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)
RIG.GIF (18KB)
Read topic starting at message #7416
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 Builtin 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
Read topic starting at message #7437
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 xcomponents:
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 ycomponents:
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.
Read topic starting at message #7440
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
Read topic starting at message #7442
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
Read topic starting at message #7446
Dave;
See Don Blair's 'Arborist Equipment', pages
175176 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
Read topic starting at message #7448
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
Read topic starting at message #7455
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)
CALCS.GIF (10KB)
Read topic starting at message #7453
>1st diagram:
>The sum of the xcomponents:
>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.
Read topic starting at message #7486
(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.
Read topic starting at message #7490
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
Read topic starting at message #7494
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
Read topic starting at message #7495
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
forcebut 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
Read topic starting at message #7507
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
>forcebut 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)
LOWER.GIF (8KB)
Read topic starting at message #7519
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(\BDC90)]  B[sin(\ADC180)= 0]
***and***
A[cos(\ADC90)]  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(7090)]  B[sin(120180)= 0]
.94A  .87B = 0
***and***
A[cos(12090)]  200[sin(9070)] 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
Read topic starting at message #7520
OOPs, here is the image:
Isaac
IMG.GIF (4KB)
Opposing Forces
Read topic starting at message #7521
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.
Read topic starting at message #7534
Joe,
I looked at the two formulas, and they seem to be the same, only set
up in different ways.
Isaac
Read topic starting at message #7537
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
Read topic starting at message #7538
I mean your formulas and
mine look similar, just that mine use calc. and yours trig.
Isaac
Read topic starting at message #7541
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
MICROSOFT EXCEL #01.GIF (44KB)
Read topic starting at message #7556
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 fullscale. If I ever do get more insight into this, I'll write
about it.
Later,
Pete
Read topic starting at message #7532
I believe that most of our
answers lie here...
Next step; tag line (I think I have it!)
Dave
TENSION CALCS.XLS (33KB)
Read topic starting at message #7539
Wow, you guys are going all
out.
The work put together on this discussion amounts to the sum total of
all written load calculations for a speed line that I have ever read.
Here is a little info from the Rescue people. This is a really good
website too.
http://www.techrescue.org/vertical/verticalref4.html
There are some good rope and rigging quizzes.
The formulas are great for use when we sit at the keyboard. Can anyone
translate them into a magnitude factor or maybe begin to make a rule
of thumb? Or, is the best we have, the charts that Don Blair pulled
together about sling loads?
Livin' @ Lat. West 93.34
Long. North 45.02
Tom
Read topic starting at message #7544
here are my thoughts on the
effect of a tag line.
I figure that the tag line acts as a line holding a frictionless
object on an inclined plane. That was simple and I think Joe and
others had the same idea.
The hard part is how this affects T1 or the parallel line. My only
thought was the first and most simplistic one that came to mind. The
force on the tagline is subtracted from T1.
Check out the Gif and see if you agree.
I also modified the xl attachment from before. Punch in any weight and
angles and it will give you the results and graph what the lines would
look like.(not exact but you get the idea)
Try estimating previous work, especially if you have photos you can
get angles from.
Dave
Livin in Bliss (don't know the coordinates)
CALCTAG.GIF (5KB)
CALCS2.XLS (17KB)
Read topic starting at message #7545
Dave,
If I wanted to know what the load was on A would I add T1+Tt?
575.87+101=686.87
Since there are two lines it seems like this is right...
If so, it shows how important a solid anchor needs to be. To reduce
the load on A another leader could be used.
Livin' dizzy
Tom
Read topic starting at message #7550
On 3/26/00 5:41:00 PM, Tom
Dunlap wrote:
>Dave,
>
>If I wanted to know what the
>load was on A would I add
>T1+Tt? 575.87+101=686.87
>
>Since there are two lines it
>seems like this is right...
There are actually 3 lines because the tag line is usually controlled
from the ground. Remember our pulley talk for before. It will not be
quite double Tt because of the angles formed. A good rough guide
though might be:
A = T1+(Tt * 1.75)
>If so, it shows how important
>a solid anchor needs to be. To
>reduce the load on A another
>leader could be used.
>
I think there still may be some slight problems with the calcs. Stay
posted...
Dave
Read topic starting at message #7551
Oops! I slipped off the
foundation of mechanical advantage.
Strong limbs just got MORE important!
Livin' @ Lat. West 93.34
Long. North 45.02
Tom
Read topic starting at message #7577
Thank you to all who have
contributed to this awesome thread. Inclement weather kept me home
today so I took the time to set up some rigging and experiment with
MA, speedline, and speedline/lowering/redirect combinations.
I haven't had the opportunity to work with speedlines much in the
field as of yet but when I do I'll be more prepared to set a system
that is much safer than if I had done so before, saving property
damage and/or life or limb.
Cheers!
Strong anchors and slack speedlines,
Gerry
Read topic starting at message #7613
Hey Dave,
When you are talking about the forces on a speedline. Are you talking
about the angle's put on the rope? Like for example, if a rope is at 0
degree's the load is at 50% on each end of the rope, and at 90
degree's, the load is at 70%, and at 120 degree's, the load is at
100%, and at 150 degree's, the load it at 200%, and at 170 degree's,
the load is at 1150%???
Brad
Topic: forces on a speedline (31 of 31)
Back again. I added another
scale to my experimental station. I'm beginning to find when a tag
line is added to the speedline
as a system, the tension at the "anchor points" increase. The tension
at B is larger than the tension when no tag line is used. The sum of
the tensions at A are larger than the tension at A without a tag line.
However, the individual tensions, that of the tag line and that of A
are smaller than the tension at A without a tag line. The numerical
values at the anchor points result from the angles made on the lines
due to anchor point and weight location. These #'s can show the
sensitivity of any changes in angle within the system. My observations
are not absolute.
If I ever figure an acceptable set of equations which reflect the
tensions in a speedline with a tag line attached, there's some serious
calculus and physics involved with bringing dynamics into the picture.
Can any one confirm or deny my observations? Has anybody had a nagging
itch they couldn't scratch? This is what this problem does to me.
Joe
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