根据我在 Math Stackexchange 上的 question:

我正在为我的 3D 图形类(class)做一个项目。该项目是用 C++ 和 OpenGL/Glut 构建的。基本上，我创建了一个水平矩形窗口， segmentation 为两个正方形。在左侧，我有一个二维坐标平面，它允许用户指向并单击并定义轮廓“曲线”。然后我需要将这条曲线围绕 Y 轴绕 n 次。

那么，有人能指导我如何使用三角函数计算连续点的 X 和 Z 值吗？例如，如果用户点击并创建点:

(1, 1, 0)

并且它们的扫描分辨率 (n) 设置为 10，然后我需要围绕 Y 轴每 36 (360/10) 度重绘该点。

我假设三角函数会在这方面对我有帮助，我的假设是否正确？如果是这样，有人可以告诉我如何计算 3D 空间中翻译点的位置吗？自从我使用 Trig 以来已经有一段时间了，我不相信我们曾经离开过 2D 空间。

编辑:我尝试使用:

x'=xcos(theta)-zsin(theta)

y'=y

z'=xsin(theta)+zcos(theta)

，根据我对 AMPerrine 的 answer 的理解，我认为它没有像我希望的那样工作:

// this is in a loop

// setup the new angle
double angle = i>0 ? (360/sweepResolutionMod)*i : 0;

angle = angle * (M_PI/180);

// for each point...
for( int i=0; i<clickedPoints.size(); i++ )
{
    // initial point, normalized
    GLfloat tempX = (clickedPoints[i].x-250)/250;
    GLfloat tempY = (clickedPoints[i].y-250)/250;
    GLfloat tempZ = 0.0;

    // log the initial point
    cout << "(" << tempX << ", " << tempY << ", 0.0) by " << angle << " radians = ";

    // generate the new point
    GLfloat newX = (tempX * cos(angle)) - (tempZ * sin(angle));
    GLfloat newY = tempY;
    GLfloat newZ = (tempX * sin(angle)) - (tempZ * cos(angle));

    // log the new point
    cout << "(" << newX << ", " << newY << ", " << newZ << ")\n";

    // render the new point
    glVertex3d(newX, newY, newZ);
}

这不会产生屏幕输出，但会产生控制台输出:

(0.048, -0.296, 0.0) by 0 radians = (0.048, -0.296, 0)
(0.376, -0.508, 0.0) by 0 radians = (0.376, -0.508, 0)
(0.72, -0.204, 0.0) by 0 radians = (0.72, -0.204, 0)
(0.652, 0.176, 0.0) by 0 radians = (0.652, 0.176, 0)
(0.368, 0.504, 0.0) by 0 radians = (0.368, 0.504, 0)

(0.048, -0.296, 0.0) by 0.628319 radians = (0.0388328, -0.296, 0.0282137)
(0.376, -0.508, 0.0) by 0.628319 radians = (0.30419, -0.508, 0.221007)
(0.72, -0.204, 0.0) by 0.628319 radians = (0.582492, -0.204, 0.423205)
(0.652, 0.176, 0.0) by 0.628319 radians = (0.527479, 0.176, 0.383236)
(0.368, 0.504, 0.0) by 0.628319 radians = (0.297718, 0.504, 0.216305)

(0.048, -0.296, 0.0) by 1.25664 radians = (0.0148328, -0.296, 0.0456507)
(0.376, -0.508, 0.0) by 1.25664 radians = (0.11619, -0.508, 0.357597)
(0.72, -0.204, 0.0) by 1.25664 radians = (0.222492, -0.204, 0.684761)
(0.652, 0.176, 0.0) by 1.25664 radians = (0.201479, 0.176, 0.620089)
(0.368, 0.504, 0.0) by 1.25664 radians = (0.113718, 0.504, 0.349989)

...

