Circle line-segment collision detection algorithm?

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被撕碎了的回忆 2020-11-22 06:38

I have a line from A to B and a circle positioned at C with the radius R.

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What is a good alg

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  •  自闭症患者
    2020-11-22 06:57

    Taking

    1. E is the starting point of the ray,
    2. L is the end point of the ray,
    3. C is the center of sphere you're testing against
    4. r is the radius of that sphere

    Compute:
    d = L - E ( Direction vector of ray, from start to end )
    f = E - C ( Vector from center sphere to ray start )

    Then the intersection is found by..
    Plugging:
    P = E + t * d
    This is a parametric equation:
    Px = Ex + tdx
    Py = Ey + tdy
    into
    (x - h)2 + (y - k)2 = r2
    (h,k) = center of circle.

    Note: We've simplified the problem to 2D here, the solution we get applies also in 3D

    to get:

    1. Expand
      x2 - 2xh + h2 + y2 - 2yk + k2 - r2 = 0
    2. Plug
      x = ex + tdx
      y = ey + tdy
      ( ex + tdx )2 - 2( ex + tdx )h + h2 + ( ey + tdy )2 - 2( ey + tdy )k + k2 - r2 = 0
    3. Explode
      ex2 + 2extdx + t2dx2 - 2exh - 2tdxh + h2 + ey2 + 2eytdy + t2dy2 - 2eyk - 2tdyk + k2 - r2 = 0
    4. Group
      t2( dx2 + dy2 ) + 2t( exdx + eydy - dxh - dyk ) + ex2 + ey2 - 2exh - 2eyk + h2 + k2 - r2 = 0
    5. 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.*
    6. And then,
      t2( _d * _d ) + 2t( _d * ( _e - _c ) ) + ( _e - _c ) * ( _e - _c ) - r2 = 0
    7. 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
    So solving the quadratic equation:

    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 ;
    }
    

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