In fact, the longer the fall, the softer it feels... Usually, when the climber asks for a "soft catch" in case of a fall, the belayer gives a bit of extra slack (or even jump) to make the fall smoother Another point: the bolts aren't usually placed in that zigzag pattern, the rope drag would make the climb impossible
Hey Paul, yeah this is a bad climbing form example. You'd never zig zag the rope because, as you said, the friction in the system would make it stupidly difficult to climb. However, for the sake of this fall factor 2 example, that's exactly what we want. The only available rope to be stretched is from the climber to the last bolt, after that it's too tangled up to stretch. And yes, you're absolutely right, a soft catch by a belayer jumping will spread the energy of the climber out over a longer distance decreasing the impact force. Thankfully this is why fall factors of 2 are very rare.
@@ciaranmcevoy9875 yeah, factor 2 is very rare when sport climbing, but more common in another scenarios like multipitch climbing... could be a topic for another video ;)
The likelihood of an F2 fall might be greater on multi pitch climbing but surely the physics won't change. Gravity doesn't know how many pitches you did before the fall.
@@Andy-ej4bb physics won't change, but on a single pitch route yo cannot fall below your belayer, which is possible when starting a pitch far from the ground...
Great video! So if the tensile force is your unknown value and I want to solve for Fmax… how can I do so if the tensile force is also required to solve for the elastic modulus? That makes two unknowns. Do I need to find a published modulus for my rope or can I simply sub out the tensile force somehow? I also thought about just using the gravitational force of the person but that may be too easy. Thanks!
Interesting video. The more rope you have the bigger the spring you have to catch you, and longer springs absorb more energy. I imagine the impact force is therefore dependent on the stiffness of the rope?
Yes that's right. The spring constant of a rope is EA/l0. However, treating that as one constant may lead us to believe that all different heights result in different forces, which we know not to be true. It's only when we break it up do we notice the height is not independent of the rope length l0. Revealing the fall factor ratio of h/l0.
Great video. Are the forces on the bolt/gear the same no matter the hight. Does the same elasticity in the rope distribute the potential energy over more time... protecting the bolt from a sudden force??
Hey Zoli, so glad you liked the video! So, in this scenario I'm talking about a basic, yet thankfully not too common, scenario of a fall factor of two. Almost simulating a bad rope line, with multiple zig-zagging anchor points, effectively stalling the rope. This means that we're assuming the only part of the rope that is available to be stretched is the rope between the climber and their last anchor point (or bolt). Therefore the force acting on the climber will be the same as the force acting on the anchor point but in the opposite direction. I believe, in a climbing scenario, since the rope would be anchored to the ground (or a belayer), the last anchor would witness twice the force that the climber experiences, but again, in this scenario the fall factor wouldn't be 2 since there'd be plenty of stretchable rope available to the climber. So, to answer your question, I believe that since the force on the climber depends on the fall factor (not the height exclusively), the force on the anchor point (bolt) depends on that also, and therefore will not be the same no matter the height, but will be the same with the same fall factors. I hope this helps answer your question, I had to get my brain thinking about the problem again haha.
Given that there is a maximum velocity that a falling object will reach I assume that a climber that falls just far enough to reach terminal velocity will have a more severe shock than a climber that falls a greater distance. Is this correct?
I don't like the asumption that exactly (and only) the rope between you and the anchor stretches. This is absolutely not typical for a climbing scenario. This would only be true if you were tied to that anchor. If you wouldn't dismiss everything in the system before the anchor, the force would absolutely depend on the height of the fall. But what's interesting and what I like about your video is, that you have proven the existence of an upper bound, which is independent of the fall height. I don't see this as counter intuitive, but it's definetly not obvious. Good job
Hey Frederik, glad you enjoyed the video. Yeah the example is completely unrealistic, no climber would have heavily zig-zagged anchors enough to mean no rope beyond their last anchor point would stretch. But, this simplifies the problem greatly to allow us to find the "fall factor". What I gathered from my reading on the topic, albeit from a physics point of view, was that the maximum impact of the force depends on the fall factor. Therefore, climbers falling from greater heights will experience the same impact force as those falling from lesser heights if their fall factors are the same. That, to me, is the counter intuitive part. I would immediately assume the greater height would have a greater impact force.
So wait, are you telling me that if I fall from 50 meters while on a normal climbing rope my spine won't come out of my ass and nail the ground below??
😂 that tickled me. The maths and physics says you should survive. There'll be more rope to stretch and absorb your fall. I know, it's so hard to fathom, and still feels odd to me.
hey, great video! what happens if you don't make the rope drag assumption? in reality, a climber might have 30m+ of rope out when they take a fall, all of which can stretch (climbers do everything they can to make sure the rope is in a straight line). what would happen is h/2 was, say 5m with 30m of rope out, vs if it was 2m....?
Hey Oliver, thanks for the kind words, glad you enjoyed it. Yeah so in the scenario you've mentioned, a 10 meter (h/2 being 5m) fall with 30m of stretchable rope would yield a fall factor of 10/30 or 1/3. This would make the term containing the fall factor ratio not contribute as much to the maximum force and bring it down by a considerable amount. If the h/2 was 2m then you'd fall 4m and so the fall factor would be 4/30 so an even easier fall. Correction: After reminding myself of the resulting equation, I saw that the fall factor is under the square root and only affects one term so it won't affect the max force as simply as I first suggested.
