举例 ②等精度法: A 2 D 2/3 30×50=3873D2=√18×30=2324D3=√50×80=63.25 求出:T=0.068,T=0.058,T=0080 A1=40°(9 选4为协调环,设定41、A2并靠档标准公差 A2=20032(9) T=√Tb-72-72=012-0.0622-0052=00886 A1=40 +0.062 40.031±0.031 A2=20002=20.026±0026 A0=0.5 0.1 0.56±0.06 A=59497±00443=595:001 A3=40.031+20.026-056=59497举例： A1 A2 A3 A0 ② 等精度法： =  = m i i i i D T D T 1 2 3 3 0 D1 = 3050 = 38.73 D2 = 1830 = 23.24 D3 = 5080 = 63.25 求出： T1=0.068，T2=0.058，T3=0.080 选A3为协调环，设定A1、A2并靠档标准公差 40 ( 9) 0.062 1 0 A IT + = 20 ( 9) 0.052 2 0 A IT + = 0.12 0.062 0.052 0.0886 2 2 2 2 2 2 1 2 T3 = T0 −T −T = − − = 0.5 0.56 0.06 20 20.026 0.026 40 40.031 0.031 0.1 2 0 0 0.052 2 0 0.062 1 0 = =  = =  = =  + + + A A A A3 = 40.031+ 20.026−0.56 = 59.497 0.0413 3 5 0.0473 59.497 0.0443 59. +  A =  = −