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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
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>> x=60*(cos(t)+t*sin(t))2 H& R$ O) c* V) `8 A0 `* D& q' g# O9 i/ ?7 E' W
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x =/ I+ U, p* y: l; [( t) |$ ^
# F9 A. x5 F8 x2 h# S60*cos(t) + 60*t*sin(t)! a" e8 H( F% n
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* e7 t1 L# w4 u; f$ H7 M>> y=60*(sin(t)-t*cos(t))2 y8 A& V# z9 V! p# s& `$ u
6 K1 e5 {& S* N- |; R2 b1 t5 yy =
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60*sin(t) - 60*t*cos(t)
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& @6 T4 K3 x3 t# ^) J>> x1=-x*cos(11/6*q)-y*sin(11/6*q)+110*cos(q)
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9 a- [. \8 I$ N: m8 Hx1 =% u, w- Q! A" I) D, G" ~. X
( M9 Q; X+ {- O, [. w& @0 M- @3 K3 @# W! \, [
0 F( ]5 C& T1 @2 a110*cos(q) - cos((11*q)/6)*(60*cos(t) + 60*t*sin(t)) - sin((11*q)/6)*(60*sin(t) - 60*t*cos(t))8 T) K+ l( [: \- V( Q9 `' g$ E \9 U7 U
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# `4 p/ V& M7 D5 U; |; |>> y1=-x*sin(11/6*q)+y*cos(11/6*q)+110*sin(q)- G3 q* t" b0 A( q1 s7 Z
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8 w: O i5 i- B! D* s3 Yy1 =' @9 _; f% k- o6 a: N& g+ D) }
% M& ~. W; Q# d: r. [
110*sin(q) + cos((11*q)/6)*(60*sin(t) - 60*t*cos(t)) - sin((11*q)/6)*(60*cos(t) + 60*t*sin(t))
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>> diff(x1,t)- @2 a6 S% ^5 G7 b' N8 D
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ans =
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- 60*t*cos((11*q)/6)*cos(t) - 60*t*sin((11*q)/6)*sin(t)" v9 m# E+ O* ?: k6 f7 P: Z- X! V C5 |# J2 a, u
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0 }- u+ A3 }7 E0 a9 M>> diff(y1,t)+ N! `! Q( i3 @% n9 o- D
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4 Z; w% V3 j$ Sans =7 W6 `+ M2 x- F( R" s
$ k: K4 R5 @0 U: x, l& }# y. ^; B1 Z9 M" u% p! a' A* o# q% d* H, F) v! X5 l! ~7 K
60*t*cos((11*q)/6)*sin(t) - 60*t*sin((11*q)/6)*cos(t)6 g8 N: I6 _$ l( a* k7 `/ m
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>> diff(x1,q)3 F5 L: n% Y! m4 a
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. [2 d# O; r9 p0 s% |" [1 lans =
0 ^3 `0 ?) N# I0 V1 w! g
; X) e) y8 d! i* x l6 ]5 s$ V% K6 s(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)
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>> diff(y1,q)3 J5 s6 T9 l" G$ S! f! T; j9 e
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+ r1 j9 [0 s9 _3 }ans = ^1 m" ^7 `3 ?
7 Y8 A+ G8 N+ v5 v$ s* a
g% k( u" Z, V0 }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+ c6 t% S" Q4 b/ R# X4 r
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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))')3 \1 k& w3 ?2 K* U' Z) Y
5 J) L5 h: l: T- N8 Vf1 =
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6 Q4 ?- b9 F) e/ `, n( R. Q& D(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)6 N: d: l' s4 d" V. Q
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>> simplify(f1)) [ m8 o C& N b2 a/ @* y4 ^7 U0 H" X" ?, K1 {
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ans =4 i' A% H% c% V; u# N# ~4 ?6 G! @# \ J9 W( l1 u
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-6600*t*(cos((5*q)/6 - t) - 1)+ K. J5 p; ?8 J$ v* t+ r2 c
8 {( y, X% H2 f# H. B& U5 o) O* s4 t/ t7 M) V0 y+ D4 V9 `+ o
-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
' a2 b# N9 @5 o+ E6 {& V$ { y1=-x*sin(11/6*q)+y*cos(11/6*q)+110*sin(q)
) F$ \* ? x0 Z8 m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))
; F1 N+ ~* ]% g3 R$ \从方程上可以看出小齿轮的方程仍为渐开线5 N. Y' \- r( |' R" j5 x+ f
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