是否可以使用OpenGL点精灵来模拟广告牌精灵？

Question

我试图在OpenGL中设置点精灵来改变距离,就像一个广告牌精灵那样,但我无法获得值GL_POINT_DISTANCE_ATTENUATION_ARB来做任何有用的事情.是否存在与给定投影匹配的值的相关性？我正在努力做甚么可能吗？

渲染使用的代码:

glPointParameterfARB  = (PFNGLPOINTPARAMETERFARBPROC)wglGetProcAddress("glPointParameterfARB");
glPointParameterfvARB = (PFNGLPOINTPARAMETERFVARBPROC)wglGetProcAddress("glPointParameterfvARB");

glClear(GL_COLOR_BUFFER_BIT|GL_DEPTH_BUFFER_BIT);
glLoadIdentity();
gluPerspective(100.0, 800.0/600.0, 0.1, 10.0);

float quadratic[] =  { 5.0f, 0.1f, 10.0f };
glPointParameterfvARB( GL_POINT_DISTANCE_ATTENUATION_ARB, quadratic );

float maxSize = 0.0f;
glGetFloatv( GL_POINT_SIZE_MAX_ARB, &maxSize );
if( maxSize > 100.0f )  maxSize = 100.0f;
glPointSize( maxSize );

glPointParameterfARB( GL_POINT_FADE_THRESHOLD_SIZE_ARB, 0.1f );
glPointParameterfARB( GL_POINT_SIZE_MIN_ARB, 0.1f );
glPointParameterfARB( GL_POINT_SIZE_MAX_ARB, maxSize );

glTexEnvf( GL_POINT_SPRITE_ARB, GL_COORD_REPLACE_ARB, GL_TRUE );

glEnable( GL_POINT_SPRITE_ARB );

glScalef(0.75,1,1);
glTranslatef(0.00,0.0,-1.0);
glScalef(0.5,0.5,0.5);
glRotatef(counter*0.1+0.5,1.0,1.0,0.0);

glBegin( GL_POINTS );

for( int i = 0; i < 100; ++i )
{
    glColor4f( i%10*0.1, i/10*0.1, 0.5, 1.0f );

    glVertex3f( i%10*0.2-1.0,i/10*0.2-1.0,
    ((i%10-5)*(i%10-5)+(i/10-5)*(i/10-5))*0.01 );
}

glEnd();

glDisable( GL_POINT_SPRITE_ARB );

Answer 1

这就是我如何让我的穷人缩小磅值的方法:

void render() {
    glEnable(GL_VERTEX_PROGRAM_POINT_SIZE_ARB);
    glHint(GL_POINT_SMOOTH_HINT, GL_NICEST);
    glEnable(GL_POINT_SPRITE);
    glActiveTexture(GL_TEXTURE0);
    glTexEnvi(GL_POINT_SPRITE, GL_COORD_REPLACE, GL_TRUE);

    /* Activate shader program here */
    /* Send pointSize to shader program */

    glBegin(GL_POINTS);
        /* Render points here */
        glVertex3f(...);
    glEnd(GL_POINTS);
}

顶点着色器:

uniform float pointSize;
void main() {
    gl_Position = ftransform();
    gl_PointSize = pointSize / gl_Position.w;
}

您可以在片段着色器中执行任何操作,但您必须自己计算颜色,光照和纹理.