Mechanistic Virtual Modeling: Coupling a Plant Simulation Model with a Three-dimensional Plant Architecture Component

Abstract

The aim of this research is to integrate plant architectural modeling or “visualization modeling” and “mechanistic” or physiologically based modeling to describe how a real plant functions using a virtual crop. Virtual crops are life-like computer representations of crops based on individual plants and including the representation of the substrate on which the plants grow. The integration of a three-dimensional expression and the mechanistic model of plant development and growth requires the knowledge of the position of the organs along the different plant axes (the topology), their sizes, their forms, and their spatial orientation. The plant simulation model simulates the topology and organ weight or length. The superposition of spatial position and the topology produces the architecture of the plant. The association between sizes and organs creates what we refer to as the plant morphological model. Both components, the architectural model and the morphology model, are detailed in this paper. Once the integration is complete, the system produces a movie-like animation that shows the plant growing. The integrated model may simulate one or several plants growing simultaneously (in parallel). Visual capabilities make the proposed system very unique as it allows users to judge the results of the simulation the same way a farmer judges the situation of the crops in real life, by visually observing the field.

Appendix

1.1 The Bending Model

1.1.1 Generality

The elasticity of a material depends on the applied forces, the section of the material, and an intrinsic characteristic named Young’s modulus [5]. For example, the elongation of a cylindrical bar can be expressed as:

$${\text{h}} - {\text{h}}_0 = \frac{1}{{\text{E}}} \times \frac{{{\text{h}}_0 }}{{\text{S}}} \times {\text{F}}$$

(16)

where: h − h ₀ is the elongation, S the section, F the traction force applied at the end of the bar, and E the Young’s modulus.

A fruiting branch can be assimilated to an embedded bar in the mainstem and submitted to a vertical force F (see below Fig. 15).

Fig. 15

Coordinate system used in the bending model

At the point of coordinates (x,y) by definition, the flexing moment M is:

$${\text{M}} = {\text{E}} \times {\text{{\rm I}}} \times \frac{{d\theta }}{{d{\text{S}}}}$$

(17)

where: E is the Young module, I is the inertia (${\text{{\rm I}}} = \frac{{\pi {\text{R}}^4 }}{4}$ in the case of a bar with circular section, where R is the radius of the section), dS a small length of the bar, dθ the angle between the tangent of two successive length portions of the bar, and $\frac{{d\theta }}{{d{\text{S}}}}$ the bending of the bar.

At the end of the bar where bending force is applied, the flexing moment is nil. At the point of (x, y) coordinates, the flexing moment M and the bending force equilibrate (see Fig. 15). This can be expressed as:

$${\text{M}} = \left( {{\text{e}} - {\text{y}}} \right) \times {\text{F}}\cos \theta _0 + \left( {{\text{h}} - {\text{x}}} \right) \times {\text{F}}\sin \theta _0 $$

(18)

$${\text{E}} \times {\text{{\rm I}}} \times \frac{{d\theta }}{{d{\text{S}}}} = \left( {{\text{e}} - {\text{y}}} \right) \times {\text{F}}\cos \theta _0 + \left( {{\text{h}} - {\text{x}}} \right) \times {\text{F}}\sin \theta _0 $$

(19)

where: F cosθ ₀ is the “compression” component and F sinθ ₀ is the “flexion” component.

By differentiation of Eq. 19 with respect to S, we obtain:

$${\text{E}} \times {\text{{\rm I}}} \times \frac{{d^2 \theta }}{{d{\text{S}}^2 }} = - {\text{F}}\cos \theta _0 \frac{{d{\text{y}}}}{{d{\text{S}}}} - {\text{F}}\sin \theta _0 \frac{{d{\text{x}}}}{{d{\text{S}}}}$$

(20)

where: $\frac{{d{\text{y}}}}{{d{\text{S}}}} = \sin \theta$ and $\frac{{d{\text{x}}}}{{d{\text{S}}}} = \cos \theta$ (see Fig. 15) so Eq. 20 can be expressed as:

$${\text{E}} \times {\text{{\rm I}}} \times \frac{{d^2 \theta }}{{d{\text{S}}^2 }} = - \left( {{\text{F}}\cos \theta _0 \sin \theta + {\text{F}}\sin \theta _0 \cos \theta } \right)$$

(21)

By multiplying Eq. 21 by $\frac{{d\theta }}{{d{\text{S}}}}$, we obtain:

$${\text{E}} \times {\text{{\rm I}}} \times \frac{{d^2 \theta }}{{d{\text{S}}^2 }} \times \frac{{d\theta }}{{d{\text{S}}}} = - \left( {{\text{F}}\cos \theta _0 \sin \theta \times \frac{{d\theta }}{{d{\text{S}}}} + {\text{F}}\sin \theta _0 \cos \theta \times \frac{{d\theta }}{{d{\text{S}}}}} \right)$$

(22)

