is
τi = κi θi
(32)
= κi ( µi – φi ) ,
(33)
where τ i is the torque of spring i , κ i is the spring constant, θ i is
the angle between segments i and i – 1 , φ i is the rest angle of
segment i – 1 , and µ i is as shown in Figure 16.
segment i
segment i
gravity
ferred
nt i−1
segme
e
st stat
φ re
pre
1
ent i−
ion
segm
direct
−φ
θ
µ
(a)
(b)
Figure 12: (a) An axis seeking to grow in an upward direction. φ is the rest angle with respect to segment i . (b) Due to gravity, segment i – 1 is rotated by angle θ with respect to it’s rest state. The resulting angle µ between segment i and i – 1 is equal to
–φ+θ.
n
or
ie
nt
at
io
al
iti
in
450
(a)
(b)
(c)
(d)
Figure 13: (a)-(c) An axis that has a tendency to grow upward at an angle of 45° with respect to the horizontal plane and that is subject to gravity. (d) The simulation of the two-dimensional model in three-dimensional space.
= ( κ1 + κ2 + … + κn )
A collection of pipes that makes up a segment can be considered as
a single pipe. In other words, there exists a spring constant κ and a
κ 1 φ 1 + κ 2 φ 2 + … + κ n φ n
*  µ i – ------------------------------------------------------------ .


κ1 + κ2 + … + κn
rest angle φ such that the multi-pipe system will behave as a single angular spring described by the equations
τi = κi θi
(34)
= κi ( µi – φi )
Comparing formulas (38) and (35), we obtain:
(35)
To show this, assume that a segment i consists of n pipes. Since
torques are additive, the resultant total torque τ i will be equal to:
τi = κ1 ( µi – φ1 ) + κ2 ( µi – φ2 ) + … + κn ( µi – φn )
(36)
= ( κ1 + κ2 + … + κn )
* ( µi – κ1 φ1 – κ2 φ2 – … – κn φn )
(37)
(38)
κi = κ1 + κ2 + … + κn ,
(39)
κ1 φ1 + κ2 φ2 + … + κn φn
- .
φ i = -----------------------------------------------------------κ1 + κ2 + … + κn
(40)
In the simulation, each time a new pipe is added to a segment the
overall spring constant and rest angle are recalculated for each segment. The change in rest angle affects the overall shape of the rod
when the external forces are removed. That is, an axis that has
been growing over a period of time, subject to external forces, will
be deformed when the forces are removed because of the change in
rest states over time.
As a segment grows radially, the mass of the segment will increase
as well. The increase in mass of a segment will be proportional to
the increase in cross-section of the segment.
segment i
segment i
segment i
new segment
radial growth
of preceding
segment
segment i−1
φ1
segment i−2
φ2
µ
(a)
Figure 14: The longitudinal
and radial growth of an axis.
When a new internode is
attached to the end of an axis,
previous internodes will grow
radially.
segment i−1
segment i−1
θ
φ
(b)
Figure 15: (a) Rest angle φ 1 of the first pipe in segment i – 1 .
(b) When an internode is added to the end of an axis, the rest
angle φ 2 of the second pipe will be equal to the current angle
between segments i and i – 1 (resulting from the total torque
applied to segment i – 1 ).
Figure 16: Graphical interpretation of θ , µ , and φ for
a single pipe.
n
tio
ien
ta
or
al
in
iti
(a)
50
0
(b)
(c)
(d)
Figure 17: (a) A branch that has a tendency to grow upwards but is subject to gravity. (b) A rod subject to gravity and having an upward
rest angle of 50° with respect to the horizontal plane. (c) A rod subject to the same conditions as rod (b) but having a smaller elasticity
modulus. (d) The simulation of the two-dimensional model in three-dimensional space.
4.3 Calculating the Spring Constant of a
New Layer
the simulation of the two-dimensional model in three-dimensional
space.
As stated above, each new pipe that is added to a segment acquires
a unique spring constant that is used in the calculation of the overall spring constant at that joint. The spring constant for an individual pipe can be calculated using Equation (30) which states that
5 BRANCHING STRUCTURES
EI
κ = ------ ,
s
(41)
where E is the elasticity modulus of the material of a segment, s
is the length of a segment, and I is the moment of area that is calculated according to one of two cases. If the segment has just been
added then the cross-section is assumed to be a full circle; the
cross-section of a segment is the sum of the cross-sections of the
pipes that make up that segment. The moment of area is calculated
using the formula [7]
πR 4
I = ---------- ,
4
(42)
5.1 A Relationship Between Branch
Widths and Branching Angles
The model of axis bending can be extended to include branching.
