An oilrig is floating on the surface. One of the mooring lines keeping the rig in place has an attached buoy at some point, in order to move the touchdown point (where the line meets the seabed) further away from the rig.

The curve will look something like this: http://www.globalmaritime.com/navalarch/mooring1.png where the discontinouation is the point where the upwards buoyancy force from the buoy is acting.

M in the equation is the weight of the line in N/m, and D and R is the horizontal and vertical distances from the rig (fairlead) to the buoy. However, I belive there must be a simpler approach than what I have attempted. (e.g. by using line distance S, instead of D and R)

Basically, I need to find an expression for the slope of the line, given that i know the line tension at the rig (fairlead), the buoyancy force of the buoy, the line weight M, the point on the line where the buoy is attached, and the total length of the line.

Think that's it

Dan

]]>2A = (1+e^B)e^C + (1+e^(-B))e^(-C)

is similar to drift-diffusion in a transistor .

We couldn't get it into closed form unless we made some assumptions.

is there anything special about

C1, MD/H and MR/H

like MR/H+C1 ≈ C1

this would be nice

]]>Dan

]]>Well, let's call MD/H "A", MR/H "B" and C1 "C"

A=cosh(C)+cosh(B+C)

A = ½(e^C + e^(-C)) + ½(e^(B+C) + e^(-B-C))

2A = e^C + e^(-C) + e^(B+C) + e^(-B-C)

2A = e^C + e^(-C) + e^B × e^C + e^(-B) × e^(-C)

2A = (1+e^B)e^C + (1+e^(-B))e^(-C)

... that is as far as I have got, and I have to go ... I just feel that with a bit of manipulation, and some hyperbolic identitities, that we may arrive at a solution.

]]>Also, if anyone has derived an expression for a catenary with an applied force (like e.g. a mooring line with an attached buoy) this would also be helpful

Dan

]]>its the same as the expansion for cosine

1 + x²/2 + x^4/ 24+ ....

]]>This could take some figuring!

First of all, you have MD/H and MR/H - are they the same?

This might help too ...

And the inverse:

If I had a bit more time I would try brute algebra on it, to see where we get to.

]]>I'm stuck with this equation at work, which im trying to solve in order to do some mooring calculations. I have a mooring line, with an attached buoy. When deriving one of the catenary equations, I end up with this:

MD/H=cosh(C1)+cosh(MR/H+C1)

Basically M, D, H and C1 are all constants and I need to find a symbolic expression for C1, i.e. C1=????

Any bright minds out there who can help me??

Dan

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