均匀与准均匀 B样条算法

2024-01-07 23:59:21

B 样条曲线的定义

p ( t ) = ∑ i = 0 n P i F i , k ( t ) p(t) = \sum_{i=0}{n} P_i F_{i, k}(t) p(t)=i=0?nPi?Fi,k?(t)

方程中 n + 1 n+1 n+1 个控制点, P i P_i Pi?, i = 0 , 1 , ? n i=0, 1, \cdots n i=0,1,?n 要用到 n + 1 n+1 n+1 k k k 次 B 样条基函数 F i , k F_{i, k} Fi,k?, i = 0 , 1 , ? ? , n i=0, 1, \cdots, n i=0,1,?,n, 节点矢量为 T = [ t 0 , t 1 , ? ? , t n + k + 1 ] T = [t_0, t_1, \cdots, t_{n+k+1}] T=[t0?,t1?,?,tn+k+1?] F i , k ( t ) F_{i, k}(t) Fi,k?(t) 是由一个称为节点矢量的非递减的参数 t t t 的序列, t 0 ≤ t 1 ≤ ? ≤ t n + k + 1 t_0 \leq t_1 \leq \cdots \leq t_{n+k+1} t0?t1??tn+k+1?所决定的 k k k 次分段多项式。

B 样条曲线划分为四种类型,均匀 B 样条曲线,准均匀B 样条曲线,分段 Bezier 曲线和非均匀 B 样条曲线。

定义域

给定 n + 1 n+1 n+1 个控制点, P i P_i Pi?, i = 0 , 1 , ? n i=0, 1, \cdots n i=0,1,?n, 相应地要求 n + 1 n+1 n+1 个 B 样条基函数 F i , k ( t ) F_{i, k}(t) Fi,k?(t) 定义一个 k k k 次 B 样条曲线,这 n + 1 n+1 n+1 k k k 次 B 样条由节点矢量 T = [ t 0 , t 1 , ? t n + k + 1 ] T = [t_0, t_1, \cdots t_{n+k+1}] T=[t0?,t1?,?tn+k+1?] 所决定。
并非这个些节点矢量所包含的 n + k + 1 n+k +1 n+k+1 个区间都在该曲线的定义域,其中两端的各 k k k 个几点区间,不能作为 B 样条曲线的定义区间。
这是因为 n + 1 n+1 n+1 个控制点中最前的 n + 1 n+1 n+1 个顶点 P i P_i Pi?, i = 0 , 1 , ? k i=0,1, \cdots k i=0,1,?k 定义了 B 样条曲线的首段,其定义区间为 t ∈ [ t k , t k + 1 ] t\in [t_k, t_{k+1}] t[tk?,tk+1?] 往后移动一个顶点 P i P_i Pi? i = 1 , 2 , ? k + 1 i=1, 2, \cdots k+1 i=1,2,?k+1 定义第二段,其定义区间为 t ∈ [ t k + , t k + 2 ] t \in [t_{k+}, t_{k+2}] t[tk+?,tk+2?] 依次类推,最后 k + 1 k+1 k+1 个顶点, P i P_i Pi?, i = n ? k , b ? k ? 1 , ? n i=n-k, b-k-1, \cdots n i=n?k,b?k?1,?n 定义最后一段,其定义区间为 t ∈ [ t n , t n + 1 ] t\in[t_n, t_{n+1}] t[tn?,tn+1?], 因此,高于零次的 k k k 次B 样条曲线的定义域为

t ∈ [ t k , t n + 1 ] t \in [t_k, t_{n+1}] t[tk?,tn+1?]

三次均匀 B 样条曲线

{ F 0 , 3 ( t ) = 1 6 ( 1 ? t ) 3 = ( ? t 3 + 3 t 2 ? 3 t + 1 ) F 1 , 3 ( t ) = 1 6 ( 3 t 3 ? 6 t 2 + 4 ) F 2 , 3 ( t ) = 1 6 ( ? 3 t 3 + 3 t 2 + 3 t + 1 ) F 3 , 3 ( t ) = 1 6 t 3 \begin{cases} F_{0,3}(t) = \frac{1}{6} (1-t)^3 = (-t^3 + 3t^2 -3t+1)\\ F_{1,3}(t) = \frac{1}{6} (3t^3 - 6t^2 +4)\\ F_{2,3}(t) = \frac{1}{6} (-3t^3 + 3t^2 + 3t +1) \\ F_{3,3}(t) = \frac{1}{6} t^3\\ \end{cases} ? ? ??F0,3?(t)=61?(1?t)3=(?t3+3t2?3t+1)F1,3?(t)=61?(3t3?6t2+4)F2,3?(t)=61?(?3t3+3t2+3t+1)F3,3?(t)=61?t3?

