> > 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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