2 sticks and a rope

sedgwick - How long is the rope?

Ok I have got it, the rope is Y metres long. There is no analytical solution to this problem, it is well known.

sedgwick, I retract the previous stement there might be an analytical solution.

The ratio is Y/f(x)
where f(x) is obtained from

X+A = (A/2)(e^(x/A) + e^(-x/A)) and A is an arbitrary number >0.

sedgwick, I am assuming that the rope behaves as a caternery curve. These are miserable problems and more often then not require a numerical soultion. I am not in the mood for hyperbolic functions today, however you may find this helpful
http://www.math.niu.edu/~rusin/known-math/95/cables

there are other sites on the web as well.

aburr: can u simplify your solution please ?

y : y - (2x)
???

Presumably the starting condition is that the rope forms a horizontal straight line. This then is the problem since it is impossible both mathematically and in the real world.

The equation for a catenary is y = a.cosh(x/a) and there is no poss1ble value of "a" which gives "y" a constant value for all values (or even a range of values) of "x".

Since the starting condition cannot be met it is therefore impossible to solve the problem.

lengore,
we are not actually interested in start position.
The position when rope touches the ground is catenary.
I used catenary more general catenary equation
y=a*cosh(x/a) + C (const a depends on the unit length weight of the rope, C depends on the off the ground heigth), and integral formula for calculating length of fnction graph:
l = integral( a to b from sqtr( 1 + (f')^2));
which gives me
Y = exp(X/2) - exp(-X/2); as you can see, const a goes away after taking derivative from a*cosh(x/a) + C.
So,
Y/X = exp(X/2)/X - exp(-X/2)/X ~ exp(X/2)/X;
since exp(-X/2)/X is small enough.

lengore,
>The equation for a catenary is y = a.cosh(x/a) and there is no poss1ble value of "a" which gives "y" a constant value for all values (or even a range of values) of "x".

how about a=0 ?
that gives y = C;
a=0 could mean there is no weight on rope and thus it stays straight.

baqi,

Please re-read the original question and you will see that the start position is necessary.  He asks for the ratio between the distance the sticks were apart at the START to the distance between the sticks when the rope just touches the ground.

I accept your more generalised equation for a catenary but you still have not solved the problem as stated.

lengore,
The ratio at start position can be anything as given by user, and don't to be hardly worked out. It is simply Y/X.
But, when you push them toward each other, X varies, while Y stayes constant. And I think, sedgwick is interested in value of X when the rope touches the ground.

sedgwick, please comment.

My reply referred to your comment before last.

Re your last comment you suggest:
        y = 0.cosh(x/0) + c  gives y = c
Not too sure about zero multiplied by infinity!

The ratio for which your equation applies is the height of the sticks to the distance between them when the rope just touches the ground.  Fine, but that was not the question.

The fact that the staring position is not representatable by the caternary is irrelevant. The question can be stated as given a rope of lenght Y suspended on sticks of height X, how far apart must the sticks be for the rope to just touch the ground. I have have not checked but there is a good chance there is not an analytical solution.

Perhaps Sedgwick could clarify the question.

Sory, in my previous calc I made an error. (wich is only a=1 case)

We have no analytical solution if we can't integrate length fully. But luckily,
f=a*cosh(x/a) + C => f'= (exp(x/a) - exp(x/a))/2
=> 1 + f'^2 = ((exp(x/a) + exp(x/a))/2 )^2
so it goes out of sqrt.
After that each of exp functions integrate easily.
Finally, Y = a(exp(X/(2a)) - exp(-X/(2a))).

Any answer must include extreme cases.
So that if Height of pole is 1 and length of rope is 2 the distance between poles at which the rope touches ground is zero giving a ratio of 2/0 which is insoluble. Since the extreme case is insoluble then the general case is probably insoluble as well!

Heh this was the Car Talk Puzzler this week

hmmm

assume this: Y > X1+X2
Y = rope length
X1, X2 = sticks hight

sorry for the misunderstanding

btw, the rope cannot be streched

Hi baqi,

Sorry, I had my blindfold on yesterday!

The start distance between the sticks is, of course, the length of the catenary so the problem can be re-expressed as the ratio of the length of the curve to the distance between the sticks when the rope just touches the ground.

Last comment from Sedgwick has introduced the variation of the two sticks being of different lengths.