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三次 B 样条的几何性质

{ p ( 0 ) = 1 6 ( p 0 + 4 p 1 + p 2 ) = 1 3 ( p 0 + p 2 2 ) + 2 3 p 1 = 1 3 p m + 2 3 p 1 p ( 1 ) = 1 6 ( p 1 + 4 p 2 + p 3 ) = 1 3 ( p 1 + p 3 2 ) + 2 3 p 2 = 1 3 p n + 2 3 p 2 \begin{cases} p(0) = \frac{1}{6}(p_0 + 4 p_1 + p_2) = \frac{1}{3} (\frac{p_0 + p_2}{2}) + \frac{2}{3}p_1 = \frac{1}{3}p_m + \frac{2}{3}p_1\\ p(1) = \frac{1}{6}(p_1 + 4 p_2 + p_3) = \frac{1}{3} (\frac{p_1 + p_3}{2}) + \frac{2}{3}p_2 = \frac{1}{3}p_n + \frac{2}{3}p_2\\ \end{cases} {p(0)=61?(p0?+4p1?+p2?)=31?(2p0?+p2??)+32?p1?=31?pm?+32?p1?p(1)=61?(p1?+4p2?+p3?)=31?(2p1?+p3??)+32?p2?=31?pn?+32?p2??

{ p ′ ( 0 ) = 1 2 ( p 2 ? p 0 ) p ′ ( 1 ) = 1 2 ( p 3 + p 1 ) \begin{cases} p'(0) = \frac{1}{2}(p_2 - p_0) \\ p'(1) = \frac{1}{2}(p_3 + p_1) \\ \end{cases} {p(0)=21?(p2??p0?)p(1)=21?(p3?+p1?)?

{ p ′ ′ ( 0 ) = p 0 ? 2 p 1 + p 2 = 2 ( p 0 + p 2 2 ? p ) = 2 ( p m ? p 1 ) p ′ ′ ( 1 ) = p 1 ? 2 p 2 + p 3 = 2 ( p 1 + p 3 2 ? p 2 ) = 2 ( p n ? p 2 ) \begin{cases} p''(0) = p_0 - 2p_1 + p_2 = 2(\frac{p_0 + p_2}{2} -p)= 2(p_m - p_1) \\ p''(1) = p_1 - 2p_2 + p_3 = 2(\frac{p_1 + p_3}{2} -p_2)= 2(p_n - p_2) \\ \end{cases} {p′′(0)=p0??2p1?+p2?=2(2p0?+p2???p)=2(pm??p1?)p′′(1)=p1??2p2?+p3?=2(2p1?+p3???p2?)=2(pn??p2?)?

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#include <QWidget>
#include <QApplication>
#include <QPainter>
#include <QPointF>
#include <QPainterPath>


const double knot[13] = {-3/6.0, -2/6.0, -1/6.0, 0.0, 1 / 6.0, 2 / 6.0, 3 / 6.0, 4 / 6.0, 5 / 6.0, 1.0, 7/ 6.0, 8/ 6.0, 9/6.0};


