So I am trying to solve a problem, I’ve done a simple solution but I want a more comprehensive answer and don’t know how to do the math. The problem involves climbing anchors and the difference between max loads of an equalized vs non equalized anchor. The two different situations, I’ll try to keep as simple as I can and still represent likely real life scenarios: 1. An anchor with 2 points perfectly equalized. 2. An anchor with 2 points but .5m of fall between anchor points, meaning that the load of a fall comes down on one anchor point until that point breaks, then there will be a .5 meter fall before the load comes down on the second anchor point. What I’m trying to figure out is how much more total force can be expected from #2 vs #1 in a fall that generates 15kn of maximum load in #1 I did the basic calculations (f=ma) and the additional force is negligible: I assumed a 100kg weight falling an arbitrary distance to generate a force of 15kn. Each anchor point breaks at 10kn. In #1 the anchor doesn’t break. In #2 the first anchor point breaks at 10kn (dissipating the force as heat in the components) and the second anchor point experienced a load of 5kn plus the force of 100kg falling .5 meters (~100n) For 5.1kn force, and the second anchor point does not break. Where I get lost is the fact that the rope stretches (and recoiles) and the disparity that that causes, which I do not understand. According to an online force calculator for climbing rope falls (That presumably factors in rope stretch) http://ferforge.tripod.com/Srt002.htm The force generated by a .5m fall is .5.6kn to 11.25kn depending on the length of rope Resulting in a force of at least 10.6kn and the second anchor point breaks. This doesn’t apply to the problem precisely because, I assume, it is calculating an acceleration from 0m/s to 1m/s with the rope unstretched. I would expect that there will be no significant additional load but I don’t know how to work it out, and the online calculator seems to suggest otherwise, I’m stuck here. I don’t understand where all the extra force can come from.