Cloth Simulation

Mass-Spring Model based

Cloth Simulation

Wanho Choi

The explicit forward Euler time integration x = dv dt a = dv dt = d dt dx dt ⎛ ⎝⎜ ⎞⎠⎟ = d2x dt 2 = M−1 f : Newton' s 2nd law v = dx dt position : velocity : acceleration : differentiation differentiation integration integration d dt x v ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = v M−1f ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 1 Δt xn+1 − xn vn+1 − vn ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = vn M−1f n ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ discretization f : force

(

)

M = m 0 0 0 m 0 0 0 m ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥

: 3 × 3 mass matrix ← 3D :(x, y,z)

The explicit forward Euler time integration d dt x v ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = v M−1f ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 1 Δt xn+1 − xn vn+1 − vn ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = vn M−1f n ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ discretization

1 Δt xn+1 − xn vn+1 − vn ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = v n+1 M−1f n+1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟

The implicit backward Euler time integration

⇐ f n+1 = f n + ∂f ∂x x n+1 − x n

(

)

+ ∂f ∂v v n+1 − v n

(

)

: the 1st order Taylor’s expansion

1 Δt xn+1 − xn Δt − v n ⎛ ⎝⎜ ⎞ ⎠⎟ = M −1 _fn + ∂f ∂x x n+1 − xn

(

)

+ ∂f ∂v xn+1 − xn Δt − v n ⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 1 Δt v n+1 − vn

(

)

= M−1 f n+1 = M−1 f n + ∂f ∂x x n+1 − xn

(

)

+ ∂f ∂v v n+1 − v n

(

)

⎡ ⎣⎢ ⎤ ⎦⎥ ⇐ v n+1 = x n+1 − xn Δt xn+1 − xn

(

)

− Δt2 vn = Δt2M−1 f n + ∂f ∂x x n+1 − xn

(

)

+ ∂f ∂v xn+1 − xn Δt − v n ⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = Δt2M−1f n + Δt2M−1 ∂f ∂x x n+1 − xn

(

)

+ ΔtM−1 ∂f ∂v x n+1 − xn

(

)

− Δt2 M−1 ∂f ∂v v n I − Δt 2M−1 ∂f ∂x ⎛ ⎝⎜ ⎞⎠⎟

(

xn+1 − xn

)

= Δt2

(

vn + M−1f n

)

Ax

= b

We don't know fn+1 at n-th frame,

but we can estimate fn+1 from fn, xn and, vn.

I − Δt2M−1 ∂f ∂x − ΔtM −1 ∂f ∂v ⎛ ⎝⎜ ⎞⎠⎟

(

xn+1 − xn

)

= Δt2 ⎡_⎣⎢vn + M−1⎛⎝⎜ f n − ∂_∂vf vn⎞⎠⎟ ⎤_⎦⎥ ⇐ ∂_∂vf = 0 by later formulation I − Δt2M−1 ∂f ∂x ⎛ ⎝⎜ ⎞⎠⎟

(

xn+1 − xn

)

= Δt2

(

vn + M−1fn

)

f

n

: forces

_x

n+1

_{: positions}

n +1 ( ) instead of n

1 Δt 3 2 x n+1 − 2xn + 1 2 x n−1 3 2 v n+1 − 2vn + 1 2 v n−1 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ = vn+1 M−1f n+1 ⎛ ⎝ ⎜ ⎞ ⎠

⎟ : the 1st order Taylor’s expansion

The semi-implicit second-order backward time integration (BDF)

1 Δt 3 2 v n+1 − 2vn + 1 2 v n−1 ⎛ ⎝⎜ ⎞⎠⎟ = M−1f n+1 = M−1 f n + ∂ f ∂x x n+1 − xn

(

)

+ ∂f ∂v v n+1 − v n

(

)

⎡ ⎣⎢ ⎤ ⎦⎥ 2Δt 3 v n+1 − 8Δt 9 v n + 2Δt 9 v n−1 = 4Δt 2 9 M −1 f n + 4Δt 2 9 M −1 ∂f ∂x x n+1 − 4Δt 2 9 M −1 ∂f ∂x x n+ 4Δt 2 9 M −1 ∂f ∂v v n+1 − 4Δt 2 9 M −1 ∂f ∂v v n ⇐ vn+1 = 1 Δt 3 2 x n+1 − 2xn + 1 2 x n−1 ⎛ ⎝⎜ ⎞⎠⎟ 2 3 3 2 x n+1 − 2xn + 1 2 x n−1 ⎛ ⎝⎜ ⎞⎠⎟ − Δt₉

(

8vn − 2vn−1

)

