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COMS W4167: Physically Based Computer Animation
Theme IV, Milestone I – Elastic Simulation
Due 22:00:00 PM Tuesday 23th Nov 2021 (EST)
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1 Introduction
In Theme IV, we’ll tackle the elastic body problem. Simulation of the elastic material is based on formulation
of an elastic energy that is quadratic in a strain that describes how the object is deformed. Using different
aspects of the deformation as our strain, we can capture different modes of deformation in the energies, and
thus obtain the forces that counter-act these modes. In this milestone we’ll examine three elastic forces:
spring force, bending force, and the constant-strain-triangle (CST) force.
2 New XML Features
This milestone introduces the following new simulation features:
- The elasticbodyspringforce feature specifies a spring force:
The i1 and i2 attributes specify between which particles this force acts; note that different than the
spring force we implemented in the first theme, this force does not require an edge between the two
endpoints. The alpha attribute specifies the material’s elastic modulus (equivalent to EA where E is
the young’s modulus and A is the area of the cross-section of the spring). The l0 attribute specifies
the rest length which means the same as the spring in theme I.
- The elasticbodybendingforce feature specifies a bending force:
The i1, i2 and i3 attributes specify between which particle this force acts. To represent a bending
deformation we need three particles, where the second one is the hinge. The alpha attribute specifies
the material’s elastic modulus in resisting the bending mode. The theta0 attribute specifies the rest
angle in radian.
- The elasticbodycstforce feature specifies a CST force:
ebyy=”0″ ebxy=”0″ xb1x=”0″ xb1y=”0″ xb2x=”1″ xb2y=”0″ xb3x=”0″ xb3y=”1″/>
1
Figure 1: The deformation map.
The i1, i2 and i3 attributes specify the three vertices of the triangle where this force acts. The
youngsmodulus and poisonratio attributes are the material parameters. ebxx, ebxy and ebyy encodes
the“resting strain”ˉ that is discussed in the next section. xb1x and xb1y specify the undeformed
position of vertex 1, and xb2x, xb2y, xb3x and xb3y are similar. - Kinematics
3.1 The Deformation Map and Deformation Gradient
The kinematics of a deforming material are represented by a deformation mapping Φ : R2 → R2 taking every
‘undeformed’material point x on the plane to its corresponding deformed position Φ(x). In general, Φ is
an arbitrary nonlinear map taking 2 coordinates denoting a position to a new set of 2 coordinates denoting
some other position.
Now we’d like to know how much this deformation is stretching, compressing or shearing the material.
Since a generic deformation is an arbitrary nonlinear map, we only look at each point locally (by examining
its infinitesimal neighborhood). Consider a point xˉ in the undeformed space, and a nearby point xˉ + δxˉ.
The deformed position are x and x + δx, and by definition of Φ we have:
x = Φ(xˉ)
x + δx = Φ(xˉ + δxˉ)
= Φ(xˉ) +?Φ(xˉ)δxˉ +O(‖δxˉ‖2)
δx = ?Φ(xˉ)δxˉ +O(‖δxˉ‖2)
Φ(xˉ) is the Jacobian of the transformation Φ evaluated at xˉ, often called the Deformation Gradient.
The high order terms in the end can be omitted because we assume δxˉ is infinitesimal.
3.2 The Strain
Now we are ready to evaluate how much lengths change after such a transformation. Under the assumption
that δxˉ is infinitesimal, the line from xˉ to xˉ+δxˉ in the undeformed space remains a line after the deformation,
2
going from x to x+ δx. Therefore the lengths of this piece of material before and after deformation are ‖δxˉ‖
and ‖δx‖ respectively. For the ease of computation we look at the change in squared length (so as to avoid
taking square roots):
δxT δx? δxˉT δxˉ = (?Φ(xˉ)δxˉ)T (?Φ(xˉ)δxˉ)? δxˉT δxˉ
= δxˉT?Φ(xˉ)T?Φ(xˉ)δxˉ? δxˉT δxˉ
= δxˉT [?Φ(xˉ)T?Φ(xˉ)? I]δxˉ
The reason we can use change in squared length instead of change in length itself as a measure of stretching
is that, to the first order, they are equivalent as long as the deformation is small. In order to make this
change in squared length a scale-invariant representation of relative stretching/compression, we extract out
the term in the middle, [?Φ(xˉ)T?Φ(xˉ)? I], and call it the strain, or .
This strain is a second-order tensor, unlike the 1D case where a single scalar fully describes how much a
spring is stretched as shown in class two weeks ago. In 2D materials can be stretched differently in different
directions, so a second-order tensor is needed. The physical meaning of this tensor is that when you hit
it on both sides with a direction δxˉ, it returns a change in squared length in that direction. In its matrix
representation, the eigenvectors of this matrix are the two directions that are stretched or compressed the
most, and the corresponding eigenvalues tell you how much the stretching/compression is in that direction.
Another interesting thing to note is that is rotation-invariant, even though ?Φ is not. As a measure
of deformation, the strain should not depend on any rigid motion such as translation and rotation. It is
obvious that ?Φ, which is the result of a spatial differential operator, stays the same under translation, but
what about rotation? To test this, we apply a constant rotation Q to any deformation Φ to obtain a new
deformation Φ?, and see how responds:
Thus we’ve convinced ourselves that our strain is invariant under any rigid motions, therefore it qualifies
as a measure of deformation.
3.3 Example Deformation
Let’s look at an example with real numbers. Consideration an affine deformation of a rectangular sheet of
rubber as in Figure 2. This rectangle is stretched along the green axis by 80%, and compressed along the
red axis by 40%. This means our deformation map Φ is multiplication by a matrix with two eigenvalues: 1.8
corresponding to the eigenvector (3, 1) (green axis), and 0.6 corresponding to the eigenvector (-1, 3) (red
axis). Therefore Φ looks like this:
Figure 2: An example deformation of square, with stretching in one direction and compression in the other.
The two arrows denote the two eigenvectors of the strain tensor with corresponding eigenvalues (λ1 and λ2)
marked on the arrows.
We can compute the strain tensor from Φ:
It is easy to verify that this 2×2 matrix has eigenvalues 2.24 and -0.64, corresponding to a 80% stretching
(1.82 1 = 2.24) and a 40% compression (0.62 ? 1 = ?0.64). - Physics
The strain tensor gives us a complete description of the deformation of the material, but it doesn’t tell us
how the material will respond. That is out of the scope of kinematics, and we have to put in some ingredients
from physics. Here in this section we’ll attempt to write down the energy that arises from the deformation,
which governs how the simulation will take place in time. For this purpose, we can proceed in either a
mechanics-oriented way, or a mathematics-oriented way.
4.1 The Mathematical Way
Imagine we have somehow obtained the energy function w of strain , we can take its Taylor expansion:
w(0) is a constant independent of and can be conveniently assume to be 0. The linear term is also - because the energy should be minimized at zero strain (the definition of“resting shape”). If we adopt
the small strain assumption, the higher order terms can be omitted since the energy is dominated by the
quadratic term. This is basically saying, no matter how complicated the material laws are, under the small
strain assumption we can always write down the energy as a quadratic function of the strain.
4.2 The Mechanical Way
Physically, the energy is equal to the product between the stress and the strain: