**General Order 95**

**Appendix C**

**Conductor Sags**

(c) Sags for unequal Spans, Level Supports and Normal Conditions

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,