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lydia said..
So lets forget the fashion and look at the physics.
A preventer rigged mid boom pulling downwards is loading the boom in a way never designed for and has the worse physics as while it may stop the boom swinging across there will be no load on the main sheet and all load on the preventer and the loads mostly in tension not compression.
Worst though is that once you do gybe, unless released you have a force aft of the centre of lateral resistance so you end up heading you head too wind which of course if also poled out means the headsail is also completely backed.
Niether the boom or the pole are designed to take these loads.
So if you do want a preventer, take it from the end of the boom to the very bow and have it able to be released immediately. That is best load wise.
Physics is just maths, and I'm a sucker for maths. Based on my calculations as below, the forces don't get to boom breaking territory at all.
Perhaps I have got the maths wrong at some point.
I've a 12 metre boat with the mast about 5m from the bow and a boom about 4m long. So a preventer rigged from the bow to the boom end with boom at 90 degrees to mast will be angled from the centre line about 39 degrees. Vector-wise 100 kg of forward force on the boom end requires 128kg of tension on the line. This is leaving aside the vertical angle of the rope (given boom isn't level with bow), as at about 11 degrees it doesn't much alter the vector equation (adds about 2 per cent, so 130 kg all up).
The beam at the widest point is 3.7metres, so 1.85 from the centre line. At the mast and chainplates (spreaders are not back swept and shrouds are in line with mast) the beam is 3m, and 2 metres for'ard of the mast is 2m. If I were to rig a preventer from the toe rail 2m for'ard of the mast, the angle of the rope (to the centre line) to the end of the boom would be 56 degrees. And vertical angle would be about 27 degrees. To exert 100 kg of forward force on the boom end would require 200kg of tension on the rope, this time allowing for the both the horizontal and vertical angles as the latter adds about 11 per cent. The preventer would be about 6 metres shorter, as it wouldn't have to travel to the bow and back.
The vector forces can be understood the other way round. That is, if the main backs, the preventer to the bow will experience 1.3 times the load on the rope as the pressure back on the boom measured at the end of the boom. Whereas the preventer rigged to the toe rail 2 metres in front of the shroud will experience 2 times the load on the rope as the pressure back on the boom measured at the end of the boom.
If the wind is 15 degrees ahead of the mainsail, and my main is an unreefed 27 square metres in 20 knots of wind, the initial back pressure on the sail is about 46kg. As the boom is a lever and the centre of effort is about one third of the boom length back from the mast, the force on the end of the boom is about 15kg. So not likely to trouble a preventer rigged to the bow nor one to the toe rail 2m ahead of the chainplate.
This sounds pretty light, and it is. The reason an uncontrolled gybe in 20 knots is more forceful is that as the boom responds to the back wind, the angle of the sail changes, and so the wind pressure on it increases. If the wind is 15 degrees ahead of the beam, by the time the boom has swung to 15 degrees from the centreline the wind force on the main is 175 kilos, and boom will be moving quite fast.
Going back to preventer loads, if the wind is 15 degrees ahead of the mainsail, and my main is a third reefed 10 square metres in 40 knots of wind, the back pressure on the sail is about 68kg. As the boom is a lever and the centre of effort is now perhaps a quarter of the boom length back from the mast (moves forward as reefs put in), the force on the end of the boom is about 16kg. So again not likely to trouble a preventer rigged to the bow nor one to the toe rail 2m ahead of the chainplate (where the force on the rope will be about 32kg).
Let suppose I rig the preventer to the vang point on the boom. Mine's about 1.2 metres from the mast. This changes the angles quite a bit and loses the leverage available from using the end of the boom.
Rigged to the bow, the angle of the preventer from the centre line is about 14 degrees. Vector-wise 100 kilos of forward force at the vang point requires 103 kg of tension on the rope.
Rigged to the toe rail 2m ahead of the chain plate the horizontal angle of the preventer to the centre line is just 6 degrees, as the vang point is only 20cm outboard of the toe rail where the preventer is rigged. The more important angle is the vertical angle of about 26 degrees from the block on the toe rail to the vang point. Together 100 kg of forward force on the vang point would require about 112 kg of tension on the rope.
If the wind is 15 degrees ahead of the mainsail, and my main is an unreefed 27 square metres in 20 knots of wind, the initial back pressure on the sail is about 46kg. As the boom is a lever and the centre of effort is about one third of the boom length back from the mast, the force on the vang point is about 64kg. Still not likely to trouble a preventer rigged to the bow nor one to the toe rail 2m ahead of the chainplate, where the force on the rope will be about 72kg.
If the wind is 15 degrees ahead of the mainsail, and my main is a third reefed 10 square metres in 40 knots of wind, the back pressure on the sail is about 68kg. As the boom is a lever and the centre of effort is perhaps a quarter of the boom length back from the mast (moves forward as reefs put in), the force on the vang point is also about 68kg. Still not likely to trouble a preventer rigged to the bow or to the toe rail 2m ahead of the chainplate (where the force on the rope will be about 76kg).
Lets suppose I'm racing and so only have two reefs in at 40 knots, for about 16 square metres of main. If the wind is 15 degrees ahead of the main, then the force on the sail will be 110kg and maybe 130kg at my vang point if the centre of effort is a bit more than 1.2 metres aft/outboard of the luff. Force on the toe rail mounted preventer might be about 145kg, and perhaps 135kg for the bow mounted preventer.
None of this equates to broken booms. Perhaps if to the 2 reef/40 knot scenario, we add another 20 degrees as the boat broaches a bit on a wave, then the wind force backwinding the sail gets up to about 240kg, and at the vang point about the same, with the load on the toe rail mounted preventer getting to 270kg. Still no drama for 12mm double braid rope. But perhaps some booms might break with the weight of 3 large people concentrated at the vang point. I've got my doubts though, since my vang has a five to one purchase and then is run to a winch, I'd be mighty surprised if it doesn't routinely exceed 300kg of force and more when winched tight.
To be clear, if the main is somehow 90 degrees to a fifty knot gust, with just 2 reefs in, the pressure on the main is well over half a tonne at 650 kg. But that's an unlikely scenario, unless the boat has lost steering and somehow ended up with the bow to the wind with the preventer holding the boom backwinded at 90 degrees to the wind. Or perhaps if there's a 180 degree wind change accompanying a squall. Or...?
Have I stuffed up the calculations somewhere?