# Conductor sags – comparing

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• Conductor sags - comparing

Sometimes the values you get for conductor sag don’t seem intuitive. You need to remember some of the maths involved.

One equation is that sag is inversely proportional to tension. $$H = \frac{mgL_s ^2}{ 8S}$$ where H is tension and S is sag.

Comparing 4B150 and 4B95, the 150 conductor is heavier and bigger.

You’d think therefore that 4B150 would sag more than 4B95. But because the breaking load (CBL) for 150 is greater than for 95, the 150 will sag less for the same stringing tension (assuming you use %CBL as the tension specifier) because sag is inversely proportional to tension.

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