Horizontal Lifeline Tension and Anchorage (OSHA 1926.502)

A horizontal lifeline MULTIPLIES the force it is asked to catch, and the multiplier is set by SAG.

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Details, formula, and sources

An 1,800 lb arrest at midspan of a 30 ft line sagging 1 ft puts over 13,500 lb into each anchor -- 7.5 times the arrest force -- which is why a cable strung between two roof-hatch handles is not a lifeline. Halve the sag and the tension roughly doubles, so the line pulled drum-tight is the dangerous one and the slack that looks sloppy is what keeps the anchors alive. It also costs clearance below, which is the counterweight. The engineered demand at OSHA's factor of two routinely exceeds the 5,000 lb per worker everyone quotes.

static midspan cable: T = W x sqrt((L/2)^2 + s^2) / (2 s); horizontal pull at each anchor H = W x L / (4 s). Anchorage demand = T x the safety factor, compared against the prescriptive 5,000 lb per employee attached; the larger governs. Inverse: the sag for a target tension is s = W (L/2) / sqrt(4 T^2 - W^2), which has no solution for a target at or below W/2.

Cable statics, with the regulatory requirements from OSHA 29 CFR 1926.502(d)(8), (d)(15), and (d)(16). A US federal regulation in the public domain, quoted directly.

29 CFR is published in full at no cost by OSHA and the eCFR.

Estimate. AHJ and licensed professional govern.

Field names used by the API: span_ft, sag_ft, arrest_force_lb, workers, safety_factor, anchorage_capacity_lb, target_tension_lb, cable_tension_lb, horizontal_pull_lb, tension_multiple, angle_deg, anchorage_demand_lb, sag_for_target_ft

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