283 
1921.] Robinson—Cable Spans with Suspension Insulators. 
This is illustrated in the accompanying diagram. 
Anchor-- 
Post 
Anchor - — 
Posh 
Fig. 1.—Diagram illustrating movement of insulators. 
The movements of the insulators are small, and they generally will be 
small. The limit of swing is the length of the insulator-string, but if the 
movement approaches that length the tension in the spans cannot be 
considered constant owing to the weight of the insulator, and the above 
method would have to be modified. 
To find the Tension in remaining Spans when the Wire in one Span 
breaks .—This is best illustrated by considering the wire to break in the 
300 ft. span when the stretch is at a tension of 13,5001b. per square 
inch. The insulators are assumed to be vertical before the wire breaks. 
Consider the 750 ft. span : the insulator will swing to the right and decrease 
the span by an amount nearly equal to the length of the insulator-string— 
i.e., 2-71 ft. 
Let 2ax = — 2'70 (for a trial). Equation (6) becomes 
750 /P t — 13,500\ _ 1 . /747-3\ 2 3 91 2 
I\ 16 X lO® - ’ ~ \ 2 / (Tx P t 3 
750 ^ 2 3-91 2 
\~Y 6^<"13,500 2 ’ 
which reduces to 
P t 3 + P t 2 7*55 X 10 4 — 5-69 X 10 12 - 0, 
whence Pt = 8,2501b. per square inch. Now, the area of 7/12 wire is 
•0585 sq. in., so that the pull in the wire is ‘0585 X 8,250 = 4831b. The 
inclination of the insulator to the vertical is 
483 
tan_1 3l-5 = 86 
so that 2 A£c = 2"71 cos 86° = 2*70 (34'51b. being one-half of the weight 
of the insulator-string). 
For the tension in the remaining spans, summing all the equations 
obtained by substituting in equation (6), 
2x (P t - 13,500) „ , 2x 3 3-91 2 
-—-|-j = 2A# -f- 
16 X 10 6 -6 P t 2 
Now, %x z = 55-3 X 10 6 and 2* = 775 X = \ 
%ax = - 1-35. 
2x 3 x 3-91 2 
6 x 13,500 2 
55-3 X 10 6 
= 268, 
and 
775 
