Mooring Catenary 1

[GrenMeera]

I need to write a computer program that simulates data in a mooring system but am unable to find much information on the shape of the catenary in a way that I can work with. I should point out that I am a skilled programmer, but have not delved into much high level engineering mathematics. I am, however, quite talented at math; I just warn you that I may not recognize terminology or a formula intrinsically.

My goal is to calculate the tension (vertical and horizontal components separated) of each mooring cable, preferably in kips. I have as variables: total length of chain (L), weight of chain in water (W), distance between mooring winch and the sea bed vertically (V), and distance from the mooring winch and the anchor horizontally (H).

At some point, the chain catenary will hit the sea bed, most likely before the anchor. All I really need is to find how much chain is in the catenary off the sea bed, and a method to factor that weight into vertical and horizontal components (which will be relative to the created angle of the catenary).

I very much appreciate any help I can get on this! Thank you for all responses.

 

[paddingtongreen]

The amount that lays on the bottom is controlled by the tension applied by the winch, I suppose, that that is the push of the tide or the wind on the craft. If there is no force on the craft the weight of the cable will pull the craft in until it is vertical, or nearly so.

What are your starting points, the "knowns" going in to find the "unknowns"?

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[GrenMeera]

I appreciate your attention to this.

There are more than one cable in the system, each creating their own tension forces, and therefore stabilizing the craft. Because of this, the craft should never be pulled to being vertical above the chain location on the sea bed (unless we have weird data, which then I suppose it's possible yet unlikely).

Outside forces, such as wind and waves, move the craft, which changes the positional data. Essentially, if this positional data can be used in an abstract manner, these external forces can be ignored(also, we are ignoring the current, as there is no realistic way to predict a current). All outside forces acting upon the craft are handled separately and move and rotate the craft accordingly.

I need to work on a per cable basis, finding the instantaneous forces for each cable to apply. This way I can total them for a final force being applied by all cables on the craft at an instantaneous time. The system already handles applying forces over time from instaneous results, so no worries about that.

The data I have is as follows:
Total length of chain (L)
Weight of chain in water (W)
Distance between mooring winch and the sea bed vertically (V) (this distance include craft heave)
Distance from the mooring winch and the anchor horizontally (H) (this distance includes craft sway and motion from outside forces)

 

[SlideRuleEra]

GrenMeera - Take a look in "USS Wire Rope Engineering Hand Book", which you can download (free .pdf) from this page of my website:

http://www.slideruleera.net/miscellaneous.html

Of course, this book addresses wire rope instead of chain... but I'll bet the math (covered in detail in the book) is essentially the same.

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[Denial]

Tricky, but interesting, problem. Making life easier for you is the fact that (I think) it is safe to assume that your chain is inextensible. Making life harder for you is the fact that the chain will not be sufficiently taut for you to be able to assume that the shape of the suspended part is parabolic. You will have to use the true chain shape equation.

Some useful properties of the chain equation:
» Equation is y = c*cosh(x/c) in a suitably chosen cartesian axis system.
» Length from x=a to x=b is L = c[sinh(b/c)-sinh(a/c)]
» The horizontal component of the chain's tension is cW at all points, where W is the chain's weight per unit length. The actual tension therefore only requires the additional knowledge of the chain's slope at any point.

Your further complication, that part of the chain might be lying along the sea bed, will force you to iterate to determine the length so lying. Obviously this lying length can not be negative. Slightly less obviously, and probably not relevant to any reasonably proportioned problem, neither can it be greater than your dimension H. (This degeneracy will arise only if L>=H+V.)

Place your origin at the anchor point. The chain now runs from (0,0) to (H,V), probably via a straight horizontal line to (B,0) then a catenary. B is the length of chain lying on the sea bed. There will be gradient continuity at (B,0), ie the catenary will be horizontal at this point.

One possible solution approach proceeds as follows.

Assume a value for B.

Determine equation for the catenary portion. This will have the form
(y-y0) = c * cosh((x-x0)/c)
with three unknowns (x0, y0 and c). Luckily(?) we have three constraints, so we can solve for the equation. The constraints are:
» Catenary passes through (B,0).
» Catenary passes through (H,V).
» Length of catenary between these two points is L-B.
Not an easy set of simultaneous equations, but far from impossible.

