我最初在3D空间储存了一种物体的原产地和方向,使用代表物体来源、前向和方向的3个病媒。
适用于对模型概览矩阵的正确转变,是使用这三种病媒的组合式转变矩阵。
翻译是三维的,但是,通过构建正确的轮流矩阵(取决于轮流的角和轴心)和将这些矩阵应用到这3个病媒中来进行轮换。
使用这一方法处理大量物体和轮用/制式矩阵堆肥造成工作瓶颈。
我不禁要问,是否有更明智、更高效的方式储存方向?
我最初在3D空间储存了一种物体的原产地和方向,使用代表物体来源、前向和方向的3个病媒。
适用于对模型概览矩阵的正确转变,是使用这三种病媒的组合式转变矩阵。
翻译是三维的,但是,通过构建正确的轮流矩阵(取决于轮流的角和轴心)和将这些矩阵应用到这3个病媒中来进行轮换。
使用这一方法处理大量物体和轮用/制式矩阵堆肥造成工作瓶颈。
我不禁要问,是否有更明智、更高效的方式储存方向?
我最初在3D空间储存了一种物体的原产地和方向,使用代表物体来源、前向和方向的3个病媒。
换言之: 页: 1 矩阵表的开放式计算器的使用只是一样的,虽然是4×4,但唯一的区别是,要素4,4总是1,要素0.3,4全部是0,第1栏是前向产,通常称为
这实际上是代表位于3D空间的物体的最简明和最直接获得的途径。 您可以采用任何一种转变方式,然后采用单一矩阵的多重复。
另一种方法是使用quaternion与一种被抵消的病媒。 但是,如果你希望你的物体能够转换成一个矩阵(或者你可以把许多翻译/轮换的配对连接起来,进行转变,但使用矩阵实际上造成较少的间接费用)。
Other than memory, what s stopping you from storing the whole 4x4 affine matrix?
更有甚者,《国际减灾战略》规定,如果阵列实现正常,则底层的行文为usually[0、0、1]
,那么你只需要储存上三个行。
储存方向可能是最自然/最有效率的方法(它应当在所有方面比前导/上调器更好)。 缩略语将包含4个价值,储存10个多模版和15个增补。
如果你特别关注空间,你可能只得到3个价值,因为你能够从其余3个单位中产生一个要素。 这将需要另外增加3个倍增效应、3个增补和一个平方根(也稍微trick,因为你们需要确保第1个要素是不可否定的......)
Quaternions transformations may be faster then Matrix on current GPGPUs where the access for global memory is much slower then to local memory. According to this tables you compute twice much but need to fetch less then half of memory.
投放有作为地方记忆储存的垂直穿透镜,这里唯一合理的是使用矩阵。
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