oh for pete's sake

[tshanefreeman]

Tom,

While I'm not trying to defend myself, I think that my description sounded a little more devistating than it actually was. The gate had come open about 3/4 to 1 revolution. The gate itself was still on 4+ treads. But it still sent a chill down my spine and created a lump in my throat. The one thing that I can say that I learned from this was that my few years of experience had conditioned me to visually inspection not a physically inspect, you now actually turning, snapping, and clicking things.

Point taken!

 

[John Paul Sanborn]

I'd have to use it for a while before saying anything, though the concept does give a V8 moment. I think the connector is a tad over built right now. They do say that this is the first incarnation and better renditions are in the works.

 

[Joe]

Originally posted by Tom Dunlap
The formula that Don Blair has promoted for free fall loads is to take the weight of the piece, multily by the fall distance and add the weight of the piece. There IS NOT a doubling. For a twelve pound saw dropping three feet...you do the math.

Tom

Hi, Tom;

Ken correcly used the Don Blair rule of thumb earlier in the thread. Ken also quoted the rule of thumb along with Don's example earlier in the thread, but later than in the 1st example.

I tried rereading some of the replies in this thread and don't see where some1 went so far off track they doubled something. I'm scratching my head here.

Joe

 

[hillbilly]

Made I'm mistaking, but wasn't the original question really:
What force is generated on the saw lanyard if the saw was
dropped l meters ?

Joe wrote:
"O.K., now the saw is dropped and falls 1 foot. Multiply the distance of fall by the weight. This gives 7 ft›lbs of energy. It is kinetic energy, energy spent."

Joe, you can't say that kinetic energy is measured in ft*lbs.
Check out the links you posted yourself.
Ft vs. meters and lbs vs kg might be a difference between countries. However kinetic energy is not.
Kinetic energy is measured in kg * m^2/s^2.
kg means kilgrams, m means meters, s means seconds.

The formula for kinetic energy is 0.5 * m * v^2.
In words that is: half the mass multiplied with the square of
the objects velocity.

Kinetic energy in a free fall could also be interpreted as the
differnece in potential energy.
KE = m * g * l
m is the mass of the object
g is the gravitational constant of the earth, approx. 9.82
l is the length of the fall.
Note that this yields the same unit as above: kg * m^2/s^2

Now an example:
A chainsaw weighing 3.5 kg (7 lbs) is being dropped 1 meter
will have the kinetic energy = 3.5 * 9,82 * 1 = 34.37
This is NOT the force that will work on the lanyard, it is infact NOT
a force at all, it is the kinetic energy that the chainsaw now posesses.

This kinetic energy must now be converted into some other form
of energy to stay at 1 meter below the original position.
It must be converted to heat energy.
How this is done depends heavily on the material of the lanyard.
How this is done also dictates the actual force that will work on
the lanyard.

This is where my knowledge in mechanics start to run out.
So I'll leave it to Joe to calculate the actual force...good luck!

 

[Joe]

Originally posted by hillbilly
Made I'm mistaking, but wasn't the original question really:
What force is generated on the saw lanyard if the saw was
dropped l meters ?

Joe wrote:
"O.K., now the saw is dropped and falls 1 foot. Multiply the distance of fall by the weight. This gives 7 ft›lbs of energy. It is kinetic energy, energy spent."

Joe, you can't say that kinetic energy is measured in ft*lbs.
Check out the links you posted yourself.
Ft vs. meters and lbs vs kg might be a difference between countries. However kinetic energy is not.
Kinetic energy is measured in kg * m^2/s^2.
kg means kilgrams, m means meters, s means seconds.

We're still cool on this. You aren't wrong and neither am I. It's just the difference of using the metric system by you and the english system by me. The link uses the metric system.

The formula for kinetic energy is 0.5 * m * v^2.
In words that is: half the mass multiplied with the square of
the objects velocity.

...which can be measured using the metric system like in the links or the english system in the U.S.A.