= 4Δt2 9 M −1_fn + 4Δt 2 9 M −1 ∂f ∂x x n+1 − 4Δt 2 9 M −1 ∂f ∂x x n+ 4Δt 9 M −1 ∂f ∂v 3 2 x n+1 − 2xn + 1 2 x n−1 ⎛ ⎝⎜ ⎞⎠⎟ − 4Δt2 9 M −1 ∂f ∂v v n xn+1 − 4 3 x n + 1 3 x n−1 − Δt 9 8v n − 2vn−1

(

)

= 4Δt2 9 M −1 fn + 4Δt 2 9 M −1 ∂f ∂x x n+1 − 4Δt 2 9 M −1 ∂f ∂x x n+ 2Δt 3 M −1 ∂f ∂v x n+1 − 8Δt 9 M −1 ∂f ∂v x n + 2Δt 9 M −1 ∂f ∂v x n−1 − 4Δt 2 9 M −1 ∂f ∂v v n ⇐ f n+1 = f n + ∂f ∂x x n+1 − x n

(

)

+ ∂f ∂v v n+1 − v n

(

)

second order time discretization

I − 4Δt 2 9 M −1 ∂f ∂x − 2Δt 3 M −1 ∂f ∂v ⎛ ⎝⎜ ⎞ ⎠⎟ x n+1 − xn − 1 3 x n − xn−1

(

)

− Δt 9 8v n − 2vn−1

(

)

= 4Δt2 9 M −1 fn − 4Δt 2 9 M −1 ∂f ∂x x n− 8Δt 9 M −1 ∂f ∂v x n + 2Δt 9 M −1 ∂f ∂v x n−1 − 4Δt 2 9 M −1 ∂f ∂v v n I − 4Δt 2 9 M −1 ∂f ∂x − 2Δt 3 M −1 ∂f ∂v ⎛ ⎝⎜ ⎞ ⎠⎟ x n+1 − I − 4Δt 2 9 M −1 ∂f ∂x x n− 2Δt 3 M −1 ∂f ∂v ⎛ ⎝⎜ ⎞ ⎠⎟ x n = 1 3 x n − xn−1

(

)

+ Δt 9 8v n − 2vn−1

(

)

+ 4Δt2 9 M −1 fn − 2Δt 9 M −1 ∂f ∂v x n + 2Δt 9 M −1 ∂f ∂v x n−1 − 4Δt 2 9 M −1 ∂f ∂v v n I − 4Δt 2 9 M −1 ∂f ∂x − 2Δt 3 M −1 ∂f ∂v ⎛ ⎝⎜ ⎞ ⎠⎟ x n+1 − xn

(

)

= 1 3 x n − xn−1

(

)

+ Δt 9 8v n − 2vn−1

(

)

− 2Δt 9 M −1 ∂f ∂v x n − xn−1

(

)

+ 4Δt2 9 M −1 fn − ∂f ∂v v n ⎛ ⎝⎜ ⎞⎠⎟

I − 4Δt 2 9 M −1 ∂f ∂x ⎛ ⎝⎜ ⎞ ⎠⎟ x n+1 − xn

(

)

= 1 3 x n − xn−1

(

)

+ Δt 9 8v n − 2vn−1

(

)

+ 4Δt2 9 M −1 fn I − 4Δt 2 9 M −1 ∂f ∂x − 2Δt 3 M −1 ∂f ∂v ⎛ ⎝⎜ ⎞ ⎠⎟ x n+1 − xn

(

)

= 1 3 x n − xn−1

(

)

+ Δt 9 8v n − 2vn−1

(

)

− 2Δt 9 M −1 ∂f ∂v x n − xn−1

(

)

+ 4Δt2 9 M −1 fn − ∂f ∂v v n ⎛ ⎝⎜ ⎞⎠⎟ ⇐ ∂ f ∂v = 0 by later formulation I − 4Δt 2 9 M −1 ∂f ∂x ⎛ ⎝⎜ ⎞ ⎠⎟ x n+1 − xn

(

)

= 1 3 x n − xn−1

(

)

+ Δt 9 8v n − 2vn−1

(

)

+ 4Δt 2 9 M −1 f n

I

− Δt

2

M

−1

∂f

∂x

⎛

⎝⎜

⎞

⎠⎟

(

x

n+1

− x

n

)

= Δt

2

(

v

n

+ M

−1

f

n

)

Ax

= b

The explicit forward Euler time integration

The implicit backward Euler time integration : 3n × 3n symmetric matrix equation

(n : the number of mesh vertices)

: We need to derive f(= force) and ∂f

Stretch Force & Jacobian

for a Linear Spring

E_ij = kij 2

(

xij − Lij

)

2 f_i (= −f_j ) = − ∂Eij ∂x_i = kij 1− L_ij x_ij ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ xij

x

_i

x

_j

x

_ij

= x

_j

− x

_i

: Hook’s law (L_ij : rest length)

