fiddling with a Poldo tackle

[knotsaver]

  You do not need the Feyman Lectures on Physics to measure, and to count, do you ? 

Xarax, I cited (for reference) Feynman lectures for the principle of conservation of energy!

 In other words, if the ends of the two straights segments, at the maximum extension, are, say, 3L apart, at the minimum extension, when those two segments will become three, they will be at 2L apart - simply because the total length l=6L of the rope has not changed , so 2 x 3L= 3 x 2L = 6L !  ( as I, too, remember, since I my elementary school service      )

 of course, by construction!

   However, you have only proved that the gain on the sum of the work done by utilizing the mechanical advantage is 33.3% ( NOT 50% ! You start from the state of the maximum extension, where the distance between the anchors is 3L, you pull, you consume work, and you end with the state of minimum extension, where the distance becomes 2L, so you go from 3L to 2L ( a 33.3% reduction), not from 2L to 3L ( a 50% reduction) ! )

The load was lifted from 0 to 1L of height, but I (or someone else) pulled the rope 2L in length (the rope in the middle of the Poldo tackle at maximum extension is zero but at minimum is 2L)

   You have not proved that the consumption of the work will be linear, throughout the transformation - that the mechanical advantage will remain constant from the start to the finish of the pulling ( although I think that, given the linearity of the arrangement of the segments before and after any tensioning, at any two distances between the anchor points, this would be easy. If the segments are not parallel to each other - which happens when we have four anchor points, not two - and the angles between them are not 0 degrees, the mechanical advantage varies. )

yes, sure, but we can consider infinitesimal displacement of the rope and we obtain that pulling 2 infinitesimal piece of rope is equivalent to lifting the load 1 infinitesimal piece of height, (then we can consider the integral of the force), so the mechanical advantage is 2:1

P.S. The Super Poldo is nice ! However, to be able to utilize its full potential, the bowline should be able to pass through the ring - so you better use a wider one ! 

or we can use eye splices!
but friction exists!
ciao,
s.

[xarax]

I cited ( for reference ) Feynman lectures for the principle of conservation of energy !

  You should conserve your energy, too ! 
   I mean, we should better try to understand and explain things as simply as possible ( especially the simple things, as this ), because I have seen that this MA thing has a bad reputation among knot tyers, as being something magical and incomprehensible. I believe you could had simplified your exposition a lot.

the rope in the middle of the Poldo tackle at maximum extension is zero but at minimum is 2L.

   Right. You could had started from this ! Specify that, what you do, is to pull the two ends of this segment, the two eyes of the two loops, apart from each other, in order to extend its length, and that you will calculate the mechanical advantage regarding THIS action. The way you had described it was ambiguous ; you had not said explicitly which segment(s) of the rope you pull, from which point(s), towards which direction(s), to shorten the tackle ! You extend the middle segment by 2L, and doing this, you shorten the tackle ( and you lift the hanged load ) by 1L. ( Tex would had told you that tight from the start...  )
   EDIT : Notice that I am talking about pulling the ends of the middle segment here, that is, pulling the eyes of the loops, NOT the line itself, from some point P. By pulling the line, from some point P, you do achieve a 4:1 mechanical advantage, indeed, because when you drag P, you drag the eye of the corresponding loop half as much, so the 2:1 mechanical advantage is duplicated.
   HOWEVER, read my next post, and have a look at the attached picture there : you should take into account friction, which is required to stabilize the mechanism and establish an initial equilibrium - or, in the "ideal" case where there is no friction, the action of an incarnated ghost-spring, or the action of two more not-annihilated hands !   

   we can consider infinitesimal displacement of ... 2 infinitesimal pieces of ... is equivalent to... infinitesimal piece of ... we can consider the integral of the...

   THAT is what I was talking about ! Speaking like this, I believe that you lose most knot tyers, who will think that this is rocket science !  I tell you this, because I am an expert in loosing readers    Simplicity is not as simple as we believe... It is veeery difficult, at least for me, to explain something as simply as possible ( but not more... ).

[knotsaver]

  You should conserve your energy, too ! 

 yes sure! I'm knotSaver, energySaver ...

Simplicity is not as simple as we believe... It is veeery difficult, at least for me, to explain something as simply as possible ( but not more... ).

