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'[EE]: Pulleys'
2001\09\21@005942 by Jinx

face picon face
(made this EE rather than OT in case there are spin-offs
for the Bot Boys)

I'm working backwards from a load to get a ballpark figure
for the size of motor required to lift it with a 3-pulley block
and tackle (which I'm making)

Everything seemd OK until I read this paragraph from a
technical book

"Ignoring friction, for a simple pulley system with equal sized
pulleys, the mechanical advantage is equal to the number of
pulleys. The ratio of the distance moved by the effort to that
moved by the load (the velocity ratio) will also equal the number
of pulleys, and is not affected by friction or the weight of the
equipment. In practice, these factors reduce the mechanical
advantage, and the efficiency of the system is calculated by
dividing the mechanical advantage by the velocity ratio"

Pretty straight-forward. To raise 10kg weight 1m with 3 pulleys
requires a 3.33kg force and 3m of rope. I plan to use bearings
rather than plain shafts so that should cut the friction down. What
I hadn't expected was the statement "equal sized pulleys". I've
got 2 large (200mm) and 1 small (110mm) pulleys. Trying to
think what difference the size would make. Any Ideas ? Looked
around a little bit but can't find a mention of this. Plenty of very
good sites about what I already knew though

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2001\09\21@045426 by Russell McMahon

picon face
IMHO, for practical purposes the result is the same for unequal sized
pulleys
There would not normally be a reason for doing this.

With equal sized pulleys the mechanical advantage is equal to the number of
ropes supporting the load. ie if there are 4 ropes then the tension in each
rope is equal to 1/4 of the load so you have to pull only 1/4 as hard on the
rope end to lift the load. You have to pull the end 4 times as far though.

With unequal sized pulleys the ropes "slope" vertical component of the force
in the ropes supports the load and the horizontal component is wasted by
opposing an equal and opposite force in the matching rope on the other side.
The rope force for N ropes is therefore higher than Load/N and you are worse
off.

Bearings can be a very good idea as you can certainly build a block and
tackle with negative advantage if you have lots of stiction.

Depending on how far you have to lift the load there may be better solutions
such as a lead screw using reasonably cheap threaded rods. Not as flexible
as rope though (pun intended).


RM


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2001\09\21@060407 by Jinx

face picon face
> With unequal sized pulleys the ropes "slope" vertical
> component of the force in the ropes supports the load
> and the horizontal component is wasted by opposing
> an equal and opposite force in the matching rope on
> the other side

OK, gotcha, it's a vector thing. So for a vertical lift keep all
sections of the rope vertical. If I have to I can build up the
smaller one to 200mm or look around for a 200mm. Picked
these up at a scrap dealer for $5. 2nd-hand block down the
road was $130, so there's a pretty good "tinker" margin.
If I found another pair of $5 200mm pulleys I'd think myself
very lucky

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2001\09\21@071344 by Russell McMahon

picon face
> > With unequal sized pulleys the ropes "slope" vertical
> > component of the force in the ropes supports the load
> > and the horizontal component is wasted by opposing
> > an equal and opposite force in the matching rope on
> > the other side
>
> OK, gotcha, it's a vector thing. So for a vertical lift keep all
> sections of the rope vertical. If I have to I can build up the
> smaller one to 200mm or look around for a 200mm. Picked
> these up at a scrap dealer for $5. 2nd-hand block down the
> road was $130, so there's a pretty good "tinker" margin.
> If I found another pair of $5 200mm pulleys I'd think myself
> very lucky

Yes - you've got it - but the result is liable to be insignificant until
size mismatch gets quite severe.
When the pulleys are a reasonable distance apart even quite large size
differences would result in smallish angles from vertical. Only as you
pulled the blocks close together would the angles get very far away from
vertical.
Vertical component will be the cosine of the deviation from vertical

   Useful Force = rope tension x cos(angle)

Even at 10 degrees cos(10) = 0.984 = almost 1.
cos(20) = 0.94.

In practice it will usually not matter but you could use the following (E&OE
:-) ) to calculate effect of pulley size mismatch at various separations.

1.below  gives the angle of the rope for a given arrangement.

2 below gives the distance between pulleys at which the rope attains a given
angle.

Back of envelope figures using simple geometry (which assume that the ropes
touch pulleys at right angles to the vertical (which is true for equal
pulleys and almost true for reasonable size mismatches) give -

1        Angle = tan^-1((n-1)*x/d)

and

2        d = ((n-1) * x) / tan(angle)

Where

   N is ratio of pulley sizes.
   d = distance between pulley centres
   angle = angle that ropes make away from vertical
   x = radius of small pulley

Set x = 1 to get answer in multiples of small pulley radius.