I will ponder further on the problem.

I have done a little research and now believe that the length of a catenary from X1 to X2 can only be found by numerical methods.

My little book says that for a very shallow curve a catenary can be approximated to a parabola and, using this approximation, the length of the curve can be derived analytically.

Assuming the rope is long enough and it touches the ground the ratios 1:1 as the triangle created by the rope would be isosceles       |\  /|
                   | \/ |
                   <z><z>

sedgwick:
Your first question states they are X meters each, but then you have X1 and X2 for the lengths later.

Is it the intent that each stick is a DIFFERENT length, or intended the same length?

Assuming the rope is long enough and it touches the ground the ratios 1:1 as the triangle created by the rope would be isosceles

       |\    /|
       | \  / |
       |  \/  |
       <z-><-z>

Drawing got messed up first time heres 2nd attempt.

The two z dimensions should be in the middle of the M shaped diagram

the ratio is Y:wouldsomeonedothis

Refer to http://www.maa.org/pubs/Calc_articles/ma028.pdf
We want to solve the following equation for d (l=Y,g=X) :
Y=c sinh((d/2c) + tanh-1 (v/Y))+ c sinh((d/2c) + tanh-1 (v/Y))

Since tanh-1 (0/Y) = 0 (v=0 because poles same height and assume same level)
(Also, tanh-1 = (log  (1+x) - log (1-x))/2)

So we have Y= 2c sinh(d/2c)
But c is unknown, so we need to solve for c:

Now, from the picture on the article above, it seems like c+g=X here, which would make c=0 since g=X, but it is hard to tell because in their example the poles are different heights.
But the equation for c/g/x (x=Y/2 -- x<>X here) is
c+g=c cosh(Y/2c)

and cosh(Y/2c)=e^(Y/2c)+e^(-Y/2c).

But it has been too long since I've worked with logs and ln and e, so I can't simply the equation.
So solve for c in c+g=c cosh(Y/2c)
The substitue for c in
Y= 2c sinh(d/2c)
and solve for d.
Anyway, the article above might help somebody get the answer if I've totally screwed up the math above.

I have given the solution but not in the form f(x) because I have not the time now to invert the relation I gave. The solution is based on a catenary. The A had to be added because if the catenary touches the ground you get into trouble with divide by zero.

The sticks at either end were originally S apart where S = length of the rope.  Assume each stick is of height H.

Assume the sticks are now 2L apart.  The equation for the curve into which the rope has drooped is a catenary for which the general formula is:

               y = b.cosh(x/a) + c

Now x ranges from ¨CL to L and at the point where the rope just touches the ground x is zero.  Cosh(0) is 1 so c must equal ¨Cb for the rope to just touch the ground.

The problem is to find the ratio of 2L to S.

The length of a curve which is described by y = f(x) is given by the expression:

          S = ¡Òsqrt (1 + f¡ä(x)^2).dx

In our case the range is x = -L to x = L

We have f(x) = b.cosh(x/a) ¨C b so f¡ä(x) = b.sinh(x/a) therefore we need to solve:

          S = ¡Òsqrt (1 + (b.sinh(x/a))^2).dx
                from x = -L to x = L

I do not believe this integral can be solved analytically.

lengore,

look at my comments.

Catenary

L : sqrt(cosh(L)^2) * tanh(L)

sedgwick

  Consider 2 poles of height X = cosh(A) - 1  a distance 2A apart with rope touching the ground at (0,0). Assuming a caternary then the ropes curve  y(x) is given by

      y(x)=cosh(x)-1


the length of the rope Y is given by

   integral(-A to A ) sqrt(1+(y'(x))^2) dx = 2sinh(A) = Y

so giving A=arcsinh(Y/2)

the ratio of distance apart to the height is given by 2A/X

         Ratio=2{arcsinh(Y/2)}/X

GwynforWeb :-)

sedgwick
  I can not do the above ( I am working too late in the morning here). I can do this though

 Consider 2 poles of height cosh(A) - 1  a distance 2A apart with rope touching the ground at (0,0). Assuming a caternary then the ropes curve  y(x) is given by

     y(x)=cosh(x)-1


the length of the rope Y is given by

  integral(-A to A ) sqrt(1+(y'(x))^2) dx = 2sinh(A)

now choose A st the ratio of the pole height to rope length is X/Y ie choose A s.t

  (cosh(A) - 1)/(2sinh(A)) = X/Y     (1)

the ratio of pole height to distance apart is

        Ratio= (cosh(A) - 1)/(2A)  


where A saisfies equation (1), which will require a numerical solution for A.              

