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UID396100
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注册时间2007-10-17
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发表于 2011-12-24 13:07:17
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来自: 中国山东青岛
工具 宏 打开VBA管理器
1 N( d2 v$ T( T( h* x9 ^ 插入模块 复制如下 运行 即得渐开线 有了渐开线 齿轮不就出来了吗9 j" J1 N8 `' f2 k6 `
Sub jkx()! }4 G9 g, k* T( Q
Rem 绘制渐开线,1 }, o5 b' F1 C
Dim d As Double '节圆直径. l7 t6 e' S, x% o# f* Q/ f5 l7 p
Dim r As Double '节圆半径9 w. T6 r0 T; r: R. G6 y
Dim A As Double '总展开角度9 Z0 ?; P' U" b) r3 u5 O& \
Dim Ai As Double '展开角度
4 q7 X. p7 H" j7 p" U R% s% S% Y. u* QDim Li As Double '展开弧长
1 v. R7 z- X/ k! N8 x% Kd = 100. m2 {' I$ R& g1 c: H
A = 360
2 ~7 Y) y" q% o- c, R4 M. qr = d / 2
9 } R/ x* B( {8 p: J" s9 vDim Pnt1(2) As Double
7 q' ~ ]$ X: O7 n( r# `Dim Pnt2(2) As Double, d3 H: c6 n) ]2 k4 U
Dim PntLst() As Double, N As Integer
9 w2 E, V, b9 K- J+ EThisDrawing.ModelSpace.AddCircle Pnt1, r" n- s' {+ { W/ o: x# O8 i
For Ai = 0 To A * Atn(1) / 45# Step Atn(1) / 45#
8 e8 `/ h3 C6 b! G- {Li = r * Ai
" }/ z; R0 R8 G6 J1 R0 N- U' IPnt1(0) = r * Sin(Ai)
( s- j5 j, R! |, e+ m* W+ c/ A& S/ V8 hPnt1(1) = r * Cos(Ai)
- s- ]' L, N* ^9 H# w0 d1 XPnt2(0) = Pnt1(0) - Li * Cos(-Ai)
1 e9 D4 e4 a& APnt2(1) = Pnt1(1) - Li * Sin(-Ai)
0 A$ I1 X C- a4 I) _! ~ThisDrawing.ModelSpace.AddLine Pnt1, Pnt25 q- T0 v ^5 O
N = N + 1
" T& b: d+ _* I+ E' O# pReDim Preserve PntLst(N * 2 - 1)1 ^+ B' C( m% u! g6 E
PntLst(N * 2 - 2) = Pnt2(0)! d, g3 k" e5 _1 E) l1 z/ e% z( J* \
PntLst(N * 2 - 1) = Pnt2(1)
8 c5 k& c9 P/ E+ [/ W) zNext
0 H' A) \3 J4 p& y* [8 D" x0 S PIf N > 1 Then
0 C+ q( b2 u1 p/ ^: c- ?- \+ WThisDrawing.ModelSpace.AddLightWeightPolyline PntLst7 A: c+ S2 }6 g) K
End If3 M2 G$ P& M, \, C* o- ?) y
End Sub |
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