2266998 发表于 2011-10-25 12:22 ' Q, N2 P8 R) l
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, ... ! H7 i) i9 V! r' d3 n' Q
方程这样的) }& |# h1 Z6 n2 M' @
xat = ((r - r * Sin(αn) - ha) * Tan(αn) - sn1) / Cos(β)5 X% w) z' W+ Y |, w- M
xbt = (-sn1 / 2 + (h - ha) * Tan(αn)) / Cos(β)
4 m; p0 ]! ^" ~, r9 m yat = r - r * Sin(αn) - ha& {- z% y; x: |+ ~/ z: H0 M, Y0 A
x1 = λa: ^ l9 f S0 W1 k* R2 j7 v3 N
y1 = yat + (λa - xat) * Atan(αn) * Cos(β)6 k1 k" p2 ^: p$ q K
ψ = (x1 + y1 * Atan(αn) * Cos(β)) / r11 B2 m" Z B0 p; }8 T5 K% [9 n
x2 = Cos(ψ) * x1 + Sin(ψ) * y1 + r1 * (Sin(ψ) - ψ * Cos(ψ))& f) ^+ j, D& D4 P7 o
y2 = -Sin(ψ) * x1 + Cos(ψ) * y1 + r1 * (Cos(ψ) + ψ * Sin(ψ))7 S# r! ]! U: F# Y- J9 o, u
8 V, A/ y2 I4 R, y) K4 u% D: e$ Z xet = xat - r * Cos(αn) / Cos(β)8 _! E N/ M; p: H! @/ o, e2 _: a
x3 = xet + r * Sin(λe) / Cos(β), g9 J$ ?% q' z; U7 ]
y3 = -ha + r - r * Cos(λe)
( ]) P- E+ ^& R: N7 K ψ1 = (x3 + y3 * Atan(αn) * Cos(β)) / r16 l3 D2 G: {& C. q' o/ W( c
x4 = Cos(ψ1) * x3 + Sin(ψ1) * y3 + r1 * (Sin(ψ1) - ψ1 * Cos(ψ1))4 _+ d. Y" \( Q# w! G5 C0 t! e
y4 = -Sin(ψ1) * x3 + Cos(ψ1) * y3 + r1 * (Cos(ψ1) + ψ1 * Sin(ψ1))1 ?$ Z' E+ a0 Z0 W) {
λa,λe是在一定的范围内,联立方程组怎么解啊0 y8 Z' o+ c5 m0 y9 @, _3 P% P
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