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設(shè)一對中心距為110的齒輪 齒數(shù)比為6:5 則大齒輪的參數(shù)方程為:x=60*(cos(t)+t*sin(t)) y=60*(sin(t)-t*cos(t)) 根據(jù)嚙合原理 利用matlab編程 4 ]* ?: a6 m! M! e; F" q
> syms t q/ S% z0 `8 {$ h3 m8 @( R! l
>> x=60*(cos(t)+t*sin(t))2 H& R$ O) c* V) `8 A0 `
8 L. ?( ]9 S+ S* l0 ?6 P% X, d" a- \
8 a5 [% \1 z5 W& C Sx =
0 Y( s. b+ N3 F; Z0 {
. x. B6 `' a( Q60*cos(t) + 60*t*sin(t)! a" e8 H( F% n, R& Q! i; K% G b( x0 T7 A
/ C3 z- [+ A, _( T2 s3 }% b6 ~7 w) M# e7 r2 H, d- a
>> y=60*(sin(t)-t*cos(t))0 W5 y4 i: X7 b1 u( @
7 h d7 ^9 L/ j0 O! V$ Py =4 [0 q! i7 U; D) I4 W9 d- G$ c" p
7 r1 a' t# V. {( H7 M
& x) T: ~( X6 E% z0 b1 G60*sin(t) - 60*t*cos(t)
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>> x1=-x*cos(11/6*q)-y*sin(11/6*q)+110*cos(q)
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) ^, W* `$ L5 S' L$ Gx1 =% u, w- Q! A" I) D, G" ~. X s. L* \5 ]4 \+ b
& @0 M- @3 K3 @# W! \, [
q1 x6 z5 |5 v0 m! O1 ~5 l* j" \110*cos(q) - cos((11*q)/6)*(60*cos(t) + 60*t*sin(t)) - sin((11*q)/6)*(60*sin(t) - 60*t*cos(t))+ K' s. w; I8 s# f, Y! _
' W* I* o" i: E. Y; J! a. T6 o: x
) U2 R" a" `6 g( G+ t! ^>> y1=-x*sin(11/6*q)+y*cos(11/6*q)+110*sin(q)- G3 q* t" b0 A( q1 s7 Z
- V V: b1 L* z! v3 s9 y1 o2 U+ O4 d! C+ [) ?" t4 z+ J1 }5 U \2 Y {
y1 =
# J {# q; K" B' k2 H2 ]
5 _8 C; \+ O+ G6 L( @2 Q* x110*sin(q) + cos((11*q)/6)*(60*sin(t) - 60*t*cos(t)) - sin((11*q)/6)*(60*cos(t) + 60*t*sin(t))
. p" F; I5 a, f" p1 W) Y
! L- P0 \/ ^, f% m>> diff(x1,t)
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ans =4 |! Y: V1 E" m! I0 o
, O3 p- Y5 F) A3 Z Y! d5 d- 60*t*cos((11*q)/6)*cos(t) - 60*t*sin((11*q)/6)*sin(t)" v9 m# E+ O* ?: k6 f7 P: Z7 R6 m0 W c0 x+ m) n) ~
" V+ y6 _ T! D& H6 i+ k/ I' U2 O$ w( \9 x! g8 M. m! C
>> diff(y1,t)+ N! `! Q( i3 @% n9 o- D" _ a0 o' O* [1 _; Q$ c o
* F3 P; E4 v4 J( Z; W3 ^0 Cans =7 W6 `+ M2 x- F( R" s# h/ T5 ]' C+ i+ Z
; B1 Z9 M" u% p! a' A* o
% c& e8 W+ q% O# g( J! m! X60*t*cos((11*q)/6)*sin(t) - 60*t*sin((11*q)/6)*cos(t)
! `6 S9 n! J/ v4 G% ]* l4 I; k. W: Q! ~
% M4 K& ~/ N4 S4 i; W _! v>> diff(x1,q)
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6 F# }% N$ ~( F: F) b$ Pans =
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(11*sin((11*q)/6)*(60*cos(t) + 60*t*sin(t)))/6 - (11*cos((11*q)/6)*(60*sin(t) - 60*t*cos(t)))/6 - 110*sin(q)
+ y# G5 g4 K- ]1 _& E4 _% M' N& ~2 ^' t9 C5 i
>> diff(y1,q)3 J5 s6 T9 l" G$ S! f! T; j9 e* k' x8 s0 Y) _
6 t" V6 ^- K A+ Q. n8 s& Y* M" j: v: f2 ?