E

_ij

L

_ij : the 1st derivative ∂f_i ∂x _j = kij x_ijx_ijT x_ijTx_ij + kij 1− L_ij x_ij ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ I − xijxij T x_ijTx_ij ⎛ ⎝⎜ ⎞ ⎠⎟ : the 2nd derivative (elastic energy) (force) (Jacobian) (3x3 symmetric matrix) ∴ ∂f ∂x ⎛

⎝⎜ ⎞⎠⎟ _i = _{j∈neighbors}

∑

_∂x∂fi_j : the Jacobian of i-th node (3x3 symmetric matrix)

proof) f_i = − ∂Eij ∂x_i = kij 1− L_ij x_ij ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ xij = k_ij

(

x_ij − L_ij

)

xij x_ij = kij 1− L_ij x_ij ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ xij f_i = − ∂Eij ∂x_i = − ∂∂x_i k_ij 2

(

xij − Lij

)

2 ⎡ ⎣⎢ ⎤ ⎦⎥ = − k_ij 2 ∂ ∂x_i

(

xij − Lij

)

2 ⎡ ⎣⎢ ⎤⎦⎥ = −kij

(

xij − Lij

)

∂ ∂x_i

(

xij − Lij

)

⎛ ⎝⎜ ⎞ ⎠⎟ = −kij

(

xij − Lij

)

∂ ∂x_i xij ⎛ ⎝⎜ ⎞ ⎠⎟ ⇐ ∂∂x_i xij = − x_ij x_ij proof) ∂fi ∂x _j = kij x_ijx_ijT x_ijTx_ij + kij 1− L_ij x_ij ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ I − xijxij T x_ijTx_ij ⎛ ⎝⎜ ⎞ ⎠⎟ ∂f_i ∂x_j = ∂ f_i ∂x_ij ∂x_ij ∂x_j = ∂ f_i ∂x_ij ∂ ∂x_j

(

x j − xi

)

= ∂ f_i ∂x_ij I = ∂ f_i ∂x_ij = ∂∂x_ij kij 1− L_ij x_ij ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ xij ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = kij ∂ ∂x_ij 1− L_ij x_ij ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ xij ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = k ∂∂x 1− L x ⎛ ⎝⎜ ⎞ ⎠⎟ x ⎡ ⎣ ⎢ ⎤ ⎦ ⎥

: Let’s omit ij index for convenience = k 1− L x ⎛ ⎝⎜ ⎞ ⎠⎟ I + x ∂∂x 1− L x ⎛ ⎝⎜ ⎞ ⎠⎟ ⎧ ⎨ ⎪ ⎩⎪ ⎫ ⎬ ⎪ ⎭⎪ T ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ∵ ∂ ∂x[ f (x)x] = f (x)I + x ∂⎡_⎣⎢∂x f (x)⎤_⎦⎥ T ⎛ ⎝⎜ ⎞ ⎠⎟ ∂ ∂x 1− L x ⎛ ⎝⎜ ⎞ ⎠⎟ = ∂∂x x − L x ⎛ ⎝⎜ ⎞ ⎠⎟ = ∂∂x xTx − L xTx ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ∂ ∂x x T x − L

(

)

xTx − x

(

Tx − L

)

∂ ∂x x T x ⎛ ⎝⎜ ⎞⎠⎟ xTx = 2x 2 xTx x T x − x

(

Tx − L

)

2x 2 xTx xTx = x − x − L

(

)

x x xTx = L x x xTx = L x x xTx ∴ ∂fi ∂x_j = k 1− L x ⎛ ⎝⎜ ⎞ ⎠⎟ I + x L x x xTx ⎛ ⎝⎜ ⎞ ⎠⎟ T ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = k 1− L x ⎛ ⎝⎜ ⎞ ⎠⎟ I + L x xxT xTx ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = k 1− L x ⎛ ⎝⎜ ⎞ ⎠⎟ I + xxT xTx − xxT xTx + L x xxT xTx ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = k 1− L x ⎛ ⎝⎜ ⎞ ⎠⎟ I + xxT xTx − 1− L x ⎛ ⎝⎜ ⎞ ⎠⎟ xxT xTx ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = k xx T xTx + 1− L x ⎛ ⎝⎜ ⎞ ⎠⎟ I − 1− L x ⎛ ⎝⎜ ⎞ ⎠⎟ xxT xTx ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = k xx T xTx + 1− L x ⎛ ⎝⎜ ⎞ ⎠⎟ I − xxT xTx ⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = k xx T xTx + k 1− L x ⎛ ⎝⎜ ⎞ ⎠⎟ I − xxT xTx ⎛ ⎝⎜ ⎞ ⎠⎟

