76 lines
3.0 KiB
Markdown
76 lines
3.0 KiB
Markdown
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# B-Spline Volumes
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B-Spline Volumes are a simple extension of B-Splines to 3 Dimensions. This is
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a straightforward adaption of the 2-Dimensional version.
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## Nomenclature
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$x,y,z$ denote space-coordinates,
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$u,v,w$ denote spline-coordinates (Between 0-1),
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$P_{ijk}$ denote the control-Points on the control-Polygon,
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$N_{i,d,\tau}(u)$ denote the value of the underlying Basis-Functions at value
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$u$ using the $i$-th Basis-Function of degree $d$ in range $\tau$.
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For our case we only care about degree-3 splines, so we omit the d furtheron.
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$\tau$ is defined statically (in each direction) with each $P$ as Position on the whole surface/volume and within [0,1].
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For a regular Control-Grid this defaults to $\tau_i = i/n$
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Given $n,m,o$ control points in $x,y,z$-direction each Point on the curve is
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defined by
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$$C(u,v,w) = \sum_{i=0}^{n-d-2} \sum_{j=0}^{m-d-2} \sum_{k=0}^{o-d-2} P_{ijk} N_{i}(u) N_j(v) N_k(w)$$
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## Calculate $u, v, w$
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Given a target-point $\textbf{p}^*$ and an initial guess $\textbf{p}=C(u,v,w)$ we
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define the error-function as:
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$$Err(u,v,w,\textbf{p}^{*}) = \textbf{p}^{*} - C(u,v,w)$$
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$$Err_x(u,v,w,\textbf{p}^{*}) = p^{*}_x - \sum_{i=0}^{n-d-2} \sum_{j=0}^{m-d-2} \sum_{k=0}^{o-d-2} {P_{ijk}}_x N_{i}(u) N_j(v) N_k(w) $$
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To solve this we derive:
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$$
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\begin{array}{rl}
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\displaystyle \frac{\partial Err_x}{\partial u} & p^{*}_x - \displaystyle \sum_{i=0}^{n-d-2} \sum_{j=0}^{m-d-2} \sum_{k=0}^{o-d-2} {P_{ijk}}_x N_{i}(u) N_j(v) N_k(w) \\
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= & \displaystyle - \sum_{i=0}^{n-d-2} \sum_{j=0}^{m-d-2} \sum_{k=0}^{o-d-2} {P_{ijk}}_x N'_{i}(u) N_j(v) N_k(w)
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\end{array}
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$$
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The other partial derivatives follow the same pattern yiedling the Jacobian:
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$$
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J(Err(u,v,w)) =
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\left(
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\begin{array}{ccc}
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\frac{\partial Err_x}{\partial u} & \frac{\partial Err_x}{\partial v} & \frac{\partial Err_x}{\partial w} \\
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\frac{\partial Err_y}{\partial u} & \frac{\partial Err_y}{\partial v} & \frac{\partial Err_y}{\partial w} \\
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\frac{\partial Err_z}{\partial u} & \frac{\partial Err_z}{\partial v} & \frac{\partial Err_z}{\partial w}
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\end{array}
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\right)
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$$
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Iterate with
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$$J(Err(u,v,w)) \cdot \Delta \left( \begin{array}{c} u \\ v \\ w \end{array} \right) = -Err(u,v,w)$$
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using Cramers rule for solving the SLE.
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## Basis-Splines and Derivatives
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The previously mentioned $N_{i,d,\tau}$ are defined recursively:
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$$N_{i,0,\tau}(u) = \begin{cases} 1, & u \in [\tau_i, \tau_{i+1}[ \\ 0, & \mbox{otherwise} \end{cases} $$
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and
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$$N_{i,d,\tau}(u) = \frac{u-\tau_i}{\tau_{i+d}} N_{i,d-1,\tau}(u) + \frac{\tau_{i+d+1} - u}{\tau_{i+d+1}-\tau_{i+1}} N_{i+1,d-1,\tau}(u) $$
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This fact can be exploited to get the derivative for an arbitrary $N$:
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$$\frac{d}{du} N_{i,d,r}(u) = \frac{d}{\tau_{i+d} - \tau_i} N_{i,d-1,\tau}(u) - \frac{d}{\tau_{i+d+1} - \tau_{i+1}} N_{i+1,d-1,\tau}(u)$$
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*Warning:* in the case of $d=1$ the recursion-formula yields a $0$ denominator, but $N$ is also $0$. The right solution for this case is a derivative of $0$
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