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jt512
May 25, 2010, 7:02 PM
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adatesman wrote: jt512 wrote: adatesman wrote: jt512 wrote: Since one can convert between pounds of force and kilonewtons by multiplying by a constant, it is difficult to see why one would suck any more than the other. Jay I suggest you give some thought to the bigger picture suckage-wise. 1km = 1000m, 1ml = 1cm^3 and g = 9.81m/s^2 (~10m/s^s) is at least an order of magnitude easer to deal with than 1mi = 5280ft, 1ft = 12in, 1gal = 128oz = 231in^3 and g=32.2ft/s^s. It's no wonder one system is preferred over the other. Sure, conversions between metric units are easier than between English units. I still don't know how many fluid ounces there are in a cup. But it's no harder to calculate the impact force of a fall in English or SI units. Jay Do you really think there's the same amount of work in the calculation when you use 10.5 meters as compared to 34 feet, 5-3/8 inches?
Do you think there's the same amount of work in the calculation when you use 3 feet or 1.093633 m?
Jay
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jt512
May 25, 2010, 7:07 PM
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chadnsc wrote: Ok, using Jays handy link: With my weight of 240 pounds, a 15 foot fall on 40 feet of rope with the rope having a force rating of 9.4 Kn and assuming a friction factor of .33 we get: Standard On climber: 5.93 On belayer: 3.95 On anchor: 9.88 Friction-adjusted On climber: 6.61 On belayer: 4.41 On anchor: 11.02 All forces in Kn.
I think you calculated the fall factor incorrectly. The fall factor should be 15/40 = 0.375.
Jay
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chadnsc
May 25, 2010, 7:13 PM
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Hmm, I must have hit a wrong number, thanks Jay.
Fall Factor: .375
Impact Force Rating: 9.4
Friction Factor: .333
Standard
On climber 5.78
On belayer 3.85
On anchor 9.63
Friction-adjusted
On climber 6.45
On belayer 4.30
On anchor 10.75
All number in Kn.
Edit to add:
One question though Jay, what is the difference between 'standard' and 'friction adjusted' and why is the friction adjusted values higher? People here keep saying that the added friction in the system lowers the forces on the climber. I'm not disputing your link; I'd just like to actually know the truth.
(This post was edited by chadnsc on May 25, 2010, 7:15 PM)
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adatesman
May 25, 2010, 7:13 PM
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jt512
May 25, 2010, 7:19 PM
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adatesman wrote: jt512 wrote: adatesman wrote: jt512 wrote: adatesman wrote: jt512 wrote: Since one can convert between pounds of force and kilonewtons by multiplying by a constant, it is difficult to see why one would suck any more than the other. Jay I suggest you give some thought to the bigger picture suckage-wise. 1km = 1000m, 1ml = 1cm^3 and g = 9.81m/s^2 (~10m/s^s) is at least an order of magnitude easer to deal with than 1mi = 5280ft, 1ft = 12in, 1gal = 128oz = 231in^3 and g=32.2ft/s^s. It's no wonder one system is preferred over the other. Sure, conversions between metric units are easier than between English units. I still don't know how many fluid ounces there are in a cup. But it's no harder to calculate the impact force of a fall in English or SI units. Jay Do you really think there's the same amount of work in the calculation when you use 10.5 meters as compared to 34 feet, 5-3/8 inches? Do you think there's the same amount of work in the calculation when you use 3 feet or 1.093633 m? Jay No. That would be the same amount of work because neither measurement requires conversion to a base unit, as it would if the measurement was 3 feet, 6 inches. Did you miss my point, or just refusing to concede that you are in fact incorrect about it being more difficult in English units?
The only reason that it is more work to do the calculation using English units in your example is that you gave the number in mixed units and fractions. But we're not talking about building a shed; we're talking about doing physics calculations. If you had expressed length in decimal feet, the calculation would have been no harder than with decimal meters.
jay
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whiskeybullets
May 25, 2010, 8:56 PM
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jt512,
You don't find a measurements system that uses the same term for mass and weight to be confusing? How many people really understand that a kg~lbm~slug and kn~lbf? I'd wager that if you ask the average person, a pound is equivalent to a kilogram, and no one knows what is equivalent to a kilonewton. Since in the US we often fail to really specify between weight and mass, and we often quote people's "body weight" in either Kg or Lbs, much confusion regarding the units does exist.
That's why so many climber's are confused about kN ratings. To most everyone, a lb measures the same quantity as a kg, even though that's not totally correct. Adatesman's link does help to clarify this, if people take the time to read it.
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jt512
May 25, 2010, 9:16 PM
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whiskeybullets wrote: jt512, You don't find a measurements system that uses the same term for mass and weight to be confusing?
