OpenGL旋转和平移

Question

我正在阅读一系列 NeHe OpenGK 教程。 Tutorial #9 做了一些花哨的事情；我什么都懂，除了我认为是整个教程的支柱的两件事。

在DrawGlScene函数中，下面这行我没看懂。

glRotatef(tilt,1.0f,0.0f,0.0f);  // Tilt The View (Using The Value In 'tilt') 

我了解该行的作用，并且在教程中也非常清楚地提到了它。但我不明白他为什么要倾斜屏幕。

另一件事是他首先倾斜屏幕，然后将其旋转一个星角，然后立即执行相反的操作。那是什么技术？什么需要倾斜？只需在星星面向用户时旋转星星。

glRotatef(star[loop].angle,0.0f,1.0f,0.0f); // Rotate To The Current Stars Angle
glTranslatef(star[loop].dist,0.0f,0.0f);    // Move Forward On The X Plane

glRotatef(-star[loop].angle,0.0f,1.0f,0.0f); // Cancel The Current Stars Angle
glRotatef(-tilt,1.0f,0.0f,0.0f);             // Cancel The Screen Tilt

如果有人告诉我引擎盖下的机制，我将非常感激。

Answer 1

我不明白他为什么要倾斜屏幕。

倾斜可以让您从另一个角度看到星星，而不仅仅是“正上方”。

另一件事是他首先倾斜屏幕，然后将其旋转一个星角，然后立即执行相反的操作。这是什么技术？

那是因为他想围绕选定平面（在本例中为 Y 平面）旋转星形，但是（！）他还希望纹理四边形面向观察者。假设他将其旋转 90 度，如果是这样，您只会看到（就像他在教程中所说的那样）一条“粗”线。

考虑这些 cmets：

// Rotate the current drawing by the specified angle on the Y axis 
// in order to get it to rotate.
glRotatef(star[loop].angle, 0.0f, 1.0f, 0.0f); 

// Rotating around the object's origin is not going to make 
// any visible effects, especially since the star object itself is in 2D.
// In order to move around in your current projection, a glRotatef()
// call does rotate the star, but not in terms of moving it "around" 
// on the screen.
// Therefore, use the star's distance to move it out from the center.  
glTranslatef(star[loop].dist, 0.0f, 0.0f);

// We've moved the star out from the center, with the specified 
// distance in star's distance. With the first glRotatef() 
// call in mind, the 2D star is not 100 % facing
// the viewer. Therefore, face the star towards the screen using 
// the negative angle value.
glRotatef(-star[loop].angle, 0.0f, 1.0f, 0.0f);

// Cancel the tilt on the X axis.
glRotatef(-tilt, 1.0f, 0.0f, 0.0f);