When a crossing span and its adjoining spans are of different
lengths it is not possible to string the conductors so as to make both the
normal tension and the loaded tension balance in the several spans. This
condition should be met by selecting a sag for the longest span not less
than that shown in the accompanying curves, charts
1
to 6
, inclusive.
The sags for the other spans should then be determined as follows:
For each span multiply the sag for the longest span by the square of the
ratio of the length of the span under consideration to that of the longest
span. The total normal tension in each of the spans will then balance and
the total tension under loaded conditions will be slightly less in the short
spans than in the longest span.
Example
Assume -
A crossing span length of 250 feet-Heavy Loading District. Adjoining spans of 300 feet and 200 feet, respectively. Conductors No. 0 AWG copper, medium–hard– drawn, stranded, bare. Sag from curve on chart 4 , for a 300–foot span is 5.30 feet.
Making the sags in the other spans proportional to the squares of their length, the sag in the 250 foot span will be,
The sag in the 200–foot span will be,
