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After many years climbing I still haven't a clue what a "Kn" is.I know what it means but what is it? for example 25 cm = 1 inch 1 Kn = ?.Anyone any idea.
In reply to Homeward: kn=kilo newton = 1000 newtons (i think.)
In reply to jimtheape: Yep it is kilo newtons but whats a kilo newton,I get all
kinds of answers like" it'll hold a car" but as its me and
not a car on the end of my rope it would be nice to know
exactly what wieght a Kn holds,,,,and no I'm not heavier
than a car!! but I know a climber who is
kinds of answers like" it'll hold a car" but as its me and
not a car on the end of my rope it would be nice to know
exactly what wieght a Kn holds,,,,and no I'm not heavier
than a car!! but I know a climber who is
In reply to Homeward:
I was curious about myself so I just looked it up for you.
1 lb = roughly 4.5 N.
So 22kN (roughly what a krab will hold) is roughly 4900 lbs, and if a man is 12 stone = 168lbs, that's 29 men.
Hope that helps! (Hope it's right too...)
I was curious about myself so I just looked it up for you.
1 lb = roughly 4.5 N.
So 22kN (roughly what a krab will hold) is roughly 4900 lbs, and if a man is 12 stone = 168lbs, that's 29 men.
Hope that helps! (Hope it's right too...)
In reply to Sarah_Clough:
& to metricise, 1 newton must be roughly equal to 100 grams, ie 10 newtons = 1 kg.
& to metricise, 1 newton must be roughly equal to 100 grams, ie 10 newtons = 1 kg.
In reply to Homeward:
A kilonewton is the dynamic equivalent to 100 kilograms in static weight. Which is the equivalent of one big bloke. that's how I think of it anyway.
A kilonewton is the dynamic equivalent to 100 kilograms in static weight. Which is the equivalent of one big bloke. that's how I think of it anyway.
In reply to Mike C:
Newtons are a unit of force, 10 Newtons is (approx) the force exerted by a mass of 1kg in earths gravity at the surface (I think?)
anyone up to correcting my recollection of school physics, please go right ahead...
> (In reply to Sarah_Clough)
> & to metricise, 1 newton must be roughly equal to 100 grams, ie 10 newtons = 1 kg.
> & to metricise, 1 newton must be roughly equal to 100 grams, ie 10 newtons = 1 kg.
Newtons are a unit of force, 10 Newtons is (approx) the force exerted by a mass of 1kg in earths gravity at the surface (I think?)
anyone up to correcting my recollection of school physics, please go right ahead...
In reply to Homeward:
It's not Kn, it's KN. 1 KN = 1000 Newtons
A Newton is the force required to accelerate a mass of one kilogram at one metre per second squared in the direction of the force.
It's not Kn, it's KN. 1 KN = 1000 Newtons
A Newton is the force required to accelerate a mass of one kilogram at one metre per second squared in the direction of the force.
On this planet, the acceleration due to gravity, g, is 9.8 metres per second squared.
This means that a mass of 1 kilogram at rest on a horizontal surface exerts a force of 9.8 Newtons vertically downwards, and, as it is at rest, there is an equal and opposite reaction from the surface.
Although it can sometimes be useful to interpret forces in Newtons in terms of the gravitational forces acting on static masses (in kilograms), it would be a mistake to do so in the situations of dynamic loading that arise in climbing.
This means that a mass of 1 kilogram at rest on a horizontal surface exerts a force of 9.8 Newtons vertically downwards, and, as it is at rest, there is an equal and opposite reaction from the surface.
Although it can sometimes be useful to interpret forces in Newtons in terms of the gravitational forces acting on static masses (in kilograms), it would be a mistake to do so in the situations of dynamic loading that arise in climbing.
In reply to Cosmic John: I think it's kN. Usually letters are capitals only when they came from someone's name.
It's defined as a unit of force equal to the force that imparts an acceleration of 1 m/sec/sec to a mass of 1 kilogram. On earth it's equal to the force exerted by a 1kg mass due to gravity.
It's defined as a unit of force equal to the force that imparts an acceleration of 1 m/sec/sec to a mass of 1 kilogram. On earth it's equal to the force exerted by a 1kg mass due to gravity.
In reply to Cosmic John:
9.81 m/s^2
9.81 m/s^2
In reply to Duma Brickhill:
Thanks for that, & Cosmic. Had an idea that Newtons were force, not weight, but minimal physics knowledge here!
Can anyone tell me why krabs used to carry kg ratings, yet now kN ratings? Is it because it is only in the last few years manufacturers have realised falling climbers are moving?
Thanks for that, & Cosmic. Had an idea that Newtons were force, not weight, but minimal physics knowledge here!
Can anyone tell me why krabs used to carry kg ratings, yet now kN ratings? Is it because it is only in the last few years manufacturers have realised falling climbers are moving?
In reply to Tom Fuller:
Yes.
What is "It" ? In the context of your post, it looks as if you mean a kilonewton. You actually define a newton. Well, sort of. As force is a vector quantity, you should add "in the direction of the force." (As I did.)
>On earth it's equal to the force exerted by a 1kg mass due to gravity.
No it is not!
There is no defined "force due to gravity", there is only acceleration due to gravity, which applies equally to all masses. As I explained in my previous post, it may sometimes be convenient to attempt grasp static forces expressed in newtons in terms of static loading in kilograms.
In climbing we are really concerned with dynamic loading, not static.
This is all pretty basic stuff. It shouldn't be that hard to find on the net, for heaven's sake.
> (In reply to Cosmic John) I think it's kN. Usually letters are capitals only when they came from someone's name.
Yes.
> It's defined as a unit of force equal to the force that imparts an acceleration of 1 m/sec/sec to a mass of 1 kilogram.
What is "It" ? In the context of your post, it looks as if you mean a kilonewton. You actually define a newton. Well, sort of. As force is a vector quantity, you should add "in the direction of the force." (As I did.)
>On earth it's equal to the force exerted by a 1kg mass due to gravity.
No it is not!
There is no defined "force due to gravity", there is only acceleration due to gravity, which applies equally to all masses. As I explained in my previous post, it may sometimes be convenient to attempt grasp static forces expressed in newtons in terms of static loading in kilograms.
In climbing we are really concerned with dynamic loading, not static.
This is all pretty basic stuff. It shouldn't be that hard to find on the net, for heaven's sake.
In reply to Mike C:
Well, from a scientific (and engineering) perspective, it's simply the more correct, more meaningful, way of defining the actual strength, in practice, of these items of gear.
As with most things, the more things are studied, the more complicated than was first imagined they tend to become.
Well, from a scientific (and engineering) perspective, it's simply the more correct, more meaningful, way of defining the actual strength, in practice, of these items of gear.
As with most things, the more things are studied, the more complicated than was first imagined they tend to become.
In reply to Cosmic John:
That certainly makes sense! Cheers.
I wonder which climbing physicist/engineer introduced the change?
> (In reply to Mike C)
> As with most things, the more things are studied, the more complicated than was first imagined they tend to become.
That certainly makes sense! Cheers.
I wonder which climbing physicist/engineer introduced the change?
In reply to Mike C:
I have an idea that it was introduced with the first specifications for "UIAA approved" certification of climbing equipment.
I'll have a quick google to see if I can confirm this.
Some old hands from Clog (later DMM) might be able to comment, if they happen to read this.
I have an idea that it was introduced with the first specifications for "UIAA approved" certification of climbing equipment.
I'll have a quick google to see if I can confirm this.
Some old hands from Clog (later DMM) might be able to comment, if they happen to read this.
I'm pretty sure that it predates the more recent wave of "CE" "EN" and "PPE" specifications, anyway.
In reply to Cosmic John:
Happy to let you do the Googling. I have a selection of old Clog krabs, circa 1980, with 2200 to 2500 kg ratings. Interestingly one which opened (recently) to allow the gate to "pop" out on loading from a 20ft + fall. It closed back to almost original shape without breaking, only the gate is on the outside now! No UIAA stamp on it.
Happy to let you do the Googling. I have a selection of old Clog krabs, circa 1980, with 2200 to 2500 kg ratings. Interestingly one which opened (recently) to allow the gate to "pop" out on loading from a 20ft + fall. It closed back to almost original shape without breaking, only the gate is on the outside now! No UIAA stamp on it.
In reply to Niall:
A small car weighs 1000 kgs, or ten big blokes. So a 2500 kN crab will hold 25 big blokes or 2.5 small cars, or the equivalent falling weight.
A small car weighs 1000 kgs, or ten big blokes. So a 2500 kN crab will hold 25 big blokes or 2.5 small cars, or the equivalent falling weight.
In reply to Alison Stockwell:
A 2500 kN krab would hold 250 small cars. It would also be a bit big to fit on your harness... A 25 kN krab would hold 2.5 small cars But this isn't actually a very helpful thing to know, since it's the forces generated during falls which are actually important, and they aren't directly equivalent to the static weight. You've got to take into account the properties of the rope, gear etc. to find out what they might be.
A 2500 kN krab would hold 250 small cars. It would also be a bit big to fit on your harness... A 25 kN krab would hold 2.5 small cars But this isn't actually a very helpful thing to know, since it's the forces generated during falls which are actually important, and they aren't directly equivalent to the static weight. You've got to take into account the properties of the rope, gear etc. to find out what they might be.
In reply to Cosmic John:
Not necessarily. Capital letters are used for the bigger multipliers: mega (M), giga (G), etc.
http://whatis.techtarget.com/definition/0,,sid9_gci499008,00.html
And capital K is 2^10, as in Kb.
> > (In reply to Cosmic John) I think it's kN. Usually letters are capitals only when they came from someone's name.
> Yes.
Not necessarily. Capital letters are used for the bigger multipliers: mega (M), giga (G), etc.
http://whatis.techtarget.com/definition/0,,sid9_gci499008,00.html
And capital K is 2^10, as in Kb.
In reply to Richard:
You're right, of course.
I have no excuse other than my irritation at the much more fundamental mistakes in the post to which I was replying.
Which is no excuse at all.
You're right, of course.
I have no excuse other than my irritation at the much more fundamental mistakes in the post to which I was replying.
Which is no excuse at all.
