几何定义
射线
射线由起点 O O O D D D t t t P P P
平面
平面可以由平面内任一点 P P P N N N P P P P ′ P^{\prime} P ′ P P ′ PP^{\prime} P P ′ N N N 0 0 0
( P − P ′ ) ⋅ N = 0 (P-P^{\prime}) \cdot N = 0 ( P − P ′ ) ⋅ N = 0 球体
已知球体的方程为 x 2 + y 2 + z 2 = R x^2+y^2+z^2 = R x 2 + y 2 + z 2 = R P ( x , y , z ) P(x,y,z) P ( x , y , z ) C C C P P P C C C R R R
( P − C ) ⋅ ( P − C ) = R 2 (P-C) \cdot (P-C) = R^2 ( P − C ) ⋅ ( P − C ) = R 2 面积和体积
面积
两向量叉乘的模即为平行四边形的面积，而任意两个二维向量组合成矩阵，其行列式的绝对值即为二者所围成的平行四边形的面积。
S = ∥ a × b ∥ = ∣ d e t ( a , b ) ∣ S = \left\| \mathbf{a} \times \mathbf{b}\right\| = |det(\mathbf{a},\mathbf{b})| S = ∥ a × b ∥ = ∣ d e t ( a , b ) ∣ 体积
由相邻的三个矢量所构成的平行六面体，其体积可以表示为三个矢量的标量三重积的绝对值。
V = ∣ a ⋅ ( b × c ) ∣ = ∣ d e t ( [ a , b , c ] ) ∣ = ∥ a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 ∥ V=|\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})|=
|det([\mathbf{a},\mathbf{b},\mathbf{c}])| =
\left\|\begin{array}{lll}
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3} \\
c_{1} & c_{2} & c_{3}
\end{array}\right\| V = ∣ a ⋅ ( b × c ) ∣ = ∣ d e t ([ a , b , c ]) ∣ = ∥ ∥ a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 ∥ ∥ 三角形重心坐标的求解
三角形 △ A B C \triangle ABC △ A BC P P P A A A △ A B C \triangle ABC △ A BC P P P △ A B C \triangle ABC △ A BC
A P = u A B + v A C A − P = u ( A − B ) + v ( A − C ) P = ( 1 − u − v ) A + u B + v C {A P}=u {A B}+v {A C}
\\
A-P=u(A-B)+v(A-C)
\\
P=(1-u-v) A+u B+v C A P = u A B + v A C A − P = u ( A − B ) + v ( A − C ) P = ( 1 − u − v ) A + u B + v C 将由三角形 △ A B C \triangle ABC △ A BC A P AP A P 0 0 0
u A B + v A C + P A = 0 u {A B}+v {A C}+{P A}=0 u A B + v A C + P A = 0 [ u v 1 ] ⋅ [ A B x A C x P A x ] = 0 \left[\begin{array}{l}
u \\
v \\
1
\end{array}\right]
\cdot
\left[\begin{array}{l}
{A B_{x}} \\
{A C}_{x} \\
{P A_{x}}
\end{array}\right]=0 ⎣ ⎡ u v 1 ⎦ ⎤ ⋅ ⎣ ⎡ A B x A C x P A x ⎦ ⎤ = 0 [ u v 1 ] ⋅ [ A B y A C y P A y ] = 0 \left[\begin{array}{l}
u \\
v \\
1
\end{array}\right]
\cdot
\left[\begin{array}{l}
{A B_{y}} \\
{A C}_{y} \\
{P A_{y}}
\end{array}\right]=0 ⎣ ⎡ u v 1 ⎦ ⎤ ⋅ ⎣ ⎡ A B y A C y P A y ⎦ ⎤ = 0 两向量点乘为 0 0 0 [ u , v , 1 ] T [u,v,1]^T [ u , v , 1 ] T [ u , v , 1 ] T [u,v,1]^T [ u , v , 1 ] T
[ A B x A C x P A x ] × [ A B y A C y P A y ] = [ u v 1 ] \left[\begin{array}{l}
{A B_{x}} \\
{A C}_{x} \\
{P A_{x}}
\end{array}\right]
\times
\left[\begin{array}{l}
{A B_{y}} \\
{A C}_{y} \\
