A new method for T-spline parameterization of complex 2D geometries

Abstract

We present a new strategy, based on the idea of the meccano method and a novel T-mesh optimization procedure, to construct a T-spline parameterization of 2D geometries for the application of isogeometric analysis. The proposed method only demands a boundary representation of the geometry as input data. The algorithm obtains, as a result, high quality parametric transformation between 2D objects and the parametric domain, the unit square. First, we define a parametric mapping between the input boundary of the object and the boundary of the parametric domain. Then, we build a T-mesh adapted to the geometric singularities of the domain to preserve the features of the object boundary with a desired tolerance. The key of the method lies in defining an isomorphic transformation between the parametric and physical T-mesh finding the optimal position of the interior nodes by applying a new T-mesh untangling and smoothing procedure. Bivariate T-spline representation is calculated by imposing the interpolation conditions on points sited both in the interior and on the boundary of the geometry. The efficacy of the proposed technique is shown in several examples. Also we present some results of the application of isogeometric analysis in a geometry parameterized with this technique.

Appendices

Appendix

Regular node case

Here we explain and proof the election of the weights of the objective function for a regular node.

Let us consider the objective function for the conformal local submesh, as shown in Fig. 18a

$$ \begin{aligned} K({\mathbf{x}})&=\sum\limits_{i=1}^M\eta_i({\mathbf{x}}).\\ \eta_i({\mathbf{x}}) &=\eta({S}_i({\mathbf{x}}))=\frac{1}{q(S_i({\mathbf{x}}))}, \quad S_i({\mathbf{x}})=A_i({\mathbf{x}}){W}^{-1}, \end{aligned} $$

where q(S _i ) is any algebraic scale-invariant quality metric for the triangle T _i . In this work, we have used mean ratio quality metric $q(S_i({\mathbf{x}}))=\frac{2\det(S_i({\mathbf{x}}))}{\|S_i({\mathbf{x}})\|^2}.$

Fig. 18

a Conformal local submesh with optimal position (x ₀,y ₀), b non-conformal local submesh with the same optimal position (x ₀,y ₀) due to modification of objective function

Note that, according to the definition of an algebraic scale-invariant quality metric

$$ \eta(\tau\;S_i({\mathbf{x}}))=\eta(S_i({\mathbf{x}})),\quad \tau \in {\mathbb{R}}^+. $$

(11)

If we consider that ${\mathbf{x}}_{0}=(x_0,y_0)$ is the optimal position of the free node, then

$$ \left\{ \begin{array}{l} \partial_x K(x_0,y_0)=\sum\limits_{i=1}^M\partial_x\eta_i({x_0,y_0})=0\\ \partial_y K(x_0,y_0)=\sum\limits_{i=1}^M\partial_y\eta_i({x_0,y_0})=0.\\ \end{array} \right. $$

(12)

Let us suppose that, for the conformal case (see Fig. 18a), the matrix A _i is given by

$$ A_i({\mathbf{x}})=\left( \begin{array}{ll} x_1-x& x_2-x \\ y_1-y&y_2-y \end{array} \right). $$

Let us consider now the non-conformal case (Fig. 18b), where each triangle $\bar{T}_i$ is a scaled version of the triangle T _i when the free node is sited in its optimal position ${\mathbf{x}}_{0}$. That is, the triangle T _i is transformed into $\bar{T}_i$ by means of the affine mapping $\bar{\mathbf{x}}={\mathbf{x}}_0+\tau_i({\mathbf{x}}-{\mathbf{x}}_0)$. The objective function $\bar{K}$ for the non-conformal case is defined as

$$ \begin{aligned} \bar{K}({\mathbf{x}})&=\sum\limits_{i=1}^M\bar{\eta}(S_i({\mathbf{x}})) =\sum\limits_{i=1}^M\eta(\bar{S}_i({\mathbf{x}})), \quad \hbox {where}\\ \bar{S}_i({\mathbf{x}})&=\bar{A}_i({\mathbf{x}}){W}^{-1}\quad \hbox{and}\\ \bar{A}_i({\mathbf{x}})&=\left( \begin{array}{ll} \bar{x}_1-x &\bar{x}_2-x \\ \bar{y}_1-y&\bar{y}_2-y \end{array} \right). \end{aligned} $$

In general, the optimal position obtained by minimizing $\bar{K}$ is different from the one obtained by minimizing K. The goal is to modify the objective function $\bar{K}$ to the optimal position of the free node be the same as in the conformal case, as illustrated in Fig. 18.