In reply to Cosmic John: more to the point, do you know where that scatty bint's got to?
In reply to Cosmic John:
Fair enough. They were pretty egregious.
> I have no excuse other than my irritation at the much more fundamental mistakes in the post to which I was replying.
Fair enough. They were pretty egregious.
In reply to Homeward:
1 kN (kilonewton) is the weight of 1kg (kilgram), that is the force applied by it (kg is a mass, not a force).
If a mass is 1 kg, then the force in newtons is 9.81kN.
(you have to x 9.81 for the conversion when considering gravity).
I believe there are about 70 kg in 11 stone, if this helps to convert from imperial mass-wise.
For forces other than gravity ie. weight, i have no idea since i have only ever used kN (ive no idea wha tthe imperial unit of force is, or how it converts).
Sorry of this reply is useless.
Si
1 kN (kilonewton) is the weight of 1kg (kilgram), that is the force applied by it (kg is a mass, not a force).
If a mass is 1 kg, then the force in newtons is 9.81kN.
(you have to x 9.81 for the conversion when considering gravity).
I believe there are about 70 kg in 11 stone, if this helps to convert from imperial mass-wise.
For forces other than gravity ie. weight, i have no idea since i have only ever used kN (ive no idea wha tthe imperial unit of force is, or how it converts).
Sorry of this reply is useless.
Si
In reply to SidH:
The Imperial unit of force is (was) the poundal, believe it or not.
That is, the force required to accelerate a mass of 1 pound at a rate of 1 foot per second squared in the direction of the force.
Obscure or what?
I'm afraid that the first line of your post, as it stands, is just wrong. It may not be exactly what you meant to say.
If I may express it this way, the "weight" of a mass of 1 kilogram is 9.81 Newtons (Not kilonewtons, as you say.)
As individuals, learning, we tend to grasp the concept of "weight" before the idea of "mass", and both of these before "force" We tend to associate all these different concepts too closely.
The strict definition of "force" involves the acceleration of defined masses, in a vacuum, in free space, and under conditions of zero gravity. Not that easy to grasp intuitively!
Really, force is force, mass is mass, and gravity is gravity.
They interact in complex ways, but they are separate, very different things. It's a bit too easy to get them confused.
The Imperial unit of force is (was) the poundal, believe it or not.
That is, the force required to accelerate a mass of 1 pound at a rate of 1 foot per second squared in the direction of the force.
Obscure or what?
I'm afraid that the first line of your post, as it stands, is just wrong. It may not be exactly what you meant to say.
If I may express it this way, the "weight" of a mass of 1 kilogram is 9.81 Newtons (Not kilonewtons, as you say.)
As individuals, learning, we tend to grasp the concept of "weight" before the idea of "mass", and both of these before "force" We tend to associate all these different concepts too closely.
The strict definition of "force" involves the acceleration of defined masses, in a vacuum, in free space, and under conditions of zero gravity. Not that easy to grasp intuitively!
Really, force is force, mass is mass, and gravity is gravity.
They interact in complex ways, but they are separate, very different things. It's a bit too easy to get them confused.
In reply to Cosmic John:
cosmic cosmic numbers and all.
what we want to know is:
Q1. a 70kg climber takes a fall of ten meters. The slack in the system = rope stretch 15%, belayer smoking fag 4%, shite belay 1%.
if this load was applied to a single point what would the load be?
Q2 blah de blah
Q3 etc. etc
cosmic cosmic numbers and all.
what we want to know is:
Q1. a 70kg climber takes a fall of ten meters. The slack in the system = rope stretch 15%, belayer smoking fag 4%, shite belay 1%.
if this load was applied to a single point what would the load be?
Q2 blah de blah
Q3 etc. etc
In reply to simmo:
1) "F*ck Me! I nearly decked it there! Let me down!"
2) Abseil for stuck gear. Fanny about for 45 minutes. End up leaving it in.
3) Go to Pub.
1) "F*ck Me! I nearly decked it there! Let me down!"
2) Abseil for stuck gear. Fanny about for 45 minutes. End up leaving it in.
3) Go to Pub.
In reply to Cosmic John:
Was the pound force, lbf not also used? Which, just to confuse things, was the force applied due to a mass of 1 pound on the surface of the earth...? Or have I got that totally wrong?
> (In reply to SidH)
>
> The Imperial unit of force is (was) the poundal, believe it or not.
>
> That is, the force required to accelerate a mass of 1 pound at a rate of 1 foot per second squared in the direction of the force.
>
> Obscure or what?
>
>
> The Imperial unit of force is (was) the poundal, believe it or not.
>
> That is, the force required to accelerate a mass of 1 pound at a rate of 1 foot per second squared in the direction of the force.
>
> Obscure or what?
>
Was the pound force, lbf not also used? Which, just to confuse things, was the force applied due to a mass of 1 pound on the surface of the earth...? Or have I got that totally wrong?
In reply to stuartf:
1 pound force would equal 32 poundals.
I don't think that the Imperial system was ever developed to the same level of consistency as the present standard system (Is it still just called "S.I."(= Systeme Internationale)?)
Just don't ask me to define Tesla, Lux or Lumens in terms of the fundamental units!
1 pound force would equal 32 poundals.
I don't think that the Imperial system was ever developed to the same level of consistency as the present standard system (Is it still just called "S.I."(= Systeme Internationale)?)
Just don't ask me to define Tesla, Lux or Lumens in terms of the fundamental units!
In reply to Cosmic John:
A kilonewton (on Earth) is approximately equal to 100 kilograms or one fat bastard.
> (In reply to Homeward)
>
> It's not Kn, it's KN.
>
is it fuk, its kN.
>
> It's not Kn, it's KN.
>
A kilonewton (on Earth) is approximately equal to 100 kilograms or one fat bastard.
In reply to DaveN:
It's FORCE NOT MASS you dimwit.
If you don't get the difference after all the above, then I just give up.
It's not fuk, It's f u c k !
Tw*t.
It's FORCE NOT MASS you dimwit.
If you don't get the difference after all the above, then I just give up.
It's not fuk, It's f u c k !
Tw*t.
In reply to DaveN: I`m going to do this in a Newtonian style as its more readily understandable.
Newtons second law of motion is
Force=mass X acceleration
Using the SI system of units mass is measured in kg. An objects mass is constant no matter where they are(i.e. moon or Jupiter etc) Th Force excerted on them due to gravity can be calculated by finding the acceleration due to gravity using Newtons law of Gravitation.
This allows the force due to gracity, or weight, to be found. The weight will vary as the acceleration is a function of distance and mass of the atracting object. Because the mass of the moon is lower than that of the Earth the acceleration due to gravity is reduced meaning that an objects weight is lower though its mass is the same as on earth. Hence astronaughts doing big jumps.
The 22kilonewtons on a crab is equivalent to a static loading of 2200 kg. However during a fall once the rope has gone tight and rope stretch starts the acceleration(or deceleration) may be more rapid than that due to gravity, going back to newtons second law allows the force on the rope to be worked out. The force on the crab is nearly double the force on the rope due to the fact that there are essentially two ropes going downward from the crab.
Hope that made some sense.
Newtons second law of motion is
Force=mass X acceleration
Using the SI system of units mass is measured in kg. An objects mass is constant no matter where they are(i.e. moon or Jupiter etc) Th Force excerted on them due to gravity can be calculated by finding the acceleration due to gravity using Newtons law of Gravitation.
This allows the force due to gracity, or weight, to be found. The weight will vary as the acceleration is a function of distance and mass of the atracting object. Because the mass of the moon is lower than that of the Earth the acceleration due to gravity is reduced meaning that an objects weight is lower though its mass is the same as on earth. Hence astronaughts doing big jumps.
The 22kilonewtons on a crab is equivalent to a static loading of 2200 kg. However during a fall once the rope has gone tight and rope stretch starts the acceleration(or deceleration) may be more rapid than that due to gravity, going back to newtons second law allows the force on the rope to be worked out. The force on the crab is nearly double the force on the rope due to the fact that there are essentially two ropes going downward from the crab.
Hope that made some sense.
In reply to Cosmic John:
I converted it to a mass on earth as I assumed the original poster does most of their climbing on earth and most people have more of an understanding of mass than of weight.
I see why you posted but I do feel that I justified what I said by saying "on earth" if you regularly climb elsewhere please make sure that your gear is capable of taking the forces involved. At least I made the distinction of being on earth unlike some of the earlier respondants. How come you didn`t correct Mike C when he said "Had an idea that Newtons were force, not weight".
> (In reply to DaveN)
>
> It's FORCE NOT MASS you dimwit.
>
Right, that is probably why I said that one kilonewton (on Earth) is approximately equal to 100 kg. I understand that kg are a measure of mass and that kN are a measure of force. I also understand that they are linked by acceleration and on earth this can, in most cases be approximated as 10 m/s^2. Though by Newtonian mechanics this varies with distance and the mass of the objects.
>
> It's FORCE NOT MASS you dimwit.
>
I converted it to a mass on earth as I assumed the original poster does most of their climbing on earth and most people have more of an understanding of mass than of weight.
I see why you posted but I do feel that I justified what I said by saying "on earth" if you regularly climb elsewhere please make sure that your gear is capable of taking the forces involved. At least I made the distinction of being on earth unlike some of the earlier respondants. How come you didn`t correct Mike C when he said "Had an idea that Newtons were force, not weight".
In reply to Cosmic John:
But only on Earth...
I believe that the Tesla is equal to 1 kg s^-1 C^-1. It's the magnetic field strength required such that a 1C charge moving at 1m/s experiences a force of 1N
> (In reply to stuartf)
>
> 1 pound force would equal 32 poundals.
>
>
> 1 pound force would equal 32 poundals.
>
But only on Earth...
> Just don't ask me to define Tesla, Lux or Lumens in terms of the fundamental units!
I believe that the Tesla is equal to 1 kg s^-1 C^-1. It's the magnetic field strength required such that a 1C charge moving at 1m/s experiences a force of 1N
In reply to stuartf:
I can quote your own post back at you :
" ... this isn't actually a very helpful thing to know"
And in the context of this thread, the same goes for DaveN's simplistic offerings.