{P A_{y}}
\end{array}\right]
=
\left[\begin{array}{l}
u \\
v \\
1
\end{array}\right] ⎣ ⎡ A B x A C x P A x ⎦ ⎤ × ⎣ ⎡ A B y A C y P A y ⎦ ⎤ = ⎣ ⎡ u v 1 ⎦ ⎤ 还可以通过直接计算的方式求出重心坐标。
α = − ( x − x B ) ( y C − y B ) + ( y − y B ) ( x C − x B ) − ( x A − x B ) ( y C − y B ) + ( y A − y B ) ( x C − x B ) β = − ( x − x C ) ( y A − y C ) + ( y − y C ) ( x A − x C ) − ( x B − x C ) ( y A − y C ) + ( y B − y C ) ( x A − x C ) γ = 1 − α − β \begin{aligned}
\alpha &=\frac{-\left(x-x_{B}\right)\left(y_{C}-y_{B}\right)+\left(y-y_{B}\right)\left(x_{C}-x_{B}\right)}{-\left(x_{A}-x_{B}\right)\left(y_{C}-y_{B}\right)+\left(y_{A}-y_{B}\right)\left(x_{C}-x_{B}\right)} \\
\beta &=\frac{-\left(x-x_{C}\right)\left(y_{A}-y_{C}\right)+\left(y-y_{C}\right)\left(x_{A}-x_{C}\right)}{-\left(x_{B}-x_{C}\right)\left(y_{A}-y_{C}\right)+\left(y_{B}-y_{C}\right)\left(x_{A}-x_{C}\right)} \\
\gamma &=1-\alpha-\beta
\end{aligned} α β γ = − ( x A − x B ) ( y C − y B ) + ( y A − y B ) ( x C − x B ) − ( x − x B ) ( y C − y B ) + ( y − y B ) ( x C − x B ) = − ( x B − x C ) ( y A − y C ) + ( y B − y C ) ( x A − x C ) − ( x − x C ) ( y A − y C ) + ( y − y C ) ( x A − x C ) = 1 − α − β 射线与几何相交
射线与球相交
已知射线和球体的定义，若二者相交，将射线定义代入球体定义后即可得到二者交点的定义。
( O + t D − C ) ⋅ ( O + t D − C ) = r 2 D ⋅ D t 2 + 2 ( O − C ) ⋅ D t + ( O − C ) ⋅ ( O − C ) − R 2 = 0 (O+tD-C) \cdot (O+tD-C) = r^2
\\
D \cdot Dt^2 + 2(O-C) \cdot Dt + (O-C) \cdot (O-C) -R^2 = 0 ( O + t D − C ) ⋅ ( O + t D − C ) = r 2 D ⋅ D t 2 + 2 ( O − C ) ⋅ D t + ( O − C ) ⋅ ( O − C ) − R 2 = 0 经过变形可以得到一个以时间 t t t t t t
a = D ⋅ D b = 2 ( O − C ) ⋅ D c = ( O − C ) ⋅ ( O − C ) − R 2 x = − b ± b 2 − 4 a c 2 a \begin{array} {l}
a = D \cdot D
\\
b = 2(O-C) \cdot D
\\
c= (O-C) \cdot (O-C) -R^2
\\ \\
x=\frac{-b \pm \sqrt{b^2 -4ac}}{2 a}
\end{array} a = D ⋅ D b = 2 ( O − C ) ⋅ D c = ( O − C ) ⋅ ( O − C ) − R 2 x = 2 a − b ± b 2 − 4 a c 设 h = b 2 h = \frac{b}{2} h = 2 b D D D
x = − h ± h 2 − a c a x = − h ± h 2 − c x=\frac{-h \pm \sqrt{h^2 -ac}}{a}
\\\\
x=-h \pm \sqrt{h^2 -c} x = a − h ± h 2 − a c x = − h ± h 2 − c 射线与平面相交
和球体相交相同，只需要将射线的定义代入平面的定义即可，且为一个简单的以时间 t t t
( O + t D − P ′ ) ⋅ N = 0 t = ( P ′ − O ) ⋅ N D ⋅ N (O+tD-P^{\prime}) \cdot N = 0
\\
\\
t = \frac{(P^{\prime} - O) \cdot N}{D \cdot N} ( O + t D − P ′ ) ⋅ N = 0 t = D ⋅ N ( P ′ − O ) ⋅ N 射线与矩形相交
给定矩形 A B C D ABCD A BC D A = P ′ A = P^{\prime} A = P ′ t t t P P P P P P P P P A B C D ABCD A BC D
