Toprope ground falls and rope stretch

Reading about ground fall injuries on toprope inspired me to derive the "effective" fall height (as perceived by the ankles) as a function of fall height: h = H * [ 1  M/m * H/(e*L) * a/(a+1) ] , where:  h = effective fall height  H = fall height  m = climber mass  M = mass used for static elongation spec (80kg)  e = static elongation (e.g. 0.08)  L = total rope length involved (2x pitch length for TR)  a = TR carabiner friction multiplier (1.6) Top graphs are for belaying from the top (a = 1). For the bottom graphs ("TR") I modeled anchor carabiner friction by pretending the belayer half of the rope has a 1.6x higher spring constant. I treat the m*e product as a single parameter to reduce the number of graphs. This assumes no slack and no tension before the fall. EDIT: worstcase effective fall height is reduced by 23x when the rope's damping (internal friction) is taken into account. 

Is the Y axis supposed to be m/s? Edit: Nevermind I get it now I think. The "effective fall height" or the equivalent fall height with no protection that would produce the same speed at impact? Is that right? 

please could we have titles on the x and y axis thanks 

Wow..... never knew that something as simple as "put a bunch of weight on the rope before you climb, just to be safe..." could be so complicated. Did you take into account that a even longer fall can be achieved if the top rope extender is also a full length climbing rope?


climbing friend, you would quit it stop it with your mathematics, and instead simply lead the climb and not toprope your way into lifetime of mediocrity and crippling fear of safe fallings! if you have this much sexual frustration for posting equations tubes of internet, perhaps you may utilize the hangboard and spreadsheetz? 

This is scary as shit since at 95 kg I am literally off the charts. I'm sticking to lead climbing. Top roping is way too fucking dangerous 

I see somebody just bought a new Macbook and is tinkering with the Numbers app. 

Vaughn wrote:Edit: Nevermind I get it now I think. The "effective fall height" or the equivalent fall height with no protection that would produce the same speed at impact? Is that right? Yes  exactly. 

Top roping is so dangerous... 

We need a Weekend Whipper: Toprope cratering. 

What's a kg? 

kevin deweese wrote: A large beer container, usually made of aluminum. 

Do climbing ropes obey Hooke's law? 

kevin deweese wrote: It's the thing that defines what a pound is. 

Without having looked into the actual maths and equations, I know there's no way this is right. Take the very first, upper left, graph using the red curve as the example. Your plot tells me that when I'm at 5m of the possible 15m climbing height that my effective fall is 0. After that it goes negative, that's a hell of a way to get up a route, "Hey climbing friend, I'm beyond 5m now and the crux is tough so I'll just keep falling until I get to the top." Any good scientist/engineer is going to test boundaries and limits for sanity before posting results or scrutinizing any further. I doubt this should be a quadratic function (or maybe it is and your scaling/boundaries are just wrong). Beyond that, I'm afraid that most folks here aren't really going to give a shit about TR falls. Now, if you factor in rope elongation to ground fall potential after falling while clipping between waistheight and headheight then you've got yourself a discussion. 

physnchips wrote: An effective fall of 0 means that you reach the ground with zero velocity. A negative fall height means you end up above the ground. Why wouldn't this be a quadratic equation? All of the forces in question (friction, spring force, gravity) are polynomial in nature with a maximum order of 2. Also, the plot you're talking about is for a 30m climb with a top belay. Read what you're criticizing, maybe? 

physnchips wrote: You have misunderstood the graph. The x axis is the total fall height. Looking at the line you've chosen, the maximum rope stretch is 5 meters and the "equivalent fall height" is 0 meters since you would have come to a complete stop at this point. It's an interesting exercise if nothing else. I think something is missing though because I find it hard to believe that under any top roping scenario you could potentially fall 8+ meters (24+ feet), and that you could have an "equivalent" fall to almost 2.5 meters (~8ft). Edit: Other folks beat me to it. 

i think 24 feet is somewhat realistic for a TR fall on a really long slingshot TR (ie two ropes, 150+ feet). if static elongation is about 8% for a rope and there is about 300 feet out... i have seen this a few times (where somebody comes off at the bottom of a long TR pitch) and they can definitely drop a ways. the creek is a pretty typical place for it. 

physnchips wrote: Dude I think it's the 2nd time I see you fail maths in a topics... 

Vaughn wrote: So all this is presumed you hit the ground and his effective fall height is the equivalent speed you hit the ground if you fell without a rope? Why not just make the yaxis momentum, or even impact force? I thought it was net fall distance, which would also make sense as a number to report and would clearly show when you crater and when you don't (isn't this what all the "recent" discussions have been about). 

Franck Vee wrote: Where's the other time? 