I can quote your own post back at you :
" ... this isn't actually a very helpful thing to know"
And in the context of this thread, the same goes for DaveN's simplistic offerings.
In reply to DaveN:
Er..., because he was right?
> (In reply to Cosmic John)
> [...]
>
>
> How come you didn`t correct Mike C when he said "Had an idea that Newtons were force, not weight".
> [...]
>
>
> How come you didn`t correct Mike C when he said "Had an idea that Newtons were force, not weight".
Er..., because he was right?
In reply to Cosmic John:
On the contrary, it is very useful to know what 1lbf is on Earth. In the aviation industry, for example, jet engines performance is often defined in terms of pound force, and often this needs to be converted into Newtons. To do this, one has to assume the equivalence between weight on earth, and force.
I just pointed out that your conversion from poundal to lbf was only valid on Earth, since you've constantly been pointing out that the conversion between Newtons and kilograms also depends on the acceleration due to gravity.
On the contrary, it is very useful to know what 1lbf is on Earth. In the aviation industry, for example, jet engines performance is often defined in terms of pound force, and often this needs to be converted into Newtons. To do this, one has to assume the equivalence between weight on earth, and force.
I just pointed out that your conversion from poundal to lbf was only valid on Earth, since you've constantly been pointing out that the conversion between Newtons and kilograms also depends on the acceleration due to gravity.
In reply to Cosmic John:
Newtons are a unit of force, weight is a force (the mass of an object multiplied by the acceleration due to gravity). He should have said "Had an idea that Newtons were force, not mass"
So to quote you back at yourself
It's FORCE NOT MASS you dimwit.
If you don't get the difference after all the above, then I just give up.
My earlier offering may appear simplistic to you as I tried to make them easy to understand. I appresiate that not everyone has a scientific/engineering background, and can only apologise for using newtonian formulae when this has been superseeded by the works of Einsteinand Hawking. Would you care to explain in greater depth with reference to the work of these two?
> (In reply to DaveN)
> [...]
>
> Er..., because he was right?
> [...]
>
> Er..., because he was right?
Newtons are a unit of force, weight is a force (the mass of an object multiplied by the acceleration due to gravity). He should have said "Had an idea that Newtons were force, not mass"
So to quote you back at yourself
It's FORCE NOT MASS you dimwit.
If you don't get the difference after all the above, then I just give up.
My earlier offering may appear simplistic to you as I tried to make them easy to understand. I appresiate that not everyone has a scientific/engineering background, and can only apologise for using newtonian formulae when this has been superseeded by the works of Einsteinand Hawking. Would you care to explain in greater depth with reference to the work of these two?
>
> " ... this isn't actually a very helpful thing to know"
>
> And in the context of this thread, the same goes for DaveN's simplistic offerings.
> " ... this isn't actually a very helpful thing to know"
>
> And in the context of this thread, the same goes for DaveN's simplistic offerings.
When the question was "what are kilo Newtons?" I thought that an explaination including how to work them out what they measured and how they vary with acceleration was reasonably useful.
In reply to stuartf:
Sorry, I was only attempting a little humour in what seemed to be becoming a slightly tedious thread.
Seriously, though, I think the strict definition of lbf would specify standard terrestrial gravity, and that the lbf unit could then be used universally without further qualification.
I think you'd agree that this is getting pretty academic, however.
The continued use of lbf in specifying jet engine thrust is a bit of a historical anomaly, similar to the use of btus in the classification of domestic boilers. It's what people are used to.
>since you've constantly been pointing out that the conversion between Newtons and kilograms also depends on the acceleration due to gravity.
On the contrary, I've been trying to persuade people NOT to make this conversion, but to adapt their approach to thinking about the forces exerted on climbing gear directly in kilonewtons.
Sorry, I was only attempting a little humour in what seemed to be becoming a slightly tedious thread.
Seriously, though, I think the strict definition of lbf would specify standard terrestrial gravity, and that the lbf unit could then be used universally without further qualification.
I think you'd agree that this is getting pretty academic, however.
The continued use of lbf in specifying jet engine thrust is a bit of a historical anomaly, similar to the use of btus in the classification of domestic boilers. It's what people are used to.
>since you've constantly been pointing out that the conversion between Newtons and kilograms also depends on the acceleration due to gravity.
On the contrary, I've been trying to persuade people NOT to make this conversion, but to adapt their approach to thinking about the forces exerted on climbing gear directly in kilonewtons.
In reply to Cosmic John:
Doe this matter? most people are familiar with the concept of mass, kilograms, and to be honest you can measure things in kilogram force anyway (on Earth approximately 9.81 Newtons). We can consider that the acceleration is pretty much constant so the conversion between force and mass is constant.
The gear manufactures could mark everything in kg-f and then more people would get a hadle on what the gear can take-statically at least.
The dynamic bit is still an issue thought and I for one have never been entirely satisfied with th approach via fall factors. The actual forces experienced with gear will vary with the decelleration of the climber which will be related to the "softness" of the catch and the properties of the rope. The speed that theuy are decelerated from will depend on the height above the last peice of gear as this gives the velocity when they are caught (using constant acceleration formulae).
> >since you've constantly been pointing out that the conversion between Newtons and kilograms also depends on the acceleration due to gravity.
>
> On the contrary, I've been trying to persuade people NOT to make this conversion, but to adapt their approach to thinking about the forces exerted on climbing gear directly in kilonewtons.
>
> On the contrary, I've been trying to persuade people NOT to make this conversion, but to adapt their approach to thinking about the forces exerted on climbing gear directly in kilonewtons.
Doe this matter? most people are familiar with the concept of mass, kilograms, and to be honest you can measure things in kilogram force anyway (on Earth approximately 9.81 Newtons). We can consider that the acceleration is pretty much constant so the conversion between force and mass is constant.
The gear manufactures could mark everything in kg-f and then more people would get a hadle on what the gear can take-statically at least.
The dynamic bit is still an issue thought and I for one have never been entirely satisfied with th approach via fall factors. The actual forces experienced with gear will vary with the decelleration of the climber which will be related to the "softness" of the catch and the properties of the rope. The speed that theuy are decelerated from will depend on the height above the last peice of gear as this gives the velocity when they are caught (using constant acceleration formulae).
In reply to Cosmic John:
No! Why do you ask.
I`m just about to look up his profile thugh.
> (In reply to DaveN)
>
> Are you related to Bruce Hooker?
>
> Are you related to Bruce Hooker?
No! Why do you ask.
I`m just about to look up his profile thugh.
In reply to DaveN: He appears to have posted several times in relation tofall factors recently.
I just thik the fall factor thing is oversimplified, and as you appear to have an understanding of physics/mechanics surely you must agree.
I know climbers who weigh 8 stone and 16 stone. If they are both above there gear and fall off they will be doing the same speeds as each othe rwhen the system catches them. The force they experience is dependent on the acceleration as well as the mass. If they decelerate at the same rate then the forces on the heavier climber and his equipment will be larger (Although his rope stretch would also be greater).
If a soft catch approach is used on the heavier climberthen the deceleration and hence forces will be reduced. Plus he`ll probably get more scared so you can tease him more in the pub later.
Fall factor is probably a good rule of thumb but I wouldn`t rely on it for every thing-at least not till I see/do a reasonable and justified mathematical proof.
I just thik the fall factor thing is oversimplified, and as you appear to have an understanding of physics/mechanics surely you must agree.
I know climbers who weigh 8 stone and 16 stone. If they are both above there gear and fall off they will be doing the same speeds as each othe rwhen the system catches them. The force they experience is dependent on the acceleration as well as the mass. If they decelerate at the same rate then the forces on the heavier climber and his equipment will be larger (Although his rope stretch would also be greater).
If a soft catch approach is used on the heavier climberthen the deceleration and hence forces will be reduced. Plus he`ll probably get more scared so you can tease him more in the pub later.
Fall factor is probably a good rule of thumb but I wouldn`t rely on it for every thing-at least not till I see/do a reasonable and justified mathematical proof.
In reply to Homeward: Just read this thread and got totally confused (and I got an A in A-level Physics). So here's my 2d's worth...
Mass is a fundamental unit and a scalor (sp?) quantity ie it has a quantity but no direction unlike velocity which has a direction. Force is also a vector, as is acceleration - the rate of change of velocity with respect to time (a scalor). The three measures are related by Newton's second law,
F = m x a
Standard units are force in Newtons (N), mass in kilograms (kg) and acceleration (a)in metres per second per second.
So a force of 1N is sufficient to cause a mass of 1kg to accelerate by metre per second per second. BUT... this is the RESULTANT force. Newton's first law states that a body (mass) at rest will remain at rest unless acted upon by an external force. Similarly a body moving with constant velocity (ie zero acceleration) will continue to move at the same velocity unless acted upon by an unbalanced (resultant) force.
To illustrate this, consider a mass on a table top. The mass is acted upon by a force, gravity. Gravity is effectively the attractive force which occurs between any to bodies with mass, in this case between the mass and the Earth. If the body were to 'fall' it would effectively be attracted to the centre of the earth; simultaneously the earth would be attracted to the centre of the mass, but because the Earth is so massive, this second point is often overlooked. The point is, the two bodies are moving towards each other. Anyway, back to the mass on the table top. Gravity acts downwards on the mass, but because Newton's third law states that every action has an equal and opposite REaction, the table top exerts a force on the mass ('upwards')equal to the size of the force on the mass ('downwards')due to gravity (the mass is stationary remember - see Newton's first law). This means the forces acting upon it are balanced.
Now consider pushing the mass from left to right along the table top. You may be able to push against it without it moving. This must mean that the forces acting upon it are also balanced. A force must therefore be acting on the body to oppose the force of you pushing it. This is friction between the mass and the table top (acting from right to left) and may depend on the size of the mass and the nature of the surfaces in contact with one another (if they're lubricated, for example). If the mass is moving, the pushing force must be greater than the resisting friction force; therefore the forces are unbalanced hence the pushing force is the resultant one.
Right.
Let's consider a karabiner with a quoted breaking strain of 22kN - remember this is a force.
A force of 22,000 Newtons will cause the krab to snap (in reality, thankfully, it will probably take more than this)
Remembering Newton's second law
F = m x a
we can convert this to what mass such a krab can be expected to hold.