∂f_i ∂x _j = kij x_ijx_ijT x_ijTx_ij + kij 1− L_ij x_ij ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ I − xijxij T x_ijTx_ij ⎛ ⎝⎜ ⎞ ⎠⎟

Geometric meaning of Jacobian

projection operator complementary projection operator X = x x a b y = a + b a = X ⋅ y

(

)

X = X X ⋅ y

(

)

= XX

( )

T y = x x xT x y = xxT xTx y = Pay Pa = xxT xTx : projection operator on x b = y − a = y − P_ay = I − P

(

_a

)

y = P_by _Pb = I − xx T

xTx : complementary projection operator on x x

Stretch Force & Jacobian

for a Triangular Spring

static triangle per frame deformed triangle (uv-coordinate) (xyz-coordinate) (u_i,v_i ) (u_j,v_j ) _{(uk ,vk )} x_k (x_k , y_k , z_k ) x_i (x_i, y_i, z_i ) x _j (x _j, y_j, z_j ) S(u,v) = (x, y,z) x _j − x_i = S_u

(

u_j − u_i

)

+ S_v

(

v_j − v_i

)

x_k − x_i = S_u

(

u_k − u_i

)

+ S_v

(

v_k − v_i

)

x _j − x_i x_k − x_i ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = u_j − u_i v_j − v_i u_k − u_i v_k − v_i ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ S_u S_v ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = u_ij v_ij u_ik v_ik ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ S_u S_v ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ S_u = ∂S ∂u , Sv = ∂ S ∂v : 3D vectors ⎛ ⎝⎜ ⎞⎠⎟ ∴ Su S_v ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = u_ij v_ij u_ik v_ik ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ −1 x _j − x_i x_k − x_i ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 1 u_ijv_ik − v_iju_ik v_ik −v_ij −u_ik u_ij ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ x _j − x_i x_k − x_i ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ S_u = vik

(

x j − xi

)

− vij

(

xk − xi

)

2A S_v = −uik

(

x j − xi

)

+ uij

(

xk − xi

)

2A

Triangle based Stretch Spring (Baraff’s 98)

∂S_u ∂x_i = v_ij − v_ik 2A ∂S_v ∂x_i = u_ik − u_ij 2A ∂S_u ∂x _j = v_ik 2A ∂S_u ∂x_k = −v_ij 2A ∂S_v ∂x _j = − u_ik 2A ∂S_v ∂x_k = u_ij 2A

3D vectors scalar values

= 1 2A v_ik −v_ij −u_ik u_ij ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ x _j − x_i x_k − x_i ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ A = areauv = u_ijv_ik − v_iju_ik 2 ⎛ ⎝⎜ ⎞ ⎠⎟

f_i = − ∂Eij ∂x_i = kij 1− L_ij x_ij ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ xij (elastic energy) (force) (Jacobian) E_t = A_t kt 2

(

Su −1

)

2 + A_t kt 2

(

Sv −1

)

2 f_i = − ∂Et ∂x_i = Atkt 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ Su ∂S_u ∂x_i + 1− 1 S_v ⎛ ⎝⎜ ⎞ ⎠⎟ Sv ∂S_v ∂x_i ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = A_t k_t ∂Su ∂x_i ∂S_u ∂x _j S_uS_uT S_uTS_u + Atkt 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ ∂S_u ∂x_i ∂S_u ∂x _j I − S_uS_uT S_uTS_u ⎛ ⎝⎜ ⎞ ⎠⎟ (elastic energy) (force) (Jacobian)

Stretch “Linear” Spring _{Stretch “Triangular” Spring}

∂f_i ∂x _j = kij x_ijx_ijT x_ijTx_ij + kij 1− L_ij x_ij ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ I − xijxij T x_ijTx_ij ⎛ ⎝⎜ ⎞ ⎠⎟ ∂f_i ∂x _j +A_tk_t ∂Sv ∂x_i ∂S_v ∂x _j S_vS_vT S_vTS_v + Atkt 1− 1 S_v ⎛ ⎝⎜ ⎞ ⎠⎟ ∂S_v ∂x_i ∂S_v ∂x _j I − S_vS_vT S_vTS_v ⎛ ⎝⎜ ⎞ ⎠⎟ E_ij = kij 2

(

xij − Lij

)

2

(L_ij : rest length)

(

t : triangle index

)

