使用OpenGL如何实现物体绕体外任意轴旋转?
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我用glRotatef(m_fAngleX,1.0f,0.0f,0.0f);然后让m_fAngleX循环加10,可以让立方体绕中心轴旋转;如果我想让立方体绕体外的任意轴旋转,我该怎么办呢?
可以这样假设你要绕轴AB旋转
1、计算把轴AB一个点平衡原点(假如我这里是B点移动到原点)所需移动量为向量xyz,把立方体也移动向量xyz
2、把立方休按A点位置旋转(因为B点已经在原点了),即把A点坐标填入glRotate后面三个参数就是
3、把物体移动回向量xyz即可 参考技术A glQuaternion m_qHeading; //四元数
float m_HeadingDegrees; //转动角度
Vector3 eye, center; //摄像机参数
Vector3 m_Position; //轴位置
Matrix matrix; //矩阵
m_qHeading.CreateFromAxisAngle(0.0f, 1.0f, 0.0f, m_HeadingDegrees); //将转动角变成四元数
m_qHeading.CreateMatrix(matrix); //将四元数矩阵化
eye *= Matrix;
center *= Matrix;
gluLookAt(eye.x + m_Position.x,eye.y + m_Position.y,eye.z + m_Position.z, center.x + m_Position.x,center.y + m_Position.y,center.z + m_Position.z,0,1,0);
其中四元数可以直接百度“glQuaternion" 可以找到矩阵化的算法,向量和矩阵相乘你也可以找到一些资料的。本回答被提问者采纳
在 OpenGL 中绘制一个绕 Y 轴旋转的 Cog
【中文标题】在 OpenGL 中绘制一个绕 Y 轴旋转的 Cog【英文标题】:Drawing a Cog which rotates around the Y axis in OpenGL 【发布时间】:2019-09-13 06:23:38 【问题描述】:如标题中所述:我想创建一个 3D 齿轮,它有 10 个围绕其中心旋转的齿(就像齿轮一样)。齿轮有方形齿,为了简单起见,齿之间有平坦的侧面 - 这个齿轮上没有曲线。
齿轮一侧应该是什么样子的可视化。请注意,角度并非 100% 完美。
根据上图,每个齿轮齿都必须是 8 边多边形,而齿之间的每条边都必须是 4 边多边形。然而,现在,牙齿绘制做了以下事情:
-
尽管未应用初始旋转,但应面向摄像机的侧齿旋转到地板中。
虽然缩放和顶点创建是使用统一的数字完成的,但牙齿的宽度是高度的两倍(它在 x、y、z 方向上按
s
缩放,并使用0
或0.5
作为顶点坐标)。
完全可重现的示例:
import java.awt.Dimension;
import javax.swing.JFrame;
import com.jogamp.opengl.GL;
import com.jogamp.opengl.GL2;
import com.jogamp.opengl.GLAutoDrawable;
import com.jogamp.opengl.GLEventListener;
import com.jogamp.opengl.awt.GLJPanel;
import com.jogamp.opengl.glu.GLU;
import com.jogamp.opengl.util.FPSAnimator;
public class SpinCog3D implements GLEventListener
JFrame jf;
GLJPanel gljpanel;
Dimension dim = new Dimension(800, 600);
FPSAnimator animator;
float rotation;
float speed;
// set up the OpenGL Panel within a JFrame
public SpinCog3D()
jf = new JFrame();
gljpanel = new GLJPanel();
gljpanel.addGLEventListener(this);
gljpanel.requestFocusInWindow();
jf.getContentPane().add(gljpanel);
jf.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
jf.setVisible(true);
jf.setPreferredSize(dim);
jf.pack();
animator = new FPSAnimator(gljpanel, 20);
rotation = 0.0f;
speed = 0.1f;
animator.start();
public static void main(String[] args)
new SpinCog3D();
public void init(GLAutoDrawable dr)
GL2 gl2 = dr.getGL().getGL2();
GLU glu = new GLU();
gl2.glClearColor(1.0f, 1.0f, 1.0f, 0.0f);
gl2.glEnable(GL2.GL_DEPTH_TEST);
gl2.glMatrixMode(GL2.GL_PROJECTION);
gl2.glLoadIdentity();