We know the value of F (22kN) and we also know the value of a. In this situation it can be considered the acceleration due to gravity. This has been measured to be 9.81 metres per second per second, although it varies across the earth's surface. For our purposes, though, let's take it to be 10 metres per second per second. We want to find the value of m.
Hence
m = F / a
m = 22,000 / 10
m = 2,200 kg.
So a krab would be expected to hold a maximum of 2,200 kg (2.2 metric tonnes (I think)) before failing. This is of course a static load ie a mass of 2,200 kg dangling off the krab on earth. It sounds like shed loads but when you consider the forces generated in a fall things don't look quite so rosy. When a mass (eg a climber) falls onto a krab and is held by the rope, it decelerates and hence exerts a force on the krab. How big this force is depends on the rate of deceleration, which in turn depends on the velocity (due to gravity) at which the mass was travelling before it started to decelerate (which depends on how far the body has fallen and therefore for how long) and the rate at which it slows down - this depends on such factors as rope stretch and rope slip - how much rope the belayer lets through the belay device. In certain situations it's surprising what forces can be generated. Even so, I won't be losing sleep over whether my krabs will break or not in a fall. It is, however, worth keeping them in good nick, especially the gates. The recent article in Summit from the BMC Technical Committee makes very interesting reading (perhaps if like me you're very sad but then that is what my degree is all about...). Basically krab is much weaker in some directions than in others, especially across the gate, and this can have major implications for certain wiregate designs which may promote cross loading. Not healthy.
Right. Think that's everything, and I think that's right, but don't quote me on it. I just hope it clears up one or two things.
Cheers
ic
Mass is a fundamental unit and a scalor (sp?) quantity ie it has a quantity but no direction unlike velocity which has a direction. Force is also a vector, as is acceleration - the rate of change of velocity with respect to time (a scalor). The three measures are related by Newton's second law,
F = m x a
Standard units are force in Newtons (N), mass in kilograms (kg) and acceleration (a)in metres per second per second.
So a force of 1N is sufficient to cause a mass of 1kg to accelerate by metre per second per second. BUT... this is the RESULTANT force. Newton's first law states that a body (mass) at rest will remain at rest unless acted upon by an external force. Similarly a body moving with constant velocity (ie zero acceleration) will continue to move at the same velocity unless acted upon by an unbalanced (resultant) force.
To illustrate this, consider a mass on a table top. The mass is acted upon by a force, gravity. Gravity is effectively the attractive force which occurs between any to bodies with mass, in this case between the mass and the Earth. If the body were to 'fall' it would effectively be attracted to the centre of the earth; simultaneously the earth would be attracted to the centre of the mass, but because the Earth is so massive, this second point is often overlooked. The point is, the two bodies are moving towards each other. Anyway, back to the mass on the table top. Gravity acts downwards on the mass, but because Newton's third law states that every action has an equal and opposite REaction, the table top exerts a force on the mass ('upwards')equal to the size of the force on the mass ('downwards')due to gravity (the mass is stationary remember - see Newton's first law). This means the forces acting upon it are balanced.
Now consider pushing the mass from left to right along the table top. You may be able to push against it without it moving. This must mean that the forces acting upon it are also balanced. A force must therefore be acting on the body to oppose the force of you pushing it. This is friction between the mass and the table top (acting from right to left) and may depend on the size of the mass and the nature of the surfaces in contact with one another (if they're lubricated, for example). If the mass is moving, the pushing force must be greater than the resisting friction force; therefore the forces are unbalanced hence the pushing force is the resultant one.
Right.
Let's consider a karabiner with a quoted breaking strain of 22kN - remember this is a force.
A force of 22,000 Newtons will cause the krab to snap (in reality, thankfully, it will probably take more than this)
Remembering Newton's second law
F = m x a
we can convert this to what mass such a krab can be expected to hold.
We know the value of F (22kN) and we also know the value of a. In this situation it can be considered the acceleration due to gravity. This has been measured to be 9.81 metres per second per second, although it varies across the earth's surface. For our purposes, though, let's take it to be 10 metres per second per second. We want to find the value of m.
Hence
m = F / a
m = 22,000 / 10
m = 2,200 kg.
So a krab would be expected to hold a maximum of 2,200 kg (2.2 metric tonnes (I think)) before failing. This is of course a static load ie a mass of 2,200 kg dangling off the krab on earth. It sounds like shed loads but when you consider the forces generated in a fall things don't look quite so rosy. When a mass (eg a climber) falls onto a krab and is held by the rope, it decelerates and hence exerts a force on the krab. How big this force is depends on the rate of deceleration, which in turn depends on the velocity (due to gravity) at which the mass was travelling before it started to decelerate (which depends on how far the body has fallen and therefore for how long) and the rate at which it slows down - this depends on such factors as rope stretch and rope slip - how much rope the belayer lets through the belay device. In certain situations it's surprising what forces can be generated. Even so, I won't be losing sleep over whether my krabs will break or not in a fall. It is, however, worth keeping them in good nick, especially the gates. The recent article in Summit from the BMC Technical Committee makes very interesting reading (perhaps if like me you're very sad but then that is what my degree is all about...). Basically krab is much weaker in some directions than in others, especially across the gate, and this can have major implications for certain wiregate designs which may promote cross loading. Not healthy.
Right. Think that's everything, and I think that's right, but don't quote me on it. I just hope it clears up one or two things.
Cheers
ic
In reply to DaveN:
Fall factors are a simplification.
In fact it's better to be a fat climber as a fat climber's body experiences less force in the event of a big fall than a thin one. This is largely due to the increased rope stretch (I think)
Petzl/Lyon research and testing has proved this.
Eat more pies!
Fall factors are a simplification.
In fact it's better to be a fat climber as a fat climber's body experiences less force in the event of a big fall than a thin one. This is largely due to the increased rope stretch (I think)
Petzl/Lyon research and testing has proved this.
Eat more pies!
In reply to Anonymous: 'the rate at which it slows down - this depends on such factors as rope stretch and rope slip - how much rope the belayer lets through the belay device'
don't forget air resistance - particuarly in the unrestrained fall stage affecting the velocity. After the point of initial rope loading I would sugest decelleration due to the rope is the dominant force.
don't forget air resistance - particuarly in the unrestrained fall stage affecting the velocity. After the point of initial rope loading I would sugest decelleration due to the rope is the dominant force.
In reply to Dave Hunter, Rock + Run:
Much as the e"at more pies" mantra is seductive, the conclusion that a fat climber experiences less force is clearly bunk. Even statically, a fat climber sees greater force from his/her weight than a stick insect does.
Much as the e"at more pies" mantra is seductive, the conclusion that a fat climber experiences less force is clearly bunk. Even statically, a fat climber sees greater force from his/her weight than a stick insect does.
In reply to GrahamD:
If the force in the rope only depends on its stretch (a reasonable assumption if it hasn't been fallen on recently) then the fat person can only experience a greater or equal force to the thin person. If there is a peak in the force/extension curve before the point that it would stop the thin person then they will both experience the same force. Otherwise the fat person will experience a greater peak force. (This assumes no slip at the belay of course, which may well not be a reasonable assumption).
Does anyone know what the force/extension (or stress/strain) curve for a rope looks like? Is it approximately linear, getting steeper just before it snaps, or is it something more complex?
If the force in the rope only depends on its stretch (a reasonable assumption if it hasn't been fallen on recently) then the fat person can only experience a greater or equal force to the thin person. If there is a peak in the force/extension curve before the point that it would stop the thin person then they will both experience the same force. Otherwise the fat person will experience a greater peak force. (This assumes no slip at the belay of course, which may well not be a reasonable assumption).
Does anyone know what the force/extension (or stress/strain) curve for a rope looks like? Is it approximately linear, getting steeper just before it snaps, or is it something more complex?
In reply to IainM:
After the point of initial rope loading I would sugest decelleration due to the rope is the dominant force.
I would agree with this. As force is mass times acceleration then the fat bastard climber will experience greater forces-unless the rope stretch is enough to decelerate him at a significantly lower rate than the thin climber.
The kinetic energy (0.5MV^2) can be equated to the energy stored in a spring (rope) which is given by 0.5ke^2, where is extension and k is the sprng constant. The velocity of the climber as the rope begins to go tight can be found using the constant acceleration formulae ((0.5xa)^0.5) which will all be turned in to stored energy via a non constant acceleration-which is the problem. (Also need to consider any potential energy as well).
The easiest thing to do would be to set up astrain gauge on a test rig when dropping the weights for rope tests.
After the point of initial rope loading I would sugest decelleration due to the rope is the dominant force.
I would agree with this. As force is mass times acceleration then the fat bastard climber will experience greater forces-unless the rope stretch is enough to decelerate him at a significantly lower rate than the thin climber.
The kinetic energy (0.5MV^2) can be equated to the energy stored in a spring (rope) which is given by 0.5ke^2, where is extension and k is the sprng constant. The velocity of the climber as the rope begins to go tight can be found using the constant acceleration formulae ((0.5xa)^0.5) which will all be turned in to stored energy via a non constant acceleration-which is the problem. (Also need to consider any potential energy as well).
The easiest thing to do would be to set up astrain gauge on a test rig when dropping the weights for rope tests.
In reply to Dave Hunter, Rock + Run:
My mistake, decelleration, not force. A heavier climber will decellerate over a longer period. A lighter climber will decellerate faster and thus suffer more.
My mistake, decelleration, not force. A heavier climber will decellerate over a longer period. A lighter climber will decellerate faster and thus suffer more.
In reply to Dave Hunter, Rock + Run:
Why is suffering related to decelleration rather than force? That seems counterintuative!
Why is suffering related to decelleration rather than force? That seems counterintuative!
In reply to stuartf:
You are in your car. You slow from 200mph to 0mph in one second (it's an impressive car). Or you slow from 200mph to 0mph over twenty minutes. Which is going to be better for your seatbelt held body?
You are in your car. You slow from 200mph to 0mph in one second (it's an impressive car). Or you slow from 200mph to 0mph over twenty minutes. Which is going to be better for your seatbelt held body?
In reply to stuartf:
The greater the deceleration the greater the force.
