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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.
9 @, `! X9 n+ H9 Y8 P. Z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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1 u% F! Q( {9 R0 P2 ^5 W# g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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8 e4 l' a/ |$ Q) G* C- Sub HT()
" J7 w4 K! @ j# Z) } - Dim Arc As AcadArc, C(2) As Double, A As Double, A1 As Double, A2 As Double, H As Double
# }! f& N4 C/ R2 f0 f - With ThisDrawing( t1 _8 c0 x1 j, ?
- Set Arc = .ModelSpace.AddArc(C, 1, 0, 0)
+ I9 d' n$ G/ h - A2 = ThisDrawing.Utility.AngleToReal(180, acDegrees) * 2
! c* f6 y! i4 H8 u, r5 P - Do
, k! x! C1 p u1 r% [ - A = (A1 + A2) / 2$ }9 d$ u+ c) T* Q
- Arc.StartAngle = (.Utility.AngleToReal(180, acDegrees) - A) / 2
! R# c; Y2 P' H! M" U. e8 I/ g - Arc.EndAngle = Arc.StartAngle + A
5 t9 T9 `+ `1 _6 w - Arc.ScaleEntity C, 888.888 / Arc.ArcLength
8 V3 ?2 g, M- c8 X1 q - H = Arc.Radius - Arc.StartPoint(1)
& t* ^5 t5 d; X# w - If H = 123.456 Or A = A1 Or A = A2 Then5 q. O, E$ F& U: d
- Exit Do
: u4 C: l t& F1 P, z# ? - ElseIf H > 123.456 Then. r# F! r* b9 n/ `
- A2 = A) _- ]$ C* @6 A) g
- Else9 s6 I" ]: m, x! ~4 I; }
- A1 = A) B! ^ [; N5 s( B5 t
- End If
: X! d# X7 m; A" K' v! i - Loop% o$ X. L v) }/ L8 \1 s) s
- .ModelSpace.AddLine Arc.StartPoint, Arc.EndPoint
/ @# L# }* o* j Y6 T" [ - End With
$ }$ R0 w; R2 N0 L - End Sub* A/ C4 f- u8 ?5 g: v$ K
复制代码 5 `' q3 l5 j3 B* g0 L
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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