Peter, This one is directed for you, or anyone else with experience in this field.

I know how to figure out the forces on a speedline, tyrolean, or traverse (call it what you will). For those who don't know, the top diagram may be a real eye opener.
200 lbs of weight can easily = 1336 lbs of tension depending on the angle of the rope.

Of course we know that rarely in arboriculture do we have lines anchored at equal heights. In 90 % of the cases, the line is anchored high in the tree and goes to a second anchor at ground level. My vector diagrams fail me in this instance (see lower diagram) Can anyone help me figure out what leg B =? Is leg A right?

How does a tag line affect A, I would think that the tag would not exceed the weight of object. Is it as simple as subtracting the weight of the object from Leg A? Does this have an effect on B?

This topic goes well with the MA thread where we talked about amplified forces in a tree.

Dave
(wishing I had a dynamometer)

Dave,

Good thread.

I have wondered about this same thing. There are some variables to consider in actual field use. For calculating the raw numbers we could say that these variables are zero.

1 Rope stretch
2 Built-in slack in line AB
3 Deflection of anchor A

Wouldn't the tagline have the 449# load added to the anchor point A? Or would it be double since the tagline is a redirect? The way that you have this setup, the redirect would not actually double the load since the angle is less than 180 degrees. At ninety degrees A would have about 70% of the load.

Livin' in MA confusion!

Tom

I did a little research tonight and came up with the following formulas:

T(1)=tension on the speedline between the weight and anchor A

0(1)=theta angle the weight makes with the speedline towards anchor A

T(2)=tension between weight and anchor B

0(2)=theta angle the weight makes with the speedline towards anchor B

mg=weight

1st diagram:
The sum of the x-components:
T(2)cos0(2)-T(1)cos0(1)=O
T(2)cos0(2)=T(1)cos0(1) {cos0(2)=cos0(1)}
therefore,T(2)=T(1)

The sum of the y-components:
T(2)sin0(2)+T(1)sin0(1)-mg=O
substitute T(2) for T(1)
T(2)sin0(2)+T(2)sin0(1)-mg=O
add mg to both sides of the equation. Also, sin0(1)=sin0(2)
T(2)sin0(2)+T(2)sin0(2)=mg
since tension T and angle 0 are "=",
2Tsin0=mg
therefore,T = mg/(2sin0)
This should be the tension on each leg of the rope.

For the 2nd diagram:
T(1)=[mgcos0(2)]/cos(0(1)-0(2))
T(2)= mg/cos(0(1)-0(2))
This was a tedious equation to come up with so I omitted the details.
(edit 03/24/00:the equations for diagram 2 DON'T WORK! edit 08/27/00: the correct equations are posted in reply 17 of this thread.)
For the 3rd diagram:
T on the tag line = mgsin(01)
I think the tag line will relieve the tension in the speedline only as the speedline becomes more vertical at this point in my analysis of the diagram. My assumption is the tension in the speedline will start to be relieved by the tag line
when the speedline is 45 degrees as the angle of the speedline approaches 90 degrees.
My reference source for these analyses was from several problems out of Schaum's outline series, Theories and Problems of College Physics, 7th ed.
Joe
GENERAL DISCLAIMER:THIS IS AN EDUCATIONAL SITE AND THE INFORMATION IS BEING PROVIDED FOR EDUCATIONAL PURPOSES AND I DISCLAIM ANY LIABILITY FOR THE USE OF THIS INFORMATION FOR ANY PURPOSES.

Joe,

Thanks for doing our homework! Reminds me of my high school physics club, The Society of Socialistic Physicists. We would get together before tests and each member would work through a separate problem and then present the solution to the group. That way we could cover more ground quicker.

I think the tag line will relieve the tension in the speedline only as the speedline becomes more vertical at this point in my analysis of the diagram. My assumption is the tension in the speedline will start to be relieved by the tag line when the speedline is 45 degrees as the angle of the speedline approaches 90 degrees.

This confirms my thought. As the tagline relieves tension, the load from the speedline is transferred to the anchor. This will increase the anchor load be some factor.