Once the equation has been established, determine its slope at (B,0). If the slope is zero, you have your answer. The chain tension you want follows from the third cable property I gave above: convert it to kips (ugh) by multiplying by a constant of your choice.

If the slope at (B,0) is negative, you need to increase your estimate of B and repeat. If it is positive, you need to decrease and repeat. My initial instinct is to use a bisection search method, because the B versus starting-slope relationship will be monotonic. But it's your call.

Good luck with this interesting little problem.

 

[GrenMeera]

Thank you both for your responses.

SlideRuleEra - That was a fairly detailed read and had some algorithms that looked promising.

The equations that seemed to match are found on page 43, and the equation specifically that looked useful was marked as (17). I could find the x value when y reaches the sea-bed by using y = h. I still do not have the tension value however, so I hit a dead end.

Denial - I apologize but I was completely unable to follow your equations. I was unsure what many of your variables represented. Unfortunately, I am a programmer first and foremost, and have some difficulty recognizing what may be a simple and well known engineering term or variable. I was particularly unable to relate what your 'c' variable represented.

 

[Denial]

The technical name of the "chain equation" is actually a "catenary". The symbol "c" is an arbitrary constant that governs the amount of curvature (or tension, or sag) in the chain. For any hanging chain you need to calculate c, just like you would have to calculate a and b if you were trying to fit a straight line (y = a*x+b) between two known points. Note that c has the units of length.

There's a heap of stuff on the catenary equation out there in cyberland. See, for example,

http://en.wikipedia.org/wiki/Catenary

in the starting few paragraphs of its section headed "mathematical description".

The formula I give above for the length is for the length along the cable.

Once you have some understanding of "c", you should be able to follow the rest of my post. If not, speak to someone with a modicum of tertiary-level mathematics. This is more a problem in mathematics than it is a problem in engineering.

 

[Qshake]

You should also check out Cable Structures by Max Irvine. It has a mooring example in the book. It's also relatively cheap since it's in paperback.

Regards,
Qshake

Eng-Tips Forums:Real Solutions for Real Problems Really Quick.

 

[GrenMeera]

Qshake - I appreciate the response. I will look into that suggestion.

Denial - Ah, I actually have read that page trying to find equations before, and it just now hit me that you quoted the Certesian equation ver batem. I suppose I didn't recognize 'c' simply because I had seen it written as 'a'. Doesn't even make a difference really what it's called, just one of those things that slipped past me.

Unfortunately, I don't believe I can accurately assume a value for 'B' in your example. However, another possibility has fallen into my lap.

Using the formulas I found in SlideRuleEra's post in the "USS Wire Rope Engineering Hand Book" pdf, on page 43, equation (24):

L = (s^2 + h^2)^(1/2) + (w^2 * s^3 * cos^(alpha)) / (24 * t^2)

L : Length
s : Horizontal distance between anchored points
h : Vertical distance between anchored points
w : Weight per foot (need to verify how this equates)
alpha : Angle between horizontal line and anchor vector
t : Horizontal tension component

I was able to find the horizontal cable component using this data. I am still unsure if I was using (w) correctly, as I tried it as weight per foot of cable, but it may need to be:

w = w' * sec(alpha)

where w' is the weight per foot of cable. I could certainly use some feedback on that.

From this, I can estimate the proper horizontal tension that would be applied across a hanging system. Ignoring friction (not always the best idea, but we're assuming the system came to rest without many external forces), the horizontal tension component should be equivalent whether or not a section of cable comes to rest on the sea-bed or not.

Please let me know if this is incorrect as well. But, to continue:

T = t + w * h

The final tension is the horizontal component plus the weight of the chain times the distance to the sea-bed. With that, I should be able to find the Vertical tension as follows:

V = (T^2 - t^2)^(1/2)

This should actually give me all the data I need! Does anybody see problems with this resultant?

 

[Denial]

Some comments on your proposed approach using Wire Rope Engineering Handbook (WREH).

(1) The "equation 24" in WREH looks reasonable for a reasonably taut chain. (I have not checked it, but would have no reason to doubt it.) But see my comment 7 below.

(2) Your transcription of eqn 24 has left off the "cubed" that should apply to the cosine function.

