将曲线方程映射到直线方程

Question

我不知道这是否可能，但我有一条曲线，我已经找到了这条曲线的方程，现在我想把这条曲线上的所有点映射到一条直线上。如何做到这一点？

曲线方程：

K3x^3 + k2x^2 + k1x + k0 = y

然后是对应的直线方程，它包含了曲线上的所有点。如果我用曲线的端点建立一个直线方程，那么如何将曲线上的所有点映射或拟合到这条直线上。

例如，我附加了一张图像，所以我们可以通过某种方程转换将这条曲线变成一条直线。谢谢

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我需要输出转换为以下图像。

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Answer 1

所以我只需要从直线方程中减去曲线方程，然后我就得到了另一个方程(方程减去的结果)，我用它把曲线上的所有点映射到直线上。

void mapCurvePointsToLine(IplImage *img, double *coeff)
{
        // algo: mapping equation = line equation - curve equation 

        int i, y;
        double m;
        struct points p;
        p.x = malloc(sizeof(*p.x) * img->widthStep);
        p.y = malloc(sizeof(*p.y) * img->widthStep);
        p.np = img->widthStep;


        for( i = 0; i < img->widthStep; i++)  {
                y = round( (pow(i,3)*coeff[3]) + (pow(i,2)*coeff[2]) + (i*coeff[1]) + coeff[0]);

                p.x[i] = i;
                p.y[i] = y;
                img->imageData[(y*img->widthStep) + i] = 255u;
        }   

        //calculate slope and interspect of line
        m =  (p.y[(p.np - 1)] - p.y[0] ) / (p.x[(p.np - 1)] - p.x[0]);
        for( i = 0; i < img->widthStep; i++)  {

                y = p.y[i] + round( (pow(i,3)* (-1 * coeff[3]) ) + (pow(i,2)* ( -1 * coeff[2])) + (i* ( m - coeff[1]) ) ) ; 
                //y = y + round( (pow(i,3)* (-1 * coeff[3]) ) + (pow(i,2)* ( -1 * coeff[2])) ) ;
                img->imageData[(y*img->widthStep) + i] = 255u;
        }   
}

我得到的结果中有一些失真，可能是因为将计算值四舍五入为整数。请看我附加的输出图像。

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Answer 2

1. 任何2D曲线都可以这样表示：

- `p(t)=p0+p1*t+p2*t*t+p3*t*t*t+...`
- where `p,p0,p1,...` are vectors
- `t` is parameter `<0,1>`
- you already have this
- `p.x(t)=x0+(x1-x0)*t+0*t*t+0*t*t*t;`
- `p.y(t)=k0+k1*t+k2*t*t+k3*t*t*t;`
- which is the same ...
- `x0,x1` are your `x` constraints

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你想要位于行p(0)+(p(1)-p(0))*u;上的q(p(t)) (p(T))(

1. )

- where `u` is parameter `<0,1>`
- so if we assume that parameters `t,u` are proportional to curve distance from start
- then `u=curve_distance(p(t),p(0))/curve_distane(p(1),p(0));`
- so write function `curve_distance` that integrates the path length at interval <0,t>
- and that is all you need

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1. curve_distance

- first you need to find the t parameter for each point
- one point is always the start point so t0=0
- the second you need to find from your parametric equations and input point coordinates
- then just write for cycle with small enough step summing the distance
- or solve it algebraically for used polynomial and compute directly in O(1)

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