F = m x a
So if a lanky streak of piss falls s/he may slow down quicker than a fat bastard, hence experiencing a greater force.
ic
The greater the deceleration the greater the force.
F = m x a
So if a lanky streak of piss falls s/he may slow down quicker than a fat bastard, hence experiencing a greater force.
ic
In reply to Dave Hunter, Rock + Run:
Does this still mean that pies are good, though ?
> My mistake, decelleration, not force. A heavier climber will decellerate over a longer period. A lighter climber will decellerate faster and thus suffer more.
Does this still mean that pies are good, though ?
In reply to Cosmic John: one would imagine that for the zero dimensional scenario being discussed mass and weight are perfectly interchangeable.
In reply to Anonymous: Just been trying this assumin that the rate of deceleration as the rope goes tight is constant (which it won`t be, I know but thats too tricky for me to do). I appear to get the result that massand acceleration are inversely proportional to each other-though I may have dropped something out somewhere I will check again.
(I`ve also assumed that the strech of the rope is linearly related to the force)
This means that as the weight increases the acceleration will decrease by the same factor meaning the force experienced should be the same-is this represented in the form of maximum force to climber in rope specs?
As stated before though this has assumed a constant deceleration so is going o be wrong but if anyone has any recorded data from drop tests berformed in labs with the decelleration rates and masses Iwould love to see it.
(I`ve also assumed that the strech of the rope is linearly related to the force)
This means that as the weight increases the acceleration will decrease by the same factor meaning the force experienced should be the same-is this represented in the form of maximum force to climber in rope specs?
As stated before though this has assumed a constant deceleration so is going o be wrong but if anyone has any recorded data from drop tests berformed in labs with the decelleration rates and masses Iwould love to see it.
In reply to Dave Hunter, Rock + Run:
But that's because the force from the seatbelt is greater in the second case, rather than because the decelleration is greater. (It must be a greater force, since the impulse is the same, and the time is shorter).
> (In reply to stuartf)
>
> You are in your car. You slow from 200mph to 0mph in one second (it's an impressive car). Or you slow from 200mph to 0mph over twenty minutes. Which is going to be better for your seatbelt held body?
>
> You are in your car. You slow from 200mph to 0mph in one second (it's an impressive car). Or you slow from 200mph to 0mph over twenty minutes. Which is going to be better for your seatbelt held body?
But that's because the force from the seatbelt is greater in the second case, rather than because the decelleration is greater. (It must be a greater force, since the impulse is the same, and the time is shorter).
In reply to DaveN:
There's no point in assuming anything about the decelleration. What you need is the Force/Extension curve for the rope. From that, you can work out such things as the acceleration (as a function of time), etc. But if it's the force that matters then you don't really need to work out what the accelerations are - all you need to do is an energy balance to find out the maximum extension of the rope, and read off the graph what the biggest force will be.
There's no point in assuming anything about the decelleration. What you need is the Force/Extension curve for the rope. From that, you can work out such things as the acceleration (as a function of time), etc. But if it's the force that matters then you don't really need to work out what the accelerations are - all you need to do is an energy balance to find out the maximum extension of the rope, and read off the graph what the biggest force will be.
In reply to DaveN:
I guess it depends too much on the factors you've stated so the only way to know for sure would be to test it scientifically (or get someone who knows what they're doing to test it... like Petzl!...).
Either way I don't need it as an excuse to eat more pies
mmmmm... pie...
damn, dribbled on keyboard...
I guess it depends too much on the factors you've stated so the only way to know for sure would be to test it scientifically (or get someone who knows what they're doing to test it... like Petzl!...).
Either way I don't need it as an excuse to eat more pies
mmmmm... pie...
damn, dribbled on keyboard...
In reply to stuartf:
I realise that-I only did it to see what the mathematical model would be like no one else was so I thought I would. I also stated all the assumptions I made as I knew there would probably be loads of inaccuracies.
Because the decelleration is non constant the only way to accuratly work this out is to test it.
With the car decelleration thing, impulse is force X time so the greater time period leads to lower force if the impulse is the same. So the second case (lower deceleration rate) gves lower not higer force. (though re reading I think it was a typo)
I realise that-I only did it to see what the mathematical model would be like no one else was so I thought I would. I also stated all the assumptions I made as I knew there would probably be loads of inaccuracies.
Because the decelleration is non constant the only way to accuratly work this out is to test it.
With the car decelleration thing, impulse is force X time so the greater time period leads to lower force if the impulse is the same. So the second case (lower deceleration rate) gves lower not higer force. (though re reading I think it was a typo)
In reply to stuartf:
You can say something about the average decelearation, though. Assumeing that the fat bloke and the thin bloke are travelling at the same speed when the rope starts to stretch (Galileo and all that), and that the rope will stretch further with fatty on it, v^2-u^2=2as means that if the distance increases, v and u (starting and finishing velocities)are the same, the (average)acceleration (read deceleration) has to decrease.
You can say something about the average decelearation, though. Assumeing that the fat bloke and the thin bloke are travelling at the same speed when the rope starts to stretch (Galileo and all that), and that the rope will stretch further with fatty on it, v^2-u^2=2as means that if the distance increases, v and u (starting and finishing velocities)are the same, the (average)acceleration (read deceleration) has to decrease.
Jeez, you lot are so out of your collective depth.
When holding a leader fall, the rope functions as an energy-absorbing system.
The kinetic energy of a falling climber is half mv squared.
When a rope arrests a falling climber all the factors interact in an extremely complex and non-linear manner.
If the load on the falling climber's tie-in point is plotted against time, the energy absorbed will be proportional to the area under this graph. If, in the case of the light and heavy climbers, the rope stays within its normal elastic operating range, then the 2 resulting graphs will be of similar shape, but of different areas. This means that the curve for the heavier climber will extend further along the time axis before reaching equilibrium, but will also have a higher peak. It's pretty daft to imagine that it would be otherwise.
All the above is of course in addition to all the static loads due to the masses involved.
The moment-to-moment modelling of the actual velocities, forces, and energies involved in an actual leader fall are far to complex for a realistic "intuitive grasp" of the situation, such as people seem to be trying to do here.
Leave it to the real experts. (And by that I don't mean me either.)
Things are always more complicated than you think.
When holding a leader fall, the rope functions as an energy-absorbing system.
The kinetic energy of a falling climber is half mv squared.
When a rope arrests a falling climber all the factors interact in an extremely complex and non-linear manner.
If the load on the falling climber's tie-in point is plotted against time, the energy absorbed will be proportional to the area under this graph. If, in the case of the light and heavy climbers, the rope stays within its normal elastic operating range, then the 2 resulting graphs will be of similar shape, but of different areas. This means that the curve for the heavier climber will extend further along the time axis before reaching equilibrium, but will also have a higher peak. It's pretty daft to imagine that it would be otherwise.
All the above is of course in addition to all the static loads due to the masses involved.
The moment-to-moment modelling of the actual velocities, forces, and energies involved in an actual leader fall are far to complex for a realistic "intuitive grasp" of the situation, such as people seem to be trying to do here.
Leave it to the real experts. (And by that I don't mean me either.)
Things are always more complicated than you think.
In reply to GrahamD: hence me not being to surprised when at the end of 4 pages of diagrams and sums the model showed that the acceleration and mass were inversely proportional meaning that the forces are the same.
Next question is how does that relate to the figure for maximum force transferred to climber quoted in rope specs.
Next question is how does that relate to the figure for maximum force transferred to climber quoted in rope specs.
In reply to Cosmic John: I understand about energy balances as I deal with them every day so used them in the model.
I stated all the assumptions I made and only moddeled it to get some idea of what would happen.
I said I`d love to see some real data as my model would be inaccurate.
I agree about the curves and I kinow and understandthat we cannot hope to imagine the rate at which the energy is dissapated in to the system.
Has anyone actually got any data?
I stated all the assumptions I made and only moddeled it to get some idea of what would happen.
I said I`d love to see some real data as my model would be inaccurate.
I agree about the curves and I kinow and understandthat we cannot hope to imagine the rate at which the energy is dissapated in to the system.
Has anyone actually got any data?
In reply to DaveN:
trace of rope load agauinst time for a factor 2 fall here, time is not quite readable though.
http://biomech.me.unr.edu/wang/abstracts/rope_drop.htm
Can`t find anything relating to different different masses and the forces experienced.
trace of rope load agauinst time for a factor 2 fall here, time is not quite readable though.
http://biomech.me.unr.edu/wang/abstracts/rope_drop.htm
Can`t find anything relating to different different masses and the forces experienced.
In reply to DaveN:
Er.., Rope manufacturers?
The UIAA ?
That's who I meant when I said "the real experts".
Er.., Rope manufacturers?
The UIAA ?
That's who I meant when I said "the real experts".
The peak force recorded in the link i just posted was 9.7 kN (approx 970 kg-f) for a 7 meter fall in factor 2 conditions for a mass of 84kg on an 11mm rope. So thats valid for up to 3.5 meters above the last peice of gear.
Gear seems to be rated to about 22kN, I`m happy with this as I weigh about 84 kilos so now have some idea of what to expect-although cosmic John may now point out my rope may be different to that used in the test.
That would be the peak force and a good belayer with a soft catch might reduce that a bit.
Gear seems to be rated to about 22kN, I`m happy with this as I weigh about 84 kilos so now have some idea of what to expect-although cosmic John may now point out my rope may be different to that used in the test.
That would be the peak force and a good belayer with a soft catch might reduce that a bit.
In reply to DaveN:
Oh this really is fun isn't it.
Are you suggesting, using your numbers, that in this fall you have a safety factor of something like 2.25 over the breaking force for the karabiner? If so you are assuming that this load on the karabiner is applied in the same manner as it were in the test on the karabiner which was undoubtedly a conventional tensile test and therefore not dynamic. This difference will decrease the safety factor.
Oh this really is fun isn't it.
Are you suggesting, using your numbers, that in this fall you have a safety factor of something like 2.25 over the breaking force for the karabiner? If so you are assuming that this load on the karabiner is applied in the same manner as it were in the test on the karabiner which was undoubtedly a conventional tensile test and therefore not dynamic. This difference will decrease the safety factor.
