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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
' N8 J8 e1 F' q" t9 C$ f, O7 {5 A 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
5 E* X- C6 U: `+ _% c$ Z1 Y" _Sub jkx()% a3 P( ]) U# ~
Rem 绘制渐开线,7 H1 ?+ \, O9 w/ l
Dim d As Double '节圆直径5 K P* L% `4 Q6 X$ {0 t
Dim r As Double '节圆半径$ f; Q k- ^: L6 j+ _
Dim A As Double '总展开角度. y0 w& i. L/ k6 Q
Dim Ai As Double '展开角度3 `" ]2 \6 t4 l; M
Dim Li As Double '展开弧长
9 k4 {, F. p; B& \ ~4 I0 Td = 1003 r3 p; L8 u. B6 P$ k1 H6 E
A = 360
) I. p8 O# B3 F4 ~1 H n, F% vr = d / 2# b& ]# ?. k2 ]+ {3 h# T7 i1 \
Dim Pnt1(2) As Double' c- P/ n$ ]& p J. d" X3 A
Dim Pnt2(2) As Double2 o3 T4 M$ w, s+ Q( U" L2 ^
Dim PntLst() As Double, N As Integer; f7 a( Z. M U! \; K- ?- D
ThisDrawing.ModelSpace.AddCircle Pnt1, r
* Q% i( E5 Y2 I/ lFor Ai = 0 To A * Atn(1) / 45# Step Atn(1) / 45#
1 a. }; j2 O2 {* Y! _Li = r * Ai( m: @% z* b1 E8 j1 o* H+ R+ n
Pnt1(0) = r * Sin(Ai)
% ~8 W; T: {& j: z. Q' P; a! KPnt1(1) = r * Cos(Ai)
. V E2 T8 ^1 d; i& APnt2(0) = Pnt1(0) - Li * Cos(-Ai)/ H" [8 X) ^# M- K
Pnt2(1) = Pnt1(1) - Li * Sin(-Ai)1 G" u, Y* \ C$ _$ E* K
ThisDrawing.ModelSpace.AddLine Pnt1, Pnt2% K- {+ Z) G3 O2 K
N = N + 1
- C% `0 x2 e3 o- @ReDim Preserve PntLst(N * 2 - 1)& W. L, e7 K8 i# |! s" T/ N
PntLst(N * 2 - 2) = Pnt2(0)
" {$ b* q7 G- r! \$ z# z4 YPntLst(N * 2 - 1) = Pnt2(1)
9 L7 |( e) A" x: K7 R! bNext
1 o( f n/ _$ [' X5 DIf N > 1 Then8 M2 q; B5 Q; Y% W
ThisDrawing.ModelSpace.AddLightWeightPolyline PntLst
. C& D& n3 L) K, |/ K: IEnd If
" w" X$ R6 N6 s; X8 A! o PEnd Sub |
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