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设一对中心距为110的齿轮 齿数比为6:5 则大齿轮的参数方程为:x=60*(cos(t)+t*sin(t)) y=60*(sin(t)-t*cos(t)) 根据啮合原理 利用matlab编程 6 K: C% P+ ?) t' l0 i( {
> syms t q
( W7 o2 s `, O q0 x>> x=60*(cos(t)+t*sin(t))2 H& R$ O) c* V) `8 A0 `7 R6 t5 [( G: m3 Q7 g5 }
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x =
. \$ T" u, S( n$ | C. ^4 \$ ~. G: w2 F6 O- Z
60*cos(t) + 60*t*sin(t)! a" e8 H( F% n
+ X* v9 Y' m! P. a+ J/ G/ C3 z- [+ A, _( T2 s3 }% b6 ~7 w' {1 q8 j' W2 V' j5 [
>> y=60*(sin(t)-t*cos(t)), X, ~# P% N; {" h5 R T5 |) C/ F
) ?5 x: v8 }" {y =
; H3 u; s4 |: |( P0 k7 r1 a' t# V. {( H7 M
" g' \; l1 z# t60*sin(t) - 60*t*cos(t)
& L" y) o5 R! k3 J
/ t4 \! V- ]7 ~! \0 z>> x1=-x*cos(11/6*q)-y*sin(11/6*q)+110*cos(q)
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( q5 M2 ] `+ d/ u. }x1 =% u, w- Q! A" I) D, G" ~. X4 L1 F- @% t T6 ?+ N. u5 n$ ]
& @0 M- @3 K3 @# W! \, [6 k ^# Q% K; t. U8 ?
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))
# R, p$ @6 O5 n: C# R( M% }& i' W* I* o" i: E. Y; J! a. T6 o: x" J) L8 m2 C" b% @' _
>> y1=-x*sin(11/6*q)+y*cos(11/6*q)+110*sin(q)- G3 q* t" b0 A( q1 s7 Z
6 |& U9 H6 \; U9 M" J1 a% G; \9 s9 y1 o2 U+ O4 d! C+ [) ?" t
4 y4 k; q4 s# ]) s( x7 V6 `y1 =
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* B& J! g) O5 K. F8 p- b3 |5 m" i110*sin(q) + cos((11*q)/6)*(60*sin(t) - 60*t*cos(t)) - sin((11*q)/6)*(60*cos(t) + 60*t*sin(t))1 N2 y" m) j3 ]: J) ~9 G
4 z" a' Z, e2 T# g" J- b>> diff(x1,t)
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ans =1 [; P) i: \4 a2 S: x
) ~% `* r# b" t2 e! K( T4 [- 60*t*cos((11*q)/6)*cos(t) - 60*t*sin((11*q)/6)*sin(t)" v9 m# E+ O* ?: k6 f7 P: Z# D( o7 ]) [; L% l3 O4 w1 W# \
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3 ^! I3 J; v/ s- q: s6 i* t>> diff(y1,t)+ N! `! Q( i3 @% n9 o- D
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ans =7 W6 `+ M2 x- F( R" s/ J* v& F( ~3 R1 `+ i) L8 w) ^
; B1 Z9 M" u% p! a' A* o
) p* D8 t& T; d( z L60*t*cos((11*q)/6)*sin(t) - 60*t*sin((11*q)/6)*cos(t)- g8 I" b* Y& \0 z1 P/ o
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>> diff(x1,q)
6 Z) J8 m9 Z1 }0 W# ?3 j! s. e# Y Z' B4 A: U- L$ L# X% o/ c$ e1 Y2 m4 n2 ]
# q+ h+ W* P: Vans =* d. Q6 v- a" S/ d$ T7 ~! M# w: Y. `# v
3 C/ E; C3 T# w4 U& B& u; c(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)
+ q! [6 I) C) |# J/ H6 |& `3 E% ]! }. B+ Y3 B
>> diff(y1,q)3 J5 s6 T9 l" G$ S! f! T; j9 e
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2 H! }3 U+ g6 u3 g. L4 g( eans =
+ z3 {0 ~( i( t ~% f7 Y8 A+ G8 N+ v5 v$ s* a* z7 N2 X) `# H; w* L
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)))/62 F+ G$ n5 ?& y9 j ~- ]% ?
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>> 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))')0 o7 E+ Y, W9 M" A
+ H4 ]1 m2 T& y% _4 Y# ]" M
f1 =5 M5 z4 L' @: Y- ?1 ^" x3 p7 `
1 o6 c" \) D/ z( H1 r; b- q3 f2 l# Z6 L( b7 K* R& c) w8 ?8 ?8 D+ r8 T* y
(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)% H# v1 h+ T5 T% b* K
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>> simplify(f1)) [ m8 o C& N b2 a/ @. f& A, ?( f3 B1 G" N( ^
8 a; ^, `7 Z* u' N' D
& Q1 [; g" v% q9 oans =4 i' A% H% c% V; u# N# ~4 ?; q2 H% O& B8 w& _) f4 l/ }
% s8 D% N) G, f/ h
-6600*t*(cos((5*q)/6 - t) - 1)! t. ?% c1 a- c5 ]6 |, m9 [4 G
8 {( y, X% H2 f# H. B& U5 o) O; U) s( E5 @- b5 V
-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& p% J; ^/ ~1 U; ?) D9 W4 Y# o
y1=-x*sin(11/6*q)+y*cos(11/6*q)+110*sin(q) 6 W! f' Y- w- }( r/ E& v
化简后得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))$ j, w3 S/ |: i1 F; \
从方程上可以看出小齿轮的方程仍为渐开线5 ^) k3 V, E3 Y9 B1 l+ E
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