4b665fa82b
URDF import demo: add COLLADA .dae file support add FiniteElementMethod demo, extracted from the OpenTissue library (under the zlib license) don't crash if loading an invalid STL file add comparison with Assimp for COLLADA file loading (disabled by default, to avoid library dependency) Gwen: disable some flags that make the build incompatible
404 lines
16 KiB
C++
404 lines
16 KiB
C++
#ifndef OPENTISSUE_DYNAMICS_FEM_FEM_COMPUTE_B_H
|
|
#define OPENTISSUE_DYNAMICS_FEM_FEM_COMPUTE_B_H
|
|
//
|
|
// OpenTissue Template Library
|
|
// - A generic toolbox for physics-based modeling and simulation.
|
|
// Copyright (C) 2008 Department of Computer Science, University of Copenhagen.
|
|
//
|
|
// OTTL is licensed under zlib: http://opensource.org/licenses/zlib-license.php
|
|
//
|
|
#include <OpenTissue/configuration.h>
|
|
|
|
namespace OpenTissue
|
|
{
|
|
namespace fem
|
|
{
|
|
namespace detail
|
|
{
|
|
|
|
/**
|
|
* Calculate B-matrices (derivatives of shape functions N, i.e B = SN).
|
|
*
|
|
* @param e10 Given the four corner points p0,p1,p2, and p3. This is p1-p0.
|
|
* @param e20 Given the four corner points p0,p1,p2, and p3. This is p2-p0.
|
|
* @param e30 Given the four corner points p0,p1,p2, and p3. This is p3-p0.
|
|
* @param volume The volume of the tetrahedron.
|
|
* @param B Upon return this array of 4 B-vectors contains the computed values.
|
|
*/
|
|
template<typename real_type, typename vector3_type>
|
|
inline void compute_B(
|
|
vector3_type const& e10,
|
|
vector3_type const& e20,
|
|
vector3_type const& e30,
|
|
real_type volume,
|
|
vector3_type * B
|
|
)
|
|
{
|
|
//
|
|
// In summary we want to calculate
|
|
//
|
|
// B = S N
|
|
//
|
|
// where
|
|
//
|
|
// B = [ B0 B1 B2 B3 ], N = [ N0 N1 N2 N3 ]
|
|
//
|
|
// Here S is the operational matrix given by
|
|
// - -
|
|
// | d/dx 0 0 |
|
|
// | 0 d/dy 0 |
|
|
// | 0 0 d/dz |
|
|
// S = | d/dy d/dx 0 |
|
|
// | d/dz 0 d/dx |
|
|
// | 0 d/dz d/dy |
|
|
// - -
|
|
// N_i's are called the shape functions and they are used to interpolate values. For
|
|
// 3D linear elastostatics these can be written as
|
|
//
|
|
// - -
|
|
// | w_i 0 0 |
|
|
// N_i = | 0 w_i 0 |
|
|
// | 0 0 w_i |
|
|
// - -
|
|
//
|
|
// The w_i's are functions in the coordinates x,y, and z and will be derived
|
|
// shortly below of here.
|
|
//
|
|
// The matrix B_i is given by:
|
|
//
|
|
// | b_i 0 0 |
|
|
// | 0 c_i 0 |
|
|
// B_i = | 0 0 d_i |
|
|
// | c_i b_i 0 |
|
|
// | d_i 0 b_i |
|
|
// | 0 d_i c_i |
|
|
//
|
|
// The entries are stored as
|
|
//
|
|
// B[i] = [ b_i c_i d_i ]
|
|
//
|
|
// In the following we go into details of deriving formulas for the interpolating
|
|
// functions N_i. From these formulas it follows how to compute the derivatives.
|
|
// Finally we show a clever way of implementing the computaitons.
|
|
//
|
|
// Using a linear polynomial for interpolating a u(x,y,z)-value at a given
|
|
// position x,y,z inside the tetrahedron
|
|
//
|
|
// u(x,y,z) = a_0 + a_1 x + a_2 y + a_3 z = P^T A
|
|
//
|
|
// where the a's are the polynomial coefficients (to be determed) and
|
|
//
|
|
// | 1 | | a0 |
|
|
// P = | x | , and A = | a1 |
|
|
// | y | | a2 |
|
|
// | z | | a3 |
|
|
//
|
|
// Say we know the u-values, u0(x0,y0,z0),...,u3(x3,y2,z3), at the four tetrahedron corner
|
|
// points, P0^T [x0 y0 z0],...,P3^T=[x3 y3 z3], then we can set up the linear system
|
|
//
|
|
// | u0 | | 1 x0 y0 z0 | | a0 |
|
|
// | u1 | = | 1 x1 y1 z1 | | a1 | = C A
|
|
// | u2 | | 1 x2 y2 z2 | | a2 |
|
|
// | u3 | | 1 x3 y3 z3 | | a3 |
|
|
//
|
|
// The C matrix is invertible (as long as the four points are in general postion, meaning
|
|
// that no point can be written as a linear combination of any of the three other points),
|
|
// so we can solve the polynomial coefficients by
|
|
//
|
|
// | u0 |
|
|
// A = C^-1 | u1 |
|
|
// | u2 |
|
|
// | u3 |
|
|
// Substitution into the polynomial interpolation formula yields
|
|
//
|
|
// | u0 |
|
|
// u(x,y,z) = P^T A = P^T C^{-1} | u1 | (*1)
|
|
// | u2 |
|
|
// | u3 |
|
|
//
|
|
// Denoting the i'th column of, P^T C^{-1}, by N_i we have u(x,y,z) = sum_i N_i u_i
|
|
//
|
|
//
|
|
// Let us now try to look at the bary-centric coordinates, w0,...,w1 of (x,y,z), these
|
|
// can also be used for interpolation
|
|
//
|
|
// | u0 | |u0|
|
|
// u(z,y,z) = w0 u0 + w1 u1 + w2 u2 + w3 u3 = [w0 w1 w2 w3] | u1 | = W^T |u1|
|
|
// | u2 | |u2|
|
|
// | u3 | |u3|
|
|
//
|
|
// From the coordinates of the four points and the condition 1 = w0 + w1 + w2 + w3 we can set
|
|
// up the linear system
|
|
//
|
|
// | 1 | | 1 1 1 1 | | w0 |
|
|
// | x | = | x0 x1 x2 x3 | | w1 |
|
|
// | y | | y0 y1 y2 y3 | | w2 |
|
|
// | z | | z0 z1 z2 z3 | | w3 |
|
|
//
|
|
//
|
|
// P = Q W
|
|
//
|
|
// Since the four points are in general postion, Q is invertible and we can solve for W
|
|
//
|
|
// W = Q^{-1} P
|
|
//
|
|
// Insertion into the barycentric interpolation formula yields
|
|
//
|
|
// |u0| |u0| |u0|
|
|
// u(x,y,z) = W^T |u1| = ( Q^{-1} P )^T |u1| = P^T Q^{-T} |u1| (*2)
|
|
// |u2| |u2| |u2|
|
|
// |u3| |u3| |u3|
|
|
//
|
|
// Comparision with the polynomial interpolation derivation we see that C = Q^T, furthermore
|
|
// we notice that w_i = N_i. So (not very surprinsingly) bary-centric interpolation is really
|
|
// just linear polynomial interpolation.
|
|
//
|
|
// Notice that for the volume, V, of the tetrahedron we have the relation: 6 V = det( C ) = det ( Q^T )
|
|
//
|
|
// When computing the stiffness matrix we are interested in derivatives of the N_i's with
|
|
// respect to the x,y and z coordinates.
|
|
//
|
|
//
|
|
// From (*1) and (*2) we see (recall we use zero-indexing)
|
|
//
|
|
// d N_i
|
|
// ----- = C^{-1}_{1,i} = Q^{-T}_{1,i} (*3a)
|
|
// d x
|
|
//
|
|
// d N_i
|
|
// ----- = C^{-1}_{2,i} = Q^{-T}_{2,i} (*3b)
|
|
// d y
|
|
//
|
|
// d N_i
|
|
// ----- = C^{-1}_{3,i} = Q^{-T}_{3,i} (*3c)
|
|
// d z
|
|
//
|
|
// Instead of actually computing the inverse of the 4x4 C or Q matrices a computational
|
|
// more efficient solution can be derived, which only requires us to solve for a 3x3 system.
|
|
//
|
|
// By Cramers rule we have
|
|
//
|
|
// (-1)^{i+j} det( C_ji)
|
|
// C^{-1}_{i,j} = ----------------------
|
|
// det(C)
|
|
//
|
|
// where det(C_ji) is the determinant of C with the j'th row and i'th column removed.