Then, by integration between w, the angle at the end of the bar, to θ with respect to S, because the flexion moment at the end of the bar is nil, we obtain:

$$\begin{array}{*{20}c} {\frac{1}{2} \times {\text{E}} \times {\text{{\rm I}}} \times {\left( {\frac{{d\theta }}{{d{\text{S}}}}} \right)}^{2} = - {\text{F}} \times {\left( {\cos \theta _{0} {\left[ { - \cos \theta } \right]}^{\theta }_{w} + \sin \theta _{0} {\left[ {\sin \theta } \right]}^{\theta }_{w} } \right)}} \\ { = {\text{F}}{\left( {\cos {\left( {\theta _{0} + \theta } \right)} - \cos {\left( {\theta _{0} + w} \right)}} \right)}} \\ \end{array} $$

(23)

Then:

$$\begin{array}{*{20}c} {d\theta = d{\text{S}} \times {\sqrt {2\frac{{\text{F}}}{{{\text{E}} \times {\text{I}}}}{\left( {\cos {\left( {\theta _{0} + \theta } \right)} - \cos {\left( {\theta _{0} + {\text{w}}} \right)}} \right)}} }} \\ {{\text{ = K}} \times {\sqrt 2 } \times {\sqrt {{\left( {\cos {\left( {\theta _{0} + \theta } \right)} - \cos {\left( {\theta _{0} + {\text{w}}} \right)}} \right)}} } \times {\text{dS}}} \\ \end{array} $$

(24)

where: ${\text{K = }}\sqrt {\frac{{\text{F}}}{{{\text{E}} \times {\text{{\rm I}}}}}}$

1.1.2 Case of small flexion

In the case of small flexion, w is small, θ is small too, and θ ₀ varies from 0 to π. Equation 24 can be expressed as:

$$d\theta = {\text{K}} \times \sqrt 2 \times \sqrt {\text{A}} \times d{\text{S}}$$

(25)

where: ${\text{A}} = \cos \left( {\theta _0 + \theta } \right) - \cos \left( {\theta _0 + {\text{w}}} \right),$

Because θ is small, we can approximate cosθ by $\left( {1 - \frac{{\theta ^2 }}{2}} \right)$ (from Maclaurin’s formula, n = 2), cosw by $\left( {1 - \frac{{{\text{w}}^2 }}{2}} \right)$, sinθ by θ (from Maclaurin’s formula, n = 2), and sin w by w. Then, A can be expressed as:

$$\begin{array}{*{20}c} {{\text{A}} = {\left( {1 - \frac{{\theta ^{2} }}{2}} \right)} \times \cos \theta _{0} - \theta \sin \theta _{0} - {\left( {1 - \frac{{{\text{w}}^{2} }}{2}} \right)} \times \cos \theta _{0} + {\text{w}}\sin \theta _{0} } \\ { = {\text{w}} \times {\left( {\sin \theta _{0} + {\text{w}}\frac{{\cos \theta _{0} }}{2}} \right)} - {\left( {\sin \theta _{0} } \right)} \times \theta - {\left( {\frac{{\cos \theta _{0} }}{2}} \right)} \times \theta ^{2} } \\ \end{array} $$

(26)

if: ${\text{a}} = {\text{w}} \times \left( {\sin \theta _0 + {\text{w}}\frac{{\cos \theta _0 }}{2}} \right)$, ${\text{b}} = - \left( {\sin \theta _0 } \right)$ and ${\text{c}} = - \left( {\frac{{\cos \theta _0 }}{2}} \right)$,

then: A = a + b × θ + c × θ ². Equation 25 can be expressed as:

$$d\theta = {\text{K}} \times \sqrt 2 \times \sqrt {a + b \times \theta + c \times \theta ^2 } \times d{\text{S}}$$

(27)

We can express:

$${\text{k}} + \left( {\sin \theta _0 } \right) \times \theta + \left( {\frac{{\cos \theta _0 }}{2}} \right) \times \theta ^2 = \left( {\alpha + \theta \times \sqrt {\frac{{\cos \theta _0 }}{2}} } \right)^2 ,$$

with:

$$2\alpha \times \sqrt {\frac{{\cos \theta _0 }}{2}} = \sin \theta \quad ,or:\quad \alpha = \frac{{\sin \theta _0 }}{{\sqrt {2\cos \theta _0 } }}\quad and\;{\text{k = }}\alpha ^2 {\text{.}}$$

Now, we have:

$$\left( {\frac{{\cos \theta _0 }}{2}} \right) \times \theta ^2 = \left( {\theta \times \sqrt {\frac{{\cos \theta _0 }}{2}} + \frac{{\sin \theta _0 }}{{\sqrt {2\cos \theta _0 } }}} \right)^2 - \left( {\frac{{\sin \theta _0 }}{{\sqrt {2\cos \theta _0 } }}} \right)^2 - \left( {\sin \theta _0 } \right) \times \theta $$

