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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.6 h$ |9 M7 k5 q: c/ {6 j' c
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.; P; V% x+ r2 `( \5 z. r
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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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$ y3 a" E6 Y0 M: L8 l8 }9 _; ~- Sub HT()# x" M$ C8 C# Z( v8 E
- Dim Arc As AcadArc, C(2) As Double, A As Double, A1 As Double, A2 As Double, H As Double1 a: ~$ l; S8 {2 \
- With ThisDrawing
/ h( A: ]: s3 J2 H) r% C) ` - Set Arc = .ModelSpace.AddArc(C, 1, 0, 0)
$ T% K, Q" l) S - A2 = ThisDrawing.Utility.AngleToReal(180, acDegrees) * 2, c, p" D. F* w/ A
- Do9 K; a6 s' R- d' L/ I' N, ^
- A = (A1 + A2) / 2
?2 d( u+ Y5 F - Arc.StartAngle = (.Utility.AngleToReal(180, acDegrees) - A) / 2
# _! _# r) t5 T, Z0 m" p) l: P - Arc.EndAngle = Arc.StartAngle + A8 I# R7 i% x$ g2 S
- Arc.ScaleEntity C, 888.888 / Arc.ArcLength
! Z7 k) g" \; }( G9 q/ k - H = Arc.Radius - Arc.StartPoint(1); t4 Z1 m' R+ g N" ]8 a- j2 k
- If H = 123.456 Or A = A1 Or A = A2 Then
' O- P; R( N& Z" Z# P: ^- i - Exit Do' ?6 s6 F+ G* @4 t# x' H0 R
- ElseIf H > 123.456 Then' R$ F0 J$ X! q4 h- @! ?' z, G
- A2 = A
9 J' t! X, c3 K- h+ V - Else
* b0 ~+ ?2 T7 F) ?4 j; J2 N - A1 = A
3 F0 {: D7 X# x1 S7 ? - End If" \6 E2 R3 Z+ |+ X
- Loop5 f, K. A8 u8 r ^/ `+ y$ b7 c K
- .ModelSpace.AddLine Arc.StartPoint, Arc.EndPoint6 w' f |3 |$ y- s( m: `
- End With
9 D( ?2 s" q" h: p; @& | - End Sub' S9 F: _4 C3 A1 L1 S: y3 ~$ X" e
复制代码 5 H& K# J, D% x; B
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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