ans =: K# D; m, V9 I3 f0 s# H
7 Y8 A+ G8 N+ v5 v$ s* a
, r6 e/ h: _# o110*cos(q) - (11*cos((11*q)/6)*(60*cos(t) + 60*t*sin(t)))/6 - (11*sin((11*q)/6)*(60*sin(t) - 60*t*cos(t)))/6
& R3 X, a1 C, M) X0 H, |: b9 p2 J+ k& K6 E/ r2 D3 h2 P9 z/ A' ]3 l K6 \/ p: B, K; t/ T! w* o7 i6 E
>> f1=sym('(110*cos(q) - (11*cos((11*q)/6)*(60*cos(t) + 60*t*sin(t)))/6 - (11*sin((11*q)/6)*(60*sin(t) - 60*t*cos(t)))/6)*(- 60*t*cos((11*q)/6)*cos(t) - 60*t*sin((11*q)/6)*sin(t))-((11*sin((11*q)/6)*(60*cos(t) + 60*t*sin(t)))/6 - (11*cos((11*q)/6)*(60*sin(t) - 60*t*cos(t)))/6 - 110*sin(q))*(60*t*cos((11*q)/6)*sin(t) - 60*t*sin((11*q)/6)*cos(t))')( D0 b5 `& L4 ]3 X+ ?7 Z
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f1 =/ h: c! R3 |+ e4 t" f0 f) A7 H+ m' z
1 o6 c" \) D/ z( H1 r; b- q3 f2 l# Z6 L/ }5 s- z/ m% F9 W2 H
(60*t*cos((11*q)/6)*cos(t) + 60*t*sin((11*q)/6)*sin(t))*((11*cos((11*q)/6)*(60*cos(t) + 60*t*sin(t)))/6 - 110*cos(q) + (11*sin((11*q)/6)*(60*sin(t) - 60*t*cos(t)))/6) + (60*t*cos((11*q)/6)*sin(t) - 60*t*sin((11*q)/6)*cos(t))*(110*sin(q) + (11*cos((11*q)/6)*(60*sin(t) - 60*t*cos(t)))/6 - (11*sin((11*q)/6)*(60*cos(t) + 60*t*sin(t)))/6)7 ?( V' ~& ^6 L
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>> simplify(f1)) [ m8 o C& N b2 a/ @
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ans =4 i' A% H% c% V; u# N# ~4 ?
7 S p7 W4 A" F% j3 D: `7 B5 t( N
; f {4 ~6 v3 p9 ]! T4 I-6600*t*(cos((5*q)/6 - t) - 1); [* ?' C7 p: E- t3 |
8 {( y, X% H2 f# H. B& U5 o) O0 [8 T/ _8 Z/ I @5 U* V( @5 _
-6600*t*(cos((5*q)/6 - t) - 1)=0 解得q=6/5*t 代入 x1=-x*cos(11/6*q)-y*sin(11/6*q)+110*cos(q) * q1 K: y3 z1 G3 m4 N- }% z2 A+ J
4 D$ T8 H5 f- H4 P% q' S7 B y1=-x*sin(11/6*q)+y*cos(11/6*q)+110*sin(q) ; J6 d" t/ ?0 [
化簡后得X1=50*(cos(1.2*t)+1.2*t*sin(1.2*t)) y=50*(sin(1.2*t)-1.2*t*cos(1.2*t))
/ h8 v+ {) S" x$ {8 k% P. |從方程上可以看出小齒輪的方程仍為漸開線# ]& ?) g" ^# o: h
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