glu.gluPerspective(60.0, 1.0, 100.0, 800.0);
public void display(GLAutoDrawable dr)
GL2 gl2 = dr.getGL().getGL2();
GLU glu = new GLU();
gl2.glMatrixMode(GL2.GL_MODELVIEW);
gl2.glLoadIdentity();
glu.gluLookAt(0.0, 200.0, 500.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0);
gl2.glClear(GL.GL_COLOR_BUFFER_BIT | GL.GL_DEPTH_BUFFER_BIT);
// Draw 1 Cog Tooth + 1 Side
drawTooth(gl2, 100.0, 1.0f, 0.0f, 0.0f, 0.0, 0.0, 1.0, 0.0, 0);
drawSide(gl2, 100.0, 0.0f, 1.0f, 0.0f, 0.0, 0.0, 1.0, 0.0, 10);
// Draw Floor
sideRotatedColorScaledFloor(gl2, 300.0, 0.0f, 0.0f, 0.0f, 90.0, 1.0, 0.0, 0.0, 0.0);
gl2.glFlush();
rotation += speed;
if (rotation > 360.9f)
rotation = 0.0f;
// draw a single side with a set color and orientation
private void drawTooth(GL2 gl2, double s, float r, float g, float b, double a, double ax, double ay,
double az, double zoffset)
gl2.glPushMatrix();
gl2.glRotated(a, ax, ay, az);
gl2.glColor3f(r, g, b);
gl2.glTranslated(0.0, 0.0, zoffset);
gl2.glScaled(s, s, s);
tooth(gl2);
gl2.glPopMatrix();
private void drawSide(GL2 gl2, double s, float r, float g, float b, double a, double ax, double ay,
double az, double zoffset)
gl2.glPushMatrix();
gl2.glRotated(a, ax, ay, az);
gl2.glColor3f(r, g, b);
gl2.glTranslated(0.0, 0.0, zoffset);
gl2.glScaled(s, s, s);
side(gl2);
gl2.glPopMatrix();
private void sideRotatedColorScaledFloor(GL2 gl2, double s, float r, float g, float b, double a, double ax, double ay,
double az, double zoffset)
gl2.glPushMatrix();
gl2.glRotated(a, ax, ay, az);
gl2.glColor3f(r, g, b);
gl2.glTranslated(0.0, 0.0, zoffset);
gl2.glScaled(s, s, s);
side(gl2);
gl2.glPopMatrix();
private void tooth(GL2 gl2)
gl2.glBegin(GL2.GL_POLYGON);
gl2.glVertex3d(-0.5, -0.5, 0.0);
gl2.glVertex3d(-0.5, 0.5, 0.0);
gl2.glVertex3d(0.5, 0.5, 0.0);
gl2.glVertex3d(0.5, -0.5, 0.0);
gl2.glVertex3d(-0.5, -0.5, 0.5);
gl2.glVertex3d(-0.5, 0.5, 0.5);
gl2.glVertex3d(0.5, 0.5, 0.5);
gl2.glVertex3d(0.5, -0.5, 0.5);
gl2.glEnd();
private void side(GL2 gl2)
gl2.glBegin(GL2.GL_POLYGON);
gl2.glVertex3d(-0.5, -0.5, 0.0);
gl2.glVertex3d(-0.5, 0.5, 0.0);
gl2.glVertex3d(0.5, 0.5, 0.0);
gl2.glVertex3d(0.5, -0.5, 0.0);
gl2.glEnd();
是的,我知道 glBegin()
和 glEnd()
已被永远弃用,但这不是重点。如何让我的牙齿正确绘制并使其与两侧正确对齐?
【问题讨论】:
您能否添加描述,或者更好的是,添加“奇怪事物”的图像?我认为这可能会使您的问题变得更好,这样我们就不必自己运行您的代码来查看问题可能是什么。 我在不使用矩阵的情况下添加了答案...您的错误很可能在哪里... 【参考方案1】:牙齿画会做各种奇怪的事情......
不是一个好的问题描述。我敢打赌,您的网格有很多间隙和重叠,因为您使用的是硬编码形状,而没有对 Cog 的实际尺寸进行任何校正。就像牙齿必须有多大才能适合n
才能完全覆盖整个圆圈......
改用参数圆方程怎么样?
