2266998 发表于 2011-10-25 12:22
$ ~9 f: O% N4 R% V0 }齿轮与滚刀位置定了,理论上的啮合点就定了,啮进、啮出就是确定的了,既然你有了共轭线,就应当有了交点, ...
" N* f4 u$ }6 g1 ]) H( {0 {* B方程这样的4 m" s3 K+ W. J2 H6 @
xat = ((r - r * Sin(αn) - ha) * Tan(αn) - sn1) / Cos(β)1 v2 Q# j7 w& w4 S
xbt = (-sn1 / 2 + (h - ha) * Tan(αn)) / Cos(β)
7 r$ J7 G+ Q2 N" y) f* B2 Q yat = r - r * Sin(αn) - ha* E) w+ Q& ~% \
x1 = λa
5 \ {) Y0 E+ L$ e0 E4 X y1 = yat + (λa - xat) * Atan(αn) * Cos(β)( M1 k# f! [, O- ~3 _% C
ψ = (x1 + y1 * Atan(αn) * Cos(β)) / r1
* U, z1 M* r' u: {0 M$ f x2 = Cos(ψ) * x1 + Sin(ψ) * y1 + r1 * (Sin(ψ) - ψ * Cos(ψ))
$ A. w2 x& y6 b) Y6 s y2 = -Sin(ψ) * x1 + Cos(ψ) * y1 + r1 * (Cos(ψ) + ψ * Sin(ψ))
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0 t$ t C' b; i+ [% R' @ xet = xat - r * Cos(αn) / Cos(β)2 D# c& s6 u& w% l8 C% g5 P, J
x3 = xet + r * Sin(λe) / Cos(β), o- i0 a/ y# `/ L: H
y3 = -ha + r - r * Cos(λe)' v# H- g/ L8 ?1 T" _ a
ψ1 = (x3 + y3 * Atan(αn) * Cos(β)) / r1
/ [5 h9 P8 x+ n U x4 = Cos(ψ1) * x3 + Sin(ψ1) * y3 + r1 * (Sin(ψ1) - ψ1 * Cos(ψ1))
4 u4 q7 u, }5 X9 Q& i/ C; W. ~5 W y4 = -Sin(ψ1) * x3 + Cos(ψ1) * y3 + r1 * (Cos(ψ1) + ψ1 * Sin(ψ1))
+ c7 ]5 z' `7 K/ J" J% G! H: hλa,λe是在一定的范围内,联立方程组怎么解啊
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