laika
Well-Known Member
On this, would 10m for mooring warps, 12m for springs sound about right?
Stretchy ropes and elementary trig.
Any questions, see me after prep and before cocoa.
Oh dear. I confess that the only times I've tied against a wall wth a big tidal range it's been with the benefit of bars and loops. Plus Fareham's finest 6th form college where I last did a maths exam more than 30 years ago didn't issue "prep" or cocoa so I'm probably going to show myself up as a dunce on a public forum but...
The problem with Gwylan's calculations would seem to be that at the top of the tide the drop you've allowed for in the vertical plane translates to an ability to wander away from the wall in the horizontal plane. Obviously this can be addressed by use of weights as others have suggested but I strongly suspect that birdseye has some rather good experience here and a slightly more complex bit of maths might take account of the stretchy properties of the 3 strand nylon we've bought because it's stretchy.
As I (with an inexperienced-in-this-department eye) see it you're not looking to rig a 3-4-5 triangle. The longer the line the better because:
- the amount a line stretches is proportional to its length
- pythagoras tells us that the hypotenuse of Gwylan's triangle (our mooring line) needs to increase by proportionally the same amount (if the line is perpendicular to the side of the boat: i.e. the triangle is not a triangle but a straight line) or less (if it is secured laterally fore/aft) and that that proportion reduces the longer the base of the triangle (i.e. the further fore/aft away form the boat you're secured).
So as we increase the initial line length (i.e. base of the triangle or distance fore and aft from the boat that we've secured), the *proportion* of its length it needs to increase to accommodate a given drop decreases.
Ideally we'd like to make the amount the line needs to increase in length to accommodate our drop less than the amount the line will comfortably stretch. That way we're snug against the wall at the top of the tide and snug (but with tighter lines) at the bottom.
If the amount our line can comfortably stretch is s (e.g. 15% would mean s=0.15, for 20% s=0.20), the depth we need to accommodate is d, then presumably the length of line we need to achieve our goal in an idealised world (where the cleat of our 2 dimensional hypothetical boat is flush against the top of the wall) is:
l = sqrt((d^2)/(s^2 + 2*s))
Which for an 8m drop and rope which stretches 15% stretch would be:
l = sqrt((8^2)/(0.15^2 + 0.3))
l = sqrt(64/0.3255)
l = sqrt(196.62)
l = a bit over 14m
Obviously that's not real world and your lines need to be longer to be tied on somewhere and your boat probably isn't 2-dimensional and I have no idea how much 3 strand nylon can comfortably stretch (ideas?) and it probably gets less stretchy the more you do this kind of thing....but is the theory along the right lines?