这将摆脱矩阵混乱,我认为会更简单。
所以我会简单地将 cog 分成三角形切片,每个切片都有几个边缘点。因此,将它们与QUAD
s 连接起来,然后将其放入for
循环中进行所有切片。
我不会在 JAVA 中编写代码,但这里使用旧的 GL api 并且没有花哨的 C++ 东西的小 C++ 示例,因此您应该能够轻松地移植它:
void glCog(float r0,float r1,float r2,float w,int n) // shaft/inner/outer radiuses, width, tooths
int i;
float a,da,x,y,c,s;
float p[6][3],q[6][3]; // slice points
// set z for slice points
a=-0.5*w; for (i=0;i<3;i++) p[i][2]=a; q[i][2]=a;
a=+0.5*w; for (i=3;i<6;i++) p[i][2]=a; q[i][2]=a;
// init first slice
q[0][0]= r0; q[5][0]= r0;
q[0][1]=0.0; q[5][1]=0.0;
q[1][0]= r1; q[4][0]= r1;
q[1][1]=0.0; q[4][1]=0.0;
q[2][0]= r2; q[3][0]= r2;
q[2][1]=0.0; q[3][1]=0.0;
// divide circle to 2*n slices
da=2.0*M_PI/float(4*n);
glBegin(GL_QUADS);
for (a=0.0,i=0;i<=n;i++)
// points on circles at angle a
c=cos(a); s=sin(a); a+=da;
x=r0*c; y=r0*s; p[0][0]=x; p[5][0]=x;
p[0][1]=y; p[5][1]=y;
x=r1*c; y=r1*s; p[1][0]=x; p[4][0]=x;
p[1][1]=y; p[4][1]=y;
x=r2*c; y=r2*s; p[2][0]=x; p[3][0]=x;
p[2][1]=y; p[3][1]=y;
// render tooth
c=cos(a); s=sin(a); a+=da;
glNormal3f(0.0,0.0,-1.0); // -Z base
glVertex3fv(p[0]);
glVertex3fv(p[2]);
glVertex3fv(q[2]);
glVertex3fv(q[0]);
glNormal3f(0.0,0.0,+1.0); // +Z base
glVertex3fv(p[3]);
glVertex3fv(p[5]);
glVertex3fv(q[5]);
glVertex3fv(q[3]);
glNormal3f(-c,-s,0.0); // shaft circumference side
glVertex3fv(p[5]);
glVertex3fv(p[0]);
glVertex3fv(q[0]);
glVertex3fv(q[5]);
glNormal3f(c,s,0.0); // outter circumference side
glVertex3fv(p[2]);
glVertex3fv(p[3]);
glVertex3fv(q[3]);
glVertex3fv(q[2]);
glNormal3f(-s,c,0.0);
glVertex3fv(p[4]);
glVertex3fv(p[3]);
glVertex3fv(p[2]);
glVertex3fv(p[1]);
glNormal3f(s,-c,0.0);
glVertex3fv(q[1]);
glVertex3fv(q[2]);
glVertex3fv(q[3]);
glVertex3fv(q[4]);
// points on circles at angle a
c=cos(a); s=sin(a); a+=da;
x=r0*c; y=r0*s; q[0][0]=x; q[5][0]=x;
q[0][1]=y; q[5][1]=y;
x=r1*c; y=r1*s; q[1][0]=x; q[4][0]=x;
q[1][1]=y; q[4][1]=y;
x=r2*c; y=r2*s; q[2][0]=x; q[3][0]=x;
q[2][1]=y; q[3][1]=y;
// render gap
c=cos(a); s=sin(a); a+=da;
glNormal3f(0.0,0.0,-1.0); // -Z base
glVertex3fv(q[0]);
glVertex3fv(q[1]);
glVertex3fv(p[1]);
glVertex3fv(p[0]);
glNormal3f(0.0,0.0,+1.0); // +Z base
glVertex3fv(q[4]);
glVertex3fv(q[5]);
glVertex3fv(p[5]);
glVertex3fv(p[4]);
glNormal3f(-c,-s,0.0); // shaft circumference side
glVertex3fv(q[5]);
glVertex3fv(q[0]);
glVertex3fv(p[0]);
glVertex3fv(p[5]);
glNormal3f(c,s,0.0); // outter circumference side
glVertex3fv(q[1]);
glVertex3fv(q[4]);
glVertex3fv(p[4]);
glVertex3fv(p[1]);
glEnd();
这里是glCog(0.1,0.5,0.6,0.1,10);
的预览:
这里是glCog(0.2,0.5,0.52,0.2,50);
的预览:
但请注意,牙齿并非完全是矩形的。齿数越少,误差越大。如果您想获得完全矩形的牙齿形状,则需要平移最外面的点而不是旋转(或校正它们的计算角度)
使用翻译来解决这个问题:
void glCog(float r0,float r1,float r2,float w,int n) // shaft/inner/outer radiuses, width, tooths
int i,j;