We assume that an axis will bifurcate into two child branches that
are smaller in girth than the parent branch. A relationship between
parent and child branch thicknesses was proposed by da Vinci
[12]. He postulated that the cross-sectional area of the parent
branch would be equal to the sum of the cross-sectional areas of
the child branches. That is,
πr 02 = πr 12 + πr 22 ,
(44)
r 02 = r 12 + r 22 .
(45)
The diameter d is twice the radius r , so the equation becomes:
where R is the radius of the circle. If the segment has grown radially, then the cross-section of the new layer is assumed to be a hollow circle, and the moment of area of the new layer is calculated
using the more general formula
π
I = --- ( R 4 – r 4 ) ,
4
(43)
d 02 = d 12 + d 22 .
(46)
A generalization of Equation (46) was proposed by Murray (1927)
to produce diverse branching architectures:
d 0α = d 1α + d 2α ,
(47)
where R is the distance from the center of the circle to the outer
edge of the layer and r is the distance from the center of the circle
to the inner edge of the layer.
where different values of α can be used to produce different forms
of branching structures. We used Equation (47) with α = 2.49 , as
suggested by Murray, to calculate the widths of segments that had
bifurcated into two child branches.
Figure 17 illustrates the longitudinal and radial growth of axes that
are subject to gravity. In Figure 17b the axis assumes an upward
rest angle of 50° with respect to the horizontal plane. Figure 17c
shows an axis that is subject to the same environment as the axis in
17b, but the elasticity modulus of the axis in 17c has a smaller
value. That is, the axis in 17c is composed of material that is more
elastic than the material of the axis in 17b. Figure 17d illustrates
Murray also found a relationship for the branching angles of the
two child branches. It was observed that if a branch bifurcates into
two branches that are equal in diameter, then the branches will
make equal angles with the line of direction of the parent branch
[5]. If a parent branch divides into two child branches with different diameters, then the thinner branch will make a larger angle
tio
n
in
0
n
in
iti
tio
ta
al
ien
or
or
ien
ta
al
iti
60
0
60
(a)
(b)
Figure 18: (a) A branching structure that initially assumes a downward rest angle of 60° with respect to the horizontal plane. (b) A
branching structure that initially assumes an upward rest angle of 60° with respect to the horizontal plane.
with the line of direction of the parent branch [12]. Murray found
the mathematical relationship to be
d 04 + d 14 – d 24
cos x = ----------------------------- ,
2d 02 d 12
(48)
d 04 – d 14 + d 24
cos y = ----------------------------- ,
2d 02 d 12
(49)
d 04 – d 14 – d 24
cos ( x + y ) = ----------------------------- ,
2d 12 d 22
(50)
where:
• d 0 , d 1 , and d 2 are the diameters of the parent branch and
the two child branches, respectively,
• x is the angle made by the branch with diameter d 1 with the
line of direction of the parent branch,
• y is the angle made by the branch with diameter d 2 with the
line of direction of the parent branch (for a derivation of these
equations see [5]).
At a branching point, the appropriate widths and branching angles
are assigned to the parent and child segments. The child branches
grow independently of each other and are not influenced by tropisms.
Figure 18 illustrates the growth of branching structures. Figure 18a
shows a structure that initially assumes a downward rest angle of
60° with respect to the horizontal plane and is subject to gravity.
In this simulation one child branch is smaller in girth than the other
child branch. Figure 18b shows a branching structure that initially
assumes an upward rest angle of 60° with respect to the horizontal
plane and is subject to gravity. Each child branch is equal in girth.
external stimulus affects a branching structure as a whole, and each
axis in a branching structure is positioned such that it is best
adapted to the conditions of it’s environment. That is, the axes in a
branching structure seek to grow in the same direction.
In our model, the tropism vector is assumed to be the same as the
initial direction of branch growth. Thus, all branches will seek to
grow in the direction that the first branch begins to grow in. The
bending of branches is simulated by slightly rotating each branch
segment in the direction of the tropism vector. The preliminary
derivation of the angle of orientation of a segment due to a tropic
influence is given below.
By definition, the torque τ due to a force F applied to a segment
s is
τ = s×F .
