...
首页> 外文期刊>ACM Transactions on Graphics >TIME/SPACE TRADEOFFS FOR POLYGON MESH RENDERING
【24h】

TIME/SPACE TRADEOFFS FOR POLYGON MESH RENDERING

机译:多边形网格渲染的时间/空间权衡

获取原文
获取原文并翻译 | 示例
   

获取外文期刊封面封底 >>

       

摘要

We investigate architectural schemes, generalizing that of existing graphics engines, supporting fast rendering of triangle meshes. A mesh defined on n vertices is rendered by sending vertices down a graphics pipeline, after which they are pushed on a stack to be popped when no longer needed. Only individual triangles whose vertices are present in the stack may be rendered. The storage cost of the mesh rendering is the size of the stack required to store mesh vertices during the rendering process. This may be significantly less than n. The time cost of the mesh rendering is the number of vertices sent down the graphics pipeline. If a large enough stack is available, it suffices to send each vertex once. If only a small stack is available, some vertices may have to be sent more than once, so a time/space tradeoff exists. With our architecture, a stack of size O(root n) is sufficient to fender any triangle mesh defined on n vertices, such that each vertex is sent only once through the graphics pipeline (time cost = n). We provide an algorithm that generates an appropriate ''rendering sequence'' of commands for any given mesh. Moreover, we show that no algorithm can do better, that is, Omega(root n) is a lower bound. Some n-vertex meshes may be rendered using a stack whose size is significantly less than O(root n). An algorithm generating a minimum-time rendering sequence requiring the minimum stack size is an open question. We provide an approximation: if it is theoretically possible to render a triangle mesh in minimum time with a stack of size S, our algorithm generates a minimum-time rendering sequence requiring a stack of size no larger than 2S log(3/2)n. If only a stack of size k is available, we provide an algorithm generating a rendering sequence requiring a stack of size no larger than k, such that at most n(1 + c/k) vertices must be sent through the pipeline, for some constant c. [References: 15]
机译:我们研究了架构方案,概括了现有图形引擎的架构方案,支持三角形网格的快速渲染。通过沿图形管线发送顶点来渲染在n个顶点上定义的网格,此后将其推入堆栈中以在不再需要时弹出。只能渲染其顶点在堆栈中存在的单个三角形。网格渲染的存储成本是在渲染过程中存储网格顶点所需的堆栈大小。这可能大大小于n。网格渲染的时间成本是沿着图形管线发送的顶点数量。如果有足够大的堆栈可用,则只需发送一次每个顶点即可。如果只有很小的堆栈可用,则可能必须多次发送某些顶点,因此存在时间/空间折衷。在我们的体系结构中,大小为O(root n)的堆栈足以抵御定义在n个顶点上的任何三角形网格,从而每个顶点仅通过图形管线发送一次(时间成本= n)。我们提供了一种算法,可为任何给定的网格生成适当的命令“渲染序列”。而且,我们证明没有算法能做得更好,即Omega(root n)是一个下限。某些n顶点网格可以使用堆栈进行渲染,该堆栈的大小显着小于O(root n)。产生需要最小堆栈大小的最小时间渲染序列的算法是一个悬而未决的问题。我们提供一个近似值:如果理论上可以在最小时间内渲染大小为S的三角形网格,则我们的算法会生成最小时间渲染序列,该序列需要大小不超过2S log(3/2)n的栈。如果只有大小为k的堆栈可用,我们提供了一种算法,该算法生成的渲染序列需要大小不大于k的堆栈,因此对于某些对象,最多只能通过管线发送n(1 + c / k)个顶点常数c。 [参考:15]

著录项

相似文献

  • 外文文献
  • 中文文献
  • 专利
获取原文

客服邮箱:kefu@zhangqiaokeyan.com

京公网安备:11010802029741号 ICP备案号:京ICP备15016152号-6 六维联合信息科技 (北京) 有限公司©版权所有
  • 客服微信

  • 服务号