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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.
& x- i3 g9 y( k2 Z u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.; }+ Q0 ?; o; k
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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
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- Sub HT()
, U/ O0 ~# z3 ? - Dim Arc As AcadArc, C(2) As Double, A As Double, A1 As Double, A2 As Double, H As Double% i2 n. p$ P4 p3 r) j0 y" u: E
- With ThisDrawing
2 K- @0 f: Q, [: w - Set Arc = .ModelSpace.AddArc(C, 1, 0, 0): @( {# p/ E7 H9 `6 i9 l K# l" } O
- A2 = ThisDrawing.Utility.AngleToReal(180, acDegrees) * 20 ~7 F. ]7 F$ i+ o
- Do. x) B3 u- Z7 s2 m) p
- A = (A1 + A2) / 2
, p. p: ]' k& E) ^4 O6 ^( L, A5 n - Arc.StartAngle = (.Utility.AngleToReal(180, acDegrees) - A) / 2
1 [1 @( U/ K6 `3 k$ d" ^ - Arc.EndAngle = Arc.StartAngle + A0 ?- H, Q) } @% r V
- Arc.ScaleEntity C, 888.888 / Arc.ArcLength5 G- i+ f3 r* Z h- c! u: {. y
- H = Arc.Radius - Arc.StartPoint(1)
8 i1 [; k, F" ^& `+ J" Z - If H = 123.456 Or A = A1 Or A = A2 Then
$ d0 z$ B' W4 O- {4 p5 I - Exit Do
$ B' O2 S5 ~! V$ n& F& `+ P0 ? - ElseIf H > 123.456 Then b% w6 i& m7 u1 M0 D/ P9 K
- A2 = A5 E9 ^5 x: @4 h% ?9 R4 y
- Else
# i6 t0 x W# p+ U - A1 = A
* B/ k/ F( y4 [) s/ a$ ]0 ~% k - End If0 o' k! t) `$ _5 c! a: O
- Loop7 p0 Z# ]/ I$ i
- .ModelSpace.AddLine Arc.StartPoint, Arc.EndPoint2 _' k# _( b" E0 U9 N' s
- End With
9 o3 D# Q1 o+ w1 s q( y - End Sub
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复制代码 , M2 C" D t" g% k* q+ k
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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