3点参数插值曲线 [英] 3 point parametric interpolation curve

问题描述

我想绘制一个3点参数样条曲线，如本网页所示。这不是一个贝塞尔曲线。积分必须都在曲线上。



http://www.doc.ic.ac.uk/~dfg/AndysSplineTutorial/Parametrics.html



我需要两个同时的方程式，x和y有一个公共变量（t = 0-1）。不知道如何用数学创造它们。任何帮助，将不胜感激。谢谢。

I would like to draw a 3 point parametric spline as demonstrated at this webpage. This is not a bezier curve. The points must all be on the curve.

"http://www.doc.ic.ac.uk/~dfg/AndysSplineTutorial/Parametrics.html"

I need two simultanious equations, x and y with a common variable(t = 0-1). Don''t know how to create them from the mathematics. Any help would be appreciated. thanks.

推荐答案

http：//www.extremeoptimization .com / QuickStart / CSharp / CubicSplines.aspx [ ^ ]



通过一组带有Bezier基元的2D点绘制平滑曲线 [ ^ ]



和一些免费的源代码示例 http://www.codeforge.com/s/ 0 / parametric-cubic-spline [ ^ ]

http://www.extremeoptimization.com/QuickStart/CSharp/CubicSplines.aspx[^]

Draw a Smooth Curve through a Set of 2D Points with Bezier Primitives[^]

and some free source examples http://www.codeforge.com/s/0/parametric-cubic-spline[^]

退出简单：提供的等式（您链接的页面）是矢量，让我们写它（为简单起见）

It is quite simple: the equation (the page you linked) provided is vectorial, let''s write it (for simplicity)

P(0)={x0,y0}={a0,b0};
P(s)={xs,ys}={a2*s*s+a1*s+a0, b2*s*s+b1*s+b0}
P(1)={x1,y1}={a2+a1+a0, b2+b1+b0}

其中P（0），P（s），P（1）是已知（0< s< 1）。

我们必须找到六个参数 a2，a1，a0，b2，b1，b0 。将计算限制为 x （y是相同的）：

Where the P(0), P(s), P(1) are known (0<s<1).
We have to find the six parameters a2,a1,a0, b2,b1,b0. Limiting computation to x (y is the same):

x0 = a0
x1 = a2+a1+x0 => a1 = x1-x0-a2
xs = a2*s*s+x1*s-x0*s-a2*s+x0 => a2 = (xs-x0+(x0-x1)*s)/(s-s*s)

a2 使用第二个等式计算 a1 的值。

现在 a2，a1，a0 （并且以相同的方式 b2，b1，b0 ）已知并且您可以计算 P（t）= {x（t），y（t ）} = {a2 * t * t + a1 * t + a0，b2 * t * t + b1 * t + b0} 每个 0< t< 1 。

（我希望它是正确的......）

With a2 value you compute a1 using the second equation.
Now a2,a1,a0 (and in the same way b2,b1,b0) are known and you may compute P(t) = {x(t), y(t)} = {a2*t*t+a1*t+a0, b2*t*t+b1*t+b0} for each 0<t<1.
(I hope it is correct...)

引用网站中的某些公式不正确。特别是 ₁ 的结果公式是错误的。



你必须明白你可以自由选择t ₁ 对于你的P ₁ ：通过改变你选择的P ₁ 的t ₁ ，曲线的形状也会有所不同。 />


假设你有三个定义点：

- 在P ₀ 开始（t = 0）

- 结束（t = 1）在P ₂

- 在开始和结束之间的点P ₁ 你还决定在 ₁ （例如0.5）。



给定这四个参数（P ₀ ，P ₂ ，P ₁ ，t ₁ ），所需的曲线系数为：

a ₀ = A

a ₁ = B - t ₁ C

a ₂ = C - B

其中

A = P ₀

B =（P ₁ - P < sub> 0 ）/（（1 - t ₁ ）t ₁ ）

C =（P ₂ - P ₀ ）/（1 - t ₁ ）



如果在2D中工作，每个系数由两个值组成，每个轴一个 - 在3D中工作，每个系数由每个轴上的3个值组成。



例如对于每个轴系数值，将点P _u 的相应轴值加上参数t ₁ 值设置为上面的公式。



然后你将参数t从0扫到1并计算要绘制的曲线点P（t）。



伪代码：

Some formulas in the referenced web site are not correct. Especially the result formula for a₁ is wrong.

You must understand that you have the freedom to chose the t₁ for your P₁: by varying t₁ for your chosen P₁, the shape of the curve will vary too.

Assuming you have three defined points:
- start (t = 0) at P₀
- end (t = 1) at P₂
- a point P₁ between start and end where you also decide for a t₁ (e.g. 0.5).

Given these four parameters (P₀, P₂, P₁, t₁), the needed curve coefficients are:
a₀ = A
a₁ = B - t₁ C
a₂ = C - B
Where
A = P₀
B = (P₁ - P₀) / ((1 - t₁) t₁)
C = (P₂ - P₀) / (1 - t₁)

If working in 2D, each coefficient consists of two values, one for each axis - working in 3D, each coefficient consists of 3 values, on per axis.

E.g. for each axis coeficient value, you set the respective axis value of the points P_u plus the parameter t₁ value into the formulas above.

Then you sweep the parameter t from 0 to 1 and calculate the curve points P(t) to draw.

pseudo code:

class Formula
{
    private Point a0;
    private Point a1;
    private Point a2;

    private Point A(Point p0, Point p1, Point p2, double t1) { return p0; }
    private Point B(Point p0, Point p1, Point p2, double t1) { double d = 1/((1-t1)*t1); return (p1-p0)*d; }
    private Point C(Point p0, Point p1, Point p2, double t1) { return d = 1/(1-t1); return (p2-p0)*d; }

    public Formula(Point p0, Point p1, point p2, double t1)
    {
       Point a = A(p0, p1, p2, t1);
       Point b = B(p0, p1, p2, t1);
       Point c = C(p0, p1, p2, t1);

       a0 = a;
       a1 = b-c*t1;
       a2 = c-b;
    }
    public Point P(double t) { return ((a2*t)+a1)*t+a0; }
}
...
var start = new Point(0,2); // p0
var end   = new Point(7,6); // p2
var p1    = new Point(2,3); // p1
double t1 = 0.5;
var Formula curve = new Formula(start, p1, end, t1);
...
int steps = 100;
double step = 1.0/(double)steps;
Point curr = p0;
foreach(int i = 1; i <= steps; ++i)
{
   Point next = curve.P(step*(double)i);
   Draw(curr, next);
   curr = next;
}

干杯

Andi

Cheers
Andi

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