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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.
7 ^: B; E" d g9 G- Y- m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.
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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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& P$ L, P( b6 P5 x0 e* _- Sub HT(): Y9 K4 L# I/ {) I1 J; a4 r
- Dim Arc As AcadArc, C(2) As Double, A As Double, A1 As Double, A2 As Double, H As Double! F. n' [. @0 A! a% H! O Y
- With ThisDrawing( [! i) S% _: ]* K: ~2 ~
- Set Arc = .ModelSpace.AddArc(C, 1, 0, 0)" U2 E7 p$ `; z1 L/ Q
- A2 = ThisDrawing.Utility.AngleToReal(180, acDegrees) * 2
$ A( Y8 n* ^; C- B/ G - Do
( t/ p( V# E' M - A = (A1 + A2) / 2
6 U6 u2 [4 J B# M. e0 E - Arc.StartAngle = (.Utility.AngleToReal(180, acDegrees) - A) / 2
' g6 G" K+ c) w# T: y4 E - Arc.EndAngle = Arc.StartAngle + A3 P3 G; m- g: M! P
- Arc.ScaleEntity C, 888.888 / Arc.ArcLength
9 u+ B3 b4 L0 O3 w! |3 `, u' P - H = Arc.Radius - Arc.StartPoint(1)$ W/ K' j0 D$ ?. y2 Y5 R( ^
- If H = 123.456 Or A = A1 Or A = A2 Then- ]# I# I9 B: |8 o1 u
- Exit Do
2 J. T( J6 i3 U: N, C- {0 v - ElseIf H > 123.456 Then1 n* }# ]" [. M, T8 v/ J
- A2 = A
! w# E; z! v8 ^3 a9 r - Else
: S* x' X8 o& P" ~, D* M - A1 = A
9 X7 R+ O c; A1 [6 Y/ C - End If
+ Z7 u7 Z! u: h& _ - Loop
0 @, Q( Z& C5 p; N6 j6 ?9 C. a4 | - .ModelSpace.AddLine Arc.StartPoint, Arc.EndPoint5 O' v0 m. J' o! ~+ b" t
- End With
9 g# a- w* g+ \1 Q6 y, s% |; F - End Sub' i- y2 K- E; y# @* C
复制代码
3 N6 h! B9 l% q0 E2 x7 E+ Q- D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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