1: rest stretch ratio

proof) f_i = − ∂Eij ∂x_i = kij 1− L_ij x_ij ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ xij

from the equation

For the force in u − direciton

By chain rule, f_i = A_tk_t 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ Su ∂S_u ∂x_i ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ let k_ij = A_tk_t, x_ij = S_u , and L_ij = 1. By chain rule, f_i = A_t k_t 1− 1 S_v ⎛ ⎝⎜ ⎞ ⎠⎟ Sv ∂S_v ∂x_i ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ let k_ij = A_tk_t, x_ij = S_v, and L_ij = 1. ∴f_i = A_tk_t 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ Su ∂S_u ∂x_i ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ + Atkt 1− 1 S_v ⎛ ⎝⎜ ⎞ ⎠⎟ Sv ∂S_v ∂x_i ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ f_i = − ∂Et ∂x_i = Atkt 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ Su ∂S_u ∂x_i + 1− 1 S_v ⎛ ⎝⎜ ⎞ ⎠⎟ Sv ∂S_v ∂x_i ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥

proof) ∂fi ∂x _j = Atkt ∂S_u ∂x_i ∂S_u ∂x _j S_uS_uT S_uTS_u + At kt 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ ∂S_u ∂x_i ∂S_u ∂x _j I − S_uS_uT S_uTS_u ⎛ ⎝⎜ ⎞ ⎠⎟ + Atkt ∂S_v ∂x_i ∂S_v ∂x _j S_vS_vT S_vTS_v + Atkt 1− 1 S_v ⎛ ⎝⎜ ⎞ ⎠⎟ ∂S_v ∂x_i ∂S_v ∂x _j I − S_vS_vT S_vTS_v ⎛ ⎝⎜ ⎞ ⎠⎟ ∵ ∂ ∂x [ f (x)x] = f (x)I + x ∂⎡_⎣⎢∂x f (x)⎤_⎦⎥ T ⎛ ⎝⎜ ⎞ ⎠⎟ = A_tk_t ∂Su ∂x_i ∂S_u ∂x _j ⋅ 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ I + Su ∂ ∂S_u 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ ⎧ ⎨ ⎪ ⎩⎪ ⎫ ⎬ ⎪ ⎭⎪ T ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = A_tk_t ∂Su ∂x_i ∂S_u ∂x _j ⋅ 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ I + Su 1 S_u S_u S_uTS_u ⎛ ⎝⎜ ⎞ ⎠⎟ T ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ∵ ∂ ∂x 1− L x ⎛ ⎝⎜ ⎞ ⎠⎟ = L x x xTx ⇒ ∂∂S_u 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ = 1 S_u S_u S_uTS_u ⎛ ⎝⎜ ⎞ ⎠⎟ = A_tk_t ∂Su ∂x_i ∂S_u ∂x _j ⋅ 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ I + 1 S_u S_uS_uT S_uTS_u ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = Atkt ∂S_u ∂x_i ∂S_u ∂x_j ⋅ 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ I + S_uS_uT S_uTS_u − S_uS_uT S_uTS_u 1 S_u S_uS_uT S_uTS_u ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = Atkt ∂S_u ∂x_i ∂S_u ∂x_j ⋅ 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ I + S_uS_uT S_uTS_u − 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ S_uS_uT S_uTS_u ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = A_tk_t ∂Su ∂x_i ∂S_u ∂x _j ⋅ S_uS_uT S_uTS_u + 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ I − S_uS_uT S_uTS_u ⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = Atkt ∂S_u ∂x_i ∂S_u ∂x_j S_uS_uT S_uTS_u + Atkt 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ ∂S_u ∂x_i ∂S_u ∂x _j I − S_uS_uT S_uTS_u ⎛ ⎝⎜ ⎞ ⎠⎟ u − direciton : ∂f_i ∂x_j = ∂ f_i ∂S_u ∂S_u ∂x _j = ∂ S_u ∂x_j ∂f_i ∂S_u = ∂ S_u ∂x_j ∂ ∂S_u Atkt 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ Su ∂S_u ∂x_i ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = Atkt ∂S_u ∂x_i ∂S_u ∂x _j ⋅ ∂∂S_u 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ Su ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ v − direciton : ∂f_i ∂x_j = Atkt ∂S_v ∂x_i ∂S_v ∂x_j S_vS_vT S_vTS_v + Atkt 1− 1 S_v ⎛ ⎝⎜ ⎞ ⎠⎟ ∂S_v ∂x_i ∂S_v ∂x _j I − S_vS_vT S_vTS_v ⎛ ⎝⎜ ⎞ ⎠⎟ in a similar way ∴ ∂fi ∂x _j = ∂ f_i ∂x_j ⎛ ⎝⎜ ⎞ ⎠⎟_u + ∂ f_i ∂x_j ⎛ ⎝⎜ ⎞ ⎠⎟ _v = Atkt ∂S_u ∂x_i ∂S_u ∂x _j S_uS_uT S_uTS_u + Atkt 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ ∂S_u ∂x_i ∂S_u ∂x_j I − S_uS_uT S_uTS_u ⎛ ⎝⎜ ⎞ ⎠⎟ + Atkt ∂S_v ∂x_i ∂S_v ∂x _j S_vS_vT S_vTS_v + Atkt 1− 1 S_v ⎛ ⎝⎜ ⎞ ⎠⎟ ∂S_v ∂x_i ∂S_v ∂x _j I − S_vS_vT S_vTS_v ⎛ ⎝⎜ ⎞ ⎠⎟

f_i = − ∂Et ∂x_i = At kt 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ Su ∂S_u ∂x_i + 1− 1 S_v ⎛ ⎝⎜ ⎞ ⎠⎟ Sv ∂S_v ∂x_i ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ f_j = − ∂Et ∂x _j = Atkt 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ Su ∂S_u ∂x _j + 1− 1 S_v ⎛ ⎝⎜ ⎞ ⎠⎟ Sv ∂S_v ∂x _j ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ f_k = − ∂Et ∂x_k = At kt 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ Su ∂S_u ∂x_k + 1− 1 S_v ⎛ ⎝⎜ ⎞ ⎠⎟ Sv ∂S_v ∂x_k ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ Forces ∂f_i ∂x _j = ∂f_j ∂x_i ⎛ ⎝⎜ ⎞ ⎠⎟ = Atkt ∂S_u ∂x_i ∂S_u ∂x _j S_uS_uT S_uTS_u + 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ I − S_uS_uT S_uTS_u ⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ + Atkt ∂S_v ∂x_i ∂S_v ∂x _j S_vS_vT S_vTS_v + 1− 1 S_v ⎛ ⎝⎜ ⎞ ⎠⎟ I − S_vS_vT S_vTS_v ⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ∂f_i ∂x_k = ∂ f_k ∂x_i ⎛ ⎝⎜ ⎞ ⎠⎟ = Atkt ∂S_u ∂x_i ∂S_u ∂x_k S_uS_uT S_uTS_u + 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ I − S_uS_uT S_uTS_u ⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ + At kt ∂S_v ∂x_i ∂S_v ∂x_k S_vS_vT S_vTS_v + 1− 1 S_v ⎛ ⎝⎜ ⎞ ⎠⎟ I − S_vS_vT S_vTS_v ⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ∂f_j ∂x_k = ∂ f_k ∂x _j ⎛ ⎝⎜ ⎞ ⎠⎟ = Atkt ∂S_u ∂x _j ∂S_u ∂x_k S_uS_uT S_uTS_u + 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ I − S_uS_uT S_uTS_u ⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ + Atkt ∂S_v ∂x _j ∂S_v ∂x_k S_vS_vT S_vTS_v + 1− 1 S_v ⎛ ⎝⎜ ⎞ ⎠⎟ I − S_vS_vT S_vTS_v ⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ Jacobians ∂S_u ∂x_i = v_ij − v_ik 2A ∂S_v ∂x_i = u_ik − u_ij 2A ∂S_u ∂x _j = v_ik 2A ∂S_u ∂x_k = −v_ij 2A ∂S_v ∂x _j = − u_ik 2A ∂S_v ∂x_k = u_ij 2A

Shear Force & Jacobian

for a Triangular Spring

Shear “Triangular” Spring

Replace

S

_u

→ S

(

_u

+ S

_v

)

S

_v

→ S

(

_u

− S

_v

)

1

→ 2

E_t = A_t kt 2

(

Su −1

)

2 + A_t kt 2

(

Sv −1

)

2 f_i = − ∂Et ∂x_i = Atkt 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ Su ∂S_u ∂x_i + 1− 1 S_v ⎛ ⎝⎜ ⎞ ⎠⎟ Sv ∂S_v ∂x_i ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = A_tk_t ∂Su ∂x_i ∂S_u ∂x _j S_uS_uT S_uTS_u + Atkt 1− 1 S_u ⎛ ⎝⎜ ⎞ ⎠⎟ ∂S_u ∂x_i ∂S_u ∂x _j I − S_uS_uT S_uTS_u ⎛ ⎝⎜ ⎞ ⎠⎟ (elastic energy) (force) (Jacobian) ∂f_i ∂x _j +A_tk_t ∂Sv ∂x_i ∂S_v ∂x _j S_vS_vT S_vTS_v + At kt 1− 1 S_v ⎛ ⎝⎜ ⎞ ⎠⎟ ∂S_v ∂x_i ∂S_v ∂x _j I − S_vS_vT S_vTS_v ⎛ ⎝⎜ ⎞ ⎠⎟ t : triangle index

(

)

1: rest stretch ratio

(

)

Caution) Shearing force may be unstable espetially using without stretching force. Also, shearing force cannot detect reverse triangle.

Bend Force & Jacobian

for a Triangular Spring

x₁ x₂ x₃ x₄ x₃₄ = x₄ − x₃ : bending element (elastic energy) E = k 2

θ

2 ₍

θ

_{: the angle between triangle x}

134 and triangle x243) n₁ n₂ n₁ = x₁₃ × x₁₄ = x

(

₃ − x₁

)

× x

(

₄ − x₁

)

n₂ = x₂₄ × x₂₃ = x

(

₄ − x₂

)

× x

(

₃ − x₂

)

cos

θ

= n1 ⋅n2 n₁ ⋅ n₂ ⇔

θ

= cos −1 n1 ⋅n2 n₁ ⋅ n₂ ⎛ ⎝⎜ ⎞ ⎠⎟

f_i = − ∂E ∂x_i = − ∂∂x_i k 2

θ

2 ⎛ ⎝⎜ ⎞⎠⎟ = −k

θ

_∂x∂

θ

_i = k

θ

sin

θ

∂ ∂x_i

(

cos

θ

)

∵ ∂ ∂x_i

(

cos

θ

)

= −sin

θ

∂

θ

∂x_i = k

θ

sin

θ

⋅ ∂∂x_i n₁ ⋅n₂ n₁ ⋅ n₂ ⎛ ⎝⎜ ⎞ ⎠⎟ ∵cos

θ

= n₁ ⋅n₂ n₁ ⋅ n₂ ∂ ∂x_i n₁ ⋅n₂ n₁ ⋅ n₂ ⎛ ⎝⎜ ⎞ ⎠⎟ = ∂ ∂x_i

(

n1 ⋅n2

)

⎡ ⎣ ⎢ ⎤ ⎦ ⎥⋅ n

(

1 ⋅ n2

)

− n

(

1 ⋅n2

)

⋅ ∂∂x i n₁ ⋅ n₂

(

)

⎡ ⎣ ⎢ ⎤ ⎦ ⎥ n₁ 2 ⋅ n₂ 2 ∵ ∂ ∂x f g ⎛ ⎝⎜ ⎞ ⎠⎟ = ∂ f ∂x g − f ∂ g ∂x g2 ∂ ∂x_i

(

n1 ⋅n2

)

= ∂ n₁ ∂x_i ⎛ ⎝⎜ ⎞ ⎠⎟ T n₂ + ∂n2 ∂x_i ⎛ ⎝⎜ ⎞ ⎠⎟ T n₁ ∂ ∂x_i

(

n1 ⋅ n2

)

= ∂ n₁ ∂x_i ⋅ n2 + n1 ⋅ ∂ n₂ ∂x_i = ∂ n₁ ∂x_i ⎛ ⎝⎜ ⎞ ⎠⎟ T n₁ n₁ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥⋅ n2 + n1 ⋅ ∂ n₂ ∂x_i ⎛ ⎝⎜ ⎞ ⎠⎟ T n₂ n₂ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ∵ ∂( u ) ∂w = ∂ u ∂w ⎛ ⎝⎜ ⎞⎠⎟ T u u = n2 n₁ ⋅ ∂ n₁ ∂x_i ⎛ ⎝⎜ ⎞ ⎠⎟ T n₁ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ + n₁ n₂ ⋅ ∂ n₂ ∂x_i ⎛ ⎝⎜ ⎞ ⎠⎟ T n₂ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ∵ ∂(u⋅v) ∂w = ∂ u ∂w ⎛ ⎝⎜ ⎞⎠⎟ T v + ∂v ∂w ⎛ ⎝⎜ ⎞⎠⎟ T u ∴f_i = k

θ

sin

θ

1 n₁ ⋅ n₂ ∂n₁ ∂x_i ⎛ ⎝⎜ ⎞ ⎠⎟ T n₂ + 1 n₁ ⋅ n₂ ∂n₂ ∂x_i ⎛ ⎝⎜ ⎞ ⎠⎟ T n₁ − n1 ⋅n2 n₁ 2 ⋅ n₂ 2 n₂ n₁ ∂n₁ ∂x_i ⎛ ⎝⎜ ⎞ ⎠⎟ T n₁ − n1 ⋅n2 n₁ 2 ⋅ n₂ 2 n₁ n₂ ∂n₂ ∂x_i ⎛ ⎝⎜ ⎞ ⎠⎟ T n₂ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥

from ∂(u × v) ∂w = u * ∂v ∂w − v * ∂u ∂w ∂n₁ ∂x_i = ∂∂x_i ⎡⎣

(

x3 − x1

)

× x

(

4 − x1

)

⎤⎦ = x

(

3 − x1

)

* ∂ ∂x_i

(

x4 − x1

)

− x

(

4 − x1

)

* ∂ ∂x_i

(

x3 − x1

)

∂n₂ ∂x_i = ∂∂x_i ⎡⎣

(

x4 − x2

)

× x

(

3 − x2

)

