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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./ C2 U1 Q" |# ^# t! e
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.2 b. [# k' T: M- `% B
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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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4 v8 ?3 e4 w3 Q/ [3 L4 f- F. [- Sub HT()
8 |8 h: M- Y( D0 B: m" I - Dim Arc As AcadArc, C(2) As Double, A As Double, A1 As Double, A2 As Double, H As Double
E: s& o* T& o3 r) { - With ThisDrawing
2 {& O9 T$ L" n- j0 H - Set Arc = .ModelSpace.AddArc(C, 1, 0, 0)
1 a* H, {3 k& v& Q. A. e/ T9 W - A2 = ThisDrawing.Utility.AngleToReal(180, acDegrees) * 2' t3 s* |, F% z: o: l
- Do
- x0 S7 G$ z4 D1 x) c - A = (A1 + A2) / 2- n8 L" R/ `( k
- Arc.StartAngle = (.Utility.AngleToReal(180, acDegrees) - A) / 2
; Q* O0 M# v0 _0 Y K - Arc.EndAngle = Arc.StartAngle + A
% l8 x7 y: q8 q+ S0 x - Arc.ScaleEntity C, 888.888 / Arc.ArcLength" j, K: Z* d; u! M
- H = Arc.Radius - Arc.StartPoint(1)' ~9 _- H" ^# d+ d V7 e
- If H = 123.456 Or A = A1 Or A = A2 Then L+ ^& A3 [, t6 b, \
- Exit Do' H) k$ R/ K( n: N2 o& f# f- U9 X
- ElseIf H > 123.456 Then
8 m. e6 R( K+ R" i4 D. H8 s4 V: ^ - A2 = A i: Y: \6 F$ ^' E* E3 I( E# Z, A) k
- Else3 j+ h; [( }, I. `9 P# n& _6 R+ O
- A1 = A5 p" ~8 C7 H; s* Y J
- End If5 Z: R+ ~' _0 r- Y2 d# [& ~& _( H
- Loop
& h5 P& c& z" P( n C) l) q$ ` - .ModelSpace.AddLine Arc.StartPoint, Arc.EndPoint
" y8 Y( J, B& y( {- t5 t1 _: g( K - End With" d% Y3 t, { I
- End Sub
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复制代码
, y4 \, |$ U0 k* b- P" R# T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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