(3) Your definition of alpha is a bit ambiguous if the lower part of the anchor chain is draped along the sea bed (which I thought was the main reason for your post in the first place).

(4) A few pages earlier in WREH, "w" is defined as the weight of the chain per unit HORIZONTAL length. And I can personally attest that this is the value you should be using. (For chains that are not adequately taut this value will not be constant but will vary with position. This is part, but only part, of the reasons I make comment 7 below.)

(5) I am not sure where you are going with your last two equations (T=… and V=…). I am now at home, and do not have access to my references. However WREH gives you all you need to calculate the tension anywhere along the chain once you have calculated "t" from eqn 24.
TrueTension = HorizontalTension * sec(InstantaneousSlope)

(6) As I outlined in my first post, if you have any possibility of the lower part of the chain lying on the sea bed, you will have to adopt some sort of iterative scheme to determine how much.

(7) WREH's equation 24 and its other related equations are based on the parabolic approximation to the catenary curve. I do not know how accurately you wish to solve this problem of yours, but I have serious doubts about the wisdom of using the parabolic approximation if there is any chance of chain lying along the sea bed. I suggest you do some hand calculations for the typical geometries and loadings you anticipate, for both the parabolic and the catenary equation forms, to see what sort of errors you could get if you use the parabolic approximation.

However I note you say you are a programmer. Therefore I assume you would write yourself a program rather than do hand calculations. I certainly would, probably using Excel. (No arguments from other Eng-Tippers about whether a spreadsheet counts as a "program", please.) Once you've done this to compare the accuracies, you've already cracked the catenary approach, so use it anyway.

And as a programmer, can you be sure of the "typical geometries and loadings" that will be thrown at your finished program? If you must use the parabolic approximation in a program whose use you will have no control over, build in some checks against "out of approved envelope" data being fed to the program. Otherwise give the approximation the flick.

Sorry to finish on a negative note, but I know what I would do were I in your shoes.

 

[GrenMeera]

Denial - I truly appreciate your attention to this matter.

I will respond numerically.

(2) Yes, I did forget to write cubed here, it was a typo and a good catch.

(3) The WREH defines alpha a bit more detailed than I, however mathematically, it should not matter. I never need the deflection of any point after it has hit the sea-bed. I know the deflection... it will be the sea depth. So I can trust alpha for the shape of the curve as it approaches the sea-bed fairly accurately. All it represents is the distances along the tangential line between anchors, which is static.

(4) I did have doubts about 'w'. Would the above equation converting weight along chain to horizontal component, w * sec(alpha), be more appropriate?

(5) I scrapped the equations due to chain lying on the sea-bed after I found the horizontal component. This is because, despite not laying taught and free, if friction is small enough, the horizontal components should equate to about the same. It's the final tension and the vertical tension that would be off due to the sea-bed. Please correct me if I am way off about this, but my accuracy doesn't need to be THAT precise, just precise enough to apply forces to move the craft, or balance the craft accordingly (multiple mooring lines), and show in a digital display that somebody can eyeball and feel comfortable with. I can't estimate data fully, due to sea-depth being variable, the shape of the catenary could be drastically different.

The two equations I used instead are found in a power point that was E-mailed to me:

http://www.google.com/url?sa=t&source=web&cd=1&ved=0CBIQFjAA&url=http://

http://www.pe.tamu.edu/schubert/pub...N2fIscZxM61NOEpaA&sig2=NBnw3G33kF_dmJkVKDKKig

(7) There is a finite set of data to expect, yes. I have already limited the data within this set, but thank you for the suggestions.

No problem on the "negativity". I asked on a forum so that I can get feedback. I appreciate when professionals have criticisms much more so than when they full out agree. You can get much better feedback that way.

 

[Denial]

On (4), yes, use w = w' * sec(alpha) to get the weight per unit of horizontal distance.

On (5), I am not in a position to comment on what is "adequate".

Happy programming.

 

[chicopee]

I read your question sometime ago but I just got around to develop a possible solution. See my attachment for the solution. It is an example taken from one of my old textbooks and I expanded the problem to take into consideration a receeding tide. Several years ago I worked on a catenary problem and I developed a converging solution for the value of "C" so if interested I'll post the decision diagram for your program.