Cloth

Particle Connection

One way to connect the particles is to add a spring between for each neighbour of a particle. Then using bending springs to connect a particle with a particle two rows or columns over in the lattice.

Based on the cloth in Fig. 2.

Symplectic Euler:

$$\begin{align*} \mathfrak{M}v^{(k + 1)} &= \mathfrak{M} v^{(k)} + hf^{(k)}\\ \begin{pmatrix} M_0 & 0\\ 0 & M_1 \end{pmatrix} \begin{pmatrix} v_0^{(k + 1)}\\ v_1^{(k + 1)} \end{pmatrix} &= \begin{pmatrix} M_0 & 0\\ 0 & M_1 \end{pmatrix} \begin{pmatrix} v_0^{(k)}\\ v_1^{(k)} \end{pmatrix} + h \begin{pmatrix} \phantom{-}E\left(\frac{\ell - L}{\ell}\right)\Delta x^{(k)}\\ -E\left(\frac{\ell - L}{\ell}\right)\Delta x^{(k)} \end{pmatrix} \end{align*}$$

Implicit Euler:

$$\begin{align*} (\mathfrak{M} - h^2 \mathfrak{K}) v^{(k + 1)} &= \mathfrak{M}v^{(k)} + hf^{(k)}\\ \mathfrak{K} &= \frac{\mathrm{d}f}{\mathrm{d}x}\\ &= \frac{\mathrm{d}}{\mathrm{d}x} \begin{pmatrix} \phantom{-}E\left(\frac{\ell - L}{\ell}\right)\Delta x^{(k)}\\ -E\left(\frac{\ell - L}{\ell}\right)\Delta x^{(k)} \end{pmatrix}\\ &= \frac{\mathrm{d}}{\mathrm{d}x} \left\{ E \frac{\ell - L}{\ell} \begin{pmatrix} -\mathbf{I}_3 & \phantom{-}\mathbf{I}_3\\ \phantom{-}\mathbf{I}_3 & -\mathbf{I}_3 \end{pmatrix} \begin{pmatrix} x_0\\ x_1 \end{pmatrix} \right\}\\ &= \begin{pmatrix} -k_s & \phantom{-}k_s\\ \phantom{-}k_s & -k_s \end{pmatrix}\\ k_s &= \frac{E}{\ell^2}\bigg(\frac{\ell - L}{\ell} \overbrace{\left(\Delta x^T \Delta x\right)}^{\textrm{inner product}} + \left(1 - \frac{\ell - L}{L}\right)\overbrace{\left(\Delta x \Delta x^T\right)}^{\textrm{outer product}}\bigg) \end{align*}$$

FIXME: "Figure 3: A 3-particle cloth."

$$\begin{align*} \htmlStyle{color: var(\-\-bs\-red);}{f_{01}^{s}} &= E \frac{\ell_{01} - L_{01}}{\ell_{01}} \Delta x_{01}, \Delta x_{01} = x_1 - x_0\\ \htmlStyle{color: var(\-\-bs\-green);}{f_{12}^{s}} &= E \frac{\ell_{12} - L_{12}}{\ell_{12}} \Delta x_{12}, \Delta x_{12} = x_2 - x_1 \end{align*}$$

Remember that we have equal and opposite forces: $\htmlStyle{color: var(\-\-bs\-red);}{f_{01}^{s}} = -\htmlStyle{color: var(\-\-bs\-red);}{f_{10}^{s}}.$

Next we compute the total force acting upon each particle and combine it into the block vector $f,$

$$\left. \begin{align*} f_0(x_0, x_1, x_2) &\mathrel{+}= \htmlStyle{color: var(\-\-bs\-red)}{f_{01}^s}\\ f_1(x_0, x_1, x_2) &\mathrel{+}= \htmlStyle{color: var(\-\-bs\-red)}{f_{10}^s}\\ f_1(x_0, x_1, x_2) &\mathrel{+}= \htmlStyle{color: var(\-\-bs\-green)}{f_{12}^s}\\ f_2(x_0, x_1, x_2) &\mathrel{+}= \htmlStyle{color: var(\-\-bs\-green)}{f_{21}^s} \end{align*} \right\}_{\mathbb{R}^{9 \times 1}}\mkern-24mu = f = \begin{pmatrix} m_1\,\vec{g} + \htmlStyle{color: var(\-\-bs\-red)}{f_{01}^s}\phantom{-f_{00}^s}\\ m_2\,\vec{g} - \htmlStyle{color: var(\-\-bs\-red)}{f_{01}^s} + \htmlStyle{color: var(\-\-bs\-green)}{f_{12}^s}\\ m_3\,\vec{g} - \htmlStyle{color: var(\-\-bs\-green)}{f_{12}^s}\phantom{-f_{00}^s} \end{pmatrix} = \begin{pmatrix} f_0 \\ f_1 \\ f_2 \end{pmatrix}$$

