UG GRIP建模源代码实例_方向盘

xcb · 发表于 2014-3-24 21:30

原帖地址：http://www.ugufun.com/?p=128

fangxiangpan.rar (1.27 KB, 下载次数: 57)

这是个简单的建模实例，用了回转体、拉伸、以及布尔运算。

entity/ln(100),cr(100),pt(100),obj(100),ent(10),csys1,csys2+ x+ c+ ]/ t  ~4 F, q9 }
number/mat1(12): g: O/ s( T; e! `% W# _: r
data/d,600
7 q4 p1 M/ a$ C, J. hback:) R3 [; x4 @, b) s3 H; Z
param/'输入方向盘平均直径','方向盘平均直径(mm)',d,rsp
. [7 ~) a/ ~0 fjump/back:,cancel:,ok:,rsp
% y! D% D2 i( _6 I) G/ ]ok:+ _2 `3 q& p3 h; Q) L& `
$$尺寸判断7 H# w! a& Z* ^) y0 ~
ifthen/d<500 or d>1000
! \! }1 k( M9 Y  W2 Z+ w1 gmessg/'尺寸超出范围（500-1000），请重新输入'4 Q8 e# H4 G4 ^
jump/back:9 V, h2 {/ F( K7 U
else
# T& K. b0 I( c/ F& j) Qjump/l1:% V3 ?0 D$ c6 E3 v0 E
endif
% q3 G& b' b- l: P+ g9 h" i( Xl1:8 P8 A1 M" V' P! \
$$外圈5 ?9 [. V8 z' n& o/ h% I8 V6 w  N
r=d/30; H$ o: ~* S' q4 j2 C; V  C( L  a
pt(1)=point/0,7*d/15
& z8 v& ^7 E4 d2 _1 o1 D# X' Jcr(1)=ellips/pt(1),r,1.5*r
3 _1 `! A9 A4 b; r$ i8 Oent(1)=solrev/cr(1),origin,0,0,0,atangl,360,axis,1,0,0
& m( f) `, e' z1 M* X$$旋转坐标
8 @5 @$ C" Y7 oln(1)=line/0,0,-1,0,0,1
+ c- Z5 t$ b% V# G. Rln(2)=line/0,-1,0,0,1,0
; o- z- M- A/ y+ I3 u5 X) x7 K* lcsys1=csys/ln(1),ln(2): W& |- n3 U! f3 I9 b
&wcs=csys1! U* i. C) s1 c' K6 ?7 j- Q
$$辐条/ B( J; u4 R$ B% {; F+ k) _
pt(2)=point/0,d/20,r( N7 B5 h/ L7 P* S5 A
pt(3)=point/7*d/30,d/10,r8 O) V5 c, m% g# Q! E" [
pt(4)=point/7*d/15,d/10,r
! X3 T+ O9 {: r# K7 Kcr(2)=circle/pt(4),pt(3),pt(2)8 h2 H" t0 t0 s, _4 T4 _
pt(5)=point/0,0% _# H  ^4 X/ D* X
ln(3)=line/pt(5),atangl,-60
4 h4 h: @, x  c6 @& J0 Smat1=matrix/mirror,ln(3)
  c1 h- E% t1 k: t: e% ?- qcr(3)=transf/mat1,cr(2)
: X2 R, g2 X/ A6 Rpt(6)=point/endof,xlarge,cr(3)
3 l) m4 V: V4 I/ U2 Uln(4)=line/pt(2),pt(6)# c) ?# T6 t) b# W
pt(7)=point/0,0,r8 g% ~/ H- {9 L" m& @3 a. e
obj(1)=fillet/cr(2),ln(4),center,pt(7),radius,50
+ l$ `1 C# P5 {: ?obj(2)=fillet/ln(4),cr(3),center,pt(7),radius,50) X' x& `, y0 j' j" w% L
pt(8)=point/pt(4),delta,0,-d/10,0
* R4 B  r& ~* Hpt(9)=point/pt(7),delta,d*(cosf(60))/8,-d*(sinf(60))/8,0
2 s- R) \8 F& x2 G5 Qpt(10)=transf/mat1,pt(8)" J  C) p8 G' K: S0 k
cr(4)=circle/pt(8),pt(9),pt(10)4 C$ z7 L4 L. K: C% S1 C" C! I, X
pt(11)=point/endof,xsmall,cr(3)
1 v+ N8 P( B" F! @/ jln(5)=line/pt(11),pt(10)
3 x. O2 Q- R/ p% tln(6)=line/pt(8),pt(4)7 c6 ~% ]8 p, P
ent(2)=solext/cr(2..4),obj(1..2),ln(4..6),height,2*r,axis,0,0,-1
# \" ^% Y+ B  I; Dblend/ent(2),radius,10/ c* N5 }, Q. M
$$旋转坐标
; P1 F# B% B/ b2 o& Z* q: G5 kln(7)=line/pt(5),perpto,ln(3)0 _% g" ?2 I2 l) A7 p% z: m
csys2=csys/ln(7),ln(3)
/ d( h0 U! ^/ ~5 F' c&wcs=csys2
5 V8 |$ Q. O8 E$$生成联轴部分4 [. f! N7 k/ s3 B5 {1 c2 c: x
ent(3)=solcyl/origin,0,0,-1.3*r,height,2.6*r,diamtr,d/5,axis,0,0,1, z3 m2 Z: F3 z- s/ }4 u) G
blend/ent(3),radius,10- P( ^0 ^1 h4 g% g+ ^, N2 x9 R: ?
cr(5)=circle/0,0,-1.3*r,d/15
( p) i& m2 }; v, ~cr(6)=circle/0,0,-1.3*r,d/30: s% M2 R) S0 a# F2 \
ent(4)=solext/cr(5..6),height,4*r,axis,0,0,-1& b9 h$ r2 e. ]6 q  m% t( U' K
ent(5)=unite/ent(1),with,ent(2..4)  b0 U7 ^' ?/ i. U8 ?
$$隐藏多余实体& _. ~4 K: y) \$ K' z5 }8 z
blank/cr(1..6),ln(1..7),pt(1..11)$ e! |" V6 E+ ?: D2 r$ U* W9 _9 J
$$上色
8 B( L. R' {9 V/ p  z&color(ent(5))=&gray- C7 E# ?' Q6 B7 c
cancel:- y3 O  R$ W3 b: X2 }" H' v. y+ C) C
halt

努力活着 · 发表于 2014-9-25 22:41

大哥，厉害哦

pengfei2010 · 发表于 2017-1-22 19:10

好厉害学习了

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