(0.048, -0.296, 0.0) by 6.28319 radians = (0.048, -0.296, -1.17566e-17)
(0.376, -0.508, 0.0) by 6.28319 radians = (0.376, -0.508, -9.20934e-17)
(0.72, -0.204, 0.0) by 6.28319 radians = (0.72, -0.204, -1.76349e-16)
(0.652, 0.176, 0.0) by 6.28319 radians = (0.652, 0.176, -1.59694e-16)
(0.368, 0.504, 0.0) by 6.28319 radians = (0.368, 0.504, -9.0134e-17)

我不确定这里到底发生了什么，但我正在努力弄清楚它，所以请不要以为我在试图获得双重声誉或其他任何东西，我只是真的卡住了。

编辑 2:这是我的透视 subview 的整个显示例程:

void displayPersp(void)
{
    glClear(GL_COLOR_BUFFER_BIT);

    glMatrixMode (GL_MODELVIEW);
    glLoadIdentity ();  

    gluLookAt (-2.0, 1.0, 1.0, 0.0, 0.0, 0.0, 0.0, -1.0, 0.0);

    // draw the axis
    glBegin(GL_LINES);
    // x
    glVertex3f(500.0, 0.0, 0.0);
    glVertex3f(-500.0, 0.0, 0.0);
    // y
    glVertex3f(0.0, -500.0, 0.0);
    glVertex3f(0.0, 500.0, 0.0);
    // z
    glVertex3f(0.0, 0.0, -500.0);
    glVertex3f(0.0, 0.0, 500.0);

    glEnd(); 

    cout << endl;

    // loop as many number of times as we are going to draw the points around the Y-Axis
    for( int i=0; i<=sweepResolutionMod; i++ )
    {
        cout << endl;

        // setup the new angle
        double angle = i>0 ? (360/sweepResolutionMod)*i : 0;

        angle = angle * (M_PI/180);

        // for each point...
        for( int i=0; i<clickedPoints.size(); i++ )
        {
            GLfloat tempX = (clickedPoints[i].x-250)/250;
            GLfloat tempY = (clickedPoints[i].y-250)/250;
            GLfloat tempZ = 0.0;

            cout << "(" << tempX << ", " << tempY << ", 0.0) by " << angle << " degrees = ";

            GLfloat newX = (tempX * cos(angle)) - (tempZ * sin(angle));
            GLfloat newY = tempY;
            GLfloat newZ = (tempX * sin(angle)) - (tempZ * cos(angle));

            cout << "(" << newX << ", " << newY << ", " << newZ << ")\n";

            glVertex3d(newX, newY, newZ);
        }

        // the following was my old solution, using OpenGL's rotate(), but that
        // didn't allow me to get back the new point's coordinates.

        /*
        glRotatef(angle, 0.0, 1.0, 0.0);

        // draw a line?
        if( clickedPoints.size() > 1 )
        {
            glBegin(GL_LINE_STRIP);

            for(int i=0; i<clickedPoints.size(); i++ )
            {     
                glVertex3f((clickedPoints[i].x-250)/250, (clickedPoints[i].y-250)/250, 0.0);     
            }

            glEnd();
        }

        // everyone gets points
        glBegin(GL_POINTS);

        for(int i=0; i<clickedPoints.size(); i++ )
        { 
            glVertex3f((clickedPoints[i].x-250)/250, (clickedPoints[i].y-250)/250, 0.0);     
        }

        glEnd();
        */
    }


    glutSwapBuffers();
}

编辑 3:这是一个糟糕的插图，说明了我需要做什么。我知道视角似乎不对，但我试图获得的是右侧 subview 中的绿色“水平”(这是使用上面注释掉的 glRotatef() 代码):

最终编辑(为了子孙后代!):

在与一位大学老师讨论了一些线性代数之后，这是我最终得到的结果:

void displayPersp(void)
{
   glClear(GL_COLOR_BUFFER_BIT);

   gluLookAt (-2.0, 1.0, 1.0, 0.0, 0.0, 0.0, 0.0, -1.0, 0.0);

   glMatrixMode (GL_MODELVIEW);
   glLoadIdentity ();  

   // draw the axis
   glBegin(GL_LINES);
     // x
     glVertex3f(500.0, 0.0, 0.0);
     glVertex3f(-500.0, 0.0, 0.0);
     // y
     glVertex3f(0.0, -500.0, 0.0);
     glVertex3f(0.0, 500.0, 0.0);
     // z
     glVertex3f(0.0, 0.0, -500.0);
     glVertex3f(0.0, 0.0, 500.0);

   glEnd(); 

   cout << endl;

   double previousTheta = 0.0;

   for( int i=0; i<=sweepResolutionMod; i++ )
   {
     double theta = i>0 ? (360/sweepResolutionMod)*i : 0;

     theta = theta * (M_PI/180);

     if( clickedPoints.size() > 1 )
     {
       // the 'vertical' piece
       glBegin(GL_LINE_STRIP);

       for(int i=0; i<clickedPoints.size(); i++ )
       {     
         // normalize
         GLfloat tempX = (clickedPoints[i].x-250)/250;
         GLfloat tempY = (clickedPoints[i].y-250)/250;
         GLfloat tempZ = 0.0;

         // new points
         GLfloat newX = ( tempX * cos(theta) ) + ( tempZ * sin(theta) );
         GLfloat newY = tempY;
         GLfloat newZ = ( tempZ * cos(theta) ) - ( tempX * sin(theta) );

         glVertex3f(newX, newY, newZ);     
       }

       glEnd();

       // the 'horizontal' piece
       if( previousTheta != theta )
       {
         glBegin(GL_LINES);

         for(int i=0; i<clickedPoints.size(); i++ )
         {     
           // normalize
           GLfloat tempX = (clickedPoints[i].x-250)/250;
           GLfloat tempY = (clickedPoints[i].y-250)/250;
           GLfloat tempZ = 0.0;

           // new points
           GLfloat newX = ( tempX * cos(theta) ) + ( tempZ * sin(theta) );
           GLfloat newY = tempY;
           GLfloat newZ = ( tempZ * cos(theta) ) - ( tempX * sin(theta) );

           // previous points
           GLfloat previousX = ( tempX * cos(previousTheta) ) + ( tempZ * sin(previousTheta) );
           GLfloat previousY = tempY;
           GLfloat previousZ = ( tempZ * cos(previousTheta) ) - ( tempX * sin(previousTheta) );

           // horizontal component           
           glVertex3f(newX, newY, newZ);     
           glVertex3f(previousX, previousY, previousZ);     
         }

         glEnd();
       }
     }

     previousTheta = theta;
   }

   glutSwapBuffers();
}

最佳答案

编辑 2:好的，我明白你遇到的问题了——这是一个我忘记的限制(所以我之前发布的代码是完全错误的，根本无法工作)。问题是您不允许在 glBegin/glEnd 对之间调用 glRotate - 如果您这样做，它将设置一个错误标志，不再绘制。

这确实意味着您几乎必须自己处理旋转。幸运的是，这比您尝试实现的要简单一些:

static const double pi = 3.1416;

for (int point=0; point<NUM_POINTS; point++) {
    glBegin(GL_LINE_STRIP);
    for (double theta = 0.0; theta < 2.0 * pi; theta += pi/6.0) {
        double x = cos(theta);
        double z = sin(theta);
        glVertex3d(points[point][0]*x, points[point][1], -1.0-points[point][0]*z);
    }
    glEnd();
}   

按原样，此代码沿 Z 轴使用 -1.0 作为旋转中心。您显然可以将其移动到您希望的位置，但剪裁平截头体之外的任何内容显然都不会显示。

另请注意，要获得线框，您必须分别绘制“垂直”和“水平”线，因此代码如下所示:

for (int point=0; point<NUM_POINTS; point++) {
    glBegin(GL_LINE_STRIP);
    for (double theta = 0.0; theta < 2.0 * pi; theta += pi/6.0) {
        double x = cos(theta);
        double z = sin(theta);
        glVertex3d(points[point][0]*x, points[point][1], -1.0 - points[point][0]*z);
    }
    glEnd();
}

for (double theta = 0.0; theta < 2.0 * pi; theta += pi/6.0) {
    glBegin(GL_LINE_STRIP);
    for (int point=0; point<NUM_POINTS; point++) {
        double x = cos(theta);
        double z = sin(theta);
        glVertex3d(points[point][0]*x, points[point][1], -1.0 - points[point][0]*z);
    }
    glEnd();
}

关于c++ - OpenGL - 围绕 Y 轴旋转 'Curve'，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/7904281/