You mean, like kilograms of mass and kilograms of force?
Jay
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jt512
May 25, 2010, 9:54 PM
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chadnsc wrote: Hmm, I must have hit a wrong number, thanks Jay. Fall Factor: .375 Impact Force Rating: 9.4 Friction Factor: .333 Standard On climber 5.78 On belayer 3.85 On anchor 9.63 Friction-adjusted On climber 6.45 On belayer 4.30 On anchor 10.75 All number in Kn. Edit to add: One question though Jay, what is the difference between 'standard' and 'friction adjusted' and why is the friction adjusted values higher? People here keep saying that the added friction in the system lowers the forces on the climber. I'm not disputing your link; I'd just like to actually know the truth.
Read through this thread and/or skim through this paper. If you still have questions, let me know.
Jay
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adatesman
May 25, 2010, 10:06 PM
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adatesman
May 25, 2010, 10:07 PM
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jt512
May 25, 2010, 10:09 PM
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adatesman wrote: jt512 wrote: whiskeybullets wrote: jt512, You don't find a measurements system that uses the same term for mass and weight to be confusing? You mean, like kilograms of mass and kilograms of force? Jay From your second link: In reply to: The kilogram-force has never been a part of the International System of Units
I didn't say it had been.
Jay
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clc
May 25, 2010, 10:12 PM
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from my limited physics in university 10 years ago.
kg=mass and Newtons=force {f=mass,kg X 9.8m/s^2}
9.8m/s^2 is the acceleration due to gravity on earth at sea level. It decrease as the distance from earth increases.
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adatesman
May 25, 2010, 10:14 PM
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dugl33
May 25, 2010, 10:16 PM
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definition of newton: a unit of force equal to the force that imparts an acceleration of 1 m/sec/sec to a mass of 1 kilogram
1 kn = 1000 newtons
1.) The gravitational constant on earth is 9.8 m/sec/sec
2.) so, it takes 9.8 newtons to counteract gravity and just hold 1 kilogram of mass in place, dig it?
1000 newtons / 9.8 m/s^2 = 102.04 kg
So, 1 KN = 102.04 kgs * 2.2 lbs per kg = 225 lbs
easy conservative approximation is to double the kn number and add two zeros
e.g. 22 kn is about 4400 lbs. (more precisely 4950 lbs)
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gunkiemike
May 25, 2010, 10:24 PM
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jt512 wrote: Thanks to gravity being a constant, you don't need to know the mass of a falling object to calculate the impact force on a piece of climbing gear. Jay
So you're saying the impact force is independent of the mass of the falling object. Man, I hope I don't drop a biner in a FF 2 fall - it'll rip my anchor! OTOH, I guess it's OK to let my 150 kg partner whip onto that RP.
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curt
May 25, 2010, 10:39 PM
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chadnsc wrote: One question though Jay, what is the difference between 'standard' and 'friction adjusted' and why is the friction adjusted values higher? People here keep saying that the added friction in the system lowers the forces on the climber. I'm not disputing your link; I'd just like to actually know the truth.
Oh well, I'll answer this anyway. Friction adjusted forces are higher because the friction over the top piece prevents the portion of the rope on the belayer side (from the top piece to the belayer) from fully sharing in the absorption of energy. So, it's not correct to say that added friction lowers the force on the climber--but it is correct to say that added friction lowers the force on the belayer.
Curt
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shockabuku
May 25, 2010, 10:44 PM
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adatesman wrote: jt512 wrote: adatesman wrote: jt512 wrote: whiskeybullets wrote: jt512, You don't find a measurements system that uses the same term for mass and weight to be confusing? You mean, like kilograms of mass and kilograms of force? Jay From your second link: In reply to: The kilogram-force has never been a part of the International System of Units I didn't say it had been. Jay Yet you somehow thought it a valid example of how units are confusing in the Metric system?
The "metric system" and the SI are not exactly the same thing.
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jt512
May 25, 2010, 10:48 PM
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adatesman wrote: jt512 wrote: adatesman wrote: jt512 wrote: whiskeybullets wrote: jt512, You don't find a measurements system that uses the same term for mass and weight to be confusing? You mean, like kilograms of mass and kilograms of force? Jay From your second link: In reply to: The kilogram-force has never been a part of the International System of Units I didn't say it had been. Jay Yet you somehow thought it a valid example of how units are confusing in the Metric system?
I never said that units in the metric system are confusing. The point I was trying to make was that the kilogram is commonly used to express both weight and mass, even though it is not an SI unit of weight. I have owned carabiners whose rated strengths were marked in kilograms. When I step on a European bathroom scale, it gives me my weight in kilograms. The international freight company DHL has an online pounds-to-kilograms weight converter. Bicycle wheel spoke tensions are stated in kg-f (see here, for example). And so on.