Actually I think (cosh(A) - 1)/2sinh(A) = X/Y  can be solved by substituting 2sinh(A)=sqrt(cosh((A))^2 - 1), squaring both sides obtaining a quadratic in cosh(A). Uggh! I am going to bed.  

This is simpler than I thought

(cosh(A) - 1)/2sinh(A) = X/Y  

can be solved by substituting 2sinh(A)=sqrt(cosh((A))^2 - 1)  and collecting/cancelling terms to give

  cosh(A)  =(1+k*k)/(1-k*k)

where k= 2X/Y

so the solution is

   
         Ratio= (cosh(A) - 1)/(2A)  

      where

         cosh(A)  =(1+k*k)/(1-k*k)  and  k= 2X/Y

I tested it numerically on a few curves and it is right.

  The trick was not to keep the height of the poles constant and vary the distance apart so that the caternary equation changes because the integal for the length is nasty. But to keep the curve constant and vary the distance apart and the pole height (poles always touching the curve) until the ratio of distance apart to length is X/Y and then read off the pole height.

I should point out that I am assuming a "rope" which means it give little resistance to be being bent so that y'' = y. If you do not you have dimensional problems

There is a major difficulty here. The rope does take the shape of a catenary BUT the exact shape of the catenary depends on the mass of the rope. A heavy rope will fall closer to the poles than a lighter one, hence the poles will have to be closer together at the end of the problem.
This fact also means that
y(x)=cosh(x)-1
is not the general formula for the catenary which all ropes will take.
If one lets w be the weight per unit length and T be the tension at the lowest point the answer will be
-
Ratio = (2T/wY)arctanh (2X/Y)

-
One cannot obtain the weight per unit length or any other properties of the rope from the initial state of the problem because, as has been pointed out, that state is unphysical (Tension infinite)
-
If the rope is VERY heavy,
Ratio = 1-(2X/Y)
-
If rope is VERY light (and a fly is sitting on the center)
Ratio = sqr(1-(2X/Y)^2)
-
(The above from geometry, If rope is heavy it stays as close to the poles as possible, if light it stays as far from the poles as possible.)

In physics you are suppose to assume the weight of the rope is negligible.  It makes things easier. ;-)

baqi,

I have re-read your notes but surely

   for y = a.cosh(x/a) we get dy/dx = a.sinh(x/a)

we then get f' = sqrt(1 + a^2.(exp(x/a) - exp(x/a))^2)

hence the problem of integration.

baqi,

Ignore my last comment, blindfold on again

aburr,
 
 You are right but my analysis does in fact genralise to a.cosh(x/a)

Lengore
  y = a.cosh(x/a) we get dy/dx = sinh(x/a)


A general solution?

Consider 2 poles of height a.cosh(A/a) - a  a distance 2A apart with rope touching the ground at (0,0). Assuming a caternary then the ropes curve  y(x) is given by ( a being some constant)

    y(x)=a.cosh(x)-a


the length of the rope Y is given by

 integral(-A to A ) sqrt(1+(y'(x))^2) dx = 2sinh(A/a)

now choose A st the ratio of the pole height to rope length is X/Y ie choose A s.t

    (a.cosh(A/a) - a)/(2sinh(A/a)) = X/Y

ie  (cosh(A/a) - 1)/(2sinh(A/a)) = aX/Y     (1)

the ratio of pole height to distance apart is

       Ratio= (a.cosh(A/a) - a)/(2A)  


where A saisfies equation (1),

Equation (1) can be solved by substituting 2sinh(A)=sqrt(cosh((A))^2 - 1), squaring both sides and collecting/cancelling terms to give (taking the non-degenerate root)

 cosh(A/a)  =(1+k*k)/(1-k*k)

where k= 2aX/Y

so the solution is

 
        Ratio= a(cosh(A/a) - 1)/(2A)  

     where

        cosh(A/a)  =(1+k*k)/(1-k*k)  and  k= 2aX/Y


which when terms are collected gives

     Ratio = 1/{(1-k*k)*arccosh((1+k*k)/(1-k*k))}

     where k= 2aX/Y


(I am prone to mistakes in my algebra but I think this is correct.)

