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#1 2005-09-22 18:31:20

Danster
Member
Registered: 2005-09-22
Posts: 9

Hyperbolic equation...

Hey,

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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#2 2005-09-22 20:26:55

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,551

Re: Hyperbolic equation...

I will trust you on the formula, seems reasonable anyway as a simple catenary has the formula  y = a cosh(x/a)

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.


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#3 2005-09-22 20:34:47

Atled
Member
Registered: 2005-08-22
Posts: 9

Re: Hyperbolic equation...

cosh(x) can be approximated with a taylor expansion.

its the same as the expansion for cosine

1  + x²/2 + x^4/ 24+ ....

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#4 2005-09-22 21:06:33

Danster
Member
Registered: 2005-09-22
Posts: 9

Re: Hyperbolic equation...

Cheers for ur time guys. MD/H and MR/H are not the same, and I guess Im still stuck trying to solve for C1. It has been a while since I used the taylor expression, so I can't see how it would simplify in providing an expression for C1...

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 smile

Dan

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#5 2005-09-22 22:33:09

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,551

Re: Hyperbolic equation...

I have a few minutes ...

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.


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#6 2005-09-22 23:03:19

Danster
Member
Registered: 2005-09-22
Posts: 9

Re: Hyperbolic equation...

Thanks, I'm looking forward to the continuation. I tried to do some moves but it got messy!

Dan

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#7 2005-09-23 08:57:10

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,551

Re: Hyperbolic equation...

Anyone want to take this further?


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#8 2005-09-23 09:24:01

kylekatarn
Member
Registered: 2005-07-24
Posts: 445

Re: Hyperbolic equation...

this is not as easy as it seems.

sol11dw.th.jpg

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#9 2005-09-25 18:57:24

Danster
Member
Registered: 2005-09-22
Posts: 9

Re: Hyperbolic equation...

Wow, that is alot more complex than I thought. Tanks!

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#10 2005-09-25 22:18:12

Atled
Member
Registered: 2005-08-22
Posts: 9

Re: Hyperbolic equation...

This setup

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

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#11 2005-09-27 19:44:51

Danster
Member
Registered: 2005-09-22
Posts: 9

Re: Hyperbolic equation...

Well, here is the complete, static, problem:

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 smile

Dan

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