Kinetic energy in a free fall could also be interpreted as the
differnece in potential energy.
KE = m * g * l
m is the mass of the object
g is the gravitational constant of the earth, approx. 9.82
l is the length of the fall.
Note that this yields the same unit as above: kg * m^2/s^2

Now an example:
A chainsaw weighing 3.5 kg (7 lbs) is

In the english system we use the slug or I think also pound mass as the unit of mass. Since converting from lbs(force) to kilograms(mass) is easier tonight, 1 kg = 2.205 lbs (approximately) so a 7 lb saw has a mass of 3.175 kg.

1 lb = 4.45 N = .454 kg.

1 ft = .3048 m

Joe

 

[Tom Dunlap]

Ok, Joe and Hillbilly,

Now that the physics is working, let us know what it's going to feel like to a climber who were to drop that seven pound saw, three feet. Let's say that the lanyard is made out of a material that has no stretch. One of the high mod fibers available and there is no break-away. Would I be heading off to my chiropractor after dropping the saw? Would I be passing blood from internal organ damage? Or would I be wishing for a little bit of bungie in the lanyard?

Thanks!

Tom

 

[hillbilly]

Joe, sorry if I came on too hard on you, but you were pretty
bald a couple of posts back, and you still didn't come up with
an answer.

My mistake about the lbs, I could have calculated the
lbs to kg a bit more accurate than by roughly dividing
by 2.

The english system uses lbs to measure force as well as mass !?
It sure is a strange system!
I still can't see how you can ignore the gravitational constant
of the earth when doing the kinetic energy calculations.

No material is truly static, even steel stretches, in fact
steel has pretty good stretching capabilities, but not
compared to rope. Of course you know this.
This really has to do with spring forces and the characteristics
of the lanyard material. I would love to see a rope technichinan
come on the board and clear this out.
I rest my case as I find my mechanics knowledge hinders
me. I also feel that I'm really just messing this thread up
with my metric system.

How I wish your countries could convert to
the metric system.
1 foot = 12 inches !?
1 yard = 3 feet = 36 inches !?
How about 1 meter = 100 centimeters = 1000 millimeters
and 1 cubic meter = 1000 liters


Joe, I would love to see the resulting force and I will try
to keep my metric mouth shut

 

[Stumper]

No doubt about it Hillbilly, the metric system is much more logical and manageable overall. Unfortunately, whatever is familiar seems easier so we have resisted the conversion to the better system. The gravitational constant is there in the english system, it just isn't always made obvious. The falling object and foot pounds calculation is one of the rare 'easy' ones on the english system-a fact I wasn't aware of prior to this thread. A typical projectile energy calculation in the ES would be Projectile weight in grains (7000 grains per pound) x velocity in feet per second squared. . divide result by 450240 = energy in ft lbs. The 450240 number is the place where the gravitatonal constant is hidden-it was used in calculating that number so that several steps can be skipped. The constant is there in Joe's simple calculation because the english unit of force ( the foot pound) is actually based upon the force of a pound FALLING one foot. The gravitational constant was included from the beginning.

 

[John Paul Sanborn]

Soooo, since the saw falls 3 feet, the work is 21 ft/lbs. So this is waht the climber feels jerk on his lanyard?

So I drop a 15# saw on a 5 foot lanyard I feel 75# of force?

If I drop a 100 # chunk 60 ft there is 600# of force in the impact?

 

[Stumper]

100# x 60'= 6000 OUCH!

 

[The Climber]

Thank God my lanyard isn't 60' long!

 

[Joe]

Originally posted by hillbilly
Joe, sorry if I came on too hard on you, but you were pretty
bald a couple of posts back, and you still didn't come up with
an answer.

I'm not sure how to interpret this statement. I'm pretty bald on all my posts, a little smokey too.

My mistake about the lbs, I could have calculated the
lbs to kg a bit more accurate than by roughly dividing
by 2.

I figured it was another poster who wasn't accustomed to the english system of measure. I offered the unit conversions as a courtesy so the thread could move forward and we could see our similarities.

The english system uses lbs to measure force as well as mass !?
It sure is a strange system!

If the post is reread, it will be shown the unit of mass is the slug. I think the term slug is being phased out in favor of the term "pound mass".
I'm not sure of this.

I still can't see how you can ignore the gravitational constant
of the earth when doing the kinetic energy calculations.

The pound is a unit of force. The gravitational constant is there.

