简体繁体 English

OpenGL，剔除视图外的对象

[英]OpenGL, Culling objects that are outside the view

原文 2018-12-01 17:09:49 1 1 c++/ opengl/ viewport/ bounding-box

In my case I wanna render 50,000 or more cubes that are distributed randomly inside a large bounding box, I don't want using instancing right now so I have to render each cube, I wanna improve the performance by culling out the cubes that are outside the camera view. 就我而言，我想渲染在一个大边界框中随机分布的50,000个或更多的多维数据集，我现在不想使用实例化，因此我必须渲染每个多维数据集，我想通过剔除外面的多维数据集来提高性能相机视图。 I have a camera class that has two matrices view and projection , each cube has its own bounding box, so I am planning to check each frame if the camera view bounding box contains the center of each cube if yes call its draw function if not ignore it. 我有一个具有两个矩阵view和projection的相机类，每个多维数据集都有自己的边界框，因此我打算检查每个帧，如果相机视图边界框包含每个多维数据集的中心（如果是），请调用其draw函数（如果不忽略）它。

I have for view matrix 3 vectors eye, target and up, and for projection width, height, near, far and FOV. 我有3个向量，分别是眼，目标和上眼，以及投影的宽度，高度，近，远和FOV。

so I have two questions: 所以我有两个问题：

Is this a right scenario? 这是正确的方案吗？ I will calculate the camera view boumding box each frame then test each cube. 我将计算每个帧的相机视图合并框，然后测试每个立方体。
How can I calculate the camera bounding box each frame? 如何计算每帧的相机边框？

1 个解决方案

I got an idea from here how_to_check_if_vertex_is_visible_for_user that worked fine with me. 我从这里获得了一个想法， how_to_check_if_vertex_is_visible_for_user对我来说很好用。

multiplying the projection view matrix of the camera by the any point in the 3D space the visible ones should be between [-1,1]. 将相机的投影视图矩阵乘以3D空间中的任意点，可见的点应在[-1,1]之间。