临界转差率sn和最大电磁转矩Tm ≈+ 2+(x1+x2) X,+x m,pU ±mDC 4/++√x2+(x+x2)2]4(x+x2) 过载能力兄r= 75 临界转差率 sm 和最大电磁转矩Tm N m T T T = ' 1 2 ' 2 ' 2 1 2 2 1 ' 2 ( ) x x r r x x r s m + + + = 4 ( ) 4 [ ( ) ] ' 1 1 2 2 1 1 ' 2 1 2 2 1 1 1 2 1 1 f x x m pU f r r x x m pU Tm + + + + = 过载能力