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
( Q2 n3 P% ]& x7 t! g; A4 ?number/mat1(12)" V6 ?: v4 U+ p9 P* O9 u  c3 v
data/d,6008 G5 w' l  V" t+ b) q. ?+ Y
back:, D  ^5 {( g6 m# ~' g
param/'输入方向盘平均直径','方向盘平均直径(mm)',d,rsp
- z' s3 N# |& `' d2 s2 Z; R( J( _jump/back:,cancel:,ok:,rsp* ~0 O# K  H8 G. \8 W1 ]; r1 l% [
ok:
+ Y! H5 n$ W( @% z  O$$尺寸判断* x/ Z3 M: j' t, R
ifthen/d<500 or d>1000: I" l. J: U6 q/ m; j* ?$ a/ w+ G1 W" n
messg/'尺寸超出范围（500-1000），请重新输入'  _9 I+ S: C1 a" P; f
jump/back:4 @$ n$ F" J# M; m9 |1 u) n
else
2 K$ p% a" j/ [0 Tjump/l1:0 @+ V  [8 p5 P- S' R% w9 d
endif
6 u% x! M+ n, Q% b. hl1:3 w0 ]! b! p. A+ Q. g
$$外圈
0 Y1 r0 h' c2 o5 @; c( ^$ _/ A/ kr=d/307 K  `* i* _8 a; N( m3 V4 u4 T
pt(1)=point/0,7*d/15
% k6 F" g8 E1 X+ _cr(1)=ellips/pt(1),r,1.5*r
& X- _3 ?# h* c( ^* rent(1)=solrev/cr(1),origin,0,0,0,atangl,360,axis,1,0,0' h8 o6 ?$ ^( L
$$旋转坐标
' f* o, T! j& C+ U. e$ O3 bln(1)=line/0,0,-1,0,0,1' b( r% U' n0 }: Q5 ?6 G
ln(2)=line/0,-1,0,0,1,04 ^/ p" J7 P+ S5 h, f
csys1=csys/ln(1),ln(2)3 o2 |9 E0 d9 z- f, s* E, Z
&wcs=csys1
8 p2 d* w4 D# z$$辐条( B1 c' d2 V' Z: x3 e
pt(2)=point/0,d/20,r: ?# ]! r+ H. J
pt(3)=point/7*d/30,d/10,r
5 w& ~8 b& \7 w! C" lpt(4)=point/7*d/15,d/10,r( A- X' W  R! }, _
cr(2)=circle/pt(4),pt(3),pt(2)% Q2 V) x8 E5 ?
pt(5)=point/0,0& R2 C3 N  d0 ]
ln(3)=line/pt(5),atangl,-60, s) i, g3 c& n  o2 _. M3 Q  M' h
mat1=matrix/mirror,ln(3)
5 e3 u/ @9 A, F: D, K! e5 gcr(3)=transf/mat1,cr(2)
+ J+ n' B) X1 _5 l; H  j4 |pt(6)=point/endof,xlarge,cr(3)* _2 h( F) z- O$ B2 g7 e
ln(4)=line/pt(2),pt(6)
6 _% A: e  ?" o, Y- x) Z/ dpt(7)=point/0,0,r( ]% I0 p. _" D+ |( s
obj(1)=fillet/cr(2),ln(4),center,pt(7),radius,50
, ], O) s0 ~+ t2 [( g; e5 Cobj(2)=fillet/ln(4),cr(3),center,pt(7),radius,50
, j! U  V+ U- Bpt(8)=point/pt(4),delta,0,-d/10,0; e! a( @5 g" F$ V9 K9 ^! M3 W
pt(9)=point/pt(7),delta,d*(cosf(60))/8,-d*(sinf(60))/8,05 N5 N) A% ~9 Q6 I
pt(10)=transf/mat1,pt(8)
: j: s# [! K3 E0 N. x2 V) P8 U, Gcr(4)=circle/pt(8),pt(9),pt(10)
" b# A8 Y' i8 O- Z% apt(11)=point/endof,xsmall,cr(3). Q( A4 |) Q& `% Q, ]& \
ln(5)=line/pt(11),pt(10)
8 P1 c4 g: u6 K4 l: fln(6)=line/pt(8),pt(4)
1 r) {2 z7 F" W$ \$ Aent(2)=solext/cr(2..4),obj(1..2),ln(4..6),height,2*r,axis,0,0,-1  H, _' t  d5 r2 P1 S
blend/ent(2),radius,10
* z+ v! z) ^7 V3 ]$$旋转坐标
% @- A0 g6 e% h6 \! J0 X7 f* bln(7)=line/pt(5),perpto,ln(3)
8 L  A3 Q" P: R5 ^, E0 c; Hcsys2=csys/ln(7),ln(3); D" L2 v/ e; O% H. u2 S# V
&wcs=csys2' w+ E3 I2 W/ J% }4 b
$$生成联轴部分+ `0 D0 d8 X/ ^5 {, y. b* o1 @
ent(3)=solcyl/origin,0,0,-1.3*r,height,2.6*r,diamtr,d/5,axis,0,0,18 s  g0 h2 j* A
blend/ent(3),radius,10
& J! L8 _& x& ]& |# `; X0 C# W+ Xcr(5)=circle/0,0,-1.3*r,d/150 O' ]! e/ H/ L" U
cr(6)=circle/0,0,-1.3*r,d/30
" d( U& M' a4 ]! Oent(4)=solext/cr(5..6),height,4*r,axis,0,0,-15 T( N1 k% G# _8 X
ent(5)=unite/ent(1),with,ent(2..4)
' Z& i. a/ M/ S) z/ X: G8 D+ {$$隐藏多余实体
) Z/ w6 |, S8 g) Xblank/cr(1..6),ln(1..7),pt(1..11)% v. l- W  @( w- P5 n9 t
$$上色
) P% i; N+ w5 |( m+ M&color(ent(5))=&gray3 W/ p9 p" b: Y& r& X
cancel:. \5 U& D; z+ Y/ k9 ~
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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