TA貢獻1946條經(jīng)驗 獲得超4個贊
E
L
C
r
d
f
P=E+t*d
Px
Py
注意：我們將問題簡化為2D，我們得到的解決方案也適用于3D
得到：
擴張
x2
2
2
2
2 = 0
塞子
x
x
y
y
x
x )2
x
x
2
y
y )2
y
y
2
2 = 0
ex2
x
x
2dx2
x
x
2
y2
y
y
2dy2
y
y
2
2 = 0
群
t2
x2
y2
xdx
ydy
x
y
x2
y2
x
y
2
2
2 = 0
最后，
t2
2 = 0
然后,
t2
2 = 0
讓f=_e-_c
t2
2 = 0
t2
float a = d.Dot( d ) ;
float b = 2*f.Dot( d ) ;
float c = f.Dot( f ) - r*r ;
float discriminant = b*b-4*a*c;
if( discriminant < 0 )
{
// no intersection
}
else
{
// ray didn't totally miss sphere,
// so there is a solution to
// the equation.
discriminant = sqrt( discriminant );
// either solution may be on or off the ray so need to test both
// t1 is always the smaller value, because BOTH discriminant and
// a are nonnegative.
float t1 = (-b - discriminant)/(2*a);
float t2 = (-b + discriminant)/(2*a);
// 3x HIT cases:
// -o-> --|--> | | --|->
// Impale(t1 hit,t2 hit), Poke(t1 hit,t2>1), ExitWound(t1<0, t2 hit),
// 3x MISS cases:
// -> o o -> | -> |
// FallShort (t1>1,t2>1), Past (t1<0,t2<0), CompletelyInside(t1<0, t2>1)
if( t1 >= 0 && t1 <= 1 )
{
// t1 is the intersection, and it's closer than t2
// (since t1 uses -b - discriminant)
// Impale, Poke
return true ;
}
// here t1 didn't intersect so we are either started
// inside the sphere or completely past it
if( t2 >= 0 && t2 <= 1 )
{
// ExitWound
return true ;
}
// no intn: FallShort, Past, CompletelyInside
return false ;
}
TA貢獻1812條經(jīng)驗 獲得超5個贊
ACABADD.DCR
TA貢獻1824條經(jīng)驗 獲得超6個贊
// compute the euclidean distance between A and B
LAB = sqrt( (Bx-Ax)2+(By-Ay)2 )
// compute the direction vector D from A to B
Dx = (Bx-Ax)/LAB
Dy = (By-Ay)/LAB
// the equation of the line AB is x = Dx*t + Ax, y = Dy*t + Ay with 0 <= t <= LAB.
// compute the distance between the points A and E, where
// E is the point of AB closest the circle center (Cx, Cy)
t = Dx*(Cx-Ax) + Dy*(Cy-Ay)
// compute the coordinates of the point E
Ex = t*Dx+Ax
Ey = t*Dy+Ay
// compute the euclidean distance between E and C
LEC = sqrt((Ex-Cx)2+(Ey-Cy)2)
// test if the line intersects the circle
if( LEC < R )
{
// compute distance from t to circle intersection point
dt = sqrt( R2 - LEC2)
// compute first intersection point
Fx = (t-dt)*Dx + Ax
Fy = (t-dt)*Dy + Ay
// compute second intersection point
Gx = (t+dt)*Dx + Ax
Gy = (t+dt)*Dy + Ay
}
// else test if the line is tangent to circle
else if( LEC == R )
// tangent point to circle is E
else
// line doesn't touch circle舉報