我无法使用C ++中的OpenGL库函数glutSolidSphere()创建一个简单的3D球体。
这是我试过的:
#include<GL/glu.h>
void display()
{
glClear(GL_COLOR_BUFFER_BIT);
glColor3f(1.0,0.0,0.0);
glLoadIdentity();
glutSolidSphere( 5.0, 20.0, 20.0);
glFlush();
}
void myInit()
{
glClearColor(1.0,1.0,1.0,1.0);
glColor3f(1.0,0.0,0.0);
glMatrixMode(GL_PROJECTION);
glLoadIdentity();
gluOrtho2D(0.0,499.0,0.0,499.0);
glMatrixMode(GL_MODELVIEW);
}
void main(int argc,char **argv)
{
qobj = gluNewQuadric();
glutInit(&argc,argv);
glutInitDisplayMode(GLUT_SINGLE|GLUT_RGB);
glutInitWindowSize(500,500);
glutCreateWindow("pendulum");
glutDisplayFunc(display);
myInit();
glutMainLoop();
}
答案 0 :(得分:72)
在OpenGL中,您不创建对象,只需绘制它们即可。绘制完成后,OpenGL不再关心您发送的几何图形。
glutSolidSphere
只是向OpenGL发送绘图命令。然而,它并没有什么特别之处。因为它与GLUT联系在一起我不会使用它。相反,如果你的代码中确实需要一些球体,那么为自己创造一个怎么样?
#define _USE_MATH_DEFINES
#include <GL/gl.h>
#include <GL/glu.h>
#include <vector>
#include <cmath>
// your framework of choice here
class SolidSphere
{
protected:
std::vector<GLfloat> vertices;
std::vector<GLfloat> normals;
std::vector<GLfloat> texcoords;
std::vector<GLushort> indices;
public:
SolidSphere(float radius, unsigned int rings, unsigned int sectors)
{
float const R = 1./(float)(rings-1);
float const S = 1./(float)(sectors-1);
int r, s;
vertices.resize(rings * sectors * 3);
normals.resize(rings * sectors * 3);
texcoords.resize(rings * sectors * 2);
std::vector<GLfloat>::iterator v = vertices.begin();
std::vector<GLfloat>::iterator n = normals.begin();
std::vector<GLfloat>::iterator t = texcoords.begin();
for(r = 0; r < rings; r++) for(s = 0; s < sectors; s++) {
float const y = sin( -M_PI_2 + M_PI * r * R );
float const x = cos(2*M_PI * s * S) * sin( M_PI * r * R );
float const z = sin(2*M_PI * s * S) * sin( M_PI * r * R );
*t++ = s*S;
*t++ = r*R;
*v++ = x * radius;
*v++ = y * radius;
*v++ = z * radius;
*n++ = x;
*n++ = y;
*n++ = z;
}
indices.resize(rings * sectors * 4);
std::vector<GLushort>::iterator i = indices.begin();
for(r = 0; r < rings; r++) for(s = 0; s < sectors; s++) {
*i++ = r * sectors + s;
*i++ = r * sectors + (s+1);
*i++ = (r+1) * sectors + (s+1);
*i++ = (r+1) * sectors + s;
}
}
void draw(GLfloat x, GLfloat y, GLfloat z)
{
glMatrixMode(GL_MODELVIEW);
glPushMatrix();
glTranslatef(x,y,z);
glEnableClientState(GL_VERTEX_ARRAY);
glEnableClientState(GL_NORMAL_ARRAY);
glEnableClientState(GL_TEXTURE_COORD_ARRAY);
glVertexPointer(3, GL_FLOAT, 0, &vertices[0]);
glNormalPointer(GL_FLOAT, 0, &normals[0]);
glTexCoordPointer(2, GL_FLOAT, 0, &texcoords[0]);
glDrawElements(GL_QUADS, indices.size(), GL_UNSIGNED_SHORT, &indices[0]);
glPopMatrix();
}
};
SolidSphere sphere(1, 12, 24);
void display()
{
int const win_width = …; // retrieve window dimensions from
int const win_height = …; // framework of choice here
float const win_aspect = (float)win_width / (float)win_height;
glViewport(0, 0, win_width, win_height);
glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);
glMatrixMode(GL_PROJECTION);
glLoadIdentity();
gluPerspective(45, win_aspect, 1, 10);
glMatrixMode(GL_MODELVIEW);
glLoadIdentity();
#ifdef DRAW_WIREFRAME
glPolygonMode(GL_FRONT_AND_BACK, GL_LINE);
#endif
sphere.draw(0, 0, -5);
swapBuffers();
}
int main(int argc, char *argv[])
{
// initialize and register your framework of choice here
