Block and Tackle Efficiencies

Not long ago a friend was looking to upgrade the mainstay1 block and tackle system on his sailboat.

Mainsheet Upgrade

The question was if the proposed system would provide the anticipated reduction in force. It was an interesting question that, while seemingly straight forward, does have a couple of gotchas.

Block (pulley) and tackle (rope) calculations are usually pretty simple, with the mechanical advantage (MA) idealized as:

MA = \frac{F_B}{F_A} = n

where FA is the input force, FB is the load, and n rope sections.

Calculating the total force of multiple non-colocated blocks using the same tackle presented a fun challenge that requires one to take into account the individual location of the blocks and the forces transfered.

I did not draw every single link of the block and tackle, but in general the problem looks something like this:

diagram

Considering the moment arm:

F_{L} = (F_{A} \times A) + (F_{B} \times B)

We devise these relations:

F_{A} = n_{AD} \times F_{P} \times \arctan\left ( \frac{\left |  A-D \right |}{h}\right )

and

F_{B} = n_{BC} \times F_{P} \times \arctan\left ( \frac{\left |B-C\right |}{h}\right ) + n_{BD} \times F_{P} \times \arctan\left ( \frac{\left |B-D\right |}{h}\right )

where nxy is the mechanical advantage coefficient at a given point, x, with respect to another point, y; FP is the force exerted on the tackle (which must be uniform throughout!).

Using some assumptions regarding the lengths and relative positions of the blocks greatly simplifies the calculations, and we can plug ‘n chug from there:

F_{L}=F_{P}\cdot \left ( \frac{n_{AD}}{2} + \frac{n_{BC}}{4} + \frac{n_{BD}}{4}\right )\cdot \cos{(\arctan(\frac{l}{8h}))}

As expected, there is a loss of useful force due to the ropes not being normal to the boom and that causes the boom height/length ratio to become an interesting variable in these calculations. You lose 10% of your power with a 1:3.87 ratio, 25% with a 1:7.06 ratio, and 50% with a 1:13.86 ratio.

To model a direct input (with no block and tackle), I assume that the force was applied at a point between the two blocks, A and B, and normal to the boom.

The increase from no block and tackle system to the current system (single boom aft block, A; double boom traveling block, B) is:

\frac{F_{P}\cdot \left ( \frac{2}{2} + \frac{2}{4} + \frac{2}{4}\right )\cdot \cos{(\arctan(\frac{l}{8}))}}{\frac{3\cdot F_{P}}{8}} = \frac{16}{3}\cdot \cos{(\arctan(\frac{l}{8h}))}

…assuming the boom heigh/length ratio is 7, the mechanical advantage is 1:4.01.

Moving from no block and tackle to the the proposed system (double boom aft block, A; triple boom traveling block, B) is:

\frac{F_{P}\cdot \left ( \frac{4}{2} + \frac{3}{4} + \frac{3}{4}\right )\cdot \cos{(\arctan(\frac{l}{8}))}}{\frac{3\cdot F_{P}}{8}} = \frac{28}{3}\cdot \cos{(\arctan(\frac{l}{8h}))}

…again, assuming the boom heigh/length ratio is 7, the mechanical advantage is now 1:7.02.

The mechanical advantage from current system to proposed system is: 1:1.75

\frac{F_{P}\cdot \left ( \frac{4}{2} + \frac{3}{4} + \frac{3}{4}\right )\cdot \cos{(\arctan(\frac{l}{8h}))}}{F_{P}\cdot \left ( \frac{2}{2} + \frac{2}{4} + \frac{2}{4}\right )\cdot \cos{(\arctan(\frac{l}{8h}))}} = \frac{3.5}{2} =  1:1.75

It’s not quite the 1:2 advantage originally thought, but it’s close.

Epilogue:

The bonus gotcha occurred when said friend opted to install a double boom aft block (A) and a triple deck traveler block (D), but keep the boom traveler block (B) as a double. Essentially, putting an extra “loop” just between the boom aft block (A) and deck traveler block (D).

The imbalance of tension on the deck traveler block caused it to experience shear stress and bind on the traveler rail in ways it was not designed to — not good.

Converting the boom traveler block (B) to a triple and the deck traveler block (D) to a quad equalized the tension.

Problem solved.


  1. rope from the top of the main-mast to the foot of the fore-mast on a sailing ship 

And All I Ask is a Tall Ship and a Star to Steer Her By

Sea Fever
By John Masefield

I must go down to the seas again
to the lonely sea and sky
And all I ask is a tall ship
and a star to steer her by
And the wheel’s kick and the wind’s song
and the white sail’s shaking
And a gray mist on the sea’s face,
and a gray dawn breaking.

I must go down to the seas again
for the call of the running tide
Is a wild call and a clear call
That may not be denied
And all I ask is a windy day
with the white clouds flying
And the flung spray and the blown spume
and the sea-gulls crying.

I must go down to the seas again
to the vagrant gypsy life
To the gull’s way and the whale’s way
where the wind’s like a whetted knife
And all I ask is a merry yarn
from a laughing fellow-rover
And quiet sleep and a sweet dream
when the long trick’s over.

One of my favorite things about being back in Seattle are the opportunities to get out on the water. While we do have a speedboat of our own, I don’t think it get’s much better then sailing. At some point in time, I’d really enjoying sailing around the world – or at least part of the world. Although such an adventure will have to wait until I can get a boat of my own and a crew.

In the meantime, I’m fortunate to have a friend, Peter, who has a sailboat. And thus we went sailing on Monday and again on Tuesday (for the bonus round):


Remember all those pictures of your parents that you look at? This picture reminds me of one of those. In fact, I’d call this picture of Staples iconic.

All Images: Copyright 2008 Andrey Marchuk