No material is truly static, even steel stretches, in fact
steel has pretty good stretching capabilities, but not
compared to rope. Of course you know this.

I wouldn't be so sure of this. Some high modulus fibers are pretty stiff.

This really has to do with spring forces and the characteristics
of the lanyard material. I would love to see a rope technichinan
come on the board and clear this out.

It really has to do with a lot of things. What we are discussing in this thread is the tip of the iceberg. I'm thankful I know even a little. I doubt if a rope technician would ever show up and dump on a public site like this one. It would be nice if 1 did show up and dump.

I rest my case as I find my mechanics knowledge hinders
me. I also feel that I'm really just messing this thread up
with my metric system.

My mechanics knowledge hinders me also. There are people here who can handle the metric system and people who cannot. Perhaps the exposure of the metric system will at least let those who aren't familiar with it know it when they see it.

How I wish your countries could convert to
the metric system.
1 foot = 12 inches !?
1 yard = 3 feet = 36 inches !?
How about 1 meter = 100 centimeters = 1000 millimeters
and 1 cubic meter = 1000 liters

I guess you "are" familiar with the english system of measure. You weren't trying to pull my leg into thinking you weren't were you? It worked.

Joe, I would love to see the resulting force and I will try
to keep my metric mouth shut

What force are you asking me to calculate?

This is NOT the force that will work on the lanyard, it is infact NOT a force at all, it is the kinetic energy that the chainsaw now posesses.

Do you want me to calculate the kinetic energy force?

This kinetic energy must now be converted into some other form
of energy to stay at 1 meter below the original position. It must be converted to heat energy. How this is done depends heavily on the material of the lanyard. How this is done also dictates the actual force that will work on the lanyard.

Really? How does this work?

This is where my knowledge in mechanics start to run out.

Me too!

So I'll leave it to Joe to calculate the actual force...good luck!

What is it I need to calculate that you can't? I still don't understand what I'm supposed to calculate?

Joe

 

[ramanujan]

the actual force that a climber will feel when he drops his saw will be much less than any simple calculation gives.

Final Velocity=sqrt(2 x 9.8m/s² x distance)


sqrt(2 x 9.8m/s² x 1)=4.427m/s


so the saw would be moving 4.427m/s (assuming, unreasonably, that the saw fell straight down, didn't bump anything during it's fall, nor did it tumble.)

4.427m/s=9.9mph

now for the force felt by the climber.

Force=Mass x Acceleration

In this case the acceleration is in fact deacceleration as the saw is slowing down, and we'll assume it's at a constant rate. One more thing we'll assume is that it takes the saw about 8 centimeters (3.15 inches) to stop once it starts slowing down (due to stretch of the lanyard, swing of the saw, movement of the climbers belt, etc.)


(Final Velocity)²=(Initial Velocity)² + (2)(Acceleration)(Distance)

Acceleration=(4.427²)/((2)(0.8))=11.81m/s²

Force=Mass x Acceleration

Force= (3.18kg)(11.81m/s²)=37.56kgm/s²=0.369 kN=82.8 lbs

 

[Joe]

Originally posted by ramanujan
(Final Velocity)²=(Initial Velocity)² + (2)(Acceleration)(Distance)

Acceleration=(4.427²)/((2)(0.8))=11.81m/s²

My hat comes off for you sir, "but"

4.427² = 19.598 m²/s²

1 meter = 100 cm then 8 cm = 0.08 m

2 × 0.08 m = .16 m

(19.598 m²/s²)/ .16 m = 122.49 m/s²

Force=Mass x Acceleration

Force= (3.18kg)(11.81m/s²)=37.56kgm/s²=0.369 kN=82.8 lbs

(3.18 kg)(122.49 m/s²) = 389.5 kgm/s²

Isn't kgm/s² the equivalence of a N(newton)?

So 389.5 kgm/s² = 389.5 N = .3895 kN

then (389.5 N / 1 N) × 0.2248 lbs = 87.6 lbs ?

It's a minor thing but only in the sense of computational correctness.

It was great to read this post!

Joe

 

[TheTreeSpyder]

Wow, am i understanding that a 7# saw dropping 4' hits with 82.8# of force?