Livin' @ Lat. West 93.34
Long. North 45.02

Tom

Tom and others,
I enjoy pursuing these sort problems since they are within the bounds of my
current level of knowledge. There are other factors involved with the physics
of the tension in the speedline which haven't been covered in the analyses.
For anyone who decides to use these formulas take precautions and use them as a reference for an educated guess as to what
the forces are when actually determining tensions on your lines. For those who
own the book "On Rope", see pages 66 and 67. There is a small discussion about the forces on the type of rope configuration
as in example "1" of Dave S.'s GIF. This is considered advanced rigging and requires a higher level of knowledge to understand, thus, the word of caution
to those who are getting their feet wet
as they begin to see how the applications
of these formulas can help them.

Joe

Dave;

See Don Blair's 'Arborist Equipment', pages
175-176 and the NAA's 'Rigging for Removal
Workbook' page 70 (also by Don Blair).

I assume that the first part of the GIF
diagram was just to introduce this set up of
vector forces and that the main thrust of the
question is how much force is generated on
the legs, tagline and anchor of a speedline?
I do not have any specific numbers to offer
(I leave that to Pete or Joe), but I can
offer a way to (in some situations)
dramatically decrease the amount of tension
on the rope and on anchor A (which I will
call the lowering point).

A speedline is used because either 1) there
is something directly under A that prevents
you from lowering the load straight down; or
2) you want to move the load quickly to where
it can be processed. B is typically the base
of a tree or some other heavy, solid object
on the ground. If B is a tree I generally
look up and try to find a good solid crotch
that still gives the same line of pull, but
decreases the angle between A and B. The
speedline is put through this crotch and this
becomes anchor B. The line is not tensioned
until after the load is cut (this is often
the case with a speedline, but when the
anchor is on the ground it creates a lot more
force on the system because of the wide angle
between A and B) As the tagline lets the
load slide down the 'speedline' B is
tensioned to direct the load to the landing
zone.

Once you get the hang of how this works
you'll see that B does not have to be
directly over the drop zone (processing
area), it merely has to form a straight line
with the anchor and the drop zone. The
higher B is the more it helps to share the
force of the load and the less force there is
on A.

This is similar to a transfer line which I
tried to describe in an earlier thread, but it
seemed that I was not able to explain it well
enough without a diagram.

Thanks for the neat illustrations. Could you
explain how its done?

Mahk

tree tree tree tr

Mahk,

Good idea with the redirect at the anchor or B point.

When I have set up this system, I will set a large diameter block instead of a natural crotch. This keeps the speedline rope from tearing any of the bark off of the tree while tensioning.

I learned this the hard way, oops...

Livin' @ Lat. West 93.34
Long. North 45.02

Tom

God I love doing this stuff!

Joe, Check out the diagram attached. I tried to apply your formulas but it just doesn't seem right. I would have almost believed it if the figures T1 and T2 were switched around.


Dave

Looking forward to getting on to the tag line!

(still wishing I had a dynamometer)

>1st diagram:
>The sum of the x-components:
>T(2)cos0(2)-T(1)cos0(1)=O
>T(2)cos0(2)=T(1)cos0(1)
>{cos0(2)=cos0(1)}
>therefore,T(2)=T(1)

I do understand some of the equations but I am a very visual person (hence all my diagrams) Do you have any ideas how to put the equations into vector diagrams. I find this much simpler for me and probably others.

By the way Mahk, I do the diagrams on corel draw.

(edit between 03/24/00 and 03/25/00:keep following this thread. I solved this part of the speedline problem.(Arms up in the air, hands in a fist, stretched above my head in victory!))
Sorry Fellas, I SCREWED UP!!! I apologize
to you especially Dave S.. The equation
representing Illustration 2 of Dave S's
GIF is totally wrong. I set up an experiment at home and found the data from these equations didn't match the experimental values I came up with from doing the experiment. I also made a serious mistake by applying a subtraction formula to arrive at the final equations. However, I did work through the math again and came up with a couple of equations which at least gave values that paralleled the data results. I found that
the tension at A was higher than the experimental weight throughout the experiment. I also found that the tension at B was lower than the experimental weight. Sorry again fellas and Dave S., I thought I had it right.
Joe
P.S. equations for illustrations 1 and 3 are correct.