P = O + t D x m i n < x p < x m a x y m i n < y p < y m a x z m i n < z p < z m a x P = O+tD
\\ \\
x_{min} < x_p < x_{max}
\\
y_{min} < y_p < y_{max}
\\
z_{min} < z_p < z_{max} P = O + t D x min < x p < x ma x y min < y p < y ma x z min < z p < z ma x 射线与三角形相交
假设点 P P P △ A B C \triangle ABC △ A BC P P P A P , B P , C P AP, BP, CP A P , BP , CP P P P
还可以使用三角形所在平面内任意一点 P P P α + β + γ = 1 \alpha + \beta + \gamma = 1 α + β + γ = 1 P P P ( α , β , γ ) (\alpha,\beta,\gamma) ( α , β , γ ) P P P
P ( x , y ) = α A + β B + γ C ’ α + β + γ = 1 P(x,y) = \alpha A + \beta B + \gamma C’
\\\\
\alpha + \beta + \gamma = 1 P ( x , y ) = α A + βB + γ C ’ α + β + γ = 1 或者以三角形 △ A B C \triangle ABC △ A BC P P P x x x y = y p y = y_p y = y p P P P
Möller–Trumbore 算法统一了射线与平面相交和确定一点是否在三角形内，即在 t 时刻射线到达的点 P P P P P P
O + t D = α A + β B + ( 1 − α − β ) C O + t D = α A + β B + C − α C − β C O − C = α ( A − C ) + β ( B − C ) − t D O+ tD = \alpha A + \beta B + (1 - \alpha - \beta)C
\\\\
O + tD = \alpha A + \beta B + C - \alpha C - \beta C
\\\\
O - C = \alpha (A - C) + \beta(B - C) - tD O + t D = α A + βB + ( 1 − α − β ) C O + t D = α A + βB + C − α C − βC O − C = α ( A − C ) + β ( B − C ) − t D 令 S = O − C , E 1 = A − C , E 2 = B − C S = O - C, E_1 = A - C, E_2 = B - C S = O − C , E 1 = A − C , E 2 = B − C
− t D + α E 1 + β E 2 = S [ − D E 1 E 2 ] [ t α β ] = S -tD + \alpha E_1 + \beta E_2 = S
\\\\
\left[
\begin{array}{ccc}
-D & E 1 & E 2
\end{array}
\right]
\left[
\begin{array}{l}
t \\
\alpha \\
\beta
\end{array}
\right]
=S − t D + α E 1 + β E 2 = S [ − D E 1 E 2 ] ⎣ ⎡ t α β ⎦ ⎤ = S 由克拉默法则可解未知数，未知数的结果是行列式的比值，而行列式可以转换为标量三重积形式。其中标量三重积可以交换点积和叉积的顺序不变，初等变换变换正负号。
t = det ( [ S E 1 E 2 ] ) det ( [ − D E 1 E 2 ] ) t=
\frac
{\operatorname{det}
\left(
\left[
\begin{array}{ccc}
S & E_{1} & E_{2}
\end{array}
\right]
\right)
}
{\operatorname{det}
\left(
\left[
\begin{array}{ccc}
-D & E_{1} & E_{2}
\end{array}
\right]
\right)
} t = det ( [ − D E 1 E 2 ] ) det ( [ S E 1 E 2 ] ) t = ( S × E 1 ) ⋅ E 2 ( − D × E 1 ) ⋅ E 2 ) = ( S × E 1 ) ⋅ E 2 − D ⋅ ( E 1 × E 2 ) = ( S × E 1 ) ⋅ E 2 E 1 ⋅ ( D × E 2 ) t =
\frac
{(S \times E_1) \cdot E_2}
{(-D \times E_1) \cdot E_2)}
=
\frac
{(S \times E_1) \cdot E_2}
{-D \cdot (E_1 \times E_2)}
=
\frac
{(S \times E_1) \cdot E_2}