|
|
//
|
|
// Now defined
|
|
//
|
|
// e10 = P1 - P0
|
|
// e20 = P2 - P0
|
|
// e30 = P3 - P0
|
|
//
|
|
// and the matrix E
|
|
//
|
|
// | e10^T | | (x1-x0) (y1-y0) (z1-z0) |
|
|
// E = | e20^T | = | (x2-x0) (y2-y0) (z2-z0) |
|
|
// | e30^T | | (x3-x0) (y3-y0) (z3-z0) |
|
|
//
|
|
// Then
|
|
//
|
|
// det(C) = det(E) and C^{-1}_{i+1,j+1} = E^{-1}_{i,j} (*4)
|
|
//
|
|
// This can shown by straightforward computation, immediately is is verified that
|
|
//
|
|
// {-1}^((i+1)+(j+1)) = {-1}^(i+j)
|
|
//
|
|
// So we have to show
|
|
//
|
|
// det( C_(i+1,j+1)) det( E_(ij))
|
|
// --------------------- = --------------
|
|
// det(C) det(E)
|
|
//
|
|
// assume det(C)= det(E) (left for reader as exercise:-) then inorder to prove
|
|
// second half of (*4) we have to show
|
|
//
|
|
// det( C_(i+1,j+1)) = det( E_(ij))
|
|
//
|
|
// A total of 9 cases exist, for here we will show a single case and leave
|
|
// the remaining cases for the reader, use i=1 and j= 0, implying
|
|
//
|
|
// det( C_(12)) = det( E_(01))
|
|
//
|
|
// So
|
|
// - -
|
|
// | 1 x0 z0 | - -
|
|
// det | 1 x2 z2 | = det | (x2-x0) (z2-z0)|
|
|
// | 1 x3 z3 | | (x3-x0) (z3-z0)|
|
|
// - - - -
|
|
//
|
|
// Which is trivially true. So we have
|
|
//
|
|
// | . ... | | . ... |
|
|
// C^{-1} = | . E^{-1} | ; Q^{-T} = | . E^{-T} |
|
|
//
|
|
// As can be seen from (*3) we do not have all the values needed since
|
|
// the first columns of C^{-1} and Q^{-T} are missing. To remedy this problem we can use
|
|
// the condition of the bary-centric coordinates
|
|
//
|
|
// w0 = 1 - w1 - w2 - w3
|
|
//
|
|
// Taking the derivate wrt. x,y, and z yields
|
|
//
|
|
// d N_0
|
|
// ----- = 1 - E^{-1}_01 - E^{-1}_02 - E^{-1}_03 (*5a)
|
|
// d x
|
|
//
|
|
// d N_0
|
|
// ----- = 1 - E^{-1}_11 - E^{-1}_12 - E^{-1}_13 (*5b)
|
|
// d y
|
|
//
|
|
// d N_0
|
|
// ----- = 1 - E^{-1}_21 - E^{-1}_22 - E^{-1}_23 (*5c)
|
|
// d z
|
|
//
|
|
//
|
|
// and for i>0 we have
|
|
//
|
|
// d N_i
|
|
// ----- = E^{-1}_0i (*6a)
|
|
// d x
|
|
//
|
|
// d N_i
|
|
// ----- = E^{-1}_1i (*6b)
|
|
// d y
|
|
//
|
|
//
|
|
// d N_i
|
|
// ----- = E^{-1}_2i (*6c)
|
|
// d z
|
|
//
|
|
// The notation b_i = d N_i / dx, c_i = d N_i / dy, and d_i = d N_i / dz is used. Furthermore
|
|
// all the derivatives are returned as four B-vectors, where
|
|
//
|
|
// B[i] = [ b_i c_i d_i]
|
|
//
|
|
//
|
|
//
|
|
// The above may seem as a mathematical trick, so let us try to derive our
|
|
// equations a litlle differently, but before doing so we must first develop
|
|
// some equations for the volume of a tetrahedron.
|
|
//
|
|
// | e10^T |
|
|
// 6 V = det(E) = | e20^T | = e10 \cdot (e20 \times e30)
|
|
// | e30^T |
|
|
//
|
|
// where e10 = p1-p0, e20 = p2-p0 and so on.
|
|
// In general given the right-hand order i,j,k, and m of the nodes we
|
|
// write vol(i,j,k,m) = eji \cdot (eki \times emi) / 6
|
|
//
|
|
// Barycentric coordinates are infact the volume weighted weights coresponding to
|
|
// the four tetrahedra lying inside the tetrahedron and having apex at the
|
|
// point p=[x,y,z]^T and bases equal to the 4 triangular faces of the enclosing
|
|
// tetrahedron. To realize this we will examine bary-centric coordinates
|
|
// in the 2D case, that is the case of the Triangle. In 2D area corresponds
|
|
// to the volume, so given a trianle and a point p inside it
|
|
//
|
|
// + 2
|
|
// /|
|
|
// / |
|
|
// / |
|
|
// / |
|
|
// / + p|
|
|
// / |
|
|
// / |
|
|
// 0 +-------+ 1
|
|
//
|
|
// Let the area weighted weights be defined as
|
|
//
|
|
// w_0 = area(1,2,p) / area(0,1,2)
|
|
// w_1 = area(0,P,2) / area(0,1,2) = area(2,0,P) / area(0,1,2)
|
|
// w_2 = area(0,1,p) / area(0,1,2)
|
|
//
|
|
// It is intuitive to see that if p moves towards the i'th corner point then
|
|
// the nominator of w_i goes towards area(0,1,2). This means that w_i -> 1
|
|
// while w_j ->0 for j\neq i.
|
|
//
|
|
// Having introduced the barycentric coordinates as the area weighted weights
|
|
// it is quite intuitive to see that we also must have
|
|
//
|
|
// 1 = \sum_i w_i
|
|
//
|
|
// In 3D the barycentric coordinates are the volume weighted weights. That is
|
|
// given the right handed order 0,1,2 and 3 of the corner points, we have by
|
|
// analogy to the 2D case
|
|
//
|
|
// vol(0,1,2,p)
|
|
// w_3 = ------------ = (p-p0) \codt ( e10 \times \e20 ) / 6V
|
|
// V
|
|
//
|
|
// vol(0,1,3,p)
|
|
// w_2 = ------------ = (p-p0) \codt ( e10 \times \e30 ) / 6V
|
|
// V
|
|
//
|
|
// vol(0,2,3,p)
|
|
// w_1 = ------------ = (p-p0) \codt ( e20 \times \e30 ) / 6V
|
|
// V
|
|
//
|
|
// vol(1,3,2,p)
|
|
// w_0 = ------------ = (p-p0) \codt ( e31 \times \e21 ) / 6V
|
|
// V
|
|
//
|
|
// But since we allready know w_3, w_2 and w_1 it is faster to compute w_0 as
|
|
//
|
|
// w_0 = 1 - w_1 - w_2 - w_3
|
|
//
|
|
// Recal from previous that we seek the derivatives of w_i's wrt. the coordinaes when
|
|
// we compute the B functions. Let us write the [x y z] coordinates as [x_0 x_1 x_2] then
|
|
//
|
|
// d w_3
|
|
// ------- = ( e10 \times \e20 )_{x_j} / 6V
|
|
// d x_j
|
|
//
|
|
// That is the j'th coordinate of the cross product divided by 6 times the volume. Similar for
|
|
//
|
|
// d w_2
|
|
// ------- = ( e10 \times \e30 )_{x_j} / 6V
|
|
// d x_j
|
|
//
|
|
// d w_1
|
|
// ------- = ( e20 \times \e30 )_{x_j} / 6V
|
|
// d x_j
|
|
//
|
|
// and finally
|
|
//
|
|
// d w_0 d w_i
|
|
// ------- = 1 - sum_i>0 -------
|
|
// d x_j d x_j
|
|
//
|
|
// Our formulas may seem differnt from those in (*6), however
|
|
//
|
|
// d w_k
|
|
// ------- = ( eji \times \emi )_{x_j} / 6V = E_{-1}_{k,j}
|
|
// d x_j
|
|
//
|
|
// That is the adjoint of E_(j,k) is given by ( eji \times \emi )_{x_j} / 6.
|
|
//
|
|
//
|
|
//
|
|
|
|
real_type div6V = 1/(6.0*volume);
|
|
|
|
B[1](0) = (e20(2) * e30(1) - e20(1) * e30(2)) * div6V; // b0 = -det(E_11), (where E_ij is the submatrix of E with row i and column j removed)
|
|
B[2](0) = (e10(1) * e30(2) - e10(2) * e30(1)) * div6V; // b1 = det(E_12)
|
|
B[3](0) = (e10(2) * e20(1) - e10(1) * e20(2)) * div6V; // b2 = -det(E_13)
|
|
B[0](0) = -B[1](0) - B[2](0) - B[3](0); // b3 = -b0 - b1 - b2
|
|
|
|
B[1](1) = (e20(0) * e30(2) - e20(2) * e30(0)) * div6V; // c0 = det(E_21)
|
|
B[2](1) = (e10(2) * e30(0) - e10(0) * e30(2)) * div6V; // c1 = -det(E_22)
|
|
B[3](1) = (e10(0) * e20(2) - e10(2) * e20(0)) * div6V; // c2 = det(E_23)
|
|
B[0](1) = -B[1](1) - B[2](1) - B[3](1); // c3 = -c0 - c1 - c2
|
|
|
|
B[1](2) = (e20(1) * e30(0) - e20(0) * e30(1)) * div6V; // d0 = -det(E_31)
|
|
B[2](2) = (e10(0) * e30(1) - e10(1) * e30(0)) * div6V; // d1 = det(E_32)
|
|
B[3](2) = (e10(1) * e20(0) - e10(0) * e20(1)) * div6V; // d2 = -det(E_33)
|
|
B[0](2) = -B[1](2) - B[2](2) - B[3](2); // d3 = -d0 - d1 - d2
|
|
}
|
|
|
|
} // namespace detail
|
|
} // namespace fem
|
|
} // namespace OpenTissue
|
|
|
|
//OPENTISSUE_DYNAMICS_FEM_FEM_COMPUTE_B_H
|
|
#endif
|