I did work through
>the math again and came up
>with a couple of equations
>which at least gave values
>that paralleled the data
>results.

Do you have these new equations?

I'm looking forward to trying them out!

Dave

Sorry Dave,
I'd much rather work the physics and
mathematics out and see to it the equations better reflect the results of the data I gather through crude methods of experimentation before I give them out. It beats the risk of embarrassment.

Here's a tip on estimating limb weights in case you didn't know it that requires you find the center of mass of the limb. You can do this on the ground after the limb is removed.

1) Balance the limb and mark its center of mass.

2)Cut the limb at the point of balance and measure the length and the diameter of both sides of the part with the least amount of brush. Hopefully,the part you measure has no brush.

3)apply the data to this equation:
2((pi/12)times(D^2+Dd+d^2)lw)
this is the equation of the volume of the frustum of a cone where,
D=larger diameter of the limb
d=smaller diameter of the limb
l=length of the limb
w=weight of the wood you are estimating per cubic foot from a wood weight table.
^2=squared or to the power of 2 for those who don't know the notation

example:you want to know the weight of an oak limb you just lowered.
You've found the center of mass of the limb and cut the limb in half. You've measured the length as 20 ft. the diameters as 4 and 6 inches.
then,
2times(pi/12)times((.5^2)+(.5)(.333)+(.333^2))times(20)times(63)=348 lbs. (approx.)
notice the diameter values were converted to decimal form before solving the equation. Also, you multiply the equation by 2 because there were 2 parts to the limb and only one part was measured. The question was how much does the limb weigh.

If the ends of the limb are of the same diameter, or if you want to know the weight of a trunk section where the diameter of each end are the same, apply this formula: the equation of the volume of a cylinder
(pi/4)times(d^2)lw where,
d=diameter
l=length
w= wood weight per cubic foot

For the estimation of a trunk section of oak whose length is 10 ft. with a diameter of 3 ft.
(pi/4)times(3^2)times(10)times(63)=4453 lbs. (approx.)
These equations can be applied with a scientific calculator without knowing the
math. But be certain the numbers are correct and when working with inches they're converted into decimal form, which will be in feet.

To convert inches into its decimal equivalent, divide the # of inches by 12. example: 3 inches divided by 12 inches=(3/12)=.25 feet. I like to gather measurements from time to time in the field and take them home for analysis later to keep from holding up production. I also like to use oak as a reference when working with other trees
since wood weights vary and oak is the heavier of the different types of wood.
Joe


GENERAL DISCLAIMER: THIS IS AN EDUCATIONAL SITE AND THIS INFORMATION IS BEING PROVIDED FOR EDUCATIONAL PURPOSES AND I DISCLAIM ANY LIABILITY FOR USE OF THIS INFORMATION FOR ANY PURPOSES

Dave and or Joe;

Could you clarify and explain (or hypothesize
on) a couple of points for me?

On 3/18/00 1:16:00 PM, Dave Spencer wrote:

In 90 % of the
>cases, the line is anchored
>high in the tree and goes to a
>second anchor at ground level.
>

I assume that this refers to a speedline
setup.


>How does a tag line affect A,
>I would think that the tag
>would not exceed the weight of
>object. Is it as simple as
>subtracting the weight of the
>object from Leg A? Does this
>have an effect on B?

I also assume that in this case the tag line
is also the 'haulback' line and, as its
illustrated, is used to control the descent
of the load.
>
>This topic goes well with the
>MA thread where we talked
>about amplified forces in a
>tree.
>


What about the force that is created using a
lowering line and a tag line to pull the load
to a landing zone that is not directly under
the lowering point. That is, suppose you do
not use a speed line, but just a lowering
line. There is some obstacle directly under
the lowering point so you have to use a tag
line to pull the load sideways rather than
letting it go straight to the ground. Will
this increase the force on the line? I would
guess that the force would be the same unless
the people pulling on the tag line jerk on
the load, in which case it would increase the
force--but by how much? And if a truck is
used to pull the tag line will the movement
create more force than a static line that is
simply pre tensioned?