(51)
τ = sF sin β ,
(52)
Thus
where β is the angle between the force and the segment. When a
force is applied to the end of a segment, the segment rotates about
its end point by an angle α (see Figure 19a). The torque value is
then equal to
τ = sF sin ρ ,
where ρ is the angle between the force and the segment in its equilibrium state after the initial rotation of α .
A spring connected to a segment exerts a torque τ̂ according to the
rotational analogue of Hooke’s law
τ̂ = – κ α
5.2 Branching Structures Influenced by
Tropisms
Tropisms must be handled slightly differently for branching structures than they were handled for individual axes. In general, an
(53)
,
(54)
where κ is the spring constant. In the equilibrium, we have
τ + τ̂ = 0 , thus
sF sin ρ – κα = 0 .
(55)
F
β
F
s+ s
final orientation
ϕ
F
F
ρ
final orientation
s
α
ϕ1
F
s
ϕ
ϕ0
initial orientation
final orientation
s
initial orientation
initial orientation
(a)
(b)
(c)
Figure 19: (a) Graphical interpretation of the force F applied to a segment s , and the angles β , α , and ρ . (b) A change of variables where β = ϕ and where a small change in the length ∆s of a segment results in a small change in the rotation ∆ϕ of the segment. (c) Graphical interpretation of angles ϕ 0 and ϕ 1 .
Rearranging Equation (55) and recognizing that ρ = – β + α ,
where positive rotations are taken to be in the counterclockwise
direction, the equation becomes:
–s F
sF
α = ------ sin ρ = --------- sin ( β – α ) .
κ
κ
(56)
It is desired to make the final equation a function of the elasticity
λ of the spring. Equation (57) yields:
dλ
------ = k .
ds
(63)
Using Equation (63), the following substitution
By allowing
sF
λ = ------ = sk ,
κ
(57)
dϕ dλ
dϕ
dϕ
------ = ------ ------ = k -----dλ ds
dλ
ds
(64)
can be made into Equation (62) to yield:
where k is the rotational elasticity per unit length and λ describes
how much the spring will bend, Equation (56) can be written as:
α = – λ sin ( β – α ) = – sk sin ( β – α ) .
dϕ
k ------ = – k sin ϕ ,
dλ
(65)
dϕ
------ = – sin ϕ ,
dλ
(66)
–d ϕ
dλ = ----------- .
sin ϕ
(67)
(58)
Given the elasticity of the spring λ and the initial orientation β of
the segment with respect to the force, Equation (58) will be solved
for the resulting orientation angle α . Initially, rename the angle
β = ϕ . Given a constant force applied to the end of a segment, a
small change in the length of the segment ∆s will result in a small
change in the rotation of the segment ∆ϕ (see Figure 19b). Equation (58) becomes:
∆ϕ = – ∆sk sin ( ϕ – ∆ϕ ) ,
(59)
∆ϕ
------- = – k sin ( ϕ – ∆ϕ ) .
∆s
(60)
Integrate both sides:
∫ dλ
ϕ 1 dϕ
= – ∫ ----------- ,
ϕ 0 sin ϕ
(68)
where ϕ 0 is the initial angle between the segment and the force
and ϕ 1 is the final angle between the segment and the force as a
result of the applied force (see Figure 19c). This yields:
As the change in length of the segment decreases, the resulting
rotation of the segment decreases as well. Therefore, taking the
limit as ∆s → 0 yields:
∆ϕ
lim  ------- = – k sin ( ϕ – ∆ϕ ) ,

∆s → 0 ∆s
dϕ
------ = – k sin ϕ .
ds
(61)
(62)
ϕ
– λ = ln  tan ---

2
ϕ1
,
(69)
ϕ0
ϕ
ϕ
– λ = ln  tan -----1- – ln  tan -----0- ,



2
2
(70)
tropism vector
tropism vector
final orientation
final orientation
initial orientation
initial orientation
(a)
(b)
Figure 20: (a) The final orientation of a segment with a small elasticity value. (b) The final orientation of a segment with a relatively
large elasticity value.
tan ( ϕ 1 ⁄ 2 )
– λ = ln  -------------------------- .
 tan ( ϕ 0 ⁄ 2 )
of a segment that is influenced by a tropism can be calculated. For
a relatively small elasticity value, a segment will be slightly rotated
in the direction of the tropism vector; given a large elasticity value,
the segment will almost reach the desired orientation, that is, it will
be oriented in approximately the same direction as the tropism vector (see Figure 20).
