Android地图v2折线交叉点

Question

我有一个圆和一条折线，我想知道交点的协调

Answer 1

user3591040非常棘手....

我希望你有关于圆和线的数据。

E

你想找到这一点

**

1. 考虑这张图片 我们将尝试检测任何碰撞

**

碰撞检测算法：尝试实现此算法

以

L是光线的起点， C是光线的终点， r是您正在测试的球体的中心 d = L - E ( Direction vector of ray, from start to end ) f = E - C ( Vector from center sphere to ray start ) 是该球体的半径 计算：

P = E + t * d

然后通过..找到交叉点 堵漏：

Px = Ex + tdx
Py = Ey + tdy

这是一个参数方程式：

(x - h)2 + (y - k)2 = r2
(h,k) = center of circle.

到

x2 - 2xh + h2 + y2 - 2yk + k2 - r2 = 0

注意：我们在这里将问题简化为2D，我们得到的解决方案也适用于3D 得到：

展开

x = ex + tdx
y = ey + tdy
( ex + tdx )2 - 2( ex + tdx )h + h2 + ( ey + tdy )2 - 2( ey + tdy )k + k2 - r2 = 0
Explode
ex2 + 2extdx + t2dx2 - 2exh - 2tdxh + h2 + ey2 + 2eytdy + t2dy2 - 2eyk - 2tdyk + k2 - r2 = 0
Group
t2( dx2 + dy2 ) + 2t( exdx + eydy - dxh - dyk ) + ex2 + ey2 - 2exh - 2eyk + h2 + k2 - r2 = 0
Finally,
t2( _d * _d ) + 2t( _e * _d - _d * _c ) + _e * _e - 2( _e*_c ) + _c * _c - r2 = 0
*Where _d is the vector d and * is the dot product.*
And then,
t2( _d * _d ) + 2t( _d * ( _e - _c ) ) + ( _e - _c ) * ( _e - _c ) - r2 = 0
Letting _f = _e - _c
t2( _d * _d ) + 2t( _d * _f ) + _f * _f - r2 = 0
So we get:
t2 * (d DOT d) + 2t*( f DOT d ) + ( f DOT f - r2 ) = 0

插头

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 ;
  }

所以求解二次方程：

\G