282 
The N.Z. Journal of Science and Technology. 
[Jan. 
As regards tension, the stretch behaves as if it were a span of length 2X, 
which may be called the equivalent span for the stretch. 
As an example of use of this formula consider a transmission-line of 
copper, No. 7/12 S.W.G. size, with suspension insulators of weight 691b.. 
and length 2 ft. 8 Jin., maximum stress in the wire'25,000 lb. per square 
inch under a wind load of 301b. per square foot at a minimum tempera¬ 
ture of 20° F., loading-ratio q = 2*24 8 = 3’91. Let one.stretch consist of 
spans of lengths 400 ft., 600 ft., 550 ft., 300 ft., 750 ft., in that order. The 
equivalent span is found to be 587 ft., as shown below:— 
895-3 x 10 6 
2,600 
587. 
2x = 2 a/^- 3 = = V 
22x 
(See table below, columns 1 and 2.) 
To get the tension when the wind drops we can use either the graphs 
in Mr. Parry’s pamphlet, using 587 ft. as the length of the span, or 
equation (10) may be used. Substituting in this equation we have 
L 3 
+ P t s 
293-5 2 x 2-24 2 x 391 2 X 16 X 10 6 
: 6 X 25,000 2 
293'5 2 X 3-91 2 X 16 X 10 6 
r* ^ 5 
25,000 
Pt 3 
+ P/ 3,200 — 3-51 X 10 12 = 0. 
Whence Pt = 14,200 lb. per square inch. 
The changes in lengths of the individual spans can be calculated 
from equation (6), and hence the movements of the insulators. From 
equation (6), neglecting the term 71 
6 P t 2 
we have 
Ax — 
X 2 § 2 
6 
x 
E 
( 11 ) 
Now, Ax and x refer to the half-spans. For convenience in workings 
translating into terms of the span by substituting 2 Ax — A a and 2x — a 
in the above equation and reducing, we have 
Aa = a 3 X L96 Xl0~ 9 — a X 6*75 X 10~ 4 . 
The values of 
A a calculated from this equation are as follows:— 
1 
Span. 
Span 3 . 
A a. 
Movement of Insulator. 
Feet. 
Feet. 
Inches. 
400 
64-0 x 10 6 
-0-145 
— 0-145 
1-7 
600 
216-0 X 10 6 
+ 0*019 
-0-126 
1-5 
550 
166 4 x 10 6 
- 0-045 
-0-171 
2-0 
300 
27-0 X 10 fl 
-0-150 
-0-321 
3-8 
750 
421-9 X 10 6 
+ 0-321 
o-ooo 
2 2x == 2,600 
S(2x) 3 = 895-3 X 10 6 
