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发表于 2010-8-5 12:52:09
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来自: 中国辽宁铁岭
这个图没有几何画法.AutoCAD的约束也不能约束弧长.只能用近似画法.
8 S# u3 L. C$ W& g& k: A比较实用的方法一个是计算,即先算出结果再按结果画图.比如用EXCEL的单变量求解,算出弧的角度约为65.41920502度,按此角度画出任意半径的弧后,再缩放到合乎要求的尺寸.. n: T" H8 C1 v3 _( x6 a% s
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还有一个方法就是编程.实质上是用计算机的迭代运算代替人工尝试的逼近法,或者说是一种自定义的约束.下面是画本图的VBA代码" V3 Q6 h, X7 Z" B
+ A: d, W0 K+ Z# Y7 [2 ?, `6 M! ~0 l- Sub HT()
' ]2 g) B" `% t- s - Dim Arc As AcadArc, C(2) As Double, A As Double, A1 As Double, A2 As Double, H As Double1 s" L- w; d& n, @/ G4 e( A" H
- With ThisDrawing+ M5 ~- k* V$ b+ Q: o* p3 ]
- Set Arc = .ModelSpace.AddArc(C, 1, 0, 0)% v* {4 a, ^6 I' ?. A
- A2 = ThisDrawing.Utility.AngleToReal(180, acDegrees) * 2
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- A = (A1 + A2) / 2$ f" n5 k( U5 T& S( P, [$ f* E
- Arc.StartAngle = (.Utility.AngleToReal(180, acDegrees) - A) / 22 A3 P. s; ~3 ^' J% E
- Arc.EndAngle = Arc.StartAngle + A
- Q0 T( V7 E% \ Q, b, N+ n - Arc.ScaleEntity C, 888.888 / Arc.ArcLength
m9 l2 [4 ]+ E$ D - H = Arc.Radius - Arc.StartPoint(1)
$ }, O. j, d8 V& ~' y/ M - If H = 123.456 Or A = A1 Or A = A2 Then
" T5 S. a8 @8 ?/ D9 O3 G- S - Exit Do
4 i8 E4 k8 q, I0 @5 ~, F - ElseIf H > 123.456 Then
( V' w3 r m) f2 w) t9 v3 E/ ~: O - A2 = A- b# x& E# o; P3 h1 A
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- A1 = A/ w8 B- ~: M. [/ I V
- End If' `; m% K8 a+ }' R
- Loop
! y3 W$ l! l O1 q - .ModelSpace.AddLine Arc.StartPoint, Arc.EndPoint
8 i* b. Y. u0 q; R! b' V - End With
/ m' w. P% I z! l - End Sub
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复制代码
# j7 H) }& g: Q4 j1 o( h还有其它方法也可以近似地画出这个图来,欢迎大家进一步讨论. |
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