我正试图找到四舍五入座座座座座座座座座标,座落在四舍五入内(我想说几句能够穿透)。 我知道四点(A、B、C、D)的坐标。
我目前正在做的是,我发现四丁(最低和最高×, y,z坐标为A、B、C、D)的捆绑箱,然后通过捆绑箱内的所有各点 lo。 对于每一个此类点,我计算出灯座坐标(http://en.wikipedia.org/wiki/Barycentric_coordination_system_%28mathematics%29#Barycentric_coordinations_on_tetrahedra)。 Wikipedia的等值,如果该点位于四舍五入内,则检查(如果任何金心坐标为负数或大于1,则点为内线)。
是否有更好的办法这样做? 目前,大约1/6个机会是,我正在测试的点(从捆绑箱中)确实位于四舍五入内,因此,我认为我太过了太多不必要的计算。
我正在与一份四轮驱动的清单合作,这份清单是我通过四轮驱动的。 我没有使用任何外部图书馆。
另一项改进想法:
如果对Z-axis(即x=4, y=6)的“棒”法则通过四舍二入。 如果没有,任何(x=4, y=5,z)的数值都不得在内。
Else发现,rod杆在四氯丁二醇的边缘(通过发现构成四环ron边缘的 plan在哪里)。
Say these planes intersect at z=1.3 and z= 10.04. Then you know all points (4,5, 2) to (4,5,10) are inside.
对所有x和 y数值的答复。
这在实践中应当更快,因为它将拯救你们1个住所。
你的做法是正确的。 有一些可能的选择,可能值得选择,或者不取决于要求。 例如:
There is an easier way to check if a given point is inside or outside of the tetrahedron. It amounts to checking the which half-space the point belongs to with respect to each of the 4 sides of the tetrahedron:
每个方面由3点界定(A、B、C)。 然后是正常的(C-A)x(B-A)(该产品是平面病媒的交叉产品)。 如果该坐标为(a,b,c),则平面方程式为
要点是,你可以预先为四舍五入的每一方提出计划要求,以及相当于在最多4次评价(每边一次)中对某个特定点进行核查的标书,每笔标书有3个多重复和2个增补。
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