(71)
Solving for the final angle of the segment yields:
tan ( ϕ 1 ⁄ 2 )
e – λ = ------------------------- ,
tan ( ϕ 0 ⁄ 2 )
(72)
ϕ
ϕ 1 = 2 atan   tan -----0- e –λ .

2 
(73)
In the branching model, each time a new segment is added to an
axis, the tropism angle ϕ 1 is incorporated into the initial orientation of the segment. Figure 21 illustrates the growth of branching
structures that are influenced by tropisms. Figure 21a shows a
structure that seeks to grow in an upward direction of 45° with
respect to the horizontal and is subject to a weak tropic influence.
The structure in Figure 21b also seeks to grow in an upward direction of 45° with respect to the horizontal plane, but the structure is
subject to a strong tropic influence. Figure 21c shows a structure
Given the initial orientation ϕ 0 of a segment with respect to a tro-
(a)
al
0
iti
45
in
in
iti
al
or
or
ie
ie
nt
nt
at
at
io
io
n
n
pism vector and the elasticity λ of the spring, the final orientation
initial orientation
450
(b)
(c)
Figure 21: (a) A structure that seeks to grow in an upward direction of 45° with respect to the horizontal plane and is subject to a
weak tropic influence. (b) A structure that also seeks to grow in an upward direction of 45° with respect to the horizontal plane but
is subject to a strong tropic influence. (c) A structure that seeks to grow in a horizontal direction and is subject to a tropic influence
that is of intermediate strength.
that seeks to grow in a horizontal direction and is subject to a
tropic influence that is of intermediate strength.
[3] Hart, J. W., 1990. Plant Tropisms and Other Growth Movements. Unwin Hymann Ltd.
6 CONCLUSION
[4] Holton, M., 1994. Strands, Gravity and Botanical Tree Imagery. Computer Graphics Forum, 13(1): 57-67.
In this paper tropisms and biomechanics were combined to yield a
model of the growth and curvature of an axis in two-dimensions.
The following cases were considered: the growth of an axis that is
tangentially loaded, the growth of an axis that is influenced by tropisms with or without taking radial growth into account, and the
development of branching structures.
[5] MacDonald, N., 1983. Trees and Networks in Biological Models. John Wiley & Sons Ltd.
An obvious area of future research is the extension of our model
into three-dimensions. When considering the displacement of each
segment, the three dimensional extension must take into account
the branch’s three degrees of rotational freedom (as opposed to one
degree of rotational freedom that was dealt with in the two-dimensional case). Another open problem is the incorporation of the biomechanical phenomena described in this paper into developmental
models of plant architecture expressed using L-Systems [9].
[7] Niklas, K. J., 1992. Plant Biomechanics - An Engineering
Approach to Plant Form and Function. The University of Chicago Press Ltd.
7 ACKNOWLEDGMENTS
We would like to acknowledge Bruno Moulia for interesting discussions that led to this work. We gratefully acknowledge the
NSERC graduate scholarship and research grant, which supported
in part this research.
[6] Murray, C. D., 1927. A Relationship Between Circumference
and Weight in Trees and its Bearing on Branching Angles. J.
Gen. Physiol., 10: 725-729.
[8] Press, W. H., Teukolsky, S. A., Vetterling, W. T., & Flannery,
B. P., 1992. Numerical Recipes in C ‘The art of scientific programming’ 2nd Edition. Cambridge University Press.
[9] Prusinkiewicz, P., & Lindenmayer, A., 1990. The Algorithmic
Beauty of Plants. Springer-Verlag.
[10]Shinozaki, K., Yoda, K., Hozumi, K. & Kira, T., 1964. A
Quantitative Analysis of Plant Form-The Pipe Model Theory.
I. Basic Analysis. Jap. J. Ecol., 14: 97-105.
[11]Silk, W., 1984. Quantitative Descriptions of Development.
Ann. Rev. Plant Physiol, 35: 479-518.
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[2] Fournier, M., Bailleres, H., & Chanson, B., 1994. Tree Biomechanics: Growth, Cumulative Prestresses, and Reorientations.
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[12]Stevens, P. S., 1974. Patterns in Nature. Little, Brown and
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[13]Strasburger, E., Noll, F., Schenck, H. & Karsten, G., 1908. A
Text-Book of Botany. Macmillan & Company Ltd.
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