Replacing in Eq. 26, one obtains:

$${\text{A}} = \left( {{\text{w}} \times \sqrt {\frac{{\cos \theta _0 }}{2}} + \frac{{\sin \theta _0 }}{{\sqrt {2\cos \theta _0 } }}} \right)^2 - \left( {\theta \times \sqrt {\frac{{\cos \theta _0 }}{2}} + \frac{{\sin \theta _0 }}{{\sqrt {2\cos \theta _0 } }}} \right)^2 $$

(28)

In the last expression, we can recognize that A is expressed in function of a constant a and a variable x. Then A is in the form of: A  =  a ² − x ², so Eq. 27 can be expressed as:

$$d\theta = {\text{K}} \times \sqrt 2 \times \sqrt {{\text{a}}^2 - {\text{x}}^2 } \times d{\text{S,}}$$

(29)

where:

$${\text{a}} = {\text{w}} \times \sqrt {\frac{{\cos \theta _0 }}{2}} + \frac{{\sin \theta _0 }}{{\sqrt {2\cos \theta _0 } }},$$

(30)

$${\text{x}} = \theta \times \sqrt {\frac{{\cos \theta _0 }}{2}} + \frac{{\sin \theta _0 }}{{\sqrt {2\cos \theta _0 } }}$$

(31)

In Eq. (31), x varies from $\frac{{\sin \theta _0 }}{{\sqrt {2\cos \theta _0 } }}$ to ${\text{w}} \times \sqrt {\frac{{\sin \theta _0 }}{2}} + \frac{{\sin \theta _0 }}{{\sqrt {2\cos \theta _0 } }}$, because θ varies from 0 to w. By differentiation with respect to θ, Eq. 31 gives:

$$\begin{aligned} & {\text{d}}x = \sqrt {\frac{{\cos \theta _0 }}{2}} {\text{d}}\theta \\ & {\text{d}}\theta = \frac{{{\text{d}}x}}{{\sqrt {\frac{{{\text{cos}}\theta _0 }}{2}} }}. \\ \end{aligned} $$

When replacing dθ by dx in Eq. 27, we obtain:

$$\begin{array}{*{20}l} {{d{\text{x}} = {\sqrt {\frac{{\cos \theta _{0} }}{2}} } \times {\text{K}} \times {\sqrt 2 } \times {\sqrt {{\text{a}}^{2} - {\text{x}}^{2} } } \times d{\text{S}},} \hfill} \\ {{\quad \quad \quad \quad \;\frac{{d{\text{x}}}}{{{\sqrt {{\text{a}}^{2} - {\text{x}}^{2} } }}} = {\sqrt {\cos \theta _{0} } } \times {\text{K}} \times d{\text{S}}} \hfill} \\ \end{array} $$

(32)

The length of the branch is h (see Fig. 15) so the integration with respect to S of Eq. 32 gives:

$$\int\limits_{\frac{{\sin \theta _0 }}{{\sqrt {2\;\cos \theta _0 } }}}^{{\text{w}}\sqrt {\frac{{\cos \theta _0 }}{2}} + \frac{{\sin \theta _0 }}{{\sqrt {2\;\cos \theta } _0 }}} {\frac{{d{\text{x}}}}{{\sqrt {{\text{a}}^2 - {\text{x}}^2 } }} = \sqrt {\cos \theta _0 } \times {\text{K}} \times {\text{h}}} $$

(33)

By definition:

$$\int\limits_{{\text{x}}1}^{{\text{x}}2} {\frac{{{\text{dx}}}}{{\sqrt {{\text{a}}^2 - {\text{x}}^2 } }} = \arcsin \frac{{{\text{x}}2}}{{\text{a}}}} - \arcsin \frac{{{\text{x}}1}}{{\text{a}}}\;\;,and\;because\;\;{\text{a}} = {\text{w}} \times \sqrt {\frac{{\cos \theta _0 }}{2}} + \frac{{\sin \theta _0 }}{{\sqrt {2\cos \theta _0 } }},$$

then, Eq. 33 can be expressed as:

$$\arcsin 1 - \arcsin \left( {\frac{{\frac{{\sin \theta _0 }}{{\sqrt {2\cos \theta _0 } }}}}{{{\text{w}} \times \sqrt {\frac{{\cos \theta _0 }}{2}} + \frac{{\sin \theta _0 }}{{\sqrt {2\cos \theta _0 } }}}}} \right) = \sqrt {\cos \theta _0 } \times {\text{K}} \times {\text{h}}$$

(34)

Solving for w gives:

$${\text{w}} = \frac{{\sin \theta _0 \left( {1 - \cos \left( {\sqrt {\cos \theta _0 } \times {\text{K}} \times {\text{h}}} \right)} \right)}}{{\cos \theta _0 \times \cos \left( {\sqrt {\cos \theta _0 } \times {\text{K}} \times {\text{h}}} \right)}}$$

(35)