

11032019, 12:49

#16

Moderator
Join Date: Mar 2009
Location: Denmark (Winter), Helsinki (Summer); Cruising the Baltic Sea this year!
Boat: CutterRigged Moody 54
Posts: 33,556

Re: Possibly Original Thought About Chain Catenary  or  The Myth of the Bar Tight C
I'm very glad you've chimed in. I respect your thorough, scientific approach to such questions and I'm sure you will contribute a lot to this thread.
Quote:
Originally Posted by thinwater
. . .
c. At 30 knots or so the chain lifts off the bottom, depending on scope, chain weight and water depth. In shallow water it happens early, in deep water, hardly at all. Another way to think about this is that it is not so much the scope, as the to total amount of chain out; once you reach 250+, it's a lot more weight and better leverage.

Great; that's exactly what I'm talking about. If the chain is hardly lifting off the bottom, then catenary is still working. How to quantify it, though?
Quote:
Originally Posted by thinwater
d. Then both the angle AND the amount of energy absorbed declines, gradually, until at 4055 knots (again, depends on the depth and scope, so don't try to pin me down) the angle at the bottom is between that off 100% rope and flat, and the energy dissipation is effectively zero. Though not straight, the difference between straight and the curve you have is only inches.

We absolutely agree, that it is not a question of the chain being absolutely straight, which will never exist (as SailorBoy pointed out ). But there is a point, as you say, when so much of the curve has been pulled out, and the force curve is so steep, that the chain is behaving more like a bar, and is not doing any work any more either to keep the angle of pull on the anchor shank down, nor to absorb energy. THAT point is what we are interesting in quantifying, so I WILL pin you down
Quote:
Originally Posted by thinwater
The math is complex,

I know
Quote:
Originally Posted by thinwater
particularly when you consider the dynamic effects. I've measured and I've dived when it was blowing. The true is in the middle and hard to nail down. But something the math will quickly show you is that with G70 chain you can fit twice as much in the locker, and that will ALWAYS hold better. I think that is unarguable.

Well, it is certainly inarguable, that you will get more effect out of the same mass of chain, spread out over a greater length. That is pretty much axiomatic. But you assume here unlimited swing room  which does not normally exist. The area covered by a swinging vessel increases as a function of pi times rode length squared, and you very quickly get into untenable territory when you get over maybe 100 meters of chain, which happens to be about the most anyone carries, whether it's heavy chain or light chain.
So if you recognize that limitation, then you are not necessarily given the choice to double the scope by lightening the chain. For example, I have 100 meters of 1/2" chain  I can't gain anything except some weight out of the bow, by going to lighter chain. I already have the heaviest anchor I can handle, and 100 meters is about as much as I would ever want to have out, from the point of view of swinging room.
Quote:
Originally Posted by thinwater
There is a common area, with medium depth water and moderate weather, where heavy chain allows less scope and might be helpful. Thus the popularity of heavy chain; it is a good answer in 2530 knots, in 15 feet of water, in a crowded anchorage. In 60 knots and enough room, not so much.

I don't know about that last sentence. Actually, I doubt it. We need numbers. Also, and very importantly, "enough room" is not a simple take whatever you need proposition. "Enough room", beyond a certain point, does not grow on trees. In any kind of sheltered space you are unlikely to have room to swing to more than 100 meters of chain of whatever weight.
Quote:
Originally Posted by thinwater
The breaking strength argument is off the point for three reasons:
*The rode better not operate above the WLL, or it won't last long.
*No, it does not take nearly that much force for the difference to become academic. * Using a proper snubber you should be several times below the WLL. If not, the anchor will drag.