Jay
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jt512
May 25, 2010, 10:51 PM
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gunkiemike wrote: jt512 wrote: Thanks to gravity being a constant, you don't need to know the mass of a falling object to calculate the impact force on a piece of climbing gear. Jay So you're saying...
I have noticed that when someone begins a post with the phrase "so you're saying" what follows is inevitably something I did not say.
Jay
(This post was edited by jt512 on May 25, 2010, 10:51 PM)
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adatesman
May 25, 2010, 10:51 PM
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jt512
May 25, 2010, 10:54 PM
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adatesman wrote: jt512 wrote: I never said that units in the metric system are confusing. The point I was trying to make was that the kilogram is commonly used to express both weight and mass, even though it is not an SI unit of weight. I have owned carabiners whose rated strengths were marked in kilograms. When I step on a European bathroom scale, it gives me my weight in kilograms. The international freight company DHL has an online pounds-to-kilograms weight converter. Bicycle wheel spoke tensions are stated in kg-f (see here, for example). And so on. Jay So what you're really saying is...
See my previous post.
Jay
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roadstead
May 25, 2010, 11:10 PM
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adatesman wrote: whiskeybullets wrote: Adatesman's link does help to clarify this, if people take the time to read it. Not my link; Roadstead posted it.
I posted it for the OP...what you boys did after that is
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ptlong
May 25, 2010, 11:17 PM
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curt wrote: chadnsc wrote: One question though Jay, what is the difference between 'standard' and 'friction adjusted' and why is the friction adjusted values higher? People here keep saying that the added friction in the system lowers the forces on the climber. I'm not disputing your link; I'd just like to actually know the truth. Oh well, I'll answer this anyway. Friction adjusted forces are higher because the friction over the top piece prevents the portion of the rope on the belayer side (from the top piece to the belayer) from fully sharing in the absorption of energy. So, it's not correct to say that added friction lowers the force on the climber--but it is correct to say that added friction lowers the force on the belayer.
Curt is correct, but this might seem confusing given that Jay's calculator shows larger forces for both the climber and belayer for the friction-adjusted model. The reason for this is that it's being compared to a "standard" model that also includes friction. In a true no-friction model the force on the belayer would be higher than both the standard and friction-adjusted models.
Confused? Jay's paper covers this rigorously. But a simpler explanation of the differences between the models would be useful. Perhaps Jay could include one on his calculator page.
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jt512
May 25, 2010, 11:25 PM
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roadstead wrote: adatesman wrote: whiskeybullets wrote: Adatesman's link does help to clarify this, if people take the time to read it. Not my link; Roadstead posted it. I posted it for the OP...what you boys did after that is
Threads asking about what a kilonewton is should be exactly this long:
225 lb.
Yet every time the question comes up a dozen people have to bring up the distinction between weight (or force) and mass, and that kg is a measure of mass and kN is a measure of force. At that point someone (usually me), feels compelled to explain that there is a simple mathematical relationship between kg and kN, which allows for the sensible and common practice of expressing force (or weight) in kg. On average, only about one or two of the weight–mass pedants understands this, even though it is obvious from their own equations; and thus they argue endlessly about definitions they've memorized but haven't thought very much about. At some point, a parallel discussion about the difference between force and energy (which actually is important in understanding impact forces on gear) emerges, and then all hell breaks loose.
Jay
(This post was edited by jt512 on May 26, 2010, 7:27 PM)
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jt512
May 25, 2010, 11:34 PM
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ptlong wrote: curt wrote: chadnsc wrote: One question though Jay, what is the difference between 'standard' and 'friction adjusted' and why is the friction adjusted values higher? People here keep saying that the added friction in the system lowers the forces on the climber. I'm not disputing your link; I'd just like to actually know the truth. Oh well, I'll answer this anyway. Friction adjusted forces are higher because the friction over the top piece prevents the portion of the rope on the belayer side (from the top piece to the belayer) from fully sharing in the absorption of energy. So, it's not correct to say that added friction lowers the force on the climber--but it is correct to say that added friction lowers the force on the belayer. Curt is correct, but this might seem confusing given that Jay's calculator shows larger forces for both the climber and belayer for the friction-adjusted model. The reason for this is that it's being compared to a "standard" model that also includes friction. In a true no-friction model the force on the belayer would be higher than both the standard and friction-adjusted models. Confused? Jay's paper covers this rigorously. But a simpler explanation of the differences between the models would be useful. Perhaps Jay could include one on his calculator page.
I plan to add some online documentation to the calculator in the near future. In the meantime, some of the obvious questions about the models are answered in this thread, where there is also a link to the paper.
Jay
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