c++ - 如何在OpenGL中绘制连接两点的圆柱体

Question

我有两个点，每个点都有自己的 X 和 Y 值，并且它们具有相同的 Z 值。

我想要一个函数在这两点之间绘制圆柱体。

Answer 1

要构建具有两个给定点的圆柱体，您需要 vector 分析。您正在构建两个垂直 vector ，将它们添加到每个点并按正弦/余弦乘以半径进行缩放。 它接受所有点(旧代码有一个错误，因为它错过了长度的 sqrt() )。 现在它可以正常工作并使用 gl 例程绘制圆柱体；我在JOGL中测试了它。 为了更快地绘制，将firstPerpVector/secondPerpVector/points变量移动到私有(private)最终数组字段并在开始时初始化它们。

Java代码:

  public float[] getFirstPerpVector(float x, float y, float z) {
  float[] result = {0.0f,0.0f,0.0f};
  // That's easy.
  if (x == 0.0f || y == 0.0f || z == 0.0f) {
    if (x == 0.0f)
      result[0] = 1.0f;
    else if (y == 0.0f)
      result[1] = 1.0f;
    else
      result[2] = 1.0f;
  }
  else {
    // If xyz is all set, we set the z coordinate as first and second argument .
    // As the scalar product must be zero, we add the negated sum of x and y as third argument
    result[0] = z;      //scalp = z*x
    result[1] = z;      //scalp = z*(x+y)
    result[2] = -(x+y); //scalp = z*(x+y)-z*(x+y) = 0
    // Normalize vector
    float length = 0.0f;
    for (float f : result)
      length += f*f;
    length = (float) Math.sqrt(length);
    for (int i=0; i<3; i++)
      result[i] /= length;
  }
  return result;
}

public void drawCylinder(GL gl, float x1, float y1, float z1, float x2, float y2, float z2) {
  final int X = 0,
            Y = 1,
            Z = 2;
  // Get components of difference vector
  float x = x1-x2,
        y = y1-y2,
        z = z1-z2;
  float[] firstPerp = getFirstPerpVector(x,y,z);
  // Get the second perp vector by cross product
  float[] secondPerp = new float[3];
  secondPerp[X] = y*firstPerp[Z]-z*firstPerp[Y];
  secondPerp[Y] = z*firstPerp[X]-x*firstPerp[Z];
  secondPerp[Z] = x*firstPerp[Y]-y*firstPerp[X];
  // Normalize vector
  float length = 0.0f;
  for (float f : secondPerp)
    length += f*f;
  length = (float) Math.sqrt(length);
  for (int i=0; i<3; i++)
    secondPerp[i] /= length;

  // Having now our vectors, here we go:
  // First points; you can have a cone if you change the radius R1

  final int ANZ = 32;  // number of vertices
  final float FULL = (float) (2.0f*Math.PI),
              R1   = 4.0f; // radius
  float[][] points = new float[ANZ+1][3];
  for (int i=0; i<ANZ; i++) {
    float angle = FULL*(i/(float) ANZ);

    points[i][X] = (float) (R1*(Math.cos(angle)*firstPerp[X]+Math.sin(angle)*secondPerp[X]));
    points[i][Y] = (float) (R1*(Math.cos(angle)*firstPerp[Y]+Math.sin(angle)*secondPerp[Y]));
    points[i][Z] = (float) (R1*(Math.cos(angle)*firstPerp[Z]+Math.sin(angle)*secondPerp[Z]));
  }
  // Set last to first
  System.arraycopy(points[0],0,points[ANZ],0,3);

  gl.glColor3f(1.0f,0.0f,0.0f);
  gl.glBegin(GL.GL_TRIANGLE_FAN);
  gl.glVertex3f(x1,y1,z1);
  for (int i=0; i<=ANZ; i++) {
    gl.glVertex3f(x1+points[i][X],
                  y1+points[i][Y],
                  z1+points[i][Z]);
  }
  gl.glEnd();

  gl.glBegin(GL.GL_TRIANGLE_FAN);
  gl.glVertex3f(x2,y2,z2);
  for (int i=0; i<=ANZ; i++) {
    gl.glVertex3f(x2+points[i][X],
                  y2+points[i][Y],
                  z2+points[i][Z]);
  }
  gl.glEnd();

  gl.glBegin(GL.GL_QUAD_STRIP);
  for (int i=0; i<=ANZ; i++) {
    gl.glVertex3f(x1+points[i][X],
                  y1+points[i][Y],
                  z1+points[i][Z]);
    gl.glVertex3f(x2+points[i][X],
                  y2+points[i][Y],
                  z2+points[i][Z]);
  }
  gl.glEnd();      
}