![fellow fellow - พรุ่งนี้ไม่มีใครรู้ feat. INK WARUNTORN [OFFICIAL MV]](http://i.ytimg.com/vi/QP6JyjYx_W0/mqdefault.jpg)
In fact, the longer the fall, the softer it feels... Usually, when the climber asks for a "soft catch" in case of a fall, the belayer gives a bit of extra slack (or even jump) to make the fall smoother
Another point: the bolts aren't usually placed in that zigzag pattern, the rope drag would make the climb impossible
Hey Paul, yeah this is a bad climbing form example. You'd never zig zag the rope because, as you said, the friction in the system would make it stupidly difficult to climb. However, for the sake of this fall factor 2 example, that's exactly what we want. The only available rope to be stretched is from the climber to the last bolt, after that it's too tangled up to stretch. And yes, you're absolutely right, a soft catch by a belayer jumping will spread the energy of the climber out over a longer distance decreasing the impact force. Thankfully this is why fall factors of 2 are very rare.
@@ciaranmcevoy9875 yeah, factor 2 is very rare when sport climbing, but more common in another scenarios like multipitch climbing... could be a topic for another video ;)
Ooo interesting! Might have to do a bit of reading on that style of climbing. Cheers Paul.
The likelihood of an F2 fall might be greater on multi pitch climbing but surely the physics won't change. Gravity doesn't know how many pitches you did before the fall.
@@Andy-ej4bb physics won't change, but on a single pitch route yo cannot fall below your belayer, which is possible when starting a pitch far from the ground...
Great video! So if the tensile force is your unknown value and I want to solve for Fmax… how can I do so if the tensile force is also required to solve for the elastic modulus? That makes two unknowns. Do I need to find a published modulus for my rope or can I simply sub out the tensile force somehow? I also thought about just using the gravitational force of the person but that may be too easy. Thanks!
Interesting video. The more rope you have the bigger the spring you have to catch you, and longer springs absorb more energy. I imagine the impact force is therefore dependent on the stiffness of the rope?
Yes that's right. The spring constant of a rope is EA/l0. However, treating that as one constant may lead us to believe that all different heights result in different forces, which we know not to be true. It's only when we break it up do we notice the height is not independent of the rope length l0. Revealing the fall factor ratio of h/l0.
Great video. Are the forces on the bolt/gear the same no matter the hight. Does the same elasticity in the rope distribute the potential energy over more time... protecting the bolt from a sudden force??
Hey Zoli, so glad you liked the video!
So, in this scenario I'm talking about a basic, yet thankfully not too common, scenario of a fall factor of two. Almost simulating a bad rope line, with multiple zig-zagging anchor points, effectively stalling the rope. This means that we're assuming the only part of the rope that is available to be stretched is the rope between the climber and their last anchor point (or bolt). Therefore the force acting on the climber will be the same as the force acting on the anchor point but in the opposite direction. I believe, in a climbing scenario, since the rope would be anchored to the ground (or a belayer), the last anchor would witness twice the force that the climber experiences, but again, in this scenario the fall factor wouldn't be 2 since there'd be plenty of stretchable rope available to the climber.
So, to answer your question, I believe that since the force on the climber depends on the fall factor (not the height exclusively), the force on the anchor point (bolt) depends on that also, and therefore will not be the same no matter the height, but will be the same with the same fall factors.
I hope this helps answer your question, I had to get my brain thinking about the problem again haha.
@@ciaranmcevoy9875 that does make sense, thanks for taking the time to answer :-)
Given that there is a maximum velocity that a falling object will reach I assume that a climber that falls just far enough to reach terminal velocity will have a more severe shock than a climber that falls a greater distance. Is this correct?
gave me flashbacks to my physics 1 class:)
Happy flashbacks I hope haha
I don't like the asumption that exactly (and only) the rope between you and the anchor stretches. This is absolutely not typical for a climbing scenario. This would only be true if you were tied to that anchor. If you wouldn't dismiss everything in the system before the anchor, the force would absolutely depend on the height of the fall. But what's interesting and what I like about your video is, that you have proven the existence of an upper bound, which is independent of the fall height. I don't see this as counter intuitive, but it's definetly not obvious. Good job
Hey Frederik, glad you enjoyed the video. Yeah the example is completely unrealistic, no climber would have heavily zig-zagged anchors enough to mean no rope beyond their last anchor point would stretch. But, this simplifies the problem greatly to allow us to find the "fall factor". What I gathered from my reading on the topic, albeit from a physics point of view, was that the maximum impact of the force depends on the fall factor. Therefore, climbers falling from greater heights will experience the same impact force as those falling from lesser heights if their fall factors are the same. That, to me, is the counter intuitive part. I would immediately assume the greater height would have a greater impact force.
So wait, are you telling me that if I fall from 50 meters while on a normal climbing rope my spine won't come out of my ass and nail the ground below??
😂 that tickled me. The maths and physics says you should survive. There'll be more rope to stretch and absorb your fall. I know, it's so hard to fathom, and still feels odd to me.
Brilliant
I’m an specialist in two phase flow and thermalhydraulics and a climber, so your help in this is appreciated
Oh I'm so glad to hear that! I found notion of a fall factor quite counterintuitive, hence my intrigue, so I'm glad I could help.
@@ciaranmcevoy9875 brilliantly you did
hey, great video! what happens if you don't make the rope drag assumption? in reality, a climber might have 30m+ of rope out when they take a fall, all of which can stretch (climbers do everything they can to make sure the rope is in a straight line). what would happen is h/2 was, say 5m with 30m of rope out, vs if it was 2m....?
Hey Oliver, thanks for the kind words, glad you enjoyed it. Yeah so in the scenario you've mentioned, a 10 meter (h/2 being 5m) fall with 30m of stretchable rope would yield a fall factor of 10/30 or 1/3. This would make the term containing the fall factor ratio not contribute as much to the maximum force and bring it down by a considerable amount. If the h/2 was 2m then you'd fall 4m and so the fall factor would be 4/30 so an even easier fall.
Correction: After reminding myself of the resulting equation, I saw that the fall factor is under the square root and only affects one term so it won't affect the max force as simply as I first suggested.