And since English is not my mother tongue, It is veeery difficult,  for me too, to explain something as simply as possible.
Saver (io)

[knotsaver]

oooops...
I'm sorry, I made a mistake!  
but, as a Poldo tackle fan, I'm really happy, 'cause the IMA is 4:1 !!!
(see attached figure Poldo_max.min_ext.jpg)

When our force F (the red arrow in the figure) acts on the Poldo tackle, we pull the rope for a displacement of 4L in length (the 2 orange lines of the second diagram in the figure). The red point P moves from the A pulley (in the first diagram) to the middle of the central rope (in the second diagram), so we pull the rope 4L in length!!! (please try to pull the central rope of the Poldo tackle and you see how much rope you pull).
 
So we have gained a change of energy (from 0 to WxL) and our force F has done a work of Fx4L whilst we have lifted the load only by 1L in length.

Now, F x 4L = W x 1L (for the principle of conservation of energy)
and then
F = 1/4 W
and so IMA is 4:1
...
I was not able to sleep...
s.

[Dan_Lehman]

?!
We don't need to compare but to explain the Poldo
Tackle systems shown in this thread. 
--dl*
====

...
Hi Dan,
you are right, here I am.
The principle of conservation of energy helps us to solve the
(ideal) problem (i.e. only conservative forces act).
...
and so IMA is 4:1

Note: if we use both hands (simultaneously and with the same force
acting on points RH and LH in the figure
(right hand upwards, left hand downwards))
we have F_righthand + F_lefthand= 1/4 W
and so F_righthand = F_lefthand = 1/8 W

This analysis isn't what I wanted, and I think it must
be bogus, but I'll not wage that battle now.

But I can only think that neither of you two posting here
has actually tried <gasp> to USE this supposed "4:1" structure,
e.g. to raise (even) 25#!?  For I have, and with some
re-direction pulleys to convert hanging (barbell) weights
into upwards pulls at some specified points, IIRC, it took
more weight devoted to this raising than the weight raised!

Now, what I was looking for you to do was to analyze
either of these Poldo (-like) tackles at the static state
where they are at least mid-way in contraction (and show
the zig-zag structure; not fully extended, anyway)
and try to determine what the tension(s) on the parts
must be!!

Curiosity: look at figure Poldo_Super.jpg for
a super-min-extension of Poldo tackle!
s.
//
The Super Poldo is nice !
However, to be able to utilize its full potential, the bowline
should be able to pass through the ring ... !

I'll show you what I mean with regard to this "Super Poldo",
referring to your image (which has it oriented left-to-right,
pulling leftside towards right anchor hook).
Since we're going for IMA we must assume frictionless
sheaves.  (What I was aiming at in begging X. to name the
supposed using-pulleys (which are fairly efficient vs. friction)
device that these mimicked : there isn't one!)

At the load sheave (left side, 2 parts running away above
and below) one should see force split evenly over the four
parts --2 above in opposition to the 2 below.
But the uppermost >>1<< runs through the anchor sheave
to turn to a sheave-end which supports 2 parts coming to it
(from either side, i.e.); and one of those parts itself splits its
load into the 2nd internal sheave.  So, at the anchor sheave,
you have this ONE part opposed to the other three (ultimately),
and 25% vs 75% does not a stationary system make !!

.:.  In short, given freedom from friction, these devices will
simply elongate under load --there is no locking.

There are some other behavioral characteristics one can discover
when playing around with them and actual forces, such as which
parts actually move --and which ones you might have thought
should move, don't!  (E.g, try lifting your regular Poldo with just
the one hand --either one, but esp. the left one, I think--, where you
are supposed to get some ("1/8th"?) MA.  Good luck with that.

My playing with it now is with 1/4" laid coextruded PP/PE and standard
carabiners (polished aluminum).  The story promulgated in knots books
is for --believe it at your peril-- rope-through-rope sheaves (!).
I used the hard-plastic-slick cut-off necks of 2-litre drink bottles qua
sheaves and some fairly hard-slick laid PP cord to manifest a structure
that did (slowly) expand when weighted, as theory foretold.

Btw, a 4:1 IMA would imply that you moved 4 units of rope
to raise the load 1 unit; but Poldo shows moving 1.5:1.


--dl*
====

[Dan_Lehman]

oooops...
I'm sorry, I made a mistake!  
but, as a Poldo tackle fan, I'm really happy, 'cause the IMA is 4:1 !!!
(see attached figure Poldo_max.min_ext.jpg)
.
.
.
and so IMA is 4:1
...
I was not able to sleep...
s.

Were you also not able to TEST this theory?!