So a 4' drop on a calculated table would almost be @4' drop force=~12 x weight of object? For the speed @ 4' is the same regardless of weight, only resistant forces (air friction to shape of weight, landing resistance)? Then even a 10:1 SWL sounds risky to me, if it can be overrun in 4'; for, certainly as higher drops would yield a higher force multiplier.

So in topping and catching on the same spar; where the center of balance moves from rest (precut) to catch; at a change of 4', if i determine that i want to add the SWL factor in after calculation of force (83# force for 7# saw,after 4' drop x SWL of 10 = 830#+ test line). If i seek to calculate that guardedly, i have to deal with a 120SWL @ 4' drop. So the safe capacity of a 6000 test line would be 50# a 4' drop(50# x 120SWL=6000)??? @ 4' drop factor of 12x, arbor plex 5400 will break @450# load, how many cycles to failure?

i get lost in these higher calculations; but i know that the same math must work in both examples of falling saw, man or log. They all teach the same lesson.

i have googled trying to find a simple chart that says @ 1' drop force will be Y x Weight, @2' etc. If Force = Mass x Velocity; and everything falls at the same speed from the same height irregardless of weight; then the Factor of Velocity (F=MxV) should be a recognized accepted constant given per foot of drop from a given elevation (all this changes with altitude i would think)!!! i don't know if that is to simple to do or what!! Seems like putting it on a level like that would make it 3rd grade math with a fixed table?? i have faith in Don Blair's rep enough that i don't think such a formulae would be written, to be so far off in 4'!??!

i'm lost , willing to learn!

edit-Now Joe has upped the ante to ~87# ??

 

[Stumper]

Hmm... Ramanujan and Joe, There is a mistake or 2 there somewhere. The right answer in foot pounds can't change when worked in the metric system and converted to english. Among other things I noticed that R's formula uses initial velocity squared- Fine. But there is a serious boo-boo- in your working of the problem INITIAL velocity was 0.

 

[Stumper]

Incidentallly, Rananujan's point about the sensation of impact being lessened by spreading it over time/distance is correct but the total force or energy expended doesn't change. Decelleration is not a force multiplier.

 

[treeclimber165]

As the title says, "Oh, for Pete's sake!" Go find a fish scale or something, clip your 4' saw lanyard to it and drop your saw off of it. Read the highest number indicated.
I was always good at math in school, but I haven't used calculations like shown above in 20 years. Gives me a headache trying to wade through it all.

 

[TheTreeSpyder]

Perhaps someone has a dynomike-meter...........for calculculations of such moving maas? Using non-dynamic cable?

i think that out of the tree is the time for such checks on the strategies we use for decision models in the tree/field; to make sure wee are within realistic bounds. If we are loading lines and gear, with more force than we thought, possibly bringing it outside the safety perimiter i think that reflection is needed. Including deeper respect for the said devices, and work carried!

As the numbers of this force climb with length of drop. Several things become more well defined peering deeper into this world, that is before me all day. Mostly that the dynamic elasticity to absorb all this energy in a line might be more important than a dynamic/elastic component rather than the static weight numbers expressed? Making Nylon more valued than polyester for these extremes? How much more pretightening, solid supports, stronger hinges, load control etc. is important as contorllling forces become more precious and appreciated, as the numbers of dynamic forces express themselves increasing so quickly.

If this here, on this particular website for examination to our particular applications are not the place for this; then where? Where else could one of ours come to seriously look at these things, save outside of bugging a Rope lab?

i stand by that this is thee place on thee web, and these things and there implications to all we do are relevant; further from any kind of impatience of censorship than other topics previously addressed. i find these things worth our time and minds. A responsibility to understand them, and this place the best vehicle on the planet to share them. As being and increasingly looked to as this premiere place; perhaps this space has the responsibility for offering these topics in the mix; the headache is quantafiably less than overhead support failure!

 

[The Climber]

I agree, althow I haven't taken part in the discussion( because my math skills fall short of those of the scholars above ) I have been reading it a couple times a day because I am interested in the outcome. the resulting knowladge is something I can put to use in my every day life and I'll be safer because of it.