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python之片段着色器 : get distance from fragment to bezier curve

2025年05月04日9thcjp

让我们从我所在的位置开始; current state

我有一个通过 FBO 生成这个纹理的片段着色器。除了贝塞尔曲线,我几乎拥有所有我想要的图元。

我相当大的片段着色器:

testfrag = """//fragment 
uniform float thickness; //sets beam thickness 
uniform float light; //sets beam intensity 
 
uniform vec4 lines[200]; //pairs of lines; x1, y1, x2, y2 
uniform vec3 circles[200]; //circles x,y,radius 
uniform vec4 ellipses[200]; //ellipses x,y,width, heightmultiplication 
uniform int ellipselength = 1; //amount of ellipses 
uniform int linelength = 2; //amount of linepairs 
uniform int circlelength = 1; //amount of circles 
 
 
uniform vec2 polygons[200]; //polygon data, points in x,y. (0,0) points used as end marker 
uniform int polylength; //amount of polygons 
 
 
int i; //for counter variable 1 
int i2; //for counter variable 2 
float mindistance = 1.0; //distance from fragment to next segment 
 
float P2L(in vec2 p, in vec2 lp1, in vec2 lp2) //distance point to line segment 
{ 
  float squared = pow(distance(lp1, lp2), 2); //squared line length 
  float t = dot(p - lp1, lp2 - lp1) / squared; //relative position in parellel to line 
  if (t < 0.0) return distance(p, lp1); //before the line, get radius to point 
  if (t > 1.0) return distance(p, lp2); //after the line, get radius to point 
  return distance(p, lp1 + t * (lp2 - lp1)); //otherwise, get distance line to point 
} 
float P2C(in vec2 p, in vec3 circle) //point to circle 
{ 
    return abs(distance(p, circle.xy)-circle.z); //euclidian distance - radius 
} 
float P2E(in vec2 p, in vec4 ellipse) //point to ellipse 
{ 
    return abs(sqrt(pow(p.x-ellipse.x,2) + pow((p.y-ellipse.y)/ellipse.w, 2)) -ellipse.z); // similar to circle, with factor on height 
} 
 
bool PinPoly(in vec2 p, in int start, in int len) //test if point in polygon 
{ 
  int i, j;  
  bool c = false; 
  for (i = start, j = len-1+start; i < len+start; j = i++) { 
    if ( ((polygons[i].y>p.y) != (polygons[j].y>p.y)) && 
     (p.x < (polygons[j].x-polygons[i].x) * (p.y-polygons[i].y) / (polygons[j].y-polygons[i].y) + polygons[i].x) ) 
       c = !c; 
  } 
  return c; 
} //balls if I know how this works 
 
 
 
void main() 
{    
 
    vec2 pos = gl_TexCoord[0].xy; // get position on fbo 
    for (i = 0; i < linelength; i++) //test lines 
    { 
        mindistance = min(mindistance, P2L(pos, lines[i].xy, lines[i].zw)); 
    } 
    for (i = 0; i < circlelength; i++) //test circles 
    { 
        mindistance = min(mindistance, P2C(pos, circles[i])); 
    } 
    for (i = 0; i < ellipselength; i++) //test ellipses 
    { 
        mindistance = min(mindistance, P2E(pos, ellipses[i])); 
    } 
    i = 1; 
    int first; 
    while (i < polylength) //test polygons 
    { 
        //first for line segments 
        first = i-1; 
        while (polygons[i] != (0.0, 0.0)) 
        { 
            mindistance = min(mindistance, P2L(pos, polygons[i-1], polygons[i])); 
            i++; 
 
        } 
 
        mindistance = min(mindistance, P2L(pos, polygons[i-1], polygons[first])); 
        if (PinPoly(pos, first, i-first)) //then test if it is inside a polygon 
        { 
            mindistance = 0.0; 
        } 
        i += 2; //jump over the (0,0) vec2 
 
    } 
 
    gl_FragColor = light*(1.0/thickness)*(thickness-mindistance).xxxx; //set color of fragment 
} 
""" 

