Convert a quadratic bezier to a cubic one

问题

What is the algorithm to convert a quadratic bezier (with 3 points) to a cubic one (with 4 points)?

回答1:

From http://fontforge.sourceforge.net/bezier.html:

Any quadratic spline can be expressed as a cubic (where the cubic term is zero). The end points of the cubic will be the same as the quadratic's.

CP₀ = QP₀
CP₃ = QP₂

The two control points for the cubic are:

CP₁ = QP₀ + 2/3 *(QP₁-QP₀)
CP₂ = QP₂ + 2/3 *(QP₁-QP₂)

...There is a slight error introduced due to rounding, but it is unlikely to be noticeable.

回答2:

For reference, I implemented addQuadCurve for NSBezierPath (macOS Swift 4) based on Owen's answer above.

extension NSBezierPath {
    public func addQuadCurve(to qp2: CGPoint, controlPoint qp1: CGPoint) {
        let qp0 = self.currentPoint
        self.curve(to: qp2,
            controlPoint1: qp0 + (2.0/3.0)*(qp1 - qp0),
            controlPoint2: qp2 + (2.0/3.0)*(qp1 - qp2))
    }
}

extension CGPoint {
    // Vector math
    public static func +(left: CGPoint, right: CGPoint) -> CGPoint {
        return CGPoint(x: left.x + right.x, y: left.y + right.y)
    }
    public static func -(left: CGPoint, right: CGPoint) -> CGPoint {
        return CGPoint(x: left.x - right.x, y: left.y - right.y)
    }
    public static func *(left: CGFloat, right: CGPoint) -> CGPoint {
        return CGPoint(x: left * right.x, y: left * right.y)
    }
}

来源：https://stackoverflow.com/questions/3162645/convert-a-quadratic-bezier-to-a-cubic-one