FYI, I now have 37.5# suspended at your point F
--downwards bearing (gravity) force at that point
on the lower internal sheave--,
and 25# as W :: there is no movement
--that is not "4:1", 3:1, 2:1, 1.5:1, 1.1:1 even.

IF I attach some helper force (to use dead weight,
then a redirecting pulley will do, but will ameliorate
force via friction (roughly 0.66 per 'biner)), THEN one
can begin to raise the load.  But one must add this
added weight to the already present 37.5 and of course
that takes it farther away from W=25#.

But people/authors can publish raves about the mysterious benefits
of these structures, over & over!  (And the books, remember,
talk of pure rope-on-rope, nevermind even 'biners for help!)


--dl*
====

[xarax]

  As we can see in the attached picture, in the "ideal" case, to just establish a state of equilibrium, and prevent an automatic maximal elongation / extension of the tackle by itself, we have to keep pulling, at all times, each of two points on two segments of it towards opposite directions, with a force equal to the 1/4th of the load ( see the purple arrows, labelled by the letter F).( Or we have to place a spring  in between those two points, to pull them towards each other with a force F ... but then we will not be talking about a rope-made block and tackle mechanism any more, because those mechanisms do not store or use any induced energy ).Therefore, any gains we may have by utilizing its 4:1 mechanical advantage, are cut by half.

In short, given freedom from friction, these devices will simply elongate under load --there is no locking.

Correct. In the ideal case, you will really need that spring - or three, not-annihilated hands  : two hands to pull the two segments with a force equal to the 1/4th of the load, so the tackle remains in equilibrium and is not elongated / extended by itself, and, after you had established this equilibrium, one more hand to pull knotsaver s point P, and utilize the 4:1 mechanical advantage - which concerns only what is left if from the total load we subtract the 1/4th two times, that is, half of it !   

[Dan_Lehman]

and utilize the 4:1 mechanical advantage ...

There is no such 4:1 IMA !
Movement of rope from those indicated points
--nb: movement measured from final positions! (!?)--
Upwards & Downwards (indicating direction from
the points of contact) is respectively 1L & 0.5L
of a fully extended span of 3L (mid-point 1.5L)
with then double bearing (3 x 2 = 6L) of our ideal
6L of involved cord; the fully contracted (having
lifted Weight) system is 2L (x3 = 6L).
So, we move 1.5L to lift weight 1L :: 1.5 : 1, not 4:1.
Orrrr, am I mismeasuring here to count just rope --from
two moving points(!)-- and should count hand motion
(or, OTOH, rope attached to these points used qua haul lines),
in which case the Upwards movement goes from the
bottom/low sheave point to 0.5L from top,
netting 3-0.5 = 2.5L + 0.5L Downwards => 3:1 !?
(It sure doesn't feel like this!)


But I'm hoping that someone out there can put highly efficient
--even, relatively efficient compared to 'biners-- PULLEYS into
such a system and then see what happens with actual weights
attached at various places.  I think that the movements that
actually occur will be most interesting (as well as the weights
needed to get movement)!!
(I don't have such pulleys.  Maybe I can find those plastic-bottle
sheaves & associated cord with which I saw slow extension.)


--dl*
====

[xarax]

   Iff you keep pulling the two lines at the points I have shown with force F equal to the 1/4th of the load ( so the tackle reaches the state of equilibrium, and remains there ), the mechanical advantage is 4:1.
   You have not found any flaw in knotsaver s calculation. When you drag the line and move point P at a distance p,  the load is moved at a distance p/4. ( knotsaver has calculated this when point P moves distance 3L upwards, and then distance 1L downwards, that is, a total distance of 4L - and, at the same time, the total length of the tackle goes from 3L to 2L, so its lower end point, and the load attached on it, is raised/lifted 1L. )
   You can not "feel" the 4:1 mechanical advantage, because, at the same time you pull point P, you have also to pull points F, each one with a force to the 1/4th of the load.

[knotsaver]

Please look at the attached figure:
Suppose:
- rope lenght = 15 L
- pulley C blocked
- force F (in red) applied as indicated
- it doesn't matter that second diagram is at max extension
- lines have to be considered vertical
For me these are the moves (tested moves):
As the force F has been acting on the Poldo tackle, we have pulled the point P for a displacement of 4L in length  whilst pulley B has moved (downwards) by 2L in length and  the load (pulley A) was lifted only by 1L in length.
...
(to be continued)

[xarax]

   knotsaver, you do not need to say all that ! Obviously, P is at the middle of the line when the tackle is in its maximum extension, so it will remain in the middle of the line when the tackle will contract, and reach its minimum extension - the point P does not slide on the line ! 
   Dan Lehman does not want to accept it, because he "feels", in practice, that the total mechanical advantage can not be 4 : 1 - and he is right in that. You have to take into account any forces you should add, to establish a tackle in equilibrium, before you start extending or contacting it - otherwise you can not / should not calculate any mechanical advantage.

