图1
1 成形车刀截形设计的必要性
2 成形车刀截形设计新算法
- 棱形车刀
- 在△ACE中,∠CAE=90°-gf-af,∠AEC=90°+af,根据正弦定理有
CE = bi sin(90°-gf-af) sin(90°+af) - 化简为
CE= cos(gf+af) bi cosaf - 将上式代入(2)式得
Ti= cos(gf+af) bicosaf=bicos(gf+af) cosaf - 将式(4)代入上式,即得棱形车刀截形深度计算公式为
Ti=[(ri2-h2)½-a]cos(gf+af) (5) - 圆形车刀
- 如图2所示,在△ACO1中,根据余弦定理有
bi2+R2-Ri2 =cos(gf+af) 2Rbi - 从中可解出
Ri=[R2+bi2-2Rbicos(gf+af)]½ (6) - 在△ACE中,∠CAE=90°-gf-af,∠AEC=90°+af,根据正弦定理有
图2
图3
3 设计实例
- 计算棱形车刀各点廓深Ti
- 将h2、a、cos(gf+af)代入式(5)得计算公式为
Ti=[(ri2-0.30771)½-0.193454]×0.88295 - 棱形车刀各点廓深为
- T2=T1=[(3.01252-0.30771)½-0.193454]×0.88295=0.906mm
- T4=T3=[(4.96252-0.30771)½-0.193454]×0.88295=2.646mm
- T6=T5=[(6.9552-0.30771)½-0.193454]×0.88295=2.646mm
- 计算圆形车刀各点半径Ri
- 将h2、a、R、cos(gf+af)代入式(7)得计算公式为
bi=(ri2-0.30771)½-0.193454
Ri=(400+bi2-35.31790bi)½- 各中间变量bi为
- b2=b1=(3.01252-0.30771)½-0.193454=0.1265mm
- b4=b3=(4.96252-0.30771)½-0.193454=2.9969mm
- b6=b5=(6.9552-0.30771)½-0.193454=4.9983mm
- 圆形车刀各点半径为
- R2=R1=(400+1.06252-35.3179×1.0265)½=19.100mm
- R4=R3=(400+2.99692-35.3179×2.9969)½=17.411mm
- R6=R5=(400+4.99832-35.3179×4.9983)½=15.762mm
- 将h2、a、cos(gf+af)代入式(5)得计算公式为


