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