[xarax]

   There are many ways to calculate the mechanical advantage of a simple machine.
   
   One way is the "kinematic" way, which is implemented by knotsaver. In this, we examine how the shape / geometry of the mechanism varies during its extension or contraction. If we see that the point on which the load is attached moves distance X, and the point of the line which we drag, in order to lift this load, moves distance Z, then the mechanical advantage is Z / X. HOWEVER, you should do this only if you have a mechanism which is in a state of equilibrium, that is, which is not expanding or contracting by itself, even before you think to expand or contract it by yourself ! The Poldo tackle is NOT such a mechanism : In the ideal case, when there is no friction, it will expand by itself, and become maximally extended.
   In order to transform it into a mechanism which is static, and so we can then study the mechanical advantage it offers when we put it in motion ( remember, we are talking about uniform motion, NOT acceleration ! ), we have to "add" forces. The interested reader will see the two forces F, indicated by the purple arrows, in the sketch of the Poldo tackle in equilibrium shown in a previous post.
   THAT is knotsaver s omission / "mistake" : he examines the mechanical advantage of the mechanism, when it is in equilibrium, and he, correctly, finds it to be 4 : 1 - but the Poldo tackle itself, without the addition of the forces F ( or the existence of friction forces, which play the same role ), is NOT in equilicrium ! If we take into account the work consumed by the application of the forces F, or the additional work needed to overcome the equivalent friction forces, we see that the 4 : 1 mechanical advantage is reduced by half, and becomes 2 : 1.
   
   The second way is the "dynamic" way. One analyses the forces acting on each part of the mechanism, and he is assured that the ( "vector" ) sum of those forces, in each and every part of them, is zero. On each and every pulley, the forces are supposed to be counterbalanced, each of them cancels the effect of all the rest, so the pulley does not start moving to somewhere by itself ! 
    Then, AFTER one has found all those forces, he should probably add some more, which would be missing, if the mechanism by itself is not static. In the case of the Poldo tackle, due to its simplicity and symmetry, it is very easy to do this - in more complex structures, it may become very difficult... The interested reader will see, at a glance, how the F forces counterbalance the rest. ( Note : In this sketch, I should had also shown, by long green arrows, the forces acting on the tackle by the load, on the axle of the one end-pulley, and by the anchor, on the axle of the other end-pulley - but I had not enough space to do this !    The diagram was already very long, and those forces are 4 times as strong as the F forces, pointing outwards...)
    After the addition of those "counterbalancing" F forces, the calculation of the mechanical advantage is a piece of cake : If those forces are, say, 2F ( as in this case ), and the load is 4L, the mechanical advantage is the ratio of the sum of forces acting on the lines of the tackle to raise the load,  divided by the load to be lifted - that is , 4l / 2l = 2 : 1.
   Of course, this is the "ideal" case : Actually, in "real" cases, there are always friction forces acting on the mechanism, so, when its expansion or contraction takes place under load, the forces we "feel" we have to apply, to lift the load, are not the one half, only, of it... However, we always calculate ideal mechanical advantages, because friction can not be taken into account so easily : it may depend on many things, it may be not-linear, it may depend on the speed the mechanism expands or contracts, it may vary during some phases of the change of the overall shape of the mechanism, etc.

   I have seen that knot tyers are not very happy, when they want/need to calculate mechanical advantages !    So it may be more interesting, and useful for us, to just TRY an actual "real" mechanism, using free-rotating, bearing-made pulleys, and slippery Dyneema fishing lines, and MEASURE the "real' mechanical advantage by themselves. I am very interested to learn how much the 'real", measured mechanical advantage of the Poldo tackle will differ from the "ideal", 2 : 1 one...
           

[xarax]

   Just in passing, I want to notice this : We can transform the original Poldo Tackle, into a "Stabilized Poldo tackle", by adding two "counterbalancing" F loads, hanged by two more lines - and we will also need one more pulley to do this, because we should orient the second F force / load, too, towards the centre of the Earth !  . Then, in order to calculate the mechanical advantage, we have just to measure how much the centre of mass of the TOTAL load will move ( the to-be-lifted load, plus the counterbalancing loads ) when a point P from which we drag a line of the tackle moves. The interested reader would only spend a few minutes to do this - but they would be his most worth-to-had-been-spent minutes in the study of the Poldo Tackle !   
     