The matrix $\bar{A}_i({\mathbf{x}})$ is such that $\bar{A}_i({\mathbf{x}}_0)=\tau_i\;A_i({\mathbf{x}}_0)$, where τ _i is the scale factor for the triangle $\bar{T}_i$. Let us consider the following transformation of the matrix $\bar{A}_i({\mathbf{x}})$:

$$ \begin{aligned} \bar{A}_i({\mathbf{x}})&=\left( \begin{array}{ll} \bar{x}_1-x&\qquad \bar{x}_2-x \\ \bar{y}_1-y&\qquad \bar{y}_2-y \end{array} \right)\\ &=\left( \begin{array}{ll} x_0+\tau_i (x_1-x_0)-x&\qquad x_0+\tau_i (x_2-x_0)-x \\ y_0+\tau_i (y_1-y_0)-y&\qquad y_0+\tau_i (y_2-y_0)-y \end{array} \right)\\ &=\tau_i \left( \begin{array}{ll} x_1-(x_0+\frac{1}{\tau_i}(x-x_0))&\quad x_2-(x_0+\frac{1}{\tau_i}(x-x_0)) \\ y_1-(y_0+\frac{1}{\tau_i}(y-y_0))&\quad y_2-(y_0+\frac{1}{\tau_i}(y-y_0)) \end{array} \right)\\ &=\tau_i \left( \begin{array}{ll} x_1-\hat x&\qquad x_2-\hat x \\ y_1-\hat y&\qquad y_2-\hat y \end{array} \right)= \tau_i A_i(\hat x,\hat y)=\tau_i A_i(\hat{{\mathbf{x}}}), \end{aligned} $$

where $\hat{{\mathbf{x}}}=\hat{ {\mathbf{x}}}({\mathbf{x}})={\mathbf{x}}_0+ \frac{1}{\tau_i}({\mathbf{x}}-{\mathbf{x}}_0).$ Note that

$$ \hat {{\mathbf{x}}}({\mathbf{x}}_0)={\mathbf{x}}_0. $$

(13)

From the previous transformation we have:

$$ \begin{aligned} \bar{S}_i({\mathbf{x}})&=\tau_i\;S_i(\hat{{\mathbf{x}}})\quad \hbox{and}\\ \bar{\eta}_i({\mathbf{x}})&=\eta(\bar{S}_i({\mathbf{x}}))= \eta(\tau_i\;S_i(\hat{{\mathbf{x}}}))\overset{\hbox {{(11)}}}=\eta(S_i(\hat{{\mathbf{x}}}))=\eta_i(\hat{{\mathbf{x}}}). \end{aligned} $$

Deriving with respect to variable x we obtain

$$ \frac{\partial\bar{\eta}_i({\mathbf{x}}) }{\partial x}= \frac{\partial\eta_i(\hat{{\mathbf{x}}}) }{\partial x}= \frac{\partial\eta_i(\hat{{\mathbf{x}}})}{\partial \hat x} \; \frac{\partial \hat x}{\partial x}= \frac{1}{\tau_i}\;\frac{\partial\eta_i(\hat{{\mathbf{x}}}) }{\partial \hat x}, $$

(14)

and evaluating at ${\mathbf{x}}_0$ we have

$$ \begin{aligned} &\frac{\partial\bar{\eta}_i({\mathbf{x}}) }{\partial x}\Bigr|_{{\mathbf{x}}={\mathbf{x}}_0}\overset{\hbox {{(14)}}}= \frac{1}{\tau_i}\;\frac{\partial\eta_i(\hat{{\mathbf{x}}}) }{\partial \hat x}\Bigr|_{\hat{{\mathbf{x}}}=\hat{ {\mathbf{x}}}({\mathbf{x}}_0)}\overset{\hbox {{(13)}}}= \frac{1}{\tau_i}\;\frac{\partial\eta_i(\hat{{\mathbf{x}}}) }{\partial \hat x}\Bigr|_{\hat{{\mathbf{x}}}={\mathbf{x}}_0}\\ &\quad =\frac{1}{\tau_i}\;\frac{\partial\eta_i({\mathbf{x}}) }{\partial x}\Bigr|_{{\mathbf{x}}={\mathbf{x}}_0}. \end{aligned} $$