In reply to Horse:
Not actually my numbers I just found them on the web
The caribiner experiences double the force in the rope once it`s loaded as the rope goes up from climber to crab and back to belayer so two times (minus a bit of friction) the rope tension acting downwards, so the FOS gets quite small.
If so you are assuming that this load on the karabiner is applied in the same manner as it were in the test on the karabiner which was undoubtedly a conventional tensile test and therefore not dynamic. This difference will decrease the safety factor.
I understand your background to be meturlagy (spelling) so you will have more of a handle on the test methods for materials than me. I agreethat the tensile test of the crab wouldn`t be dynamic-however the maximum force in the rope was 9.7 kN and crabs are rated to 22 kN this is a measurement of force, is it the rate at which the force is applied (impulse) that will affect the safty factor?
> (In reply to DaveN)
>
> Oh this really is fun isn't it.
Actually I am beggining to enjoy it
>
> Oh this really is fun isn't it.
> Are you suggesting, using your numbers, that in this fall you have a safety factor of something like 2.25 over the breaking force for the karabiner?
Not actually my numbers I just found them on the web
The caribiner experiences double the force in the rope once it`s loaded as the rope goes up from climber to crab and back to belayer so two times (minus a bit of friction) the rope tension acting downwards, so the FOS gets quite small.
If so you are assuming that this load on the karabiner is applied in the same manner as it were in the test on the karabiner which was undoubtedly a conventional tensile test and therefore not dynamic. This difference will decrease the safety factor.
I understand your background to be meturlagy (spelling) so you will have more of a handle on the test methods for materials than me. I agreethat the tensile test of the crab wouldn`t be dynamic-however the maximum force in the rope was 9.7 kN and crabs are rated to 22 kN this is a measurement of force, is it the rate at which the force is applied (impulse) that will affect the safty factor?
In reply to Cosmic John:
...Oh, and Horse of course.
> (In reply to DaveN)
>
> Er.., Rope manufacturers?
>
> The UIAA ?
>
> That's who I meant when I said "the real experts".
>
> Er.., Rope manufacturers?
>
> The UIAA ?
>
> That's who I meant when I said "the real experts".
...Oh, and Horse of course.
In reply to Dave Hunter, Rock + Run:
This actually squares up with what an "expert" from Troll once told me. I can see that in a climbing situation, it's insignificant, but I do wonder whether I should replace the 11mm rope in my VF kit with 9mm as I only weigh 8 stone.
> A heavier climber will decellerate over a longer period. A lighter climber will decellerate faster and thus suffer more.
This actually squares up with what an "expert" from Troll once told me. I can see that in a climbing situation, it's insignificant, but I do wonder whether I should replace the 11mm rope in my VF kit with 9mm as I only weigh 8 stone.
In reply to DaveN:
For future ref it is Metallurgy, the ancient science of cooking metals
The difference is in the way in which the force or load is applied to the specimen in the tensile test usually expressed as strain rate. Most tensile tests are done at much lower rates than are encountered in a dynamic impact.
But one would imagine that krab materials are designed to show good plasticity beyond yield and would therefore stretch plenty before going pop. I think it is this plasticity that is important in dynamic situations. But here this expert runs out of puff.
For future ref it is Metallurgy, the ancient science of cooking metals
The difference is in the way in which the force or load is applied to the specimen in the tensile test usually expressed as strain rate. Most tensile tests are done at much lower rates than are encountered in a dynamic impact.
But one would imagine that krab materials are designed to show good plasticity beyond yield and would therefore stretch plenty before going pop. I think it is this plasticity that is important in dynamic situations. But here this expert runs out of puff.
In reply to All the physisists:
As a punter using the same gear and taking the same length of fall, would it be safer climbing near to the equator or to one of the poles?
Baz.
As a punter using the same gear and taking the same length of fall, would it be safer climbing near to the equator or to one of the poles?
Baz.
In reply to Horse: Thank you for the correct spelling of you noble art. I never paid that much attention to the materials part of the degree and am regretting it now. The link I but up earlier had the maximum force applied in less that quarter of a second-much faster that the tensile tester i imagine.
I guess that the gear manufacturer are carefull in their selection of materials to give good plastisity and avoid failure.
Thanks for your reply as it was really useful.
I guess that the gear manufacturer are carefull in their selection of materials to give good plastisity and avoid failure.
Thanks for your reply as it was really useful.
In reply to GrahamD:
This is true. But the force required to cause this acceleration will be greater due to the increased mass.
> (In reply to stuartf)
>
> You can say something about the average decelearation, though. Assumeing that the fat bloke and the thin bloke are travelling at the same speed when the rope starts to stretch (Galileo and all that), and that the rope will stretch further with fatty on it, v^2-u^2=2as means that if the distance increases, v and u (starting and finishing velocities)are the same, the (average)acceleration (read deceleration) has to decrease.
>
> You can say something about the average decelearation, though. Assumeing that the fat bloke and the thin bloke are travelling at the same speed when the rope starts to stretch (Galileo and all that), and that the rope will stretch further with fatty on it, v^2-u^2=2as means that if the distance increases, v and u (starting and finishing velocities)are the same, the (average)acceleration (read deceleration) has to decrease.
This is true. But the force required to cause this acceleration will be greater due to the increased mass.
In reply to Everyone,Thanks for all the input I think my question has been answered somewhere along the line,I'll be printing Dave's info about fat climbers to show my mates,I'm sure they'll think its aimed at them.Just a quick note for anyone who thinks "it'll hold a car"is good enough,if you tell a novice climber"it'll hold a car"and it'll only take one to think this! if 22kn will hold a car surely 0.5 kn will hold a climber, so I'll just buy "gear krabs" and use them to climb with.
In reply to Cosmic John: manner.
Wrong. If the load is plotted against time the area under the graph will be the impulse. This will be greater for the heavier climber, since impulse = change of momentum. If you want the energy absorbed you need to plot the force against extension of the rope. The area under that curve will be the work done against the rope, i.e. the energy absorbed by it. Whether the heavier person experiences a greater force depends on the shape of the force-extension curve. If a greater extension always results in a greater force then the heavier person will experience a greater force. But since we don't know the shape of the curve then that is pointless speculation.
>
> If the load on the falling climber's tie-in point is plotted against time, the energy absorbed will be proportional to the area under this graph. If, in the case of the light and heavy climbers, the rope stays within its normal elastic operating range, then the 2 resulting graphs will be of similar shape, but of different areas. This means that the curve for the heavier climber will extend further along the time axis before reaching equilibrium, but will also have a higher peak. It's pretty daft to imagine that it would be otherwise.
>
> If the load on the falling climber's tie-in point is plotted against time, the energy absorbed will be proportional to the area under this graph. If, in the case of the light and heavy climbers, the rope stays within its normal elastic operating range, then the 2 resulting graphs will be of similar shape, but of different areas. This means that the curve for the heavier climber will extend further along the time axis before reaching equilibrium, but will also have a higher peak. It's pretty daft to imagine that it would be otherwise.
>
Wrong. If the load is plotted against time the area under the graph will be the impulse. This will be greater for the heavier climber, since impulse = change of momentum. If you want the energy absorbed you need to plot the force against extension of the rope. The area under that curve will be the work done against the rope, i.e. the energy absorbed by it. Whether the heavier person experiences a greater force depends on the shape of the force-extension curve. If a greater extension always results in a greater force then the heavier person will experience a greater force. But since we don't know the shape of the curve then that is pointless speculation.
In reply to stuartf:
Yep. Fatties experience greater force (because they are heavier) but slower average deceleration (because they fall further).
Yep. Fatties experience greater force (because they are heavier) but slower average deceleration (because they fall further).
In reply to Alison Stockwell:
It is NOT insignificant. It has potentially serious consequences regarding how ropes are rated. The difference between 60kg and 100kg is marked in a climbing situation. Even more so in via feratta.
> (In reply to Dave Hunter, Rock + Run)
> [...]
>
> This actually squares up with what an "expert" from Troll once told me. I can see that in a climbing situation, it's insignificant, but I do wonder whether I should replace the 11mm rope in my VF kit with 9mm as I only weigh 8 stone.
> [...]
>
> This actually squares up with what an "expert" from Troll once told me. I can see that in a climbing situation, it's insignificant, but I do wonder whether I should replace the 11mm rope in my VF kit with 9mm as I only weigh 8 stone.
It is NOT insignificant. It has potentially serious consequences regarding how ropes are rated. The difference between 60kg and 100kg is marked in a climbing situation. Even more so in via feratta.
Horse:
Bit confused about what you mean by "good plasticity past yield" do you mean that once yield has been reached, the material will take a fair bit more stress before fracture?
(just doing 4th year mech eng, so this is one of our classes this year)
Bit confused about what you mean by "good plasticity past yield" do you mean that once yield has been reached, the material will take a fair bit more stress before fracture?
(just doing 4th year mech eng, so this is one of our classes this year)
In reply to Dave Hunter:
So I'm confused now. Are you saying that light people are at greater risk because everything is calibrated for heavier people?
> It is NOT insignificant. It has potentially serious consequences regarding how ropes are rated. The difference between 60kg and 100kg is marked in a climbing situation. Even more so in via feratta.
So I'm confused now. Are you saying that light people are at greater risk because everything is calibrated for heavier people?
In reply to Alasdair Fulton:
Basically, for a ductile material such as aluminium or steel, yes. Particularly if it work-hardens significantly - you'll find it all in Engineering Materials 1 (Ashby & Jones), which could well be your course textbook...
Basically, for a ductile material such as aluminium or steel, yes. Particularly if it work-hardens significantly - you'll find it all in Engineering Materials 1 (Ashby & Jones), which could well be your course textbook...
In reply to stuartf:
No, John is quite right, given his entirely reasonable assumption that the rope is operating in an elastic manner (i.e. you DO know the shape of the curve, and it is pretty much linear). I hope for your sake that you never have to experience non-elastic behaviour!
To summarise, for the same fall being held by the same kit:
Greater (peak) force is required to stop a more massive climber than a less massive climber. This also means that the kit of the more massive climber experiences greater forces and so is closer to possible failure.
