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sync
Sep 12, 2003, 6:44 PM
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In reply to: Only when climbing on Saturn.... Uranus is a whole different issue...
Hehe... but actually, the "surface" gravity on Saturn is about the same as on Earth.
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mustclimb69
Sep 12, 2003, 12:06 PM
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I am well aware of KN ratings and lb ratings. my question is how do you convert KN into LB (safety standards for Oakville Firedept, Police and rescue)
Here is my theory KN is approx 220lb of force thus KN rating x 220 will give you a rough Idea of the rating.
Problem is anything rated by LBs is divided up into tensile/breaking and safe working load (1/10 of Tensile).
So if 9mm static line is 16 KN (3520lbs) what is the tensile rating and the safe working load??
I dont think an accurate rating is possible due to one is dead weight and one is based on skock loading.
Thanks
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vertical_reality
Sep 12, 2003, 1:30 PM
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16 kN is the tensile strength. The working load is anything below that.
BTW: 1kN = 224.8 LBs force
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da5id
Sep 12, 2003, 1:43 PM
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The problem with this type of conversion is that, while both are measures of force, weight is very specifically applied to the constant force of gravity, whereas newtons, or kilonewtons, are independent of gravity. If you're dealing with just static lines, entirely static loads, etc. then such a lbs. to kN conversion works. However, even with static lines, any acceleration against gravity, such as lifting this load, would increase the force on the rope. So you're conversion factor may be correct (i haven't checked it or anything though), but just bear in mind that force and weight aren't quite exactly the same thing.
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mustclimb69
Sep 12, 2003, 2:10 PM
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vertical_reality
BTW "Approx" is an abbreviation for approximately and it is in the dictionary.
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sync
Sep 12, 2003, 2:44 PM
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In reply to: So you're conversion factor may be correct (i haven't checked it or anything though), but just bear in mind that force and weight aren't quite exactly the same thing.
Weight is a force.
The conversion is fine, as long as you don't stray too far from the surface of the Earth.
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vertical_reality
Sep 12, 2003, 4:00 PM
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In reply to: vertical_reality BTW "Approx" is an abbreviation for approximately and it is in the dictionary.
Read your second sentence. I answered your question.
Is As*hole in the dictionary?
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ninjaslut
Sep 12, 2003, 4:10 PM
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Can anyone just state the formula for calculating the Force in kN which is generated by a falling climber of weight "x" who has fallen for distance "y"? (For now, let's just exclude the effect of a dynamic rope.)
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jcinco
Sep 12, 2003, 4:52 PM
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In reply to: Can anyone just state the formula for calculating the Force in kN which is generated by a falling climber of weight "x" who has fallen for distance "y"? (For now, let's just exclude the effect of a dynamic rope.)
If you exclude the effect of a dynamic rope, then the force is infinite.
I'm bored right now, so I'll go through a little physics lesson,
A climber of mass, M, falling a distance h, will have a kinetic energy of:
KE = M*g*h , where g=9.8 m/s/s
To stop the fall, the rope needs to do "work" on the falling climber equal to his kinetic energy.
This is where things get tricky, but you can calculate the average force from how much the rope stretches. Suppose the rope stretches an amount d. The the work exerted by the rope on the faller is
Work = F * d , where F is the average force exerted by the rope.
Therefore the average for exerted by the rope is:
F = M*g*h/d
So if the rope doesn't stretch, d=0, and F is infinite.
However, to calculate the maximum force, its not that simple since the rope will not exert a constant force while it stretches. You could assume the rope behaves like a spring, but the calculation gets a little more complicated at this point. If the rope has a spring constant, k, then you can calculate the amount of stretch by using the potential enery of a spring,
PE = (1/2)*k*d^2 , and equting it to the kinetic energy of the faller to calculate the spring constant,
k=2*M*g*h/d^2
Therefore the maximum force exerted during the fall is
F_max= k*d = 2*M*g*h/d assuming the rope behaves like a spring.
Anyway, I hope that helps.
(edit: km changed to m is 9.8 m/s/s. Sorry guys, I'm an astrophysicist and we always use km/s for velocity)
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da5id
Sep 12, 2003, 5:26 PM
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In reply to: Weight is a force. The conversion is fine, as long as you don't stray too far from the surface of the Earth.
distance from the surface of the earth has nothing to do with the difference between weight and force. weight is the force of attraction between a body and the earth, or whatever gravitational body you calculate with. thats different from the force a fall exerts on a rope.
but that doesnt really matter anyway, and jcinco summed it up better than i have the energy to
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robmcc
Sep 12, 2003, 5:54 PM
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In reply to: :shock: :shock: :shock: Gravity is 9.8 km/s/s? Ouch! Might want to edit that. ;)
Yeah, and quickly, too! I'm going climbing tomorrow and I don't want to have to find a rope that can take the 1-8 meganewton impact when I peel.
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bertman
Sep 12, 2003, 6:23 PM
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damn, coldclimb got to it first :roll:
I was gonna have fun wit that one.
I'll just have to make sure I dont go skydiving again until that km/s/s thing is fixed :lol:
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curt
Sep 12, 2003, 6:24 PM
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da5id,
In reply to: distance from the surface of the earth has nothing to do with the difference between weight and force.
Also, refer to jcinco's post for the most correct model here and disregard most of the rest.
Curt
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fear
Sep 12, 2003, 6:24 PM
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In reply to: In reply to: KE = M*g*h , where g=9.8 km/s/s :shock: :shock: :shock: Gravity is 9.8 km/s/s? Ouch! Might want to edit that. ;)
Only when climbing on Saturn.... Uranus is a whole different issue...
-Fear
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sync
Sep 12, 2003, 6:43 PM
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In reply to: distance from the surface of the earth has nothing to do with the difference between weight and force. weight is the force of attraction between a body and the earth, or whatever gravitational body you calculate with. thats different from the force a fall exerts on a rope.
Yes, to compute the force a fall exerts on a rope is not just your weight. I see now what you are getting at.
However, the numerical conversion of pounds to newtons that he said is correct, and that's what I was commenting on.
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sync
Sep 12, 2003, 6:44 PM
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In reply to: Only when climbing on Saturn.... Uranus is a whole different issue...
Hehe... but actually, the "surface" gravity on Saturn is about the same as on Earth.
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