   P.S. What can I do ? I had made a new quick and dirty sketch of the "Stabilized Poldo tackle" mentioned in this post. The green arrows represent forces two times the size of the forces represented by the blue arrows, and four times the forces represented by the red and purple arrows. One can easily see that the whole system, and each "free-floating" part / pulley of it, is in equilibrium. The tackle can contract and expand freely, under load, without the consumption of work ( provided there is no friction, of course ).
   The to-be-lifted load and the "stabilising" loads are indicated by the purple circles. One can also see the green additional pulley, which can be hanged anywhere outside the tackle, and its only purpose is to re-orient the one purple line to the centre of the Earth  .

[knotsaver]

   knotsaver, you do not need to say all that !

First of all, I want to thank both of you, Dan and Xarax, because I'm beginning to understand how Poldo tackle really works. I have been so fascinated by the self-locking feature (I consider it the best feature of the tackle!) that I have never studied it in detail.
I want to post another figure to complete the understanding of the moves (in the ideal-ideal case), so please look at the figure Poldo_moves_Pa-Pb.jpg.
If we apply force FB (in blue) on the left side ("good luck!", as Dan said ) in this case we are able to move point PB by 2L in lenght whilst the load is lifted by 1L, so we have to double the force FA (in red and in brackets)
Note:

1. point PA is lifted by 4L in length but now we apply a double force (!?)
2. if we use both hands with the same force, the right hand will lift 2/3 of the load and the left only 1/3, for instance if the load is 30kg, F_righthand=F_lefthand = 5 kgf (kilogram-force ) ( a total of 10kgf) and so
- if we use only ritgh hand we should have a 4:1 IMA ...
- if use both hands we should have a 3:1 IMA
- if we use only left hand we should have a 2:1 IMA
  (a little confused!)
3. with a blocked (fixed) C pulley we are analysing the "noeud a cremmailler", I haven't understood  yet if it is equivalent to Poldo tackle
4. we should consider the feature of lowering a load too

[xarax]

I'm beginning to understand how Poldo tackle really works.
 (a little confused!)

    A little confused, indeed. 
    However, things are quite simple, provided you do not mess right and left hands, feet, ties, etc...   If you grab and pull the line which goes to the rim of the pulley, you have the 2 : 1 mechanical advantage. However, if you grab and pull the line which goes to the axle of the pulley, you have half of that - because, for a distance X a point on the line which turns around the pulley covers, a point on the line which "hangs" the pulley from its axle covers distance X / 2, so the mechanical advantage, from 2 : 1, becomes 1 : 1 - that is, it is evaporated !    Moral of the story : Do not pull the pulleys themselves !
   When you calculate the mechanical advantage, you are not concerned for the number of points from which you grab and pull the line(s) ! You are concerned only for the distance those points cover ( which is the same, of course, for all the points of the line ! ), in relation to the distance the centre of mass of the load covers at the same time. Your analysis is "kinematic", do not mess the number of  hands, amount of forces, and points of the line you grab with it ! Once the tackle is in equilibrium, it is either static or it moves with uniform velocity, which is the same thing. If your analysis is "dynamic" then you take account the amount of forces you apply - but we should not make a mixture of those two ways...
   There are NO 3 : 1 or 4 : 1 mechanical advantages in the Poldo tackle ! The ( ideal ) mechanical advantage is 2 : 1.

   ( A humble advice : "See" the simple sketch with the "dynamic" analysis I had posted - you will understand the whole thing immediately ! I believe that you have not yet understood that you can not analyse the mechanical advantage of a system that it is not in equilibrium - and which, if it is left alone, it will start moving NOT with a uniform velocity, but with an accelerating one ! )

   

the self-locking feature ( I consider it the best feature of the tackle ! )

  This "locking" is not achieved with any ingenious trick!  ! It is due to friction ( mainly in the end zig zag points ), which plays the role of the "counterbalancing" loads I had mentioned : they do not allow the line to slide from the one side to the other, although it tries... If you examine the distribution of forces, you will see that, without those counterbalancing loads and forces which stabilize the mechanism, at each end point, when the line "arrives" there, it is loaded by twice as much load as it has when it leaves from it - which would nt be possible, if there were no friction.