MickD 发表于 2006-7-31 10:17:33

可能会这样结束，米克
blockref有这些属性，但我没有；我没有blockref。（点和块项）
我可以使用selectatpoint selectionset获得blockref，但我想学习如何使用我首先拥有的
这是一个调试。打印；为插入点4,4 0旋转的2块跨矩阵给出的矩阵；1&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；0；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；1&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；0&nbsp；; &nbsp；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；0；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；1&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；0；4&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；4&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；1转置矩阵；1&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；4；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；1&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；4；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；1&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；0；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；1跨基质&nbsp；插入4,4 0旋转；0.707106781186547&nbsp&nbsp&nbsp&nbsp&nbsp；0.707106781186548&nbsp&nbsp&nbsp&nbsp&nbsp；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；0-0.707106781186548&nbsp&nbsp&nbsp&nbsp&nbsp；0.707106781186547&nbsp&nbsp&nbsp&nbsp&nbsp；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；0；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；1&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；0；4&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；2&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；1转置矩阵；0.707106781186547&nbsp&nbsp&nbsp&nbsp-0.707106781186548&nbsp&nbsp&nbsp&nbsp&nbsp；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；4；0.707106781186548&nbsp&nbsp&nbsp&nbsp&nbsp；0.707106781186547&nbsp&nbsp&nbsp&nbsp&nbsp；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；2；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；1&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；0；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；0&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp；1；来自；blockrecord到blockref
我刚开始做相反的事情
我可能把整数和double搞混了

MickD 发表于 2006-7-31 21:38:16

有些东西不&#039；t向右看，向量下方的底行应全部为0&#039；s、 这些只有&#039；正方形#039；向上移动矩阵或可能用于投影计算&#039；s、 你的平移向量应该在右边。我附上了帮助文档#039；这可能会有所帮助。他们绝对不应该&#039；无论如何，我不会从底部换到右侧；不要思考。

Bryco 发表于 2006-7-31 21:45:45

我认为你所需要的就是#039；倒置&#039；如果不进行转置，请尝试对值求反，而不是将其交换。也就是说，可以通过向量（3,2,1）移动某个对象，如果要以另一种方式移动它，则应使用取反的向量（-3，-2，-1），该向量应指向另一个方向。同样适用于旋转。

Bryco 发表于 2006-7-31 22:25:07

米克，你赢了
反转然后转置似乎起作用
谢谢你的提示
现在我&#039；我必须处理乘法公式。

Bryco 发表于 2006-8-3 23:44:38

到目前为止，这些似乎都很有效，有待讨论，我相信它们会更好[代码这就是我想要使用矩阵的原因Sub TestMultiplyMatrix()
    Dim P, Ent As AcadEntity
    Dim i As Integer
    Dim M, ContextData, TransMatrix
   
    ThisDrawing.Utility.GetSubEntity Ent, P, TransMatrix, ContextData
    ThisDrawing.ModelSpace.AddPoint P
    If VarType(ContextData) = vbEmpty Then
      MsgBox "This is not a block"
      Exit Sub
    End If
    M = InverseMatrix(TransMatrix)
   
    P = TransformPt(M, P)
    P = NearestPtOnObject(Ent, P)
    P = TransformPt(TransMatrix, P)
    ThisDrawing.ModelSpace.AddPoint P
   
End Sub 最近的ptonobject（Ent，p）函数不包括在I&#039中；我还在努力
我想；走这么远会让我从数学上有一点休息，但相反，我似乎打开了潘多拉&#039；s盒
距离椭圆最近的点是疯狂的。我甚至不知道这篇文章是在提供证据还是在提出问题。本人&#039；我倾向于使用一个提到“的词；最佳猜测；，
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查看完整版本: getsubentity中的TransMatrix是什么？