Check: Use as required -

       2:1 pulley ratio
       10 degrees angle max
       4  radii separation


1.    Angle at 4 radii sepn = tan^-1 ((2-1)*1/4) = 14 degrees

2.    Sepn at 10 degrees = ((2-1)* 1/tan(10)) = 5.7 x small radius

Plug second angle into 1st equation to check

Angle = tan^-1((2-1)*1/5.7) = 9.95 degrees  --> QED



       Russell

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2001\09\21@075901 by Olin Lathrop

face picon face
> "Ignoring friction, for a simple pulley system with equal sized
> pulleys, the mechanical advantage is equal to the number of
> pulleys. The ratio of the distance moved by the effort to that
> moved by the load (the velocity ratio) will also equal the number
> of pulleys, and is not affected by friction or the weight of the
> equipment. In practice, these factors reduce the mechanical
> advantage, and the efficiency of the system is calculated by
> dividing the mechanical advantage by the velocity ratio"
>
> What
> I hadn't expected was the statement "equal sized pulleys". I've
> got 2 large (200mm) and 1 small (110mm) pulleys. Trying to
> think what difference the size would make.

It will make no difference to the mechanical advantage.  The force in the
rope along the direction of the overall pull will be the total pull force
divided by the number of rope segments.  Note that this has nothing to do
with pulley size.  The issue with different size pulleys is that the rope
segment between such pulleys will not be exactly parallel to the overall all
pull force.  Since the force in the direction of the overall pull will be
the same, the force in the rope segment will be increased by 1/COS of the
rope angle with respect to the overall pull direction.  Since COS for small
angles is essentially 1, you can ignore this in most practical applications.
Even at 45 degrees, the force in the rope is only increased by 41%.  Unless
the size ratio is large and the pulleys get close together, this is not much
of an issue.  Your pulleys are not that different.  Your two pulley sizes
can only result in about 5% extra rope tension worst case, so don't worry
about it.


********************************************************************
Olin Lathrop, embedded systems consultant in Littleton Massachusetts
(978) 742-9014, spam_OUTolinTakeThisOuTspamembedinc.com, http://www.embedinc.com

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2001\09\21@090703 by William Jacobs

flavicon
face
Hi all,
       If the pulleys are fixed to a common shaft, they have to be the same
size.  Otherwise, any will work.

bill


Jinx wrote:
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2001\09\21@150736 by Olin Lathrop

face picon face
>         If the pulleys are fixed to a common shaft, they have to be the
same
> size.  Otherwise, any will work.

The pulleys *must* be able to rotate independently anyway.  If they are all
the same diameter than the rotation rates have to be different.


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2001\09\21@172107 by Jinx

face picon face
> With equal sized pulleys the mechanical advantage is equal
> to the number of ropes supporting the load. ie if there are 4
> ropes then the tension in each rope is equal to 1/4 of the load
> so you have to pull only 1/4 as hard on the rope end to lift the
> load. You have to pull the end 4 times as far though.

The How Stuff Works page shows that well, and they compare
the force magnification of pulleys to levers and hydraulics

http://www.howstuffworks.com/pulley.htm?printable=1

> When the pulleys are a reasonable distance apart even quite
> large size differences would result in smallish angles from vertical.
> Only as you pulled the blocks close together would the angles
> get very far away from vertical

I don't want the object to turn as it's being lifted. Using guides
isn't practical so I thought I'd arrange the pulleys in a square
(by adding a fourth pulley just for the rope to pass over) and
have a bar supported by two vertical ropes at the bottom. This
will keep all ropes vertical in a shortening rectangle as the
object is lifted, rather than a 3-pulley flattening triangle. There's
some small additional friction because of a fourth axle but
that's no big deal

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2001\09\21@202030 by Russell McMahon

picon face
> >   If the pulleys are fixed to a common shaft, they have to be the
> >   same
> >   size.  Otherwise, any will work.
>
>     The pulleys *must* be able to rotate independently anyway.  If they
are all
>     the same diameter than the rotation rates have to be different.


In PRACTICE I'm sure it is very wise to allow each pulley to rotate
independently as any slight differences in pulleys or rope will lead to rope
slip or tension changes etc which could cause losses that markedly reduce
the efficiency.

My mental picture (which may be wrong) suggests that in THEORY, with equal
size pulleys, each group of pulleys can be fixed to a common axle.
This is because the rope must maintain a constant tension and velocity
throughout and this would occur in these (theoretical) circumstances.
With care one could perhaps use a single rotating axle with spacer disks
between ropes (Teflon?) although even resisting the side forces which occur
along the axle due to the slight horizontal offset in each rope could be
enough to introduce unacceptable losses. The fact that pulley blocks that I
have seen all seem to use multiple independent pulleys suggests that this is
the result of practical experience.



       Russell

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2001\09\21@204810 by M. Adam Davis

flavicon
face
You should NOT use a pulley block in this manner with a common shaft.
If you intend to use a block to amplify the force, then the pulley
closest to the object to be lifted will move very little relative to the
pulley closest to the pulling force.

-Adam

Russell McMahon wrote:

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2001\09\22@010523 by Lee Jones

flavicon
face
>>> If the pulleys are fixed to a common shaft, they have to be
>>> the same size.  Otherwise, any will work.

>> The pulleys *must* be able to rotate independently anyway.
>> If they are all the same diameter than the rotation rates
>> have to be different.

> My mental picture (which may be wrong) suggests that in THEORY,
> with equal size pulleys, each group of pulleys can be fixed to
> a common axle.  This is because the rope must maintain a
> constant tension and velocity throughout

I'll agree with constant tension, but not with constant velocity.

Let's use the example of a 3 way force multiplication with 3
pulleys.  Two pulleys (#1 & #3) are on an axle which is suppended
from an overhead beam.  One pulley (#2) is on the top of a weight
under the beam.  Rope is attached to the object, goes up through
pulley #1, back down, through pulley #2, back up, through pulley
#3, and then to the motive force (human, winch, etc).

To lift the object 5 feet, you have to pull the end of the
rope 15 feet -- a 3 fold distance increase for a 3 fold
force multiplication.

Since the object is moving up 5 feet, 5 feet of rope passes
through pulley #1; 10 feet of rope passes through pulley #2;
and 15 feet of ropes passes through pulley #3.  So if pulley
#1 and #3 are the same diameter and on a common axle, they
cannot turn at the same rate.

If pulley #3 is 3 times the diameter of pulley #1, they could
be fixed to a common axle because they would turn at the same
angular velocity.
                                               Lee Jones

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2001\09\22@012426 by Raymond Choat

flavicon
face
I may be wrong (my name is "wrong way ray" after all) but all pulleys would
move same rate if the rope is same width its entire length. But I agree
still they should not have common shaft. If the rope got rained on and part
was in the sun and part was not, then the rope would not be the same
thickness its entire length because of swelling (wet) or shrinking (dry).
Wrong Way Ray (Raymond Choat)

{Original Message removed}

2001\09\22@044423 by Russell McMahon

picon face
Yes, I'm convinced - I'm wrong ! :-(
My error was in assuming that rope velocity must be constant at all points
on a continuous rope at constant tension.

Putting what you say in a different manner - this is wrong because each
additional stage has to pass not only its own rope but rope from all stages
before it so that the rope velocity in each stage increases as you add
additional stages because more rope has to pass per unit time.


       Russell McMahon



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2001\09\22@055252 by Peter L. Peres

picon face
The advantage of the unequal pulley is in damping pendulum oscillations
imho. I am sure that it is done deliberately sometimes. Also the pulleys
are arranged staggered in a certain way for the same reason.

Peter

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2001\09\22@072441 by Roman Black

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Raymond Choat wrote:
>
> I may be wrong (my name is "wrong way ray" after all) but all pulleys would
> move same rate if the rope is same width its entire length. But I agree
> still they should not have common shaft. If the rope got rained on and part
> was in the sun and part was not, then the rope would not be the same
> thickness its entire length because of swelling (wet) or shrinking (dry).
> Wrong Way Ray (Raymond Choat)

Sorry Ray, if the pulleys all moved at the
same rate there is no mechanical advantage
and the block and tackle doesn't work. :o)
-Roman

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2001\09\22@075219 by Russell McMahon

picon face
This can be a bit of a brain teaser if you haven't thought it through (or if
you have ;-) ).

> I may be wrong (my name is "wrong way ray" after all) but all pulleys
would
> move same rate if the rope is same width its entire length. But I agree
> still they should not have common shaft. If the rope got rained on and
part
> was in the sun and part was not, then the rope would not be the same
> thickness its entire length because of swelling (wet) or shrinking (dry).
> Wrong Way Ray (Raymond Choat)

Yes - you are wrong - but I was too at first (no great surprise ;-) ) and it
is not fully intuitive.

The rope velocity relative to pulleys which support the load - which is what
matters here, increases by a quantum amount as it passes through each load
bearing pulley !.

I was initially disturbed by this apparent breaking of the laws in such a
simple machine - clearly I was seeing something from the wrong point of
view. And that is exactly what the problem was - the frame of reference
changes for different parts of the rope. For our purposes the rope velocity
is measured relative to the pulleys as they are what are turning and we wish
to establish if they are turning at the same rate. However, the pulleys are
moving vertically as well and taking the vector sum of the various support
velocities plus the rope velocities makes sense of the system.

For a graphic demonstration of the velocity differences

- Take a pice of string about 500mm / 18 inches long.
- Tie  a loop in one end.to make a circle about 40mm  / 1.5" in diameter.
- Place loop over left index finger.
- Hold left hand about 150 mm above a door know or drawer handle etc.
- bring string down and around the knob and up again and over left index
finger again.
- slowly pull down on end of string
- Note rate that your left hand or a mark anywhere on the first piece of the
string is descending compared to the pulling hand.

The right (pulling )hand has exactly twice the velocity of the left hand.
While this result is an obvious one, actually seeing the string haveing
different velocities in two adjacent parts may be 'interesting".

Is that anything like "wrong way Mollison" (I think his name was)(who
definitely wasn't going the wrong way).


       Russell





>
> {Original Message removed}

2001\09\22@135306 by Raymond Choat

flavicon
face
After reading Lee Jones message on this, I can clearly see I am wrong. Lee
Jones put is so well that even I can see it now. Thanks Lee.
Wrong Way Ray (Raymond Choat)

{Original Message removed}

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