In a skyline rigging operation with a load of 3000 LBS, using a 1/2 inch rope weighing 0.46 LBS per foot, a span of 50 feet, and a deflection of 12 feet, what is the skyline tension to the nearest pound?

• A. 3137 LBS
• B. 2600 LBS
• C. 3500 LBS
• D. 3000 LBS

Answer: (A)

The situation is governed by how the load, the span, and the sag (deflection) set the tension in a skyline, with the rope’s own weight adding a small extra load. A practical approximation used in skylining is to split the tension into a part that supports the load across the span and a small correction for the rope’s weight:

T ≈ (W × span) / (4 × deflection) + (w × span) / 2

where W is the load, span is the distance between supports, deflection is the sag, and w is the rope weight per unit length.

Plugging in the values:

• (W × span) / (4 × deflection) = (3000 × 50) / (4 × 12) = 150,000 / 48 = 3,125

• (w × span) / 2 = (0.46 × 50) / 2 = 23 / 2 = 11.5

Total tension ≈ 3,125 + 11.5 = 3,136.5 ≈ 3,137 pounds.

Thus, the skyline tension is about 3,137 pounds. The rope’s own weight adds a small correction to the main load-based term, which is why the value is just a bit above the load alone.