The more massive climber takes longer to come to a stop, so they experience a lower (peak) deceleration.
So in answer to Alison's question about Via Ferrata - for a given fall, your kit will experience lower forces than it would for a climber of the (higher) rated weight, so there is no need to worry about the kit failing. However, you personally will experience a greater deceleration.
Does this matter? I think we might need an expert in biomechanics to answer that. As a rough indicator, you might think of your back as a strut that is liable to fail by buckling under a compressive load. For an idealised (Euler) strut the force required to induce failure increases with the square of the size of the strut. The force exerted on the climber increases to the 3/2 power of the size of the climber (size here means one dimension, such as height, assuming all other dimensions change in proportion). This means that smaller climbers will be closer to failure than larger climbers. (This is a gross approximation; usual caveats apply)
> (In reply to Cosmic John) manner.
> [...]
>
> Wrong. If the load is plotted against time the area under the graph will be the impulse. This will be greater for the heavier climber, since impulse = change of momentum. If you want the energy absorbed you need to plot the force against extension of the rope. The area under that curve will be the work done against the rope, i.e. the energy absorbed by it. Whether the heavier person experiences a greater force depends on the shape of the force-extension curve. If a greater extension always results in a greater force then the heavier person will experience a greater force. But since we don't know the shape of the curve then that is pointless speculation.
> [...]
>
> Wrong. If the load is plotted against time the area under the graph will be the impulse. This will be greater for the heavier climber, since impulse = change of momentum. If you want the energy absorbed you need to plot the force against extension of the rope. The area under that curve will be the work done against the rope, i.e. the energy absorbed by it. Whether the heavier person experiences a greater force depends on the shape of the force-extension curve. If a greater extension always results in a greater force then the heavier person will experience a greater force. But since we don't know the shape of the curve then that is pointless speculation.
No, John is quite right, given his entirely reasonable assumption that the rope is operating in an elastic manner (i.e. you DO know the shape of the curve, and it is pretty much linear). I hope for your sake that you never have to experience non-elastic behaviour!
To summarise, for the same fall being held by the same kit:
Greater (peak) force is required to stop a more massive climber than a less massive climber. This also means that the kit of the more massive climber experiences greater forces and so is closer to possible failure.
The more massive climber takes longer to come to a stop, so they experience a lower (peak) deceleration.
So in answer to Alison's question about Via Ferrata - for a given fall, your kit will experience lower forces than it would for a climber of the (higher) rated weight, so there is no need to worry about the kit failing. However, you personally will experience a greater deceleration.
Does this matter? I think we might need an expert in biomechanics to answer that. As a rough indicator, you might think of your back as a strut that is liable to fail by buckling under a compressive load. For an idealised (Euler) strut the force required to induce failure increases with the square of the size of the strut. The force exerted on the climber increases to the 3/2 power of the size of the climber (size here means one dimension, such as height, assuming all other dimensions change in proportion). This means that smaller climbers will be closer to failure than larger climbers. (This is a gross approximation; usual caveats apply)
In reply to Alison Stockwell:
Light people (children as an extreme example) suffer more in falls than heavy folk it seems. This is due to more rapid decceleration. Your internals get juggled about more. Nowt to do with calibration. More to do with the limits of UIAA testing. Obviously this is influenced by fall factor too. But via ferrata (and some IRATA work) are even higher risk than climbing situations with regard to this.
Light people (children as an extreme example) suffer more in falls than heavy folk it seems. This is due to more rapid decceleration. Your internals get juggled about more. Nowt to do with calibration. More to do with the limits of UIAA testing. Obviously this is influenced by fall factor too. But via ferrata (and some IRATA work) are even higher risk than climbing situations with regard to this.
In reply to darkinbad:
I was claiming that John was wrong in saying that energy absorbed was the area under a force-time curve. It's not - it's the area under a force-displacement curve. The first one can be seen to be wrong by dimensional reasoning:
energy: mass*length^2*time^-2 (as in 1/2 * m * v^2)
force*time: mass*length*time^-2*time = mass*length*time^-1 (same units as momentum)
force*distance: mass*length*time^-2*length = mass*length^2*time^-2
The rope can behave in an elastic manner, without behaving in a LINEAR-elastic manner. In fact, I think that it's highly likely that the rope is non-linear. An elastic band does not behave in a linear fashion, but that does not mean that it's not elastic.
This is quite probably true, although I can think of situations where an identical force could work, such as if the rope were to exert a constant force/displacement characteristic (highly unlikely, I know...)
This is quite possibly true (as far as it goes), but I'm not sure how useful it is, since it involves some fairly massive approximations. I'm particularly wary about the Euler strut one, since a spine has a lot of curvature in it already, and it's going to have couples applied as well as pure compressive forces.
> (In reply to stuartf)
> [...]
>
> No, John is quite right, given his entirely reasonable assumption that the rope is operating in an elastic manner (i.e. you DO know the shape of the curve, and it is pretty much linear). I hope for your sake that you never have to experience non-elastic behaviour!
>
> [...]
>
> No, John is quite right, given his entirely reasonable assumption that the rope is operating in an elastic manner (i.e. you DO know the shape of the curve, and it is pretty much linear). I hope for your sake that you never have to experience non-elastic behaviour!
>
I was claiming that John was wrong in saying that energy absorbed was the area under a force-time curve. It's not - it's the area under a force-displacement curve. The first one can be seen to be wrong by dimensional reasoning:
energy: mass*length^2*time^-2 (as in 1/2 * m * v^2)
force*time: mass*length*time^-2*time = mass*length*time^-1 (same units as momentum)
force*distance: mass*length*time^-2*length = mass*length^2*time^-2
The rope can behave in an elastic manner, without behaving in a LINEAR-elastic manner. In fact, I think that it's highly likely that the rope is non-linear. An elastic band does not behave in a linear fashion, but that does not mean that it's not elastic.
> To summarise, for the same fall being held by the same kit:
>
> Greater (peak) force is required to stop a more massive climber than a less massive climber. This also means that the kit of the more massive climber experiences greater forces and so is closer to possible failure.
>
>
> Greater (peak) force is required to stop a more massive climber than a less massive climber. This also means that the kit of the more massive climber experiences greater forces and so is closer to possible failure.
>
This is quite probably true, although I can think of situations where an identical force could work, such as if the rope were to exert a constant force/displacement characteristic (highly unlikely, I know...)
> The more massive climber takes longer to come to a stop, so they experience a lower (peak) deceleration.
>
> So in answer to Alison's question about Via Ferrata - for a given fall, your kit will experience lower forces than it would for a climber of the (higher) rated weight, so there is no need to worry about the kit failing. However, you personally will experience a greater deceleration.
>
> Does this matter? I think we might need an expert in biomechanics to answer that. As a rough indicator, you might think of your back as a strut that is liable to fail by buckling under a compressive load. For an idealised (Euler) strut the force required to induce failure increases with the square of the size of the strut. The force exerted on the climber increases to the 3/2 power of the size of the climber (size here means one dimension, such as height, assuming all other dimensions change in proportion). This means that smaller climbers will be closer to failure than larger climbers. (This is a gross approximation; usual caveats apply)
>
> So in answer to Alison's question about Via Ferrata - for a given fall, your kit will experience lower forces than it would for a climber of the (higher) rated weight, so there is no need to worry about the kit failing. However, you personally will experience a greater deceleration.
>
> Does this matter? I think we might need an expert in biomechanics to answer that. As a rough indicator, you might think of your back as a strut that is liable to fail by buckling under a compressive load. For an idealised (Euler) strut the force required to induce failure increases with the square of the size of the strut. The force exerted on the climber increases to the 3/2 power of the size of the climber (size here means one dimension, such as height, assuming all other dimensions change in proportion). This means that smaller climbers will be closer to failure than larger climbers. (This is a gross approximation; usual caveats apply)
This is quite possibly true (as far as it goes), but I'm not sure how useful it is, since it involves some fairly massive approximations. I'm particularly wary about the Euler strut one, since a spine has a lot of curvature in it already, and it's going to have couples applied as well as pure compressive forces.
In reply to Alasdair Fulton:
For ductile materials this deformation (plasticity) can be very large (think about pulling a blob of blu tack) the next peak on the curve is the ultimate tensile strength, this is as strong as it gets. Strictly speaking in a tensile test, certainly with ductile materials, the final failure will be a ductile tear not a fracture.
One has to be a little careful about grouping materials together and saying things like aluminium and steel are ductile, this is not always the case. For example cast aluminium has very poor elongation (plasticity) whereas forged components have good elongation.
Good background reading on these aspects of materials is:
"The New Science of Strong Materials: Or Why You Don't Fall Through the Floor (Penguin Science)
~J.E. Gordon"
There is another one about structures but I can't recall the name of it. Both are excellent publications written in a fairly light and humorous style and the laymen need not be frightened by the prospect of lots of horrible maths as there isn't much. Gordon demonstrates important material properties with everyday examples such as bias cut fabrics for dresses and bras, why early aero engineers had to use bi and tri planes because they didn't understand stiffness etc. Really excellent reads and taught me more about materials than any formula ever has.
> Horse:
>
> Bit confused about what you mean by "good plasticity past yield" do you mean that once yield has been reached, the material will take a fair bit more stress before fracture?
>
Well it can't have plasticity below yield can it, that part of the curve represents elastic behaviour. Once a material reaches the yield point it doesn't break but continues to deform, the difference is that this deformation is now permanent whereas deformation below yield is not permanent.
>
> Bit confused about what you mean by "good plasticity past yield" do you mean that once yield has been reached, the material will take a fair bit more stress before fracture?
>
For ductile materials this deformation (plasticity) can be very large (think about pulling a blob of blu tack) the next peak on the curve is the ultimate tensile strength, this is as strong as it gets. Strictly speaking in a tensile test, certainly with ductile materials, the final failure will be a ductile tear not a fracture.
One has to be a little careful about grouping materials together and saying things like aluminium and steel are ductile, this is not always the case. For example cast aluminium has very poor elongation (plasticity) whereas forged components have good elongation.
