Question

这适用于iPad应用程序，但它本质上是一个数学问题。

我需要绘制一个变化（单调增加）线宽的圆弧。在曲线的开始处，它将具有起始厚度（比如说2pts），然后厚度将平滑地增加直到弧的最终厚度（假设为12pts）。

我认为最好的方法是创建一个UIBezierPath并填充形状。我的第一次尝试是使用两个圆弧（带偏移中心），并且工作精细到90°，但弧度通常在90°和180°之间，因此接近不会削减它。

example of 90 degree arc with increasing thickness

我目前的方法是使用贝塞尔四边形或三次曲线制作一个轻微螺旋（一个从圆弧略微增长，一个略微收缩）。问题是我在哪里放置控制点，以便与圆弧（也就是形状“厚度”）的偏差是我想要的值。

约束：

形状必须能够以任意角度（彼此相差180°）开始和结束
形状的“厚度”（与圆的偏差）必须以给定值开始和结束
“厚度”必须单调增加（它不能再变大，然后再变小）
它必须看起来光滑，不会出现任何尖锐的弯曲

我也对其他解决方案持开放态度。

Answer 1

我的方法只是构造2个圆弧并填充其间的区域。棘手的一点是弄清楚这些弧的中心和半径。如果厚度不是太大，看起来相当不错。（剪切并粘贴并自行决定是否满足您的需求。）可以通过使用剪切路径来改进。

- (void)drawRect:(CGRect)rect
{
  CGContextRef context = UIGraphicsGetCurrentContext();

  CGMutablePathRef path = CGPathCreateMutable();

  // As appropriate for iOS, the code below assumes a coordinate system with
  // the x-axis pointing to the right and the y-axis pointing down (flipped from the standard Cartesian convention).
  // Therefore, 0 degrees = East, 90 degrees = South, 180 degrees = West,
  // -90 degrees = 270 degrees = North (once again, flipped from the standard Cartesian convention).
  CGFloat startingAngle = 90.0;  // South
  CGFloat endingAngle = -45.0;   // North-East
  BOOL weGoFromTheStartingAngleToTheEndingAngleInACounterClockwiseDirection = YES;  // change this to NO if necessary

  CGFloat startingThickness = 2.0;
  CGFloat endingThickness = 12.0;

  CGPoint center = CGPointMake(CGRectGetMidX(self.bounds), CGRectGetMidY(self.bounds));
  CGFloat meanRadius = 0.9 * fminf(self.bounds.size.width / 2.0, self.bounds.size.height / 2.0);

  // the parameters above should be supplied by the user
  // the parameters below are derived from the parameters supplied above

  CGFloat deltaAngle = fabsf(endingAngle - startingAngle);

  // projectedEndingThickness is the ending thickness we would have if the two arcs
  // subtended an angle of 180 degrees at their respective centers instead of deltaAngle
  CGFloat projectedEndingThickness = startingThickness + (endingThickness - startingThickness) * (180.0 / deltaAngle);

  CGFloat centerOffset = (projectedEndingThickness - startingThickness) / 4.0;
  CGPoint centerForInnerArc = CGPointMake(center.x + centerOffset * cos(startingAngle * M_PI / 180.0),
                                          center.y + centerOffset * sin(startingAngle * M_PI / 180.0));
  CGPoint centerForOuterArc = CGPointMake(center.x - centerOffset * cos(startingAngle * M_PI / 180.0),
                                          center.y - centerOffset * sin(startingAngle * M_PI / 180.0));

  CGFloat radiusForInnerArc = meanRadius - (startingThickness + projectedEndingThickness) / 4.0;
  CGFloat radiusForOuterArc = meanRadius + (startingThickness + projectedEndingThickness) / 4.0;

  CGPathAddArc(path,
               NULL,
               centerForInnerArc.x,
               centerForInnerArc.y,
               radiusForInnerArc,
               endingAngle * (M_PI / 180.0),
               startingAngle * (M_PI / 180.0),
               !weGoFromTheStartingAngleToTheEndingAngleInACounterClockwiseDirection
               );

  CGPathAddArc(path,
               NULL,
               centerForOuterArc.x,
               centerForOuterArc.y,
               radiusForOuterArc,
               startingAngle * (M_PI / 180.0),
               endingAngle * (M_PI / 180.0),
               weGoFromTheStartingAngleToTheEndingAngleInACounterClockwiseDirection
               );

  CGContextAddPath(context, path);

  CGContextSetFillColorWithColor(context, [UIColor redColor].CGColor);
  CGContextFillPath(context);

  CGPathRelease(path);  
}

Answer 2

一种解决方案可能是手动生成折线。这很简单，但它的缺点是，如果以高分辨率显示控件，则必须放大生成的点数。我不太了解iOS给你iOS / ObjC示例代码，但这里有一些python-ish伪代码：

# lower: the starting angle
# upper: the ending angle
# radius: the radius of the circle

# we'll fill these with polar coordinates and transform later
innerSidePoints = []
outerSidePoints = []

widthStep = maxWidth / (upper - lower)
width = 0

# could use a finer step if needed
for angle in range(lower, upper):
    innerSidePoints.append(angle, radius - (width / 2))
    outerSidePoints.append(angle, radius + (width / 2))
    width += widthStep

# now we have to flip one of the arrays and join them to make
# a continuous path.  We could have built one of the arrays backwards
# from the beginning to avoid this.

outerSidePoints.reverse()
allPoints = innerSidePoints + outerSidePoints # array concatenation

xyPoints = polarToRectangular(allPoints) # if needed

使用贝塞尔曲线绘制螺旋线

2 个答案: