根据我在 Math Stackexchange 上的问题:em>
我正在为我的 3D 图形课程设计一个项目.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,细分为两个正方形.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,有人能指导我如何使用三角学来计算连续点的 X 和 Z 值吗?例如,如果用户单击并创建点:
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) 度重新绘制该点.
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.
我认为三角函数会在这里帮助我是否正确?如果是这样,有人可以告诉我如何计算 3D 空间中平移点的位置吗?我已经有一段时间没有使用 Trig 了,我相信我们从未离开过 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 的答案的理解,我不认为它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 << ")
";
// render the new point
glVertex3d(newX, newY, newZ);
}
这不会产生屏幕输出,而是控制台输出:
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)
我不确定这里到底发生了什么,但我很难弄清楚,所以请不要认为我是在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 2:这是我的透视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 << ")
";
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(为了后代!):
在与大学老师讨论了一些线性代数之后,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();
}
编辑 2:好的,我明白你遇到的问题——这是我忘记的限制(所以我之前发布的代码大错特错,根本行不通).问题是你不允许在 glBegin
/glEnd
对之间调用 glRotate
—— 如果你这样做,它会设置一个错误标志,不再绘制.
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.
这确实意味着您几乎必须自己处理轮换.幸运的是,这比您尝试实现的要简单一些:
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.您显然可以将其移动到您想要的位置,但显然不会显示剪切视锥体之外的任何内容.
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.
另请注意,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();
}
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