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Personal fall arrest distances

Used on the ASP, CSP and CHST. The worked example below is computed by code and checked against the source.

Free fall ≤ 6 ft

Deceleration distance ≤ 3.5 ft

Limits for personal fall arrest systems in construction, 29 CFR 1926.502(d)(16).

Worked example

The longest fall the two limits allow before the worker stops.

6 ft + 3.5 ft = 9.5 ft

Clearance below must also cover the worker's height below the D-ring, harness stretch and a margin. 1926.502 doesn't set those as numbers; use the equipment maker's figures.

Where it appears on the exam

Source

The formula comes from

29 CFR 1926.502(d)(16)(iii)-(iv)(opens ecfr.gov in a new tab)
“(iii) be rigged such that an employee can neither free fall more than 6 feet (1.8 m), nor contact any lower level; (iv) bring an employee to a complete stop and limit maximum deceleration distance an employee travels to 3.5 feet (1.07 m); and,”

Questions

What is the formula for personal fall arrest distances?
Free fall ≤ 6 ft; Deceleration distance ≤ 3.5 ft. Limits for personal fall arrest systems in construction, 29 CFR 1926.502(d)(16).
Will this formula be given on the BCSP exam?
BCSP says the formula is included in the question whenever one is needed, unless the question is testing whether you know the formula. You can't bring a formula sheet into the exam, so learn what each formula does and when to use it.

BCSP on formulas at the exam

BCSP, What to Expect When Sitting for a BCSP Exam (item 7)(opens bcsp.org in a new tab)
“For any question requiring a mathematical formula, the formula will be included within the exam question (unless the question is specifically testing your knowledge of a formula). You will not be provided or permitted to bring a formula sheet.”

Related

Checked 2026-10-16. Spot a problem? Email hello@passcited.com.

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