Good background reading on these aspects of materials is:
"The New Science of Strong Materials: Or Why You Don't Fall Through the Floor (Penguin Science)
~J.E. Gordon"
There is another one about structures but I can't recall the name of it. Both are excellent publications written in a fairly light and humorous style and the laymen need not be frightened by the prospect of lots of horrible maths as there isn't much. Gordon demonstrates important material properties with everyday examples such as bias cut fabrics for dresses and bras, why early aero engineers had to use bi and tri planes because they didn't understand stiffness etc. Really excellent reads and taught me more about materials than any formula ever has.
In reply to Homeward:
I haven't read through this entire thread, but with the exception of a few people, it is blindingly obvious that very few people understand this simple concept enough.
Please stop using the terms stress, strain, strain rate, breaking strain, failure stress, force, peak force, impact force, etc, etc as if they are all interchangeable - to an engineer they are *all* very specific things. Different materials fail in completely different ways.
To answer the question:
A kilonewton (correct notation: kN) is a unit of *force*. It says nothing about mass. The base unit for force is the Newton, which is defined as the force required to accelerate a 1kg object by 1 m/s^2 (meter per second squared).
Force is not an easy concept to grasp because you can't really feel it. It's not helped by the fact that we grow up talking about "weight" as if it's mass. Force is very difficult to measure, we usually measure the acceleration (accelerometers, transducers), assume the mass is constant (a good approximation in almost all cases) and use F=ma.
Force alone does not tell you anything about failure without other information. What area is that force applied over? How is the force applied? Over what length of time (load conditions)? What is the material (behaviour, mechanical properties)? etc etc
If you want a handle of what a force is, a good approximation (on earth) is that a 1N (N = newton) force is equivalent to holding a mass of 0.1kg (100g), ie. divide by 10 (yes, technically 9.80665 etc etc). It is really important that this applies only to static loading conditions - dynamic situations are completely different.
So 1kN of force is roughly equivalent to holding (supporting) a mass of 100kg. This DOES NOT mean that a 100kg person will create a force of 1kN no matter what the situation!
It's not a simple subject, even the use of fall factors is a very big simplification on reality IMO.
I haven't read through this entire thread, but with the exception of a few people, it is blindingly obvious that very few people understand this simple concept enough.
Please stop using the terms stress, strain, strain rate, breaking strain, failure stress, force, peak force, impact force, etc, etc as if they are all interchangeable - to an engineer they are *all* very specific things. Different materials fail in completely different ways.
To answer the question:
A kilonewton (correct notation: kN) is a unit of *force*. It says nothing about mass. The base unit for force is the Newton, which is defined as the force required to accelerate a 1kg object by 1 m/s^2 (meter per second squared).
Force is not an easy concept to grasp because you can't really feel it. It's not helped by the fact that we grow up talking about "weight" as if it's mass. Force is very difficult to measure, we usually measure the acceleration (accelerometers, transducers), assume the mass is constant (a good approximation in almost all cases) and use F=ma.
Force alone does not tell you anything about failure without other information. What area is that force applied over? How is the force applied? Over what length of time (load conditions)? What is the material (behaviour, mechanical properties)? etc etc
If you want a handle of what a force is, a good approximation (on earth) is that a 1N (N = newton) force is equivalent to holding a mass of 0.1kg (100g), ie. divide by 10 (yes, technically 9.80665 etc etc). It is really important that this applies only to static loading conditions - dynamic situations are completely different.
So 1kN of force is roughly equivalent to holding (supporting) a mass of 100kg. This DOES NOT mean that a 100kg person will create a force of 1kN no matter what the situation!
It's not a simple subject, even the use of fall factors is a very big simplification on reality IMO.
In reply to Horse:
Sorry Horse I can only apologise-I would have apologised last night but after calling you an artist I decided to leave
> (In reply to DaveN)
>
> You were doing so well then you call it "art" so I am going home in disgust
>
> You were doing so well then you call it "art" so I am going home in disgust
Sorry Horse I can only apologise-I would have apologised last night but after calling you an artist I decided to leave
In reply to SimonG:
> (In reply to Homeward)
>
Different materials fail in completely different ways.
>
>
A rather ambiguous comment to say the very least. In reply to darkinbad:
Yes, a gross approximation, but not a bad first approximation in saying smaller people are closer to failure. Spine curvature is a big factor on spine failure mechanics, especially for certain types of injury (eg. lifting). Soft tissue injuries (especially very soft tissues like the brain) are often due to high accelerations (decelerations) and strain rates, rather than high forces or stresses. There's also the major complicating factor that children's musculoskeletal systems (and organs) are not completely developed, ie. inferior mechanical properties. Injury standards (ie. what force is enough to cause what type of injury) is still a big topic of debate - think car accidents and the amount of money poured into that kind of research.
> Does this matter? I think we might need an expert in biomechanics to answer that. As a rough indicator, you might think of your back as a strut that is liable to fail by buckling under a compressive load. For an idealised (Euler) strut the force required to induce failure increases with the square of the size of the strut. The force exerted on the climber increases to the 3/2 power of the size of the climber (size here means one dimension, such as height, assuming all other dimensions change in proportion). This means that smaller climbers will be closer to failure than larger climbers. (This is a gross approximation; usual caveats apply)
Yes, a gross approximation, but not a bad first approximation in saying smaller people are closer to failure. Spine curvature is a big factor on spine failure mechanics, especially for certain types of injury (eg. lifting). Soft tissue injuries (especially very soft tissues like the brain) are often due to high accelerations (decelerations) and strain rates, rather than high forces or stresses. There's also the major complicating factor that children's musculoskeletal systems (and organs) are not completely developed, ie. inferior mechanical properties. Injury standards (ie. what force is enough to cause what type of injury) is still a big topic of debate - think car accidents and the amount of money poured into that kind of research.
In reply to Horse:
I just wanted to get across the point that (high) force doesn't equal failure - you need to know other information.
I really didn't want to get into elastic/plastic failure, ductile vs brittle, fatigue vs static vs dynamic, metals vs ceramics vs polymers vs etc etc, especially as there's a metallurgist on the board !
> (In reply to SimonG)
> [...]
> Different materials fail in completely different ways.
> [...]
> A rather ambiguous comment to say the very least.
> [...]
> Different materials fail in completely different ways.
> [...]
> A rather ambiguous comment to say the very least.
I just wanted to get across the point that (high) force doesn't equal failure - you need to know other information.
I really didn't want to get into elastic/plastic failure, ductile vs brittle, fatigue vs static vs dynamic, metals vs ceramics vs polymers vs etc etc, especially as there's a metallurgist on the board !
In reply to SimonG:
Coward, you're no fun
> (In reply to Horse)
> [...]
>
> I just wanted to get across the point that (high) force doesn't equal failure - you need to know other information.
>
True, but not what you said.
> [...]
>
> I just wanted to get across the point that (high) force doesn't equal failure - you need to know other information.
>
> I really didn't want to get into .... especially as there's a metallurgist on the board !
Coward, you're no fun
In reply to Dave Hunter, Rock + Run (and others):
My VF kit is identical to the Camp VF set, so the rope runs through a device that looks a bit like a knuckle duster (don't know what it's called). It seems to me that if I fell on this thing, my body weight would be insufficient to pull the rope through the holes and absorb the shock as intended, and because the rope is quite short at this point I'd feel a significant shock-loading to my body. That's why I'm thinking of switching the rope to 9mm. More stretch, more likely to slip through the device, and on VF I think the chances of the rope running over anything sharp are minimal. I'd appreciate any comments as this is a complex issue and I'm looking for the best solution.
> But via ferrata (and some IRATA work) are even higher risk than climbing situations with regard to this.
My VF kit is identical to the Camp VF set, so the rope runs through a device that looks a bit like a knuckle duster (don't know what it's called). It seems to me that if I fell on this thing, my body weight would be insufficient to pull the rope through the holes and absorb the shock as intended, and because the rope is quite short at this point I'd feel a significant shock-loading to my body. That's why I'm thinking of switching the rope to 9mm. More stretch, more likely to slip through the device, and on VF I think the chances of the rope running over anything sharp are minimal. I'd appreciate any comments as this is a complex issue and I'm looking for the best solution.
In reply to Homeward:
Getting back to basics, I just had a look at my old aluminium crabs and they have between 3000 kg (Clog) and 1800 kg (Cassin) stamped on them. My modern ones have 25 kN stamped on them. I reckon both are strong enough, so the conversion factor is about 100 to give an idea of how the two ways of classifying crabs are concerned. The reason for the change was that whoever sets such standards decided to do it, that's all, no need to panic!
Fall factor: plus an interesting link for DaveN.
I will resist getting into this again, except (!) to say that a few days away allowed me to reflect on why it seemed an inadequate concept - over simplification, that is - and when I got home I had a look at the Beal site which gives examples bases on real tests carried out by some Italians which would seem to show why.
The error being basically due to the modellisation assuming that the tension in the rope is uniform, something which is clearly incorrect when you have a few runners in - so don't increase the spacing as you get higher (stop! you said you wouldn't!).
Here's the link: http://www.bealplanet.com/
... although it doesn't seem to work at the moment.
Getting back to basics, I just had a look at my old aluminium crabs and they have between 3000 kg (Clog) and 1800 kg (Cassin) stamped on them. My modern ones have 25 kN stamped on them. I reckon both are strong enough, so the conversion factor is about 100 to give an idea of how the two ways of classifying crabs are concerned. The reason for the change was that whoever sets such standards decided to do it, that's all, no need to panic!
Fall factor: plus an interesting link for DaveN.
I will resist getting into this again, except (!) to say that a few days away allowed me to reflect on why it seemed an inadequate concept - over simplification, that is - and when I got home I had a look at the Beal site which gives examples bases on real tests carried out by some Italians which would seem to show why.
The error being basically due to the modellisation assuming that the tension in the rope is uniform, something which is clearly incorrect when you have a few runners in - so don't increase the spacing as you get higher (stop! you said you wouldn't!).
Here's the link: http://www.bealplanet.com/
... although it doesn't seem to work at the moment.
In reply to Dave Hunter, Rock + Run:
Put like that, who could disagree!
The link is working now:
http://www.impact-force.info/anglais/impact.html
Put like that, who could disagree!
The link is working now:
http://www.impact-force.info/anglais/impact.html
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