Well, let's begin with some considerations:
1. the exact solution to the problem is hard to reach without using a computer, because, as Ben said, it's a catenary: an exponential function !!
2. the HYDROSTATIC force of the water has to be taken into account:
let's assume that we use a steel chain (specific weight of about 7 kg
per liter) 40 meters long, then, in the water the chain becomes
lighter due to the hydrostatic Archimedean force of the water;
so doing, in the water the chain has a specific weight of about 6 kg per
liter.
The weight difference is about 14%, that, i think, will produce
not neglectable effects on a chain 40 m long. (14% of 40 m = 6 m !!!)
3. so, why not use a little computer program that solves exactly the equatio for us ??? It will be fast and precise as we want.
A heavy cable (or a wire, or a chain), hung between two fixed points, dispose itself on a curve, called "catenary" of equation y=f(x) given in figure:
where:
- S = horizontal component of the force acting on the fixed points
- q = weight of the cable per meter
- HypCos = Hyperbolic Cosine
Some properties of the catenary:
- HypCos(0) = 1
- y(0) = S/q
- symmetry: HypCos(-x) = HypCos(x)
The length of the cable, measured between two different abscissas x1 and x2 is the following:
where:
- HypSin = Hyperbolic Sine
- ...and HypSin(0) = 0
- symmetry: HypSin(-x) = - HypSin(x)
- Pitagora's: 
As first case, we solve the catenary equation to find the unknown "y"
without taking care of the water: the catenary is an homogeneous one.
The system of equation that we've to solve is the following:
where "c" is the total length of the cable (it's a datum).
The system seems hard to be solved, but doing the position:
we obtain a simple formula to be solved with respect to the variable "Z":
The method used to find the right "Z" is the following:
- step 1
initialize x1(1) = 0 , ...so Z(1) = S/q
then we calculate the left side of the expression
usually, it is smaller than "c", so we continue...
- step 2
set x1(2) = x1(1) + deltax , ...then we calculate Z(2) from the "position"
made before
then we calculate the left side of the expression
IF it is smaller the "c"
THEN we continue again as in step 2
ELSE we jump outside the loop cause we've found the approximate solution
"deltax" is a parameter affecting the precision of the algorithm: if you use deltax = 0.1 meters , then you can obtain a precision of about 0.1 - 0.2 meters !!!
When "Z" is found, we can be easily obtain "y" from the folowing:
Now the situation is more complex: we need to consider the specific weight
of the cable, and, doing so, we have two different catenaries joined
together:
the lower part, plugged in the water, and
the higher part, ...plugged in the air
where:
- "h'" : height of the towing boat above the sea.
- "sw" : specfic weight of the material which the cable is made of.
The two different catenaries have different equations and different
"x-y reference axes"; so, in order to built-up a single system of equation,
we have to refer both catenaries to the same "x-y axes".
We have to join them in a point where the ordinate and the derivatives of the two eqs.
are the same !!!
It can be made by the "x-y" translation sketched below:
By applying the "x-y" translation above, we obtain the folowing system of equation:
It can be solved by making the "position":
and applying the previously described method on the
simpler formula:
Note that, this time, we iterate on "x0" !!!
The result "y" can be obtained by the formula:
Et voila'
Stehle Ferdinando