Mahk


tree tree tree tree tree tree tree tree tre

On 3/23/00 7:53:00 PM, Mahk Adams wrote:


>I assume that this refers to a speedline
>setup.

Yes



>I also assume that in this case the tag
>line
>is also the 'haulback' line and, as its
>illustrated, is used to control the
>descent
>of the load.

Yes


>What about the force that is created
>using a
>lowering line and a tag line to pull the
>load
>to a landing zone that is not directly
>under
>the lowering point. That is, suppose
>you do
>not use a speed line, but just a
>lowering
>line. There is some obstacle directly
>under
>the lowering point so you have to use a
>tag
>line to pull the load sideways rather
>than
>letting it go straight to the ground.

As soon as the rope breaks the vertical plane, the forces will increase.You will be able to use the same calcs we are working on to figure out these forces.

You have actually made a great example to help illustrate how much the forces are amplified; Take a good chunk of wood and as it is lowered, pull back on the tag line. As the wood is lowered, you will find it easier to pull back. This has to do with the angle created (theta in the diagram). It is impossible to make theta 180 degrees. You will reach a point close to it but the rope will have long ago broken. (see theoretical graph) So the lower theta is, the less tension on the rope. Let a chunk down halfway, stand way back, and feel the forces required to pull theta larger.

>Will
>this increase the force on the line? I
>would
>guess that the force would be the same
>unless
>the people pulling on the tag line jerk
>on
>the load, in which case it would
>increase the
>force--but by how much? And if a truck
>is
>used to pull the tag line will the
>movement
>create more force than a static line
>that is
>simply pre tensioned?
>
>Mahk

All of our diagrams and equations are figured in a point in time or (static system) As most of the angles change during a logs ride to the ground, we want to figure out the forces in a given instant.

As far as the truck goes. Remember how much force it can take to pull theta towards 180 degrees. These forces also affect the tension between the pulley and the log. So one thing we are trying to figure out in this thread is what is the force above the log if the force below is say 300 lbs for a given theta. These forces can run very high and remember that you are running the rope through a pulley at the top of a tree. (not doubling the force but close). If you are pulling with a truck, make sure you use a long rope to decrease theta as much as possible, thereby decreasing the force on the top of the tree.

Dave

(geez, even my head is spinning)

After doing some research, I have finally found what I believe to be a reasonable formula. Not only that, but it seams to work.

See fig. 1 - A and B are trees (or whatever else you want to tie yourself to) and C is the climber/log.

\ = angle in degrees
| | = absolute value (distance from zero)

A[cos(\BDC-90)] - B[sin(|\ADC-180|)= 0]

***and***

A[cos(\ADC-90)] - C[sin(90-\BDC)]- C = 0

When you solve the second equation, you can find B, and put that in the first equation, and you can find A.

SO. . . .
(fig. II)

C = 200 lbs.

A[cos(|70-90|)] - B[sin(|120-180|)= 0]
.94A - .87B = 0

***and***

A[cos(120-90)] - 200[sin(90-70)]- 200 = 0
.87A - 200*.34 - 200 = 0
.87A - 68 - 200 = 0
.87A = 268
A = 308.05 lbs.

so . . .

.94A - .87B = 0
.94*308.05 - .87B = 0
289.57=.87B
B = 332 lbs.

I am still trying to figure out how to put a fourth leg (the tag line) into the system.

