FAQs

FAQ Topics / Maximum Arrest Force

As a guide using either a lanyard or SRL system as designed what would be the theoretical maximum loading imposed on an anchor point assuming a 255lb man as a standard weight?

If normal use, the rated load per manufacturer’s specs or instructions.
It should be less than 1800 lbf for OSHA for SRL and 900 lbf for shock absorber lanyards per ANSI Z359.1. NOTE Nigel Ellis design limit is 900 lbf for all deceleration devices.

Also if falling at full extraction of SRL then the force may be enough to break the cable at 3000 lbf. If the lanyard or SRL wraps (e.g. on an angle iron) in a fall situation then the force usually is enough to cut the line especially with the newer tubular elastic lanyards and edges for SRL.

Load requirements must be calculated with regard to a particular application and a particular situation and what any particular result might mean. In other words, what is the impact on you and your organization regardless of the answer given. All factors are variable and should be designed on an engineering drawing with site-specific factors or an approved model for assumptions. For example: Two people on one anchorage – may or may not accumulate force. Distance to stop is critical and less than 3-6 inches can be fatal.

You might want to consider purchasing the EFSS software program where these theoretical issues can be tried out to your heart’s content. Chapter 7 in Introduction to Fall Protection, 3Rd Edition provides vital information in this regard.

Note: 255 lb plus 60 lbs tools exceeds the limits of most off the shelf fall systems; in other words the shock absorption may be used up, producing several thousand pounds of force.

How much arresting force is put on a person in a 4′ freefall when using a web lanyard without a shock absorber attached to a full body harness?

A simple formula for guidance is the following:
W x h = F x d Newton’s Conservation of Energy Rule

W is weight in pounds
h is height of fall from CoG in feet
F is force in pounds (lbf)
d is deceleration distance at point of impact

So if W = 200 lbs
and h = 4 ft
F = 800/d assume deceleration w/out S/A is 2″ or 2/12 ft
Then F = 4800 lbf dynamic force
I f d = 2 ft with a S/A then F = 400 lbf which is tolerable for a few millseconds

Other things like velocity calculate from 1/2 x m x v2 = m x g x h
Note: there are many more factors in force and distance measurements to be considered to track a test program.