⎤⎦ = x

(

4 − x2

)

* ∂ ∂x_i

(

x3 − x2

)

− x

(

3 − x2

)

* ∂ ∂x_i

(

x4 − x2

)

∂n₁ ∂x₁ = x

(

3 − x1

)

* ∂ ∂x₁

(

x4 − x1

)

− x

(

4 − x1

)

* ∂ ∂x₁

(

x3 − x1

)

= − x

(

3 − x1

)

* + x

(

₄ − x₁

)

* = x

(

₄ − x₃

)

* = x*₃₄ ∴ ∂n1 ∂x₁ ⎛ ⎝⎜ ⎞ ⎠⎟ T = −x*₃₄ ∂n₁ ∂x₂ = x

(

3 − x1

)

* ∂ ∂x₂

(

x4 − x1

)

− x

(

4 − x1

)

* ∂ ∂x₂

(

x3 − x1

)

= 0 − 0 = 0 ∴ ∂n1 ∂x₂ ⎛ ⎝⎜ ⎞ ⎠⎟ T = 0 ∂n₁ ∂x₃ = x

(

3 − x1

)

* ∂ ∂x₃

(

x4 − x1

)

− x

(

4 − x1

)

* ∂ ∂x₃

(

x3 − x1

)

= 0 − x

(

4 − x1

)

* = −x₁₄* ∴ ∂n1 ∂x₃ ⎛ ⎝⎜ ⎞ ⎠⎟ T = x₁₄* ∂n₁ ∂x₄ = x

(

3 − x1

)

* ∂ ∂x₄

(

x4 − x1

)

− x

(

4 − x1

)

* ∂ ∂x₄

(

x3 − x1

)

= x

(

3 − x1

)

* − 0 = x₁₃* ∴ ∂n1 ∂x₄ ⎛ ⎝⎜ ⎞ ⎠⎟ T = −x₁₃* ∂n₂ ∂x₁ = x

(

4 − x2

)

* ∂ ∂x₁

(

x3 − x2

)

− x

(

3 − x2

)

* ∂ ∂x₁

(

x4 − x2

)

= 0 − 0 = 0 ∴ ∂n2 ∂x₁ ⎛ ⎝⎜ ⎞ ⎠⎟ T = 0 ∂n₂ ∂x₂ = x

(

4 − x2

)

* ∂ ∂x₂

(

x3 − x2

)

− x

(

3 − x2

)

* ∂ ∂x₂

(

x4 − x2

)

= − x

(

4 − x2

)

* + x

(

₃ − x₂

)

* = x

(

₃ − x₄

)

* = −x*₃₄ ∴ ∂n2 ∂x₂ ⎛ ⎝⎜ ⎞ ⎠⎟ T = x*₃₄ ∂n₂ ∂x₃ = x

(

4 − x2

)

* ∂ ∂x₃

(

x3 − x2

)

− x

(

3 − x2

)

* ∂ ∂x₃

(

x4 − x2

)

= x

(

4 − x2

)

* − 0 = −x*₂₄ ∴ ∂n2 ∂x₃ ⎛ ⎝⎜ ⎞ ⎠⎟ T = x*₂₄ ∂n₂ ∂x₄ = x

(

4 − x2

)

* ∂ ∂x₄

(

x3 − x2

)

− x

(

3 − x2

)

* ∂ ∂x₄

(

x4 − x2

)

= 0 − x

(

3 − x2

)

* = −x*₂₃ ∴ ∂n2 ∂x₄ ⎛ ⎝⎜ ⎞ ⎠⎟ T = x*₂₃

f₁ = kθ sinθ 1 n₁ ⋅ n₂ ∂n₁ ∂x₁ ⎛ ⎝⎜ ⎞ ⎠⎟ T n₂ + 1 n₁ ⋅ n₂ ∂n₂ ∂x₁ ⎛ ⎝⎜ ⎞ ⎠⎟ T n₁ − n1 ⋅n2 n₁ 2 ⋅ n₂ 2 n₂ n₁ ∂n₁ ∂x₁ ⎛ ⎝⎜ ⎞ ⎠⎟ T n₁ − n1 ⋅n2 n₁ 2 ⋅ n₂ 2 n₁ n₂ ∂n₂ ∂x₁ ⎛ ⎝⎜ ⎞ ⎠⎟ T n₂ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = kθ sinθ 1 n₁ ⋅ n₂ −x34 *

( )

n₂ + 1 n₁ ⋅ n₂ ( )0 n1 − n₁ ⋅n₂ n₁ 2 ⋅ n₂ 2 n₂ n₁ −x34 *

( )

n₁ − n1 ⋅n2 n₁ 2 ⋅ n₂ 2 n₁ n₂ ( )0 T n₂ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