return 0;
}
答案 1 :(得分:20)
到目前为止,似乎没有人解决原始代码的实际问题,所以我想我会这样做,即使这个问题已经很久了。
问题最初与投影相关的球体半径和位置有关。我想你会发现问题不是太复杂。该程序实际上正常工作,只是绘制的内容很难看到。
首先,使用调用
创建正交投影gluOrtho2D(0.0, 499.0, 0.0, 499.0);
“is equivalent to calling glOrtho with near = -1 and far = 1.”这意味着viewing frustum的深度为2.因此半径大于1(直径= 2)的球体将不完全适合观察视锥体
然后是电话
glLoadIdentity();
glutSolidSphere(5.0, 20.0, 20.0);
使用,它加载模型 - 视图矩阵的单位矩阵,然后“[r]enders a sphere centered at the modeling coordinates origin of the specified radius.”意义,球体在原点渲染,(x,y,z)=(0,0,0) ),半径为5。
现在,问题有三个:
pi*5^2/499^2
,接近约1/3170),因此可能很难看到。这假设整个圆圈是在窗口区域内绘制的。但是,我们将在第2点看到它。pi*5^2/(4*499^2)
,更接近1 / 12,682nd)。这会使更难以看到。特别是因为球体渲染得非常靠近屏幕的边缘/角落,您可能不会想到它。解决方案只是改变视锥体和球体半径。例如,
gluOrtho2D(-5.0, 5.0, -5.0, 5.0);
glutSolidSphere(5.0, 20, 20);
呈现以下图片。
正如你所看到的,只有一小部分在半球为5的球体的“赤道”周围可见。(我改变了投影以用球体填充窗口。)另一个例子,
gluOrtho2D(-1.1, 1.1, -1.1, 1.1);
glutSolidSphere(1.1, 20, 20);
呈现以下图片。
上图显示了视锥体内部的更多球体,但球体仍然比观察平截头体大0.2个深度单位。正如你所看到的,球体的“冰盖”在北方和南方都缺失了。因此,如果我们希望整个球体适合具有深度2的视锥体,我们必须使半径小于或等于1.
gluOrtho2D(-1.0, 1.0, -1.0, 1.0);
glutSolidSphere(1.0, 20, 20);
呈现以下图片。
我希望这有助于某人。小心!
答案 2 :(得分:8)
我不明白datenwolf的索引生成怎么可能是正确的。但我仍然觉得他的解决方案很清楚。这是我在思考之后得到的:
inline void push_indices(vector<GLushort>& indices, int sectors, int r, int s) {
int curRow = r * sectors;
int nextRow = (r+1) * sectors;
indices.push_back(curRow + s);
indices.push_back(nextRow + s);
indices.push_back(nextRow + (s+1));
indices.push_back(curRow + s);
indices.push_back(nextRow + (s+1));
indices.push_back(curRow + (s+1));
}
void createSphere(vector<vec3>& vertices, vector<GLushort>& indices, vector<vec2>& texcoords,
float radius, unsigned int rings, unsigned int sectors)
{
float const R = 1./(float)(rings-1);
float const S = 1./(float)(sectors-1);
for(int r = 0; r < rings; ++r) {
for(int s = 0; s < sectors; ++s) {
float const y = sin( -M_PI_2 + M_PI * r * R );
float const x = cos(2*M_PI * s * S) * sin( M_PI * r * R );
float const z = sin(2*M_PI * s * S) * sin( M_PI * r * R );
texcoords.push_back(vec2(s*S, r*R));
vertices.push_back(vec3(x,y,z) * radius);
push_indices(indices, sectors, r, s);
}
}
}
答案 3 :(得分:2)
以下是代码:
glPushMatrix();
glTranslatef(18,2,0);
glRotatef(angle, 0, 0, 0.7);
glColor3ub(0,255,255);
glutWireSphere(3,10,10);
glPopMatrix();
答案 4 :(得分:2)
Datanewolf的代码几乎是正确的。我不得不反转绕组和法线,以使其与固定管道正常工作。以下对我来说可以正常打开或关闭:
std::vector<GLfloat> vertices;
std::vector<GLfloat> normals;
std::vector<GLfloat> texcoords;
std::vector<GLushort> indices;
float const R = 1./(float)(rings-1);
float const S = 1./(float)(sectors-1);
int r, s;
vertices.resize(rings * sectors * 3);
normals.resize(rings * sectors * 3);
texcoords.resize(rings * sectors * 2);
std::vector<GLfloat>::iterator v = vertices.begin();
std::vector<GLfloat>::iterator n = normals.begin();
std::vector<GLfloat>::iterator t = texcoords.begin();
for(r = 0; r < rings; r++) for(s = 0; s < sectors; s++) {
float const y = sin( -M_PI_2 + M_PI * r * R );
float const x = cos(2*M_PI * s * S) * sin( M_PI * r * R );
float const z = sin(2*M_PI * s * S) * sin( M_PI * r * R );
*t++ = s*S;
*t++ = r*R;
*v++ = x * radius;
*v++ = y * radius;
*v++ = z * radius;
*n++ = -x;
*n++ = -y;
*n++ = -z;
}
indices.resize(rings * sectors * 4);
std::vector<GLushort>::iterator i = indices.begin();
for(r = 0; r < rings-1; r++)
