Question

我正在编写摄像机运动控制器的代码，贝塞尔曲线方程用于使运动平滑。

由于我的数学能力不是很强，我使用了这个＆＃34; stackoverflow＆＃34;回答Interpolating values between interval, interpolation as per Bezier curve，感谢会员＆＃34; Daniel Wolf＆＃34;。

using System;
public class Point {
public Point(double x, double y) 
{
    X = x;
    Y = y;
}
public double X { get; private set; }
public double Y { get; private set; }
}

public class BezierCurve {
public BezierCurve(Point p0, Point p1, Point p2, Point p3) {
    P0 = p0;
    P1 = p1;
    P2 = p2;
    P3 = p3;
}

public Point P0 { get; private set; }
public Point P1 { get; private set; }
public Point P2 { get; private set; }
public Point P3 { get; private set; }

public double? GetY(double x) {
    // Determine t
    double t;
    if (x == P0.X) {
        // Handle corner cases explicitly to prevent rounding errors
        t = 0;
    } else if (x == P3.X) {
        t = 1;
    } else {
        // Calculate t
        double a = -P0.X + 3 * P1.X - 3 * P2.X + P3.X;
        double b = 3 * P0.X - 6 * P1.X + 3 * P2.X;
        double c = -3 * P0.X + 3 * P1.X;
        double d = P0.X - x;
        double? tTemp = SolveCubic(a, b, c, d);
        if (tTemp == null) return null;
        t = tTemp.Value;
    }

    // Calculate y from t
    return Cubed(1 - t) * P0.Y
        + 3 * t * Squared(1 - t) * P1.Y
        + 3 * Squared(t) * (1 - t) * P2.Y
        + Cubed(t) * P3.Y;
}

// Solves the equation ax³+bx²+cx+d = 0 for x ϵ ℝ
// and returns the first result in [0, 1] or null.
private static double? SolveCubic(double a, double b, double c, double d) {
    if (a == 0) return SolveQuadratic(b, c, d);
    if (d == 0) return 0;

    b /= a;
    c /= a;
    d /= a;
    double q = (3.0 * c - Squared(b)) / 9.0;
    double r = (-27.0 * d + b * (9.0 * c - 2.0 * Squared(b))) / 54.0;
    double disc = Cubed(q) + Squared(r);
    double term1 = b / 3.0;

    if (disc > 0) {
        double s = r + Math.Sqrt(disc);
        s = (s < 0) ? -CubicRoot(-s) : CubicRoot(s);
        double t = r - Math.Sqrt(disc);
        t = (t < 0) ? -CubicRoot(-t) : CubicRoot(t);

        double result = -term1 + s + t;
        if (result >= 0 && result <= 1) return result;
    } else if (disc == 0) {
        double r13 = (r < 0) ? -CubicRoot(-r) : CubicRoot(r);

        double result = -term1 + 2.0 * r13;
        if (result >= 0 && result <= 1) return result;

        result = -(r13 + term1);
        if (result >= 0 && result <= 1) return result;
    } else {
        q = -q;
        double dum1 = q * q * q;
        dum1 = Math.Acos(r / Math.Sqrt(dum1));
        double r13 = 2.0 * Math.Sqrt(q);

        double result = -term1 + r13 * Math.Cos(dum1 / 3.0);
        if (result >= 0 && result <= 1) return result;

        result = -term1 + r13 * Math.Cos((dum1 + 2.0 * Math.PI) / 3.0);
        if (result >= 0 && result <= 1) return result;

        result = -term1 + r13 * Math.Cos((dum1 + 4.0 * Math.PI) / 3.0);
        if (result >= 0 && result <= 1) return result;
    }

    return null;
}

// Solves the equation ax² + bx + c = 0 for x ϵ ℝ
// and returns the first result in [0, 1] or null.
private static double? SolveQuadratic(double a, double b, double c) {
    double result = (-b + Math.Sqrt(Squared(b) - 4 * a * c)) / (2 * a);
    if (result >= 0 && result <= 1) return result;

    result = (-b - Math.Sqrt(Squared(b) - 4 * a * c)) / (2 * a);
    if (result >= 0 && result <= 1) return result;

    return null;
}

private static double Squared(double f) { return f * f; }

private static double Cubed(double f) { return f * f * f; }

private static double CubicRoot(double f) { return Math.Pow(f, 1.0 / 3.0); }
}

Daniel编写的代码使用相对较小的值（小于1000）非常好，但是当我尝试使用更高的值（超过1000）时，该函数没有给出任何值。

有什么想法吗？

更新：

问题解决了

解决贝塞尔曲线

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