Damn! there is small mistake

    "  ie  (cosh(A/a) - 1)/(2sinh(A/a)) = aX/Y     (1)  "

should read

    "  ie  (cosh(A/a) - 1)/(2sinh(A/a)) = X/(aY)    (1) "


so that through out the analysis k=2X/(aY)

and clearly y(x)=a.cosh(x)-a  should read y(x)=a.cosh(x/a)-a

There is something that does not seem right about the final solution but I can not put my finger on it.  

Yes I was right small errors, I think this is correct and it makes sense

Consider 2 poles of height a.cosh(A/a) - a  a distance 2A apart with rope touching the ground at (0,0). Assuming a caternary then the ropes curve  y(x) is given by ( a being some constant)

    y(x)=a.cosh(x/a)-a


the length of the rope Y is given by

 integral(-A to A ) sqrt(1+(y'(x))^2) dx = 2a.sinh(A/a)

now choose A st the ratio of the pole height to rope length is X/Y ie choose A s.t

    (a.cosh(A/a) - a)/(2a.sinh(A/a)) = X/Y

ie  (cosh(A/a) - 1)/(sinh(A/a)) = 2X/Y     (1)

the ratio of pole height to distance apart is

       Ratio= (a.cosh(A/a) - a)/(2A)  


where A saisfies equation (1),

Equation (1) can be solved by substituting sinh(A/a)=sqrt(cosh((A/a))^2 - 1), squaring both sides and collecting/cancelling terms to give (taking  the non-degenerate root)

 cosh(A/a)  =(1+k*k)/(1-k*k)

where k= 2X/Y

so the solution is

 
        Ratio= (a.cosh(A/a) - a)/(2A)  

     where

        cosh(A/a)  =(1+k*k)/(1-k*k)  and  k= 2X/Y


which when terms are collected gives

     Ratio = a/{(1-k*k)*arccosh((1+k*k)/(1-k*k))}

     where k= 2X/Y

Typo, which when terms are collected gives

    Ratio = 1/{(1-k*k)*arccosh((1+k*k)/(1-k*k))}

 which is the same as the result I had in the first place assuming cosh(x)-1. ( and makes sense)

  From my memory of the derivation of a caternery the weight per unit length term cancels out. ie a heavy chain will give the same curve as a light one of the same length. I have never seen a heavy chain (eg a suspension bridge) bend like |_| or a light chain of the same length(eg a necklace bend) like \/. You are going to get variations from this when there is a resistive force to bending that is large relative to the weight per unit length, which can occur in some ropes. But then the curve is not a caternery  and depends on the phyical properties of the rope.
    Bending is a combination of shear and tensile strains. Relatively high resistance to tensile stress => \/ and to shear => |_|, I think). The problem is then an order of magnitude more difficult.

   I think this problem is about caterneries.

See derivation  of a catenary curve which assumes no bending force at

http://pascal7.freeyellow.com/Catenary/Hello%20everyone.htm

it is of the form    a.cosh(x/a) + b. There is also a picture of a really heavy arch.

X1,X2 reffered to the poles, however, they both have the same hight...

sould read
   Y=sinh(2)=3.626860      (rope length)

A complete solution

GwynforWeb –
“I think this problem is about catenaries." It is.
 I had difficulty following your numerical answer. You state that the poles started 2 units apart yet the rope is 3.6 units long. (Not the starting rope length which is 2 units.) If the 2 units is the final pole distance you do not have enough information to get the equation of the catenary (how did you determine that a = ½?) Your final equation did not have an a in it. Where did it go? It must be there. Different values of a give different catenaries, even for the same value of x, even if the coordinate system is referenced to the bottom of the catenary.
-
Consider a general catenary
y(x) = a cosh(x/a). here a is the distance from y = 0 to the bottom of the catenary. Shift the coordinate system so that the bottom of the rope is at 0,0, then
h(x) = a cosh(x/a) – a         ( a relationship to which you pointed me.)
Now let dh(x)= [a cosh(x/a) – a] – { a1 cosh(x/a1) – a1}
Set any two values you wish for a and a1. Plot.
