# Changes between Version 6 and Version 7 of ThinPlateShell

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Timestamp:
Dec 9, 2014, 5:25:02 PM (6 years ago)
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 v6 = Smoothing interpolation = To account for measurement noise, interpolation should involve smoothing, so that the interpolated data can deviate from the measurement at the measurement sites $r_i$. This can be performed by minimising a linear combination of the elastic energy and the quadratic deviation to the measurement values  $$E = \sum(|f({\bf r_i})-f_i|^2)+ \rho\int\left[\left(\frac{\partial^2 f}{\partial x^2}\right)^2 + 2\left(\frac{\partial^2 f}{\partial xy}\right)^2 + \left(\frac{\partial^2 f}{\partial y^2}\right)^2 \right] \textrm{d} x \, \textrm{d}y$$^ To account for measurement noise, interpolation should involve smoothing, so that the interpolated data can deviate from the measurement at the measurement sites $r_i$. This can be performed by minimising a linear combination of the elastic energy and the quadratic deviation to the measurement values The variational problem then leads to the equation $$\rho\Delta^2 f= \sum (f_i-f) \delta ({\bf r-r_i})$$ The solution is still obtained as a sum of radial basis functions $\phi({\bf|r-r_i}|)$, whose weight $S_i$ now satisfies $M_\rho*S=F$, with the matrix $$M_\rho=M+\rho I_{000}$$ where $I_{000}$ is the NxN identity matrix extended by three columns of 0.^ $E = \sum(|f({\bf r_i})-f_i|^2)+ \rho\int\left[\left(\frac{\partial^2 f}{\partial x^2}\right)^2 + 2\left(\frac{\partial^2 f}{\partial xy}\right)^2 + \left(\frac{\partial^2 f}{\partial y^2}\right)^2 \right] \textrm{d} x \, \textrm{d}y$ The variational problem then leads to the equation $\rho\Delta^2 f= \sum (f_i-f) \delta ({\bf r-r_i})$ The solution is still obtained as a sum of radial basis functions $\phi({\bf|r-r_i}|)$, whose weight $S_i$ now satisfies $M_\rho*S=F$, with the matrix $M_\rho=M+\rho I_{000}$ where $I_{000}$ is the NxN identity matrix extended by three columns of 0.^ = Spatial derivatives = Spatial derivatives of the interpolated quantity $f$ can be obtained by direct differentiation of the result.  For any function $\phi(r)$, with radial distance $r=|{\bf r-r_i}|$, $r^2=X^2+Y^2$, $\partial_X \phi=(d\phi/dr) \partial_X r$, and $\partial_X r=X/r$. This yields $\partial_X \phi=X (2 \log(r) +1)$, so that,^ Spatial derivatives of the interpolated quantity $f$ can be obtained by direct differentiation of the result.  For any function $\phi(r)$, with radial distance $r=|{\bf r-r_i}|$, $r^2=X^2+Y^2$, $\partial_X \phi=(d\phi/dr) \partial_X r$, and $\partial_X r=X/r$. This yields $\partial_X \phi=X (2 \log(r) +1)$, so that, $$\partial_x f({\bf r})=\sum S_i (2 \log|{\bf r-r_i}|+1)(x-x_i)+a_1;$$ $$\partial_y f({\bf r})=\sum S_i (2 \log|{\bf r-r_i}|+1)(y-y_i)+a_2;$$ $\partial_x f({\bf r})=\sum S_i (2 \log|{\bf r-r_i}|+1)(x-x_i)+a_1;$ $\partial_y f({\bf r})=\sum S_i (2 \log|{\bf r-r_i}|+1)(y-y_i)+a_2;$ = Sub-domains =