float a,da,dr,x,y,c,s;
float p[6][3],q[6][3]; // slice points
// divide circle to 2*n slices
da=2.0*M_PI/float(4*n);
dr=r2-r1;
// set z for slice points
a=-0.5*w; for (i=0;i<3;i++) p[i][2]=a; q[i][2]=a;
a=+0.5*w; for (i=3;i<6;i++) p[i][2]=a; q[i][2]=a;
// init first slice
q[0][0]= r0; q[5][0]= r0;
q[0][1]=0.0; q[5][1]=0.0;
q[1][0]= r1; q[4][0]= r1;
q[1][1]=0.0; q[4][1]=0.0;
x=r1+dr*cos(-da); y=dr*sin(-da);
q[2][0]= x; q[3][0]= x;
q[2][1]= y; q[3][1]= y;
glBegin(GL_QUADS);
for (a=0.0,i=0;i<=n;i++)
// points on circles at angle a
c=cos(a); s=sin(a);
x=r0*c; y=r0*s; p[0][0]=x; p[5][0]=x;
p[0][1]=y; p[5][1]=y;
x=r1*c; y=r1*s; p[1][0]=x; p[4][0]=x;
p[1][1]=y; p[4][1]=y;
c=cos(a-da); s=sin(a-da); a+=da;
x+=dr*c;y+=dr*s;p[2][0]=x; p[3][0]=x;
p[2][1]=y; p[3][1]=y;
c=cos(a); s=sin(a); a+=da;
// render tooth
glNormal3f(0.0,0.0,-1.0); // -Z base
glVertex3fv(p[0]);
glVertex3fv(p[1]);
glVertex3fv(q[1]);
glVertex3fv(q[0]);
glVertex3fv(p[1]);
glVertex3fv(p[2]);
glVertex3fv(q[2]);
glVertex3fv(q[1]);
glNormal3f(0.0,0.0,+1.0); // +Z base
glVertex3fv(p[3]);
glVertex3fv(p[4]);
glVertex3fv(q[4]);
glVertex3fv(q[3]);
glVertex3fv(p[4]);
glVertex3fv(p[5]);
glVertex3fv(q[5]);
glVertex3fv(q[4]);
glNormal3f(-c,-s,0.0); // shaft circumference side
glVertex3fv(p[5]);
glVertex3fv(p[0]);
glVertex3fv(q[0]);
glVertex3fv(q[5]);
glNormal3f(c,s,0.0); // outter circumference side
glVertex3fv(p[2]);
glVertex3fv(p[3]);
glVertex3fv(q[3]);
glVertex3fv(q[2]);
glNormal3f(-s,c,0.0);
glVertex3fv(p[4]);
glVertex3fv(p[3]);
glVertex3fv(p[2]);
glVertex3fv(p[1]);
glNormal3f(s,-c,0.0);
glVertex3fv(q[1]);
glVertex3fv(q[2]);
glVertex3fv(q[3]);
glVertex3fv(q[4]);
// points on circles at angle a
c=cos(a); s=sin(a);;
x=r0*c; y=r0*s; q[0][0]=x; q[5][0]=x;
q[0][1]=y; q[5][1]=y;
x=r1*c; y=r1*s; q[1][0]=x; q[4][0]=x;
q[1][1]=y; q[4][1]=y;
c=cos(a+da); s=sin(a+da); a+=da;
x+=dr*c;y+=dr*s;q[2][0]=x; q[3][0]=x;
q[2][1]=y; q[3][1]=y;
c=cos(a); s=sin(a); a+=da;
// render gap
glNormal3f(0.0,0.0,-1.0); // -Z base
glVertex3fv(q[0]);
glVertex3fv(q[1]);
glVertex3fv(p[1]);
glVertex3fv(p[0]);
glNormal3f(0.0,0.0,+1.0); // +Z base
glVertex3fv(q[4]);
glVertex3fv(q[5]);
glVertex3fv(p[5]);
glVertex3fv(p[4]);
glNormal3f(-c,-s,0.0); // shaft circumference side
glVertex3fv(q[5]);
glVertex3fv(q[0]);
glVertex3fv(p[0]);
glVertex3fv(p[5]);
glNormal3f(c,s,0.0); // outter circumference side
glVertex3fv(q[1]);
glVertex3fv(q[4]);
glVertex3fv(p[4]);
glVertex3fv(p[1]);
glEnd();
所以只是点 p[2],q[2],p[3],q[3]
略有变化,基面必须用更多的 QUADS 来补偿...
这里是glCog(0.2,0.5,0.6,0.2,10);
的预览:
【讨论】:
很好的插图(+1) @Rabbid76 thx ...它是由我自己的 GIF 捕获和编码器捕获的...但是有时在捕获 GL 渲染白框时最后会出现不知道为什么(我开发时没有这样做这可能是多年来 gfx 驱动程序的一些变化) 如果可以的话,我会给你+2这个答案。谢谢先生。以上是关于使用OpenGL如何实现物体绕体外任意轴旋转?的主要内容,如果未能解决你的问题,请参考以下文章