for(s = 0; s < sectors-1; s++) {
/*
*i++ = r * sectors + s;
*i++ = r * sectors + (s+1);
*i++ = (r+1) * sectors + (s+1);
*i++ = (r+1) * sectors + s;
*/
*i++ = (r+1) * sectors + s;
*i++ = (r+1) * sectors + (s+1);
*i++ = r * sectors + (s+1);
*i++ = r * sectors + s;
}
编辑:有一个关于如何绘制这个问题的问题......在我的代码中,我将这些值封装在G3DModel类中。这是我设置框架,绘制模型并结束它的代码:
void GraphicsProvider3DPriv::BeginFrame()const{
int win_width;
int win_height;// framework of choice here
glfwGetWindowSize(window, &win_width, &win_height); // retrieve window
float const win_aspect = (float)win_width / (float)win_height;
// set lighting
glEnable(GL_LIGHTING);
glEnable(GL_LIGHT0);
glEnable(GL_DEPTH_TEST);
GLfloat lightpos[] = {0, 0.0, 0, 0.};
glLightfv(GL_LIGHT0, GL_POSITION, lightpos);
GLfloat lmodel_ambient[] = { 0.2, 0.2, 0.2, 1.0 };
glLightModelfv(GL_LIGHT_MODEL_AMBIENT, lmodel_ambient);
glLightModeli(GL_LIGHT_MODEL_TWO_SIDE, GL_TRUE);
// set up world transform
glClearColor(0.f, 0.f, 0.f, 1.f);
glClear(GL_COLOR_BUFFER_BIT|GL_DEPTH_BUFFER_BIT|GL_STENCIL_BUFFER_BIT|GL_ACCUM_BUFFER_BIT);
glMatrixMode(GL_PROJECTION);
glLoadIdentity();
gluPerspective(45, win_aspect, 1, 10);
glMatrixMode(GL_MODELVIEW);
}
void GraphicsProvider3DPriv::DrawModel(const G3DModel* model, const Transform3D transform)const{
G3DModelPriv* privModel = (G3DModelPriv *)model;
glPushMatrix();
glLoadMatrixf(transform.GetOGLData());
glEnableClientState(GL_VERTEX_ARRAY);
glEnableClientState(GL_NORMAL_ARRAY);
glEnableClientState(GL_TEXTURE_COORD_ARRAY);
glVertexPointer(3, GL_FLOAT, 0, &privModel->vertices[0]);
glNormalPointer(GL_FLOAT, 0, &privModel->normals[0]);
glTexCoordPointer(2, GL_FLOAT, 0, &privModel->texcoords[0]);
glEnable(GL_TEXTURE_2D);
//glFrontFace(GL_CCW);
glEnable(GL_CULL_FACE);
glActiveTexture(GL_TEXTURE0);
glBindTexture(GL_TEXTURE_2D, privModel->texname);
glDrawElements(GL_QUADS, privModel->indices.size(), GL_UNSIGNED_SHORT, &privModel->indices[0]);
glPopMatrix();
glDisable(GL_TEXTURE_2D);
}
void GraphicsProvider3DPriv::EndFrame()const{
/* Swap front and back buffers */
glDisable(GL_LIGHTING);
glDisable(GL_LIGHT0);
glDisable(GL_CULL_FACE);
glfwSwapBuffers(window);
/* Poll for and process events */
glfwPollEvents();
}
答案 5 :(得分:2)
我喜欢硬币的答案。它易于理解并适用于三角形。然而,他的计划的索引有时超出界限。所以我在这里发布他的代码有两个微小的更正:
inline void push_indices(vector<GLushort>& indices, int sectors, int r, int s) {
int curRow = r * sectors;
int nextRow = (r+1) * sectors;
int nextS = (s+1) % sectors;
indices.push_back(curRow + s);
indices.push_back(nextRow + s);
indices.push_back(nextRow + nextS);
indices.push_back(curRow + s);
indices.push_back(nextRow + nextS);
indices.push_back(curRow + nextS);
}
void createSphere(vector<vec3>& vertices, vector<GLushort>& indices, vector<vec2>& texcoords,
float radius, unsigned int rings, unsigned int sectors)
{
float const R = 1./(float)(rings-1);
float const S = 1./(float)(sectors-1);
for(int r = 0; r < rings; ++r) {
for(int s = 0; s < sectors; ++s) {
float const y = sin( -M_PI_2 + M_PI * r * R );
float const x = cos(2*M_PI * s * S) * sin( M_PI * r * R );
float const z = sin(2*M_PI * s * S) * sin( M_PI * r * R );
texcoords.push_back(vec2(s*S, r*R));
vertices.push_back(vec3(x,y,z) * radius);
if(r < rings-1)
push_indices(indices, sectors, r, s);
}
}
}