We compute $\mathfrak{K}$ as a block matrix by first defining

$$\overbrace{K_{ij}^{s}}^{\mathbb{R}^{3 \times 3}} = \frac{E}{\ell^{2}_{ij}}\left(\frac{\ell_{ij} - L_{ij}}{\ell_{ij}} \Delta x_{ij}^T\Delta x_{ij}\mathbf{I}_3 + \left(1 - \frac{\ell_{ij} - L_{ij}}{\ell_{ij}}\right)\Delta x_{ij} \Delta x_{ij}^{T}\right),$$

then computing the total contributions by each spring

$$\begin{align*} K_{00}(x_0, x_1, x_2) &\mathrel{+}= \htmlStyle{color: var(\-\-bs\-red);}{K_{10}^{s}}\\ K_{01}(x_0, x_1, x_2) &\mathrel{+}= \htmlStyle{color: var(\-\-bs\-red);}{K_{01}^{s}}\\ K_{10}(x_0, x_1, x_2) &\mathrel{+}= \htmlStyle{color: var(\-\-bs\-red);}{K_{01}^{s}}\\ K_{11}(x_0, x_1, x_2) &\mathrel{+}= \htmlStyle{color: var(\-\-bs\-red);}{K_{10}^{s}}\\ K_{11}(x_0, x_1, x_2) &\mathrel{+}= \htmlStyle{color: var(\-\-bs\-green);}{K_{21}^{s}}\\ K_{12}(x_0, x_1, x_2) &\mathrel{+}= \htmlStyle{color: var(\-\-bs\-green);}{K_{12}^{s}}\\ K_{21}(x_0, x_1, x_2) &\mathrel{+}= \htmlStyle{color: var(\-\-bs\-green);}{K_{12}^{s}}\\ K_{22}(x_0, x_1, x_2) &\mathrel{+}= \htmlStyle{color: var(\-\-bs\-green);}{K_{21}^{s}}\\ \end{align*}$$

and writing the stiffness matrix as follows

$$\begin{align*} \mathfrak{K} &= \begin{pmatrix} \htmlStyle{color: var(\-\-bs\-red)}{K_{00}} & \htmlStyle{color: var(\-\-bs\-red)}{K_{01}} & K_{02}\\ \htmlStyle{color: var(\-\-bs\-red)}{K_{10}} & \htmlStyle{color: var(\-\-bs\-orange)}{K_{11}} & \htmlStyle{color: var(\-\-bs\-green)}{K_{12}}\\ K_{20} & \htmlStyle{color: var(\-\-bs\-green)}{K_{21}} & \htmlStyle{color: var(\-\-bs\-green)}{K_{22}} \end{pmatrix}_{\mathbb{R}^{9 \times 9}}\\ &= \begin{pmatrix} \htmlStyle{color: var(\-\-bs\-red)}{-K_{01}^{s}} & \htmlStyle{color: var(\-\-bs\-red)}{K_{01}^{s}} & \mathbf{0}\\ \htmlStyle{color: var(\-\-bs\-red)}{K_{01}^{s}} & \htmlStyle{color: var(\-\-bs\-red)}{-K_{01}^{s}} \htmlStyle{color: var(\-\-bs\-green)}{-K_{12}^{s}} & \htmlStyle{color: var(\-\-bs\-green)}{-K_{12}^{s}}\\ \mathbf{0} & \htmlStyle{color: var(\-\-bs\-green)}{-K_{12}^{s}} & \htmlStyle{color: var(\-\-bs\-green)}{K_{12}^{s}} \end{pmatrix} \end{align*}$$

Note that $K_{02} = K_{20} = \mathbf{0}$ because there are no springs directly connecting particles $x_0$ and $x_2$.

We define the mass and velocity block matrices as

$$\begin{align*} \mathfrak{M} &= \begin{pmatrix} M_0 & 0 & 0\\ 0 & M_1 & 0\\ 0 & 0 & M_2\\ \end{pmatrix}_{\mathbb{R}^{9 \times 9}}\\ v^{(k + 1)} &= \begin{pmatrix} v_0^{(k + 1)}\\v_2^{(k + 1)}\\v_3^{(k + 1)} \end{pmatrix}_{\mathbb{R}^{9 \times 1}} \end{align*}$$

Finally, solve $Ax = b$ where $A = \mathfrak{M} - h^2 \mathfrak{K},$ and $b = \mathfrak{M}v^{(k)} + hf$.

3 Particles in a Triangle

The force vector is given by

$$f = \begin{pmatrix} m_0\,\vec{g} + \htmlStyle{color: var(\-\-bs\-red)}{f_{01}} - \htmlStyle{color: var(\-\-bs\-blue)}{f_{20}}\\ m_1\,\vec{g} + \htmlStyle{color: var(\-\-bs\-green)}{f_{12}} - \htmlStyle{color: var(\-\-bs\-red)}{f_{01}}\\ m_2\,\vec{g} + \htmlStyle{color: var(\-\-bs\-blue)}{f_{20}} - \htmlStyle{color: var(\-\-bs\-green)}{f_{12}} \end{pmatrix}_{\mathbb{R}^{9 \times 1}}$$

and the stiffness matrix

$$\mathfrak{K} = \begin{pmatrix} \htmlStyle{color: var(\-\-bs\-red)}{-K_{01}}\htmlStyle{color: var(\-\-bs\-blue)}{-K_{20}} & \htmlStyle{color: var(\-\-bs\-red)}{K_{01}} & \htmlStyle{color: var(\-\-bs\-blue)}{K_{20}}\\ \htmlStyle{color: var(\-\-bs\-red)}{K_{01}} & \htmlStyle{color: var(\-\-bs\-green)}{-K_{12}} \htmlStyle{color: var(\-\-bs\-red)}{-K_{01}} & \htmlStyle{color: var(\-\-bs\-green)}{K_{12}}\\ \htmlStyle{color: var(\-\-bs\-blue)}{K_{20}} & \htmlStyle{color: var(\-\-bs\-green)}{K_{12}} & \htmlStyle{color: var(\-\-bs\-blue)}{-K_{20}} \htmlStyle{color: var(\-\-bs\-green)}{-K_{12}} \end{pmatrix}_{\mathbb{R}^{9 \times 9}}$$

where $K_{ij}$ is again computed as the total contribution of $K^{s}$ on a particle.

Finally, we solve $Ax = b$ like in the previous example.