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Old 04-29-2013
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Re: Forces on a dockline while docking....

Quote:
Originally Posted by MedSailor View Post
The coolest bit of sailboat seamanship that I was ever witness to was when my next door slip neighbors in Seattle came into their slip with their Moody (45?) ketch with a solid 40kt tailwind.

They came roaring in at full throttle and one crewmember handed me a dockline with an eye in the end and asked me politely (but with some urgency) to please place the eye on the aftmost dock cleat. I complied.

The line in question was run through a foot block at the quarter and then to the primary winch and back to the skipper at the helm. He hauled in the slack and the line stopped the boat. While still powering in forward (because the line was right at the picot point of the boat) it stayed stationary at the dock perfectly parallel to the finger pier.


I've used this technique quite a few times after witnessing this (though without as much throttle or tailwind!) and really like it. I'm planning on making a dedicated line that is the correct length for my dock, but when I got to sizing the line, I wondered, "how much force would be on this puppy?"

I used to feel competent with basic Newtonian physics, but now I'm too rusty. Seems like such a simple equation.... 30,000lb boat moving at x knots, stopped by line that stretches x inches....

Anyone able to help with the calculations? Otherwise I'll just grossly overbuild everything like I always do.
It is not a particularly difficult calculation. Google the "Work Energy Theorum" and go from there. The yacht has a kinetic energy equal to its mass x velocity. It's kinetic energy must be consumed/countered. The force necessary is calculated by taking the mass of the yacht multiplied by its acceleration. In this case, the yacht must go from its maneuvering speed as it enters the slip and the surge line is applied as a stopping force to zero speed as its bow reaches the inboard end of the slip and the surge line reaches its maximum extension. If the yacht's speed of approach is say, 5 knots, and its speed upon coming to rest is zero knots, the yacht's average speed over the distance of it's stop will have been 2-1/2 knots as it transitions from approach to landing. Assuming that the surge line extends half the length of the slip, from the end piling to amidships on the yacht, say 20 feet, and the line is nylon and hence cannot be stretched more than 20 percent of its length without rupture, or 4 feet (= .20 x 20), the yacht must transition from 8.43 ft per second (at 5 knots) to zero with an average speed of about 4.215 feet per second. With 4 feet of line "stretch" in the surge line, the yacht will traverse that stretch in, roughly, .95 seconds. The Delta V will be 8.43 feet/second with the Delta T being .95 seconds so the acceleration, which in this case has a negative sign (for slowing) is 8.88Ft/second squared. The required force to effectuate this Deceleration is the yacht's mass (= displacement/32.2 ft/sec^2) x -8.88 ft/sec^2. Let us assume that the yacht displaces say 21,000 lbs. It's mass is then 652.17 slugs and the force required to stop the yacht is about 5,793.30 lbs. The work energy is 5793.3 lbs applied over a distance of 4 feet or about 23,173.20 foot lbs. On the line side of the event, the issue becomes somewhat more complex because the amount of force needed to stretch the line 1% is decidedly different than the force needed to stretch the line 20% and it is the cumulative force moving over the length of elongation that must, in total, equal 23,170.20 ft lbs. Every increment of stretch can be described as a distance dx and the necessary energy to effectuate one such stretch or elongation as F*dx. To elongate the line the second increment, dx2, the force must be incrementally greater which may be, but is not necessarily, linearly related to the first incremental stretch unless the line is perfectly elastic, which of course it is not (tho' nylon in extremis is close). So, one is then confronted with a surge line that is applying a stopping force on the yacht that is not a steady 5,793.3 lbs but is increasing from, initially, zero, to some maximum, with the sum of (or the integral of) the forces applied over each incremental stretch of the surge line totaling 23,170.20 ft lbs from 0 feet to 4 feet. It will take a little pondering but it shouldn't be too difficult to come up with the size line needed from there. Note, however, as someone wisely pointed out previously, all of the kinetic energy of the yacht will be converted to potential energy in the line, which will surely take some time to burn off as heat, although it will, eventually.

FWIW...
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Last edited by svHyLyte; 05-02-2013 at 08:59 AM. Reason: Correct Misspellings
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