Question

我查找了一些与找到点和线的距离有关的公式。在这个页面上，我使用了例14

http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html

我有一个方法变成了这个：

+(bool) checkPointNearBetweenPointsWithPointA:(CGPoint)pointA withPointB:(CGPoint)pointB withPointC:(CGPoint)pointC withLimit:(float)limit {

  float A = pointB.x - pointA.x;
  float B = pointA.y - pointC.y;
  float C = pointA.x - pointC.x;
  float D = pointB.y - pointA.y;

  float dividend = fabs( A * B ) - ( C * D );

  float divisor = sqrt(pow(A,2) + pow(D,2));

  float distanceBetweenPointAndLine = dividend / divisor;

  if(distanceBetweenPointAndLine < limit){

    NSLog(@"distanceBetweenPointAndLine = %f",distanceBetweenPointAndLine);

    return YES;

  }


  return NO;
}

问题是，如果我通过了B点，如果线段被画成B-A，它仍然会返回YES。其他搞砸的事情发生在线的绘制角度。只是想知道我是否需要考虑其他任何事情，如果测试一个点是否接近有限线。我在网上看到的大多数例子都涉及无限长的行。

Answer 1

尝试下面的代码。线被认为存在于点A和A之间。 B（无论您如何绘制B-> A或A-> B）和C点是考虑距离的考虑点。

+ (bool) checkPointNearBetweenPointsWithPointA:(CGPoint)pointA
                                   withPointB:(CGPoint)pointB
                                   withPointC:(CGPoint)pointC
                                    withLimit:(float)limit
{

    CGFloat slopeLine = atan((pointB.y-pointA.y)/(pointB.x-pointA.x) );
    CGFloat slopePointToPointA = -1 *atan((pointC.y-pointA.y)/(pointC.x-pointA.x));

    CGFloat innerAngle = slopeLine + slopePointToPointA;
    CGFloat distanceAC = sqrtf(pow(pointC.y-pointA.y,2) + pow(pointC.x-pointA.x,2));

    CGFloat distanceBetweenPointAndLine = fabs(distanceAC * sin(innerAngle));

    NSLog(@"distanceBetweenPointAndLine = %f",distanceBetweenPointAndLine);
    NSLog(@"is exceeding limit ? %@",distanceBetweenPointAndLine > limit ? @"YES":@"NO");

    if(distanceBetweenPointAndLine < limit)
    {
        return YES;
    }
    return NO;
}

objective-c中点与有限线之间的距离

1 个答案: