OpenGL-绕Y轴旋转“曲线" [英] OpenGL - Rotate a 'Curve' About the Y-Axis
问题描述
根据我对Math Stackexchange的问题:
我正在为我的3D图形课程设计一个项目.该项目使用C ++和OpenGL/Glut构建.基本上，我创建了一个水平矩形窗口，分为两个正方形.在左侧，我有一个二维坐标平面，允许用户指向和单击并定义轮廓曲线".然后，我需要将此曲线绕Y轴缠绕n次.
I am working on a project for my 3D Graphics class. The project is built with C++ and OpenGL / Glut. Basically, I create a horizontal rectangle window, subdivided into two squares. On the left, I have a two dimensional coordinate plane, which allows the users to point and click and define a profile 'curve'. I then need to wrap this curve around the Y-axis n number of times.
那么，有谁能指导我如何使用三角函数来计算连续点的X和Z值?例如，如果用户单击并创建该点:
So, would anyone be able to guide me as to how I would use Trigonometry to calculate the X and Z values of the successive points? If for example, a user clicks and creates the point:
(1、1、0)
并将其扫描分辨率(n)设置为10，然后我需要绕Y轴每36(360/10)度重画一次该点.
And their sweep resolution (n) is set to, say, 10, then I need to redraw that point every 36 (360/10) degrees around the Y-axis.
我是否可以假设三角学在这里对我有帮助?如果是这样，有人可以请我启发一下如何计算平移点在3D空间中的位置吗?自从上手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.
Am I correct in assuming that Trigonometry will help me here? If so, can someone please enlighten me a bit as to how to calculate the location of a translated point in 3D space? It's been a while since I took Trig, and I don't believe we ever left 2D space.
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，而且我认为它不像我希望的那样工作:
, as per my understanding of AMPerrine's answer, and I don't think it worked as I'd hoped:
// 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);
}
这不会产生任何屏幕输出，但是会产生以下控制台输出:
This produces no screen output, but console output of:
(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)
我不确定这到底是怎么回事，但是我正经历着一段糟糕的时机试图弄清楚，所以请不要以为我试图获得双重声誉或其他任何东西，我只是真的卡住了.
I'm not sure what exactly is going on here, but I'm having a terrible time trying to figure it out, so please don't think I'm trying to get double reputation or anything, I'm just really stuck.
编辑:这是我的透视图子视图的整个显示例程:
EDIT 2: Here is my whole display routine for my perspective subview:
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();
}
这是一个可怕的插图，阐明了我需要做的事情.我知道透视图似乎不对，但是我试图获取的是右侧子视图中的绿色水平线"(这是使用上面注释掉的glRotatef()代码):
EDIT 3: Here is a terrible illustration that illustrates what I need to do. I know the perspective seems off, but what I'm attempting to acquire is the green 'horizontals' in the right subview (this is using the commented out glRotatef() code above):
最终编辑(为子孙后代！):
在与大学老师讨论了线性代数之后，我终于开始工作了:
Here is what I finally got working, after discussing some linear algebra with a teacher at college:
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();
}
推荐答案
好的，我看到了您遇到的问题-这是我忘记的限制(因此，我之前发布的代码完全错误，根本无法使用).问题是您不允许在glBegin/glEnd对之间调用glRotate －如果这样做，它将设置一个错误标志，并且将不再进行绘制.
Edit 2: Okay, I see the problem you're running into -- it's a limitation I'd forgotten about (so the code I'd posted previously was dead wrong and wouldn't work at all). The problem is that you're not allowed to call glRotate between a glBegin/glEnd pair -- if you do, it'll set an error flag, and no more drawing will be done.
那确实意味着您几乎必须自己处理旋转.幸运的是，这要比您尝试的要简单一些:
That does mean you pretty much have to handle the rotation yourself. Fortunately, that's a bit simpler than you've tried to make it:
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.显然，您可以将其移动到所需的位置，尽管截头锥体之外的任何内容显然都不会显示.
As-is, this code uses -1.0 along the Z axis as the center of rotation. You can obviously move that where you wish, though anything outside your clipping frustum obviously won't display.
还要注意，要获得线框，您必须分别绘制垂直"线和水平"线，因此代码如下所示:
Also note that to get a wireframe, you'll have to draw both your "vertical", and your "horizontal" lines separately, so the code will look something like this:
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();
}
这篇关于OpenGL-绕Y轴旋转“曲线"的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
