44 BELL SYSTEM TECHNICAL JOURNAL 



found by deducting the moments of inertia of Ti and Ri about this axis from 

 the moment of inertia of the entire top segment about the same axis. 



D = 4.09375" Sin a = ^-^^ = 0.16625 



r 



hi = 1.33" a = 9° 34' 11.49" 



i Z>i = 0.665" a = 0.16702554 radians 



r = 4.00" 2 a = 19° 8' 22.98" 



^2 = 16.000000 Sin^ a = 0.0045949941 



(1/2 biY- = 0.442225 Sin 2 a = 0.32787285 



Pi^ = 15.557775 Cos a = 0.98608364 



pi = 3.9443345" Sin o Cos a = 0.16393640 



d (Computation /) = 3.7625" 



di = pi-\- (D-r) - d = 0.2756" 



r, = pi- l/2di = 3.8065" 



Area Ri = h\di = 0.3665 sq. ins. 



A 1 = 0.0494 sq. ins. [Area of Ti by Formula (2) ] 



.vi = 3.9666" [By Formula (3)] 



By Computation I, IT bb = 10.2654 

 ITiBB [Formula (1)] — 0.7777 



IRiBB= ^' + i?i;-i- = 5.3126 



' 6.0903 



IITcbb = 4.1751 

 The moment of inertia of the 2 Tc areas with respect to the axis through 

 their own centers of gravity is given by 



21 Tc = IITcbb - ITcz^ 



where 



2 Tc is the area of the two Tc portions of the top segment and is given by 



2Tc = A - Ui^Ri) 



in which .4 is the area of the entire top segment as shown in Computation 

 I; and 

 where, by the principle of moments, 



Tx — TiX\ ~ Riti 



z = 



2Tc 



in which Tx, TiXi and RiTi are the moments of the areas of T, Ti and Ri, 

 respectivel}^ about the axis BB. (Tx = Ax of Computation I.) 

 Thus 



2Tc = 0.2940 sq. ins. 

 z = 3.7680" 



