Touch-enabled haptic modeling of deformable multi-resolution surfaces

Abstract

Currently, interactive data exploration in virtual environments is mainly focused on vision-based and non-contact sensory channels such as visual/auditory displays. The lack of tactile sensation in virtual environments removes an important source of information to be delivered to the users. In this paper, we propose the touch-enabled haptic modeling of deformable multi-resolution surfaces in real time. The 6-DOF haptic manipulation is based on a dynamic model of Loop surfaces, where the dynamic parameters are computed easily without subdividing the control mesh recursively. A local deforming scheme is developed to approximate the solution of the dynamics equations, thus the order of the linear equations is reduced greatly. During each of the haptic interaction loop, the contact point is traced and reflected to the rendering of updated graphics and haptics. The sense of touch against the deforming surface is calculated according to the surface properties and the damping-spring force profile. Our haptic system supports the dynamic modeling of deformable Loop surfaces intuitively through the touch-enabled interactive manipulation.

Appendices

Appendix 1: Basis functions for regular meshes

The column vector b(v,w) contains all 12 regular triangular spline basis functions:

$$\begin{aligned} {\user2{b}}^{{\rm {T}}} (v, w) &= \frac{1}{{12}}(u^{{\rm {4}}} + {2}u^{{\rm {3}}} v, u^{{\rm {4}}} + {2}u^{{\rm {3}}} w, u^{{\rm {4}}} + 2u^{3} w + 6u^{3} v + 6u^{2} vw + 12u^{2} v^{2} + 6uv^{2} w \\ & \quad + 6uv^{3} + 2v^{3} w + v^{4}, 6u^{4} + 24u^{3} w + 24u^{2} w^{2} + 8uw^{3} + w^{4} + 24u^{3} v \\ & \quad + 60u^{2} vw + 36uvw^{2} + 6vw^{3} + 24u^{2} v^{2} + 36uv^{2} w + 12v^{2} w^{2} + 8uv^{3}\\.& \quad + 6v^{3} w + v^{4}, u^{4} + 6u^{3} w + 12u^{2} w^{2} + 6uw^{3} + w^{4} + 2u^{3} v + 6u^{2} vw\\ & \quad + 6uvw^{2} + 2vw^{3}, 2uv^{3} + v^{4}, u^{4} + 6u^{3} w + 12u^{2} w^{2} + 6uw^{3} + w^{4} + 8u^{3} v\\ & \quad + 36u^{2} vw + 36uvw^{2} + 8vw^{3} + 24u^{2} v^{2} + 60uv^{2} w + 24v^{2} w^{2} + 24uv^{3} \\ & \quad + 24v^{3} w+ 6v^{4}, u^{4} + 8u^{3} w + 24u^{2} w^{2} + 24uw^{3} + 6w^{4} + 6u^{3} v \\ & \quad + 36u^{2} vw + 60uvw^{2} + 24vw^{3} + 12u^{2} v^{2} + 36uv^{2} w + 24v^{2} w^{2} + 6uv^{3} \\ & \quad+ 8v^{3} w + v^{4}, 2uw^{3} + w^{4}, 2v^{3} w + v^{4}, 2uw^{3} + w^{4} + 6uvw^{2} + 6vw^{3}\\ & \quad + 6uv^{2} w + 12v^{2} w^{2} + 2uv^{3} + 6v^{3} w + v^{4}, w^{4} + 2vw^{3}), \\ \end{aligned}$$

where (v,w) ∈ Ω and u = 1 − v − w. The basis functions in b(v,w) are arranged according to the order of the local control vertices shown in Fig. 13.

Fig. 13

Local control vertices corresponding to the basis functions

Appendix 2: Subdivision matrices and picking matrices

The K × K subdivision matrix A and the M × K extended subdivision matrix ${{\bar{\user2{A}}}}$ have the following block structures:

$${\user2{A}} = {\left({\begin{array}{*{20}c} {{\user2{S}}}& {{\mathbf{0}}} \\ {{{\user2{S}}_{{11}}}}& {{\user2{S}}} \\ \end{array} _{{12}}} \right)}, \quad {\bar{\user2{A}}} = {\left({\begin{array}{*{20}c} {{\user2{S}}}& {{\mathbf{0}}} \\ {{{\user2{S}}_{{11}}}}& {{{\user2{S}}_{{12}}}} \\ {{{\user2{S}}_{{21}}}}& {{{\user2{S}}_{{22}}}} \\ \end{array}} \right)},$$

where K =  N + 6 and M =  N + 12. The sub-matrix S is an (N + 1) × (N + 1) matrix:

where ${a_{N} = 1 - \alpha _{N}, b_{N} = \alpha _{N} /N, c = {3}/{8}, d = {1}/{8}}$ and ${\alpha _{N} = ({5}/{8}) - {{(3 + 2\cos (2\pi /N))^{2}}}/{{64}}}.$ The other four sub-matrices are given as follows:

$${\user2{S}}_{{11}} = \frac{1}{{16}}{\left({\begin{array}{*{20}c} {2}& {6}& {0}& {0}& {\cdots}& {0}& {0}& {6} \\ {1}& {{10}}& {1}& {0}& {\cdots}& {0}& {0}& {1} \\ {2}& {6}& {6}& {0}& {\cdots}& {0}& {0}& {0} \\ {1}& {1}& {0}& {0}& {\cdots}& {0}& {1}& {{10}} \\ {2}& {0}& {0}& {0}& {\cdots}& {0}& {6}& {6} \\ \end{array}} \right)},\quad {\user2{S}}_{{12}} = \frac{1}{{16}}{\left({\begin{array}{*{20}c} {2}& {0}& {0}& {0}& {0} \\ {1}& {1}& {1}& {0}& {0} \\ {0}& {0}& {2}& {0}& {0} \\ {1}& {0}& {0}& {1}& {1} \\ {0}& {0}& {0}& {0}& {2} \\ \end{array}} \right)},$$

$${\user2{S}}_{{21}} = \frac{1}{8}{\left({\begin{array}{*{20}c} {0}& {3}& {0}& {0}& {{\ldots}}& {0}& {0}& {1} \\ {0}& {3}& {0}& {0}& {{\ldots}}& {0}& {0}& {0} \\ {0}& {3}& {1}& {0}& {{\ldots}}& {0}& {0}& {0} \\ {0}& {1}& {0}& {0}& {{\ldots}}& {0}& {0}& {3} \\ {0}& {0}& {0}& {0}& {{\ldots}}& {0}& {0}& {3} \\ {0}& {0}& {0}& {0}& {{\ldots}}& {0}& {1}& {3} \\ \end{array}} \right)},\quad {\user2{S}}_{{22}} = \frac{1}{8}{\left({\begin{array}{*{20}c} {3}& {1}& {0}& {0}& {0} \\ {1}& {3}& {1}& {0}& {0} \\ {0}& {1}& {3}& {0}& {0} \\ {3}& {0}& {0}& {1}& {0} \\ {1}& {0}& {0}& {3}& {1} \\ {0}& {0}& {0}& {1}& {3} \\ \end{array}} \right)}.$$

The picking matrices P ₁, P ₂ and P ₃ are defined by introducing three permutation vectors as below:

$$ \begin{aligned} {\user2{q}}^{{\rm {1}}} &= (3, 1, N + 4, 2, N + 1, N + 9, N + 3, N + 2, N + 5, N + 8, N + 7, N + 10),\\ {\user2{q}}^{{\rm {2}}} &= (N + 10, N + 7, N + 5, N + 2, N + 3, N + 6, N + 1, 2, N + 4, N, 1, 3),\\ {\user2{q}}^{{\rm {3}}} &= (1, N, 2, N + 1, N + 6, N + 3, N + 2, N + 5, N + 12, N + 7, N + 10, N + 11). \end{aligned} $$

Each P _k is a 12 × M matrix, and its jth row is filled with zeros except for a one in the (q ^k) _j th column, where j = 1, 2, ..., 12, and k = 1, 2, 3.

Appendix 3: Eigenstructure of the subdivision matrix

The eigenstructure of the subdivision matrix A is denoted by ${{\left({{\varvec\Lambda}, {\user2{U}}} \right)}},$ where $\varvec\Lambda$ is the diagonal matrix containing all eigenvalues of A, and U is an invertible matrix whose columns are the corresponding eigenvectors. Firstly, the eigenstructures of the sub-matrices S and S ₁₂ (denoted by ${{\left({{\varvec{\Sigma}}, {\user2{U}}_{0}} \right)}}$ and ${{\left({{\varvec{\Delta}}, {\user2{W}}_{1}} \right)}},$ respectively) can be computed:

$${\varvec{\Sigma}} = \left\{\begin{aligned} {\hbox{diag}}(1,\mu _{2}, \mu _{3}, \mu _{3}, \ldots, \mu _{{(N + 3)/2}}, \mu _{{(N + 3)/2}})\enspace {\hbox{if}}\,N\,{\hbox{is odd,}} \\{\hbox{diag}}(1,\mu _{2}, \mu _{3}, \mu _{3}, \ldots, \mu _{{N/2 + 1}}, \mu _{{N/2 + 1}}, \tfrac{1}{8})\enspace {\hbox{if}}\, N\,{\hbox{is even,}} \\ \end{aligned} \right.$$

$${\user2{U}}_{0} = \left\{\begin{aligned}\,& {\left({{\user2{u}}_{0}, {\user2{u}}_{1}, {\user2{v}}_{1}, {\user2{w}}_{1}, {\user2{v}}_{2}, {\user2{w}}_{2}, \ldots, {\user2{v}}_{{(N - 1)/2}}, {\user2{w}}_{{(N - 1)/2}}} \right)},\,{\hbox{if}}\,N\, {\hbox{is}}\, {\hbox{odd}}; \\& {\left({{\user2{u}}_{0}, {\user2{u}}_{1}, {\user2{v}}_{1}, {\user2{w}}_{1}, {\user2{v}}_{2}, {\user2{w}}_{2}, \ldots, {\user2{v}}_{{N/2 - 1}}, {\user2{w}}_{{N/2 - 1}}, {\user2{v}}_{{N/2}}} \right)},\, {\hbox{if}}\, N\,{\hbox{is}} \,{\hbox{even}}, \\ \end{aligned} \right.$$

$${\varvec{\Delta}} = {\hbox{diag}}\left(\frac{1}{8},\frac{1}{8},\frac{1}{8},\frac{1}{{16}}, \frac{1}{{16}}\right), \quad {\user2{W}}_{1} = {\left({\begin{array}{*{20}c} {0}& {{- 1}}& {1}& {0}& {0} \\ {1}& {{- 1}}& {1}& {0}& {1} \\ {1}& {0}& {0}& {0}& {0} \\ {0}& {0}& {1}& {1}& {0} \\ {0}& {1}& {0}& {0}& {0} \\ \end{array}} \right)},$$

where

$$ \begin{aligned} {\user2{u}}_{0} &= {\left({1, 1, \cdots, 1} \right)}^{{\rm {T}}},\quad {\user2{u}}_{1} = {\left({- 8\alpha _{N} /3, 1, 1, \ldots, 1} \right)}^{{\rm {T}}},\\ {\user2{v}}_{k} &= {\left({0}, {\hbox{C(0),}}\;{\hbox{C}}(k),\;{\hbox{C2}}k), \ldots , \hbox{C}((N- 1)k) \right)}^{{\rm {T}}},\\ {\user2{w}}_{k} &= {\left({0}, {\hbox{S(0),}}\;{\hbox{S(}}k), {\hbox{S(2}}k), \ldots , \hbox{S}((N- 1)k) \right)}^{{\rm {T}}},\\ {\mu}_{1} &= 1, {\mu}_{2} = \frac{5}{8}-\alpha _{N}, {\mu}_{3} = f(1), \ldots, {\mu}_{N + 1} = f(N - 1),\\ {\hbox{C}}(k) &= \cos (2\pi k/N), \quad {\hbox{S}}(k) = \sin (2\pi k/N), \quad f(k) = \frac{3}{8} + \frac{2}{8}\cos (2\pi k/N), \end{aligned}$$

and k = 1,2,..., N−1. Then, one can obtain the eigenstructure of A as follows:

$$\varvec\Lambda = {\left({\begin{array}{*{20}c} {{\varvec{\Sigma}}}& {{\mathbf{0}}} \\ {{\mathbf {0}}}& {{\varvec{\Delta}}} \\ \end{array}} \right)}, \quad {\user2{U}} = {\left({\begin{array}{*{20}c} {{{\user2{U}}_{\text{0}}}}& {{\mathbf{0}}} \\ {{{\user2{U}}_{\text{1}}}}& {{{\user2{W}}_{\text{1}}}} \\ \end{array}} \right)},$$

by solving the unknown block U ₁ column by column from the linear system of equations

$${\user2{U}}_{1} {\varvec{\Sigma}} - {\user2{S}}_{{12}} {\user2{U}}_{1} = {\user2{S}}_{{11}} {\user2{U}}_{0}.$$