Let me know if this works,

Isaac

OOPs, here is the image:

Isaac

Thanks for jumping in on this one with us Isaac, but I found the correct solution tonight as you were posting. The solution lied in the solution to illustration one where the forces at the anchors are equal.
I wonder how many people saw this and refused to speak up about it. Here are those 2 elusive equations we've been searching for:

theta 1=(01)
theta 2=(02)
T1=tension at the top of the tree
T2=tension at the bottom of the speedline

A point I need to make when one applies these equations is where to find the proper angles so the equations give the proper results. One needs to draw a coordinate axis(a "+" sign) through the bend in the rope where the weight hangs. The angles are found by measuring them from the horizontal axis and these angles are acute. One will be positive(01) and one will be negative(02). Each angle reflects the force of the anchor its pointing towards. Don't forget to use the "-" sign when plugging the angles into the equations.

T1=(mgcos(02))/[(sin(01)cos(02))+(sin(02)cos(01)]

T2=(mgcos(01))/[(sin(02)cos(01))+(sin(01)cos(02)]

mg=200
(01)=30degrees
(02)=-10degrees

T1=(200cos(-10))/[(sin(30)cos(-10)+(sin(-10)cos(30))]=576 lb. (approx.)

T2=(200cos(30))/[(sin(-10)cos(30))+(sin(30)cos(-10))]=506 lb. (approx.)

Remember:These forces are represented in a static situation where no acceleration,
friction, rope stretch etc. are factored into the model. Otherwise this situation is not dynamic.

Joe
GENERAL DISCLAIMER:THIS IS AN EDUCATIONAL SITE AND THE INFORMATION PROVIDED IS FOR EDUCATIONAL PURPOSES AND I DISCLAIM ANY LIABILITY FOR USE OF THIS INFORMATION FOR ANY PURPOSE.

Joe,

I looked at the two formulas, and they seem to be the same, only set up in different ways.

Isaac

Isaac,
Are you referring to the 2 formulas I
presented and you think there should be only one equation? Or are you referring to your formulas as looking the same as my formulas except I used trigonometric functions. I don't understand what your
comment refers to.
Joe

I mean your formulas and mine look similar, just that mine use calc. and yours trig.

Isaac

Some people could not load the attachment so I have included a gif as well.

The purpose of this whole exercise was not to develop a formula to figure out exactly what forces are involved when you load a speed line. It was to educate people to the fact that the forces can be much greater than the weight of the object. I only hope we have saved someone from a serious speedline accident. (either having the rope break, or the top of the tree)

I will post my theory on the tag line as soon as I get a chance. Thank you to everyone who participated in this thread. It has been the most fun thread this board has ever had. (opinion of a nerd) I'll probably shed a tear when we lose it to the archives.

Dave

On 3/26/00 7:53:00 AM, Dave Spencer wrote:
>The purpose of this whole
>exercise was not to develop a
>formula to figure out exactly
>what forces are involved when
>you load a speed line. It was
>to educate people to the fact
>that the forces can be much
>greater than the weight of the
>object.

Even I don't climb with a calculator, so I have to second Dave's thoughts here. What is most important is understanding the physics enough so that you can predict changes in a rig, not the absolute forces. I think the general trends people have calculated seem to be OK, and I like the diagrams. And, certainly, the forces are way more than the weight of the log you are moving. I've heard of people pulling trees out of the ground with a speedline (granted it was a small tree and should not have been used as an anchor, but *still*).

That being said, I don't think the numbers are correct. Many people have asked me to try my hand at this calculation. As much as I have tried, I don't yet understand it to my own satisfaction, and don't want to pass off bad information. I also don't want to speak ill of others, but I was not that impressed with the treatment of speedline forces in the NAA rigging workbook (the video is a whole other story). There were equations thrown around, but I don't think anyone really understood the physics behind their application.

Someplace early on in the thread, someone mentioned the initial rope tension and stretch (engineers would talk about rope stiffness, the ability to resist stretch). This is the key to solving for the forces in a speedline. Problem is that it now becomes more complicated than just vector diagrams; there are differential equations involved. Just as important, though are some experiments to validate the equations. Joe mentioned he did some of this, but I'd also be interested in doing it full-scale. If I ever do get more insight into this, I'll write about it.

Later,
Pete

I believe that most of our answers lie here...

Next step; tag line (I think I have it!)

Dave