{E_1 \cdot (D \times E_2)} t = ( − D × E 1 ) ⋅ E 2 ) ( S × E 1 ) ⋅ E 2 = − D ⋅ ( E 1 × E 2 ) ( S × E 1 ) ⋅ E 2 = E 1 ⋅ ( D × E 2 ) ( S × E 1 ) ⋅ E 2 令 S 1 = D × E 2 , S 2 = S × E 1 S_1 = D \times E_2, S_2 = S \times E_1 S 1 = D × E 2 , S 2 = S × E 1 α , β \alpha,\beta α , β
t = S 2 ⋅ E 2 E 1 ⋅ S 1 ， α = S ⋅ S 1 E 1 ⋅ S 1 ， β = D ⋅ S 2 E 1 ⋅ S 1 t =
\frac
{S_2 \cdot E_2}
{E_1 \cdot S_1}
，
\alpha =
\frac
{S \cdot S_1}
{E_1 \cdot S_1}
，
\beta =
\frac
{D \cdot S_2}
{E_1 \cdot S_1} t = E 1 ⋅ S 1 S 2 ⋅ E 2 ， α = E 1 ⋅ S 1 S ⋅ S 1 ， β = E 1 ⋅ S 1 D ⋅ S 2 [ t α β ] = 1 S 1 ⋅ E 1 [ S 2 ⋅ E 2 S 1 ⋅ S S 2 ⋅ D ] \left[\begin{array}{l}
t \\
\alpha \\
\beta
\end{array}\right]=\frac{1}{S_{1} \cdot E_{1}}\left[\begin{array}{c}
S_{2} \cdot E_{2} \\
S_{1} \cdot S \\
S_{2} \cdot D
\end{array}\right] ⎣ ⎡ t α β ⎦ ⎤ = S 1 ⋅ E 1 1 ⎣ ⎡ S 2 ⋅ E 2 S 1 ⋅ S S 2 ⋅ D ⎦ ⎤ 验证时间 t t t
t m i n < t < t m a x α + β < 1 , ( α > 0 , β > 0 ) t_{min}<t<t_{max}
\\ \\
\alpha + \beta < 1 , (\alpha > 0, \beta > 0) t min < t < t ma x α + β < 1 , ( α > 0 , β > 0 ) 射线与包围盒相交
二维矩形可由两个互相平行的对边组成，在二维矩形所在平面上的射线必定会与两个对边相交。若射线经过二维矩形，则射线经过两对边的时间必定有重叠的部分。可以说只有射线和两个对边都第一次相交了才是真正地进入了矩形，只要有一个对边第二次相交了就已经离开了矩形。
将二维矩形的情形推广到三维的包围盒，包围盒是由三个互相平行的对面组成的。已知平面定义是由平面内任一点 P P P N N N t t t t t t
t = ( P ′ − O ) ⋅ N D ⋅ N t =
\frac{(P^{\prime} - O) \cdot N}{D \cdot N}
t = D ⋅ N ( P ′ − O ) ⋅ N [ t x t y t z ] = P ′ − O D
\begin{bmatrix}
t_x \\
t_y \\
t_z
\end{bmatrix}
=
\frac
{P^{\prime} - O}
{D} ⎣ ⎡ t x t y t z ⎦ ⎤ = D P ′ − O 分别将最小点坐标和最大点坐标代入 P ′ P^{\prime} P ′ T m i n , T m a x T_{min}, T_{max} T min , T ma x
在二维情形我们已经知道只有射线和两个对边都第一次相交了才是真正地进入了矩形，只要有一个对边第二次相交了就已经离开了矩形。三维情形下也是同样的道理，选取最小时间矢量 T m i n T_{min} T min t e n t e r t_{enter} t e n t er T m a x T_{max} T ma x t e x i t t_{exit} t e x i t
t e x i t > 0 t e x i t + e p s > t e n t e r t_{exit}>0
\\
t_{exit} + eps > t_{enter} t e x i t > 0 t e x i t + e p s > t e n t er 浮点值精度问题
由于浮点数精度的问题，当 t e x i t t_{exit} t e x i t t e n t e r t_{enter} t e n t er t e x i t t_{exit} t e x i t t e n t e r t_{enter} t e n t er t e x i t t_{exit} t e x i t e p s i l o n = 0.0001 epsilon = 0.0001 e p s i l o n = 0.0001
除零问题
在除以光线方向 D D D D D D 0 0 0 D D D 0 0 0 t e x i t + e p s > t e n t e r t_{exit} + eps > t_{enter} t e x i t + e p s > t e n t er false 的。
如果 P ′ − O P^{\prime}-O P ′ − O D D D 0 0 0 NaN (1.#IND00) 表示这不是数。NaN 和任何数进行任何比较结果都是 false 。
几何碰撞
二维多边形碰撞
分离轴法
未完待续......