You are certainly right about this. I took breaking strength just as the extreme example  if there is still effective catenary at breaking strength, then it is absolutely unrecognizably clear, that my original thesis is right and "the chain will always get bar tight in extreme conditions" is a total myth. A much more realistic test is WLL, and that is a much lower bar for my theory. WLL for G40 12mm chain is 2 tonnes.
Quote:
Originally Posted by thinwater
I suggest you take a length of chain and a comealong (chain hoist), and stretch a chain between two trees. Do you really believe it takes 4 tons to get the chain pretty straight? I don't think you believe that, but try it. You will also learn the stretching that last bit is so hard there is not shock absorption; tje stretch/stress relationship is not linear, like rope.
Seriously, stretch a chain at the boatyard. You will find it enlightening.

Well, I have quite a good feel and understanding for this, which is clearly true. But truth will be in the numbers. The catenary curve for a given set of parameters will tell us the effective angle of pull on the anchor, and the force curve will show us whether there is still any effective elasticity in the rode or not. That is the €64 question, and the numbers won't lie.
__________________
"You sea! I resign myself to you also . . . . I guess what you mean,
I behold from the beach your crooked inviting fingers,
I believe you refuse to go back without feeling of me;
We must have a turn together . . . . I undress . . . . hurry me out of sight of the land,
Cushion me soft . . . . rock me in billowy drowse,
Dash me with amorous wet . . . . I can repay you."
Walt Whitman



11032019, 13:23

#17

cruiser
Join Date: Nov 2007
Location: Probably in an anchorage or a boatyard..
Boat: Ebbtide 33' steel cutter
Posts: 5,030

Re: Possibly Original Thought About Chain Catenary  or  The Myth of the Bar Tight C
Quote:
Originally Posted by Dockhead
Unfortunately I have not been able to find a calculator for this, and I get a little lost in the formulae. If someone with better engineering chops than me, and likewise curious about this question, would like to help me out with this, I would be most grateful.

This was from playing around in Python >
Catenary force plot
Not completely sure of the maths but think it's pretty close, enough for some web discussion anyway
Should give some dimensions for when the last link of the chain next to the anchor just lifts. Chain weight & force on the boat are adjustable down the bottom.
Also another online calc here>
https://www.spaceagecontrol.com/calccabm.htm
On thing to bear in mind, say there's lots of wind, enough to produce a force which straightens out the catinary quite a lot, then a gust gets the bow and blows the boat off,  so if you have a play around in the spaceage site, you'll see a substantial increase in force only stretches the catinary a short distance if it's already shallow. So the boat gets pulled up quick and the anchor gets a big increase in load. Or if you have a decent snubber it should slow the boat own over a greater distance with less force on the hook. (f=ma).
The interesting equation imho (one of them) is energy = 1/2 * mass * velocity squared. So looks like if you can slow down the boat veering off by half you can reduce the energy which needs to go somewhere by a factor of 4.
As an aside, playing around with an arduino and a load cell out of a bust kitchen scales, the accuracy is outstanding, down to less than a gram for a 5Kg scale , so it seems likely that making a batterypowered load cell sending data over wifi should be possible & cheap  some sort of setup with the snubber onboard & loadcell measuring halfway along the length. Similar concept to a rope tension meter  anyway, that would be like data christmas during a blow, try various options and look at the numbers.
I reckon keeping the boat speed down must help a lot to reduce the force on the hook, and obviously a good snubber to help increase the distance it takes to stop the boat when if shoots off.



11032019, 13:37

#18

Moderator
Join Date: Mar 2009
Location: Denmark (Winter), Helsinki (Summer); Cruising the Baltic Sea this year!
Boat: CutterRigged Moody 54
Posts: 33,556

Re: Possibly Original Thought About Chain Catenary  or  The Myth of the Bar Tight C
Quote:
Originally Posted by conachair
This was from playing around in Python >
Catenary force plot
Not completely sure of the maths but think it's pretty close, enough for some web discussion anyway
Should give some dimensions for when the last link of the chain next to the anchor just lifts. Chain weight & force on the boat are adjustable down the bottom.
Also another online calc here>
https://www.spaceagecontrol.com/calccabm.htm
On thing to bear in mind, say there's lots of wind, enough to produce a force which straightens out the catinary quite a lot, then a gust gets the bow and blows the boat off,  so if you have a play around in the spaceage site, you'll see a substantial increase in force only stretches the catinary a short distance if it's already shallow. So the boat gets pulled up quick and the anchor gets a big increase in load. Or if you have a decent snubber it should slow the boat own over a greater distance with less force on the hook. (f=ma).
The interesting equation imho (one of them) is energy = 1/2 * mass * velocity squared. So looks like if you can slow down the boat veering off by half you can reduce the energy which needs to go somewhere by a factor of 4.
As an aside, playing around with an arduino and a load cell out of a bust kitchen scales, the accuracy is outstanding, down to less than a gram for a 5Kg scale , so it seems likely that making a batterypowered load cell sending data over wifi should be possible & cheap  some sort of setup with the snubber onboard & loadcell measuring halfway along the length. Similar concept to a rope tension meter  anyway, that would be like data christmas during a blow, try various options and look at the numbers.
I reckon keeping the boat speed down must help a lot to reduce the force on the hook, and obviously a good snubber to help increase the distance it takes to stop the boat when if shoots off.

That's really tantalizing, but I can't make sense of the Pythongenerated graphs (where is the kg/f needed to lift the last link?), nor can I make that calculator work. Do you erase the formulae and type in the values? Doing that, I get "no results". If you can explain how to use that calculator, I will be very grateful!
Or you can plug in 20000N (about 2000kg/f), 100, 3.3, and 9.81, and tell us what it says. The amount of sag will tell us a lot.
__________________
"You sea! I resign myself to you also . . . . I guess what you mean,
I behold from the beach your crooked inviting fingers,
I believe you refuse to go back without feeling of me;
We must have a turn together . . . . I undress . . . . hurry me out of sight of the land,
Cushion me soft . . . . rock me in billowy drowse,
Dash me with amorous wet . . . . I can repay you."
Walt Whitman



11032019, 13:43

#19

Moderator
Join Date: Mar 2009
Location: Denmark (Winter), Helsinki (Summer); Cruising the Baltic Sea this year!
Boat: CutterRigged Moody 54
Posts: 33,556

Re: Possibly Original Thought About Chain Catenary  or  The Myth of the Bar Tight C
Quote:
Originally Posted by Kelkara
my days of solving this kind of maths analytically are over ... but it's fairly easy to produce a numerical approximation.
The catenary curve for chain of 3.3 Kg/m and 8480Kgf horizontal tension will always look the same no matter how deep the water, we just have to find on which part of the curve your scenario of 100m of chain in 30m depth sits.
I plotted the catenary every m using the formula y=(exp(ax)+exp(ax))/2a where a=density/tension = 3.3/8480. ... Then looked for the point 100m further up the chain and checked to see where it was 30m higher ... then found the angle at the anchor at this point.
The biggest approximation is that I used a horizontal force of the chains breaking strength, not the force along the chain, but I don't think it makes much difference at this extreme pull.
If my calculation is correct (no guarantees) then the answer to your question is that the chain will be 99.995% straight and pulling at an angle of 16.3 degrees on the anchor.
so it's not pulling horizontally, and it doesn't have much snub to give.

Seems like a solid method, and a good approximation would be gold at this stage, but I do disagree about the depth of the water  the depth of the water, or vertical displacement between the two ends of the chain, has a big effect on the force vectors on the chain so should change the catenary quite a bit.
So by 99.995% straight do you mean that there is 0.005% of 5mm (!) of sag along a 100m length of chain? Really? I have not figured out the math, so can't argue with you, but that does not seem intuitively possible. If it is really true, then my theory is way wrong.
__________________
"You sea! I resign myself to you also . . . . I guess what you mean,
I behold from the beach your crooked inviting fingers,
I believe you refuse to go back without feeling of me;
We must have a turn together . . . . I undress . . . . hurry me out of sight of the land,
Cushion me soft . . . . rock me in billowy drowse,
Dash me with amorous wet . . . . I can repay you."
Walt Whitman



11032019, 13:51

#20

cruiser
Join Date: Nov 2007
Location: Probably in an anchorage or a boatyard..
Boat: Ebbtide 33' steel cutter
Posts: 5,030

Re: Possibly Original Thought About Chain Catenary  or  The Myth of the Bar Tight C
Quote:
Originally Posted by Dockhead
That's really tantalizing, but I can't make sense of the Pythongenerated graphs (where is the kg/f needed to lift the last link?),

Other way round, there are 2 sliders t the bottom, default to 2.3Kg/M chain weight (10mm chain  https://jimmygreen.com/content/81ca...dweightguide) and 230Kg force on the boat.
Then the top graph is the water depth & chain out needed so just the last link would lift. So with 10mm chain, 230Kg force  in 10m of water with 45.87m of chain out the last would have just lifted. In perfect maths world
Quote:
nor can I make that calculator work. Do you erase the formulae and type in the values? Doing that, I get "no results". If you can explain how to use that calculator, I will be very grateful!

Ooops, just copied the link without checking, looks like it's broken



11032019, 14:03

#21

Registered User
Join Date: Oct 2015
Posts: 2,007

Re: Possibly Original Thought About Chain Catenary  or  The Myth of the Bar Tight C
Quote:
Originally Posted by Suijin
Big anchor, more rode, use a snubber. And not some 6’ toy to protect your windlass...a proper long stretchy snubber of 30’ or more.

Like Suijin said! Worrying about the catanary is a fool’s errand. Just use a PROPER snubber with 4 or 5 feet of useable stretch and put your spreaddsheet to sleep.
I see so many boats with silly little “snubbers” of 4 or 6 feet in total length. That a waste of time.



11032019, 14:03

#22

Registered User
Join Date: Oct 2016
Location: Vancouver Island
Boat: Hullmaster 27
Posts: 1,021

Re: Possibly Original Thought About Chain Catenary  or  The Myth of the Bar Tight C
Quote:
Originally Posted by Dockhead
Seems like a solid method, and a good approximation would be gold at this stage, but I do disagree about the depth of the water  the depth of the water, or vertical displacement between the two ends of the chain, has a big effect on the force vectors on the chain so should change the catenary quite a bit.

We are talking about different things here ... all catenary curves have the same shape, changing the weight and tension of the chain will scale the curve, but not change the geometry ... but the weight and tension were given in your initial question ... the chain between boat and anchor will be just part of that catenary curve ... you need to find two points on that curve that are 100m apart along the curve and 30m apart vertically ... there will only be one point ... change the 30m to a different water depth, and the result will be on a different part of the curve ... We can plot this point in a graph of scope against angle of pull on the anchor ... but remember this is for your specified chain pulling at breaking strength.
Quote:
So by 99.995% straight do you mean that there is 0.005% of 5mm (!) of sag along a 100m length of chain? Really? I have not figured out the math, so can't argue with you, but that does not seem intuitively possible. If it is really true, then my theory is way wrong.

that was the value calculated by taking the ratio of the straight line distance between the anchor and boat, and the length of chain ... 8.5 tonnes of force is a pretty serious pull.



11032019, 14:09

#23

Moderator
Join Date: Mar 2009
Location: Denmark (Winter), Helsinki (Summer); Cruising the Baltic Sea this year!
Boat: CutterRigged Moody 54
Posts: 33,556

Re: Possibly Original Thought About Chain Catenary  or  The Myth of the Bar Tight C
Quote:
Originally Posted by Kelkara
We are talking about different things here ... all catenary curves have the same shape, changing the weight and tension of the chain will scale the curve, but not change the geometry ... but the weight and tension were given in your initial question ... the chain between boat and anchor will be just part of that catenary curve ... you need to find two points on that curve that are 100m apart along the curve and 30m apart vertically ... there will only be one point ... change the 30m to a different water depth, and the result will be on a different part of the curve ... We can plot this point in a graph of scope against angle of pull on the anchor ... but remember this is for your specified chain pulling at breaking strength.
that was the value calculated by taking the ratio of the straight line distance between the anchor and boat, and the length of chain ... 8.5 tonnes of force is a pretty serious pull.

5mm?
Well, if you say so  obviously you have a much deeper understanding of the math than I do. Well, if that's really so, it blows up my "at breaking strength" hypothesis.
What about 2000kg/f?
And what angle of pull do you get?
__________________
"You sea! I resign myself to you also . . . . I guess what you mean,
I behold from the beach your crooked inviting fingers,
I believe you refuse to go back without feeling of me;
We must have a turn together . . . . I undress . . . . hurry me out of sight of the land,
Cushion me soft . . . . rock me in billowy drowse,
Dash me with amorous wet . . . . I can repay you."
Walt Whitman



11032019, 14:12

#24

Registered User
Join Date: Oct 2016
Location: Vancouver Island
Boat: Hullmaster 27
Posts: 1,021

Re: Possibly Original Thought About Chain Catenary  or  The Myth of the Bar Tight C
double checking my numbers ....



11032019, 14:24

#25

Moderator
Join Date: Mar 2009
Location: Denmark (Winter), Helsinki (Summer); Cruising the Baltic Sea this year!
Boat: CutterRigged Moody 54
Posts: 33,556

Re: Possibly Original Thought About Chain Catenary  or  The Myth of the Bar Tight C
Quote:
Originally Posted by billknny
Like Suijin said! Worrying about the catanary is a fool’s errand. Just use a PROPER snubber with 4 or 5 feet of useable stretch and put your spreaddsheet to sleep.
I see so many boats with silly little “snubbers” of 4 or 6 feet in total length. That a waste of time.

Well, I do rig a snubber  10 meters of heavy nylon  in really bad conditions, but I'm still not sure whether it's really needed or not, and I'm curious. The snubber is surely a good thing, but that doesn't mean that understanding the other stuff is a "fool's errand". Snubber also has nothing to do with the angle of pull on the anchor.
__________________
"You sea! I resign myself to you also . . . . I guess what you mean,
I behold from the beach your crooked inviting fingers,
I believe you refuse to go back without feeling of me;
We must have a turn together . . . . I undress . . . . hurry me out of sight of the land,
Cushion me soft . . . . rock me in billowy drowse,
Dash me with amorous wet . . . . I can repay you."
Walt Whitman



11032019, 17:20

#26

Registered User
Join Date: Oct 2016
Location: Vancouver Island
Boat: Hullmaster 27
Posts: 1,021

Re: Possibly Original Thought About Chain Catenary  or  The Myth of the Bar Tight C
I still think I'm calculating it right.
This graph shows the same 100m of 3.3Kg/m chain in 30m of water (3.3:1 scope) with a horizontal pull of 500, 1000, 2000, 4000 and 8000 Kg horizontal force (there will also be 330Kg vertical force). The chain will take the shape of the curve between the dots. 510Kg is the point at which the last link lifts off the bottom and starts to pull up on the anchor, at which point there is only 1.5m of "slack" left in the catenary for snubbing.
Also because I was interested. for the same 100m of chain I plotted the angle of pull at the anchor after the last link starts to lift for 3:1, 4:1, 5:1 and 7:1 scope.
I also plotted how straight the chain is in the same situation ... This seems to be predominately controlled by the force, and not the scope ... these graphs are only for once the anchor starts lifting, and there is essentially no "snub" left in the catenary.