Therefore, the derivatives of the objective function $\bar{K}$ at the point (x ₀,y ₀) satisfy

$$ \left\{ \begin{array}{l} \partial_x \bar{K}(x_0,y_0)=\sum\limits_{i=1}^M\partial_x\bar{\eta}_i({x_0,y_0})= \sum\limits_{i=1}^M \frac{1}{\tau_i}\;\partial_x\eta_i({x_0,y_0})\\ \partial_y \bar{K}(x_0,y_0)=\sum\limits_{i=1}^M\partial_y\bar{\eta}_i({x_0,y_0})= \sum\limits_{i=1}^M \frac{1}{\tau_i}\;\partial_y\eta_i({x_0,y_0}),\\\end{array} \right. $$

and

$$ \left\{ \begin{array}{l} \sum\limits_{i=1}^M\tau_i\;\partial_x\bar{\eta}_i({x_0,y_0})= \sum\limits_{i=1}^M\partial_x\eta_i({x_0,y_0})\overset{\hbox {{(12)}}}=0\\ \sum\limits_{i=1}^M\tau_i\;\partial_y\bar{\eta}_i({x_0,y_0})= \sum\limits_{i=1}^M \partial_y\eta_i({x_0,y_0})\overset{\hbox {{(12)}}}=0.\\\end{array} \right. $$

So we have that (x ₀,y ₀) is the minimum for the weighted objective function defined as $\bar{K}_{\tau}(x,y)=\sum\nolimits_{i=1}^M\tau_i\;\bar{\eta}_i(x,y).$

This result is valid for a local submesh formed by any number of triangles and for any algebraic scale-invariant quality metric.

Applying the result to a T-mesh (see Fig. 19), and taking into account that all triangles of a cell have the same scale factor, we have that the weighted objective function $\bar{K}_{\tau}$ is

$$ \bar{K}_{\tau}(x,y)=\tau_1\sum\limits_{i=1}^3\bar{\eta}_i(x,y)+\tau_2\sum\limits_{i=4}^6\bar{\eta}_i(x,y) +\tau_3\sum\limits_{i=7}^9\bar{\eta}_i(x,y)+\tau_4\sum\limits_{i=10}^{12}\bar{\eta}_i(x,y). $$

Fig. 19

An example of T-mesh and its scale factors for each cell. a Conformal local submesh with optimal position (x ₀,y ₀); b T-mesh with different cell size surrounding the free node and the same optimal position (x ₀,y ₀) obtained from the weighted objective function

Note that for our quadtree-structured balanced T-mesh all possible scale factors for a free node have relation 1:2:4.

Thus, the definition of the weighted objective function given by Eq. 4 is justified, where the weights τ ₁, τ ₂, τ ₃ and τ ₄ take values 1, 2 or 4.

Hanging node case

To determine the weights of the objective function for the hanging node case, we pose the following problem. For the given physical local submesh, as shown in Fig. 20a, we have to find the appropriate weights τ ₁ and τ ₂ such that the optimal position of free node, according to the weighted objective function K _τ , will produce an orthogonal local mesh, as shown in Fig. 20b. To do that, we evaluate the derivatives of the objective function K _τ and force these derivatives to be equal to zero at (x ₀,y ₀). Objective function K _τ is defined as

$$ K_{\tau}(x,y)=\tau_1\sum\limits_{i=1}^3\eta_i(x,y)+\tau_1\sum\limits_{i=4}^6\eta_i(x,y)+ \tau_2\sum\limits_{i=7}^{11}\eta_i(x,y). $$

The derivatives of the function $\eta({\bf S}(x,y))=\frac{\|S\|^2}{2\det(S)}$ are

$$ \begin{aligned} \partial_x\eta &= \eta(S)\left[ \frac{\langle\partial_xS, S \rangle}{\|S\|^2} -\frac{\partial_x \det(S)}{2\det(S)}\right],\\ \partial_y\eta &= \eta(S)\left[ \frac{\langle\partial_yS, S \rangle}{\|S\|^2} -\frac{\partial_y \det(S)}{2\det(S)}\right], \end{aligned} $$

where the inner product $\langle\cdot,\cdot\rangle$ is defined as $\langle A,B\rangle={\rm Tr}(A^{\rm T},B)$.

Fig. 20

Modification of the objective function to obtain, when possible, the orthogonality of the physical local submesh with hanging node. a Physical mesh and the optimal position of free node for the unweighted objective function. b The optimal position (x ₀,y ₀) of free node for the weighted objective function

The derivative at (x ₀,y ₀) with respect to y is equal to zero due to the symmetry of the problem, and the derivative with respect to x is given by $\partial_x\eta(x_0,y_0)=\frac{(b^2-a^2)(5 \tau_2-8 \tau_1)}{2a^2 b}$.

Thus, the point (x ₀,y ₀) will be the minimum, independently of values a and b, if $\tau_2=\frac{8}{5} \tau_1$. The weights we have used in this work are τ ₁ = 1 and $\tau_2=\frac{8}{5}$.

It should be pointed out that the case of hanging node is a bit specific and its treatment is not completely analogous to the treatment of the regular node case.