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Hopefully this explains why a fixed anchor point at the top of the mast is different to a halyard running around a sheave and back down to deck level.

Consider the following (very rough!) sketches…

In 1A we use a turning block to double the force we apply when dragging a weight along the floor. We can clearly see that removing the block and fixing the rope directly to the weight as in 1B halves the mechanical advantage.

In 2A we have transposed the exact same system vertically. Instead of a weight, the block is now pulling downwards on the ceiling. Again it has the effect of (approximately) doubling the force applied to the running end of the rope. Again we see in 2B that removing the block halves the mechanical advantage.

In 3A the block is no longer anchored to the ceiling, but instead to a sheave at the top of a mast. Everything else remains the same, and whatever force we apply to the rope produces (approximately) double that force applied downwards to the top of the mast. 3B demonstrates again that removing the block and anchoring the rope at the masthead halves the downward force on the mast.

Note that it doesn’t matter whether the tail of the halyard is cleated halfway down the mast or led via a block at the bottom and into the boat – whatever tension is in that rope between it’s effective bearing points (either sheve to sheave or sheave to cleat) will be transferred to the mast as a compressive load. The system discussed earlier in the thread removes this load as there is no longer any running load on the halyard where it runs up the mast.

Here is a more detailed analysis of the forces acting on a typical rig:

Assume that angle “a” is the angle between the shrouds and the mast – lets say 15 degrees, and “b” is the angle between the forestay and the mast – say 30 degrees. We now apply 400lbs tension onto a conventional job halyard.

This gives a direct compressive load of 400lbs on the mast between the mast foot sheave and masthead sheave. There is a also an indirect compressive load due to the tension in the forestay, as shown in the right hand vector diagram. Rj is the forestay jib tension (400lbs), which can be resolved into a horizontal component Xj of 200lbs, and a vertical component Yj of 346lbs.

For the mast to stay where it is, the horizontal component must be resisted by the shrouds. Therefore they must exert an equal backward force of 200lbs, shown as Xs in the left hand vector diagram. This in turn produces a vertical component Ys of 746lbs.

So the total compressive force on the mast is halyard tension + Yj + Ys = 400 + 346 + 746 = 1492lbs.

But if we anchor the halyard at the top of the mast so that only the luff wire provides the tension, then we can remove the 400lb halyard compression and the total compressive force acting on the mast drops to 1092 lbs, a reduction of 27%.

(sorry for any drafting errors – typed in haste!)