11032019, 19:09

#27

Writing FullTime Since 2014
Join Date: Nov 2008
Location: Deale, MD
Boat: PDQ Altair, 32/34
Posts: 9,367

Re: Possibly Original Thought About Chain Catenary  or  The Myth of the Bar Tight C
Honestly, the math and engineering are so complex as to defy reasonable analysis. To the factors that have been listed, add sailing at anchor, the drag created by the vessel itself, and the chaotic nature of waves, gust, and veers. Just forget it!
I'm an engineer, an I know too many variables when I see them. As a result, I simply got a load cell and measured forces with chain, rope, various snubbers, and even Dyneema, in a variety of weather. I highly recommend it. Math can show you some trends, but if you want to know, for YOUR boat, just collect the data. Someone else's boat will be different, but there will be trends.
As for the angle of the chain at the bottom, if you do the math you can get VERY close if you know the angle leaving the bow and the scope. That really is just math. For example, if the angle is the same as the scope, it's straight (it wont be). If the angle is 180% of that, the chain is close to lifting. Not exact, because it depends on scope, but close enough for the field.
With mixed rode it is more complicated.
How do you figure out the rode angle? You know the height of the roller and you can put marks on the snubber. Then it's trig, since you know opposite and hypotenuse.



11032019, 20:15

#28

Registered User
Join Date: Jul 2005
Location: Bellingham
Boat: Outbound 44
Posts: 9,319

Re: Possibly Original Thought About Chain Catenary  or  The Myth of the Bar Tight C
Thinwater
What load cell are you using? How is it rigged, to the chain, to the snubber?



11032019, 20:58

#29

Writing FullTime Since 2014
Join Date: Nov 2008
Location: Deale, MD
Boat: PDQ Altair, 32/34
Posts: 9,367

Re: Possibly Original Thought About Chain Catenary  or  The Myth of the Bar Tight C
Quote:
Originally Posted by Paul L
Thinwater
What load cell are you using? How is it rigged, to the chain, to the snubber?

Nothing special. I have used many.
You rig according to what you want to measure. Generally I attached to the rode with a Spectra loop and prusik hitch. Often the anchor point was the midships cleat, with a Dyneema extension (no stretch added to distort tests).



11032019, 21:59

#30

Registered User
Join Date: Sep 2012
Location: Washington State
Boat: Colvin, Saugeen Witch (Aluminum), 34'
Posts: 2,243

Re: Possibly Original Thought About Chain Catenary  or  The Myth of the Bar Tight C
I could not leave well enough alone so I cobbled up a test designed to measure the drag of "bar tight" chain moving through water (perpendicular to its long axis).
Apparatus: 1 foot of 3/8" BBB chain with links welded to simulate a tight chain. Small metal rod (with very thin plywood "feathers") welded to midpoint. Feathers will ensure that the chain remains horizontal while freefalling in water.
Apparatus was released (dropped) into water of 17 foot depth. Time to reach bottom averaged 4.5 seconds (terminal velocity was reached almost instantaneously as time varied only .2 seconds between releasing at zero speed and dropping into water at high speed)
17 feet/4.5 seconds = 3.77 ft/second. or 2.2 knots (coincidentally similar to Thinwater's observed speed of anchor rode lifting).
At terminal velocity (freefall) the drag of the body will be equal to it's weight. One foot of 3/8 BBB weighs 1.6 lb/ft or 1.4 lb/ft in water.
Using Dockhead's original scenario's length of 100 meters of chain, and guesstimating that 1/2 the length (50 meters) to be moving at an average of 2.2 knots (upward), I calculate that the drag of the chain through the water to be about 230 lbs. (164 feet X 1.4). This drag will act similarly to a kellet (momentarily, while the chain is moving upward).
The pulling force required to counter act this 230 lbs. of chain drag will vary depending upon the amount of catenary. Without doing any math, it is easy to see that the force required will be many times more (10 times?) than the chain drag as the catenary becomes small.
I stand by my assertion that the hydrodynamic drag of chain lifting (during gusts and waves) plays a significant role in absorbing an anchored boat's energy. The exact amount is far too complicated to compute (as Thinwater mentioned, above).
Steve





Thread Tools 
Search this Thread 


Display Modes 
Rate This Thread 
Linear Mode


Posting Rules

You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts
HTML code is Off




Advertise Here
Recent Discussions 












Vendor Spotlight 




