Games Webinar Notes

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GAMES 201 Notes:

1. Matrix Operation (P8: 24min)

\(F' = (I+\tau C) F \rightarrow \frac{\partial F'}{\partial C} = \tau F^\mathsf{T}\) is derived from the fact:
\(F' = CF \rightarrow F'^\mathsf{T} = F^\mathsf{T} C^\mathsf{T} ,\) where \(\frac{\partial F'^\mathsf{T}}{\partial C^\mathsf{T}}= F^\mathsf{T}\).
It looks imediately derivation on that slide is wrong, but we might as well consider the whole differential chain. Similarly, \(\frac{\partial C^\mathsf{T}}{\partial v^\mathsf{T}}= (x^\mathsf{T})^\mathsf{T} = x\). Then,
\(\frac{\partial F'^\mathsf{T}}{\partial v^\mathsf{T}}=\frac{\partial F'^\mathsf{T}}{\partial C^\mathsf{T}}\frac{\partial C^\mathsf{T}}{\partial v^\mathsf{T}}= F^\mathsf{T}x\)
This continues forward towards \(\psi\), thence we get the expression of \((\frac{\partial \psi}{\partial v})^\mathsf{T}\). It computes the nodal force and is actually a vactor, so it doesn’t harm if it is transposed.

GAMES 103 Notes:

1. Energy of Spring System (Matrix equation with chain manipulation)

\[E = \frac{1}{2} ([\frac{\partial \phi}{\partial x}] [x])^\mathsf{T} [K] ([\frac{\partial \phi}{\partial x}] [x]) = \frac{1}{2} [x]^\mathsf{T} ( [\frac{\partial \phi}{\partial x}] ^\mathsf{T} [K] [\frac{\partial \phi}{\partial x}] ) [x]\]

Note that \(E\) is the total energy, a scalar. \([K]\) and \([ \frac{\partial \phi}{\partial x} ]\) are 2d matrices, while \([\phi]\), \([x]\) and \([F]\) are vectors.

\[[F] = \frac{\partial E}{\partial x} = ( [\frac{\partial \phi}{\partial x}]^\mathsf{T} [K] [\frac{\partial \phi}{\partial x}] )[x] = [\frac{\partial \phi}{\partial x}]^\mathsf{T} [K] [\phi]\]

2. Constrained Dynamics (P7: 58 min)

\[\left[ \begin{array}{cc} M & -\Delta t J^\mathsf{T} \\ \Delta t J & C \end{array} \right] \left[ \begin{array}{c} v' \\ \lambda ' \end{array} \right]= \left[ \begin{array}{c} Mv \\ -\phi \end{array} \right] \tag{dual variable}\]

This gives

\[\begin{cases} Mv' &-&\Delta t J^\mathsf{T} \lambda ' &= &Mv \\ \Delta t J v' &+& C \lambda ' &= &-\phi \end{cases},\]

and further

\[Mv' = Mv - \Delta t J^\mathsf{T} C^{-1} \phi - \Delta t^2 J^\mathsf{T} C^{-1} J v' = Mv + f \Delta t - \Delta t^2 J^\mathsf{T} C^{-1} J v'.\]

If implicit time integration,

\[\begin{cases} \Delta v = a' \Delta t \Rightarrow Mv' = Mv + f' \Delta t\\ \Delta x = v' \Delta t. \end{cases}\]

So

\[\begin{aligned} & f' \Delta t - f \Delta t & ≟ & - \Delta t^2 J^\mathsf{T} C^{-1} J v'\\ \Rightarrow ~ & f'- f & ≟ & -J^\mathsf{T} C^{-1} J \Delta x \\ \Rightarrow ~ & \frac{\partial f}{\partial x} & ≟ & -J^\mathsf{T} C^{-1} J \end{aligned}\]

but

\[\begin{aligned} f &= - J^\mathsf{T} C^{-1} \phi , \\ \frac{\partial f}{\partial x} &= -\frac{\partial J^\mathsf{T}}{\partial x } C^{-1} \phi -J^\mathsf{T} C^{-1} \frac{\partial \phi}{\partial x} = \frac{\partial J^\mathsf{T}}{\partial x } \lambda -J^\mathsf{T} C^{-1} J . \end{aligned}\]

Therefore, it needs to add the extra term to let

\[Mv' = Mv - \Delta t J^\mathsf{T} C^{-1} \phi - \Delta t^2 J^\mathsf{T} C^{-1} J v' + \Delta t^2 \frac{\partial J^\mathsf{T}}{\partial x } \lambda v',\]

and this induces back to

\[\left[ \begin{array}{cc} M - \Delta t^2 \frac{\partial J^\mathsf{T}}{\partial x } \lambda & -\Delta t J^\mathsf{T} \\ \Delta t J & C \end{array} \right] \left[ \begin{array}{c} v' \\ \lambda ' \end{array} \right]= \left[ \begin{array}{c} Mv \\ -\phi \end{array} \right].\]

Only in this way does it equal first-order implicit time integration.