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dokumentation/B-Spline-Volumes.md

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+# B-Spline Volumes
+
+B-Spline Volumes are a simple extension of B-Splines to 3 Dimensions. This is
+a straightforward adaption of the 2-Dimensional version.
+
+## Nomenclature
+
+$x,y,z$ denote space-coordinates,  
+$u,v,w$ denote spline-coordinates (Between 0-1),  
+$P_{ijk}$ denote the control-Points on the control-Polygon,  
+$N_{i,d,\tau}(u)$ denote the value of the underlying Basis-Functions at value
+$u$ using the $i$-th Basis-Function of degree $d$ in range $\tau$.
+
+For our case we only care about degree-3 splines, so we omit the d furtheron.
+$\tau$ is defined statically (in each direction) with each $P$ as Position on the whole surface/volume and within [0,1].
+For a regular Control-Grid this defaults to $\tau_i = i/n$
+
+Given $n,m,o$ control points in $x,y,z$-direction each Point on the curve is
+defined by
+
+$$C(u,v,w) = \sum_{i=0}^{n-d-2} \sum_{j=0}^{m-d-2} \sum_{k=0}^{o-d-2} P_{ijk} N_{i}(u) N_j(v) N_k(w)$$
+
+## Calculate $u, v, w$
+
+Given a target-point $\textbf{p}^*$ and an initial guess $\textbf{p}=C(u,v,w)$ we
+define the error-function as:
+
+$$Err(u,v,w,\textbf{p}^{*}) = \| \textbf{p}^{*} - C(u,v,w) \|_2^2$$
+
+As the error is just the sum of the components
+
+$$Err(u,v,w,\textbf{p}^{*}) = Err_x(u,v,w,\textbf{p}^{*}) + Err_y(u,v,w,\textbf{p}^{*}) + Err_z(u,v,w,\textbf{p}^{*})$$
+
+we just take one axis into account, as the others are nearly identical. So we yield
+
+$$Err_x(u,v,w,\textbf{p}^{*}) = \left( p^{*}_x - \sum_{i=0}^{n-d-2} \sum_{j=0}^{m-d-2} \sum_{k=0}^{o-d-2} {P_{ijk}}_x N_{i}(u) N_j(v) N_k(w) \right)^2 $$
+
+To solve this we derive:
+
+$$
+\begin{array}{rl}
+        \displaystyle \frac{\partial}{\partial u} & \left( p^{*}_x - \displaystyle \sum_{i=0}^{n-d-2} \sum_{j=0}^{m-d-2} \sum_{k=0}^{o-d-2} {P_{ijk}}_x N_{i}(u) N_j(v) N_k(w) \right)^2 \\
+                                  = & \displaystyle - \sum_{i=0}^{n-d-2} \sum_{j=0}^{m-d-2} \sum_{k=0}^{o-d-2} {P_{ijk}}_x N'_{i}(u) N_j(v) N_k(w) \\
+                                    & \cdot \left( p^{*}_x - \displaystyle \sum_{i=0}^{n-d-2} \sum_{j=0}^{m-d-2} \sum_{k=0}^{o-d-2} {P_{ijk}}_x N_{i}(u) N_j(v) N_k(w) \right) \\
+                                    & \cdot 2
+\end{array}
+$$
+
+The other partial derivatives follow the same pattern.
+
+## Basis-Splines and Derivatives
+
+The previously mentioned $N_{i,d,\tau}$ are defined recursively:
+
+$$N_{i,0,\tau}(u) = \begin{cases} 1, & u \in [\tau_i, \tau_{i+1}[ \\ 0, & \mbox{otherwise} \end{cases} $$
+
+and
+
+$$N_{i,d,\tau}(u) = \frac{u-\tau_i}{\tau_{i+d}} N_{i,d-1,\tau}(u) + \frac{\tau_{i+d+1} - u}{\tau_{i+d+1}-\tau_{i+1}} N_{i+1,d-1,\tau}(u) $$
+
+This fact can be exploited to get the derivative for an arbitrary $N$:
+
+$$\frac{d}{du} N_{i,d,r}(u) = \frac{d}{\tau_{i+d} - \tau_i} N_{i,d-1,\tau}(u) - \frac{d}{\tau_{i+d+1} - \tau_{i+1}} N_{i+1,d-1,\tau}(u)$$
+
+*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 $0$

dokumentation/B-Spline-Volumes2.md

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