在pyglet环境中大致是这样调用的:

def on_draw(self): 
    if not self.inited: 
        vert = shader.Shader(shader.texvertex) 
        frag = shader.Shader(shader.testfrag) 
        self.TexProgram = shader.Program([vert, frag]) 
        with self.TexProgram: 
            self.TexProgram["lines[0]"] = (0.25, 0.5, 0.75, 0.5) 
            self.TexProgram["lines[1]"] = (0.0, 0.0, 1.0, 1.0) 
            self.TexProgram["circles[0]"] = (0.5, 0.5, 0.1) 
            self.TexProgram["ellipses[0]"] = (0.75, 0.25, 0.1, 2.0) #x,y, width, multiplicator on height 
            self.TexProgram["polygons[0]"] = (0.1, 0.9) 
            self.TexProgram["polygons[1]"] = (0.2, 0.9) 
            self.TexProgram["polygons[2]"] = (0.2, 1) 
            self.TexProgram["polygons[3]"] = (0.1, 1) 
            self.TexProgram["polygons[4]"] = (0,0) 
            self.TexProgram["polygons[5]"] = (0.1, 0.1) 
            self.TexProgram["polygons[6]"] = (0.2, 0.1) 
            self.TexProgram["polygons[7]"] = (0.2, 0.2) 
            self.TexProgram["polygons[8]"] = (0.1, 0.2) 
            self.TexProgram["polygons[9]"] = (0,0) 
            self.TexProgram["polylength"] = 8 
            self.TexProgram["light"] = 1.05 
            self.TexProgram["thickness"] = 0.02 
        self.inited = True 
 
    self.clear() 
    else: 
        with self.TexProgram: 
            self.fbo.bind_texture() 
            self.fbo.clear() 
            t = (0.0, 0.0, 1.0, 0.0, 1.0, 1.0, 0.0, 1.0) 
            x1 = 0 
            x2 = w 
            y1 = 0 
            y2 = h 
 
            glColor4d(1,1,1,1) 
            glEnable (GL_TEXTURE_2D) 
            glBegin(GL_QUADS) 
            glTexCoord2d(t[0], t[1]); glVertex2d(x1, y1) 
            glTexCoord2d(t[2], t[3]); glVertex2d(x2, y1) 
            glTexCoord2d(t[4], t[5]); glVertex2d(x2, y2) 
            glTexCoord2d(t[6], t[7]); glVertex2d(x1, y2) 
            glEnd() 

贝塞尔曲线通常有 3 个以上的点,一切都在二维空间中。 在我看来,有两种方法可以解决这个问题。

1:

Generate points on curve. 
get closest curve point. 
generate points around curve point. 
get closest point of those. 
rince and repeat last steps until satisfactory precision is reached. 

2: 想出一个公式/算法来分析距离。我尝试了这个,但没有找到任何数量的贝塞尔曲线点的解决方案,只有一个常数。

我更喜欢第二种方法的解决方案。 我可以自己想出第一种方法的解决方案,但如果有人知道任何资源来查找它,这样我就不会浪费时间,无论如何都会有所帮助。

请您参考如下方法:

我不建议对任意贝塞尔曲线解析计算距离,因为贝塞尔曲线的每个点都取决于每个结,到曲线的距离也是如此。 (即使没有这个,您当前的实现似乎也非常耗时并且无法扩展。)

我可以推荐两种计算任意形状的距离图(或 Voronoi 图)的方法:

  1. Fast computation of generalized Voronoi diagrams... :对于每条线(线段),您绘制一个“帐篷”(两个像屋顶一样排列的四边形)并从上方渲染场景 - 深度缓冲区是您的距离图。可以在着色器中或通过调整帐篷高度来调整光束厚度和强度。对于每个点,您都渲染一个圆锥体。 编辑:额外的几何体也可以在几何着色器中生成。

  2. Jump Flooding :为此,您需要设置乒乓渲染方案并渲染到多个缓冲区。我跳过算法,因为它解释起来有点冗长,请看论文。

这两种算法都适用于大量形状。