double BasisFunctionValue(double t, int i, int k)
{
    double val1, val2, val;
    if (k == 0)
    {
        if ((t >= knot[i]) && t < knot[i + 1])
        {
            return 1.0;
        }
        else
        {	// 其它
            return 0.0;
        }
    }
    if (k > 0)
    {
        if (t < knot[i] || t > knot[i + k + 1]) {
            return 0.0;		// 其它
        }
        else
        {
            double coffcient1, coffcient2;	// 凸组合系数1 凸组合系数 2
            double denominator = 0.0;		// 分母
            denominator = knot[i + k] - knot[i];
            if (denominator == 0.0)
            {
                // 约定 0/0 = 0
                coffcient1 = 0.0;
            }
            else
            {
                coffcient1 = (t - knot[i]) / denominator;	// 计算的第一项
            }
            denominator = knot[i + k + 1] - knot[i + 1];	// 递推公式第二项分母
            if (denominator == 0.0)
            {
                // 约定 0/0 = 0
                coffcient2 = 0.0;
            }
            else
            {
                coffcient2 = (knot[i + k + 1] - t) / denominator;	// 递推公式第二项
            }
            val1 = coffcient1 * BasisFunctionValue(t, i, k - 1);	// 递推公式第一项的只
            val2 = coffcient2 * BasisFunctionValue(t, i+1, k - 1);	// 递推公式第二项的只
            val = val1 + val2;	// 基函数的值
        }
    }
    return val;
}




void drawBSplineCure(QPainter* painter, const std::vector<QPointF>& P)
{
    // Set line color
    QColor lineColor(0, 0, 255);
    // Set point color
    QColor pointColor(255, 0, 0);

    QPainterPath bezierPath;
    QPen pen(lineColor);
    pen.setWidth(2);  // Set the line width as needed
    painter->setPen(pen);

    QPointF center(900, 600);  // Center coordinates

    int k = 3;  // Degree of the B-spline curve

    for (int i = k; i <= P.size() - k; ++i)
    {
        double tStep = 0.01;
        for (double t = 0.0; t <= 1.0; t += tStep)
        {
            QPointF p(0, 0);  // Discrete point
            for (int j = 0; j < P.size(); ++j)
            {
                double BValue = BasisFunctionValue(t, j, k);
                p += P[j] * BValue;
            }

            if (t == 0.0)
            {
                bezierPath.moveTo(p + center);
            }
            else
            {
                bezierPath.lineTo(p + center);

            }


            painter->setPen(pointColor);
            painter->setBrush(Qt::NoBrush);
            painter->drawEllipse(p + center, 5, 5);

        }
    }

    painter->drawPath(bezierPath);
}



void drawControlPolygon(QPainter* painter, std::vector<QPointF> P)
{
    QColor lineColor(0, 0, 0);
    QColor pointColor(0, 0, 255);  // Blue color for points

    QPen polyLinePen(lineColor);
    painter->setPen(polyLinePen);


    QBrush pointBrush(pointColor);
    painter->setBrush(pointBrush);

    QPointF center(900, 600);

    QVector<QPointF> shiftedPoints;
    std::transform(P.begin(), P.end(), std::back_inserter(shiftedPoints),
                   [center](const QPointF& point) { return point + center; });


    painter->drawPolyline(shiftedPoints.data(), shiftedPoints.size());


    for (const QPointF& point : shiftedPoints)
    {
        painter->drawEllipse(point, 5, 5);
    }

}


std::vector<QPointF> getControlPoints(){
    std::vector<QPointF> controlPoints = {
        QPointF(-600, -50),
        QPointF(-500, 200),  // 控制点
        QPointF(-160, 250),
        QPointF(-250, -300),
        QPointF(160, -200),  // 控制点
        QPointF(200, 200),
        QPointF(600, 180),
        QPointF(700, -60),   // 控制点
        QPointF(500, -200)
    };
    return controlPoints;
}



class MyWidget : public QWidget {
public:
    MyWidget(QWidget* parent = nullptr) : QWidget(parent) {
        setFixedSize(1800, 1200);
    }

protected:
    void paintEvent(QPaintEvent* event) override {
        Q_UNUSED(event);

        QPainter painter(this);
        painter.setRenderHint(QPainter::Antialiasing, true);

        std::vector<QPointF> controlPoints = getControlPoints();
        drawBSplineCure(&painter, controlPoints);
        drawControlPolygon(&painter, controlPoints);
    }




public:
    int n = 8;
    int k = 3;



};


int main(int argc, char* argv[]) {
    QApplication app(argc, argv);

    MyWidget widget;
    widget.show();

    return app.exec();
}

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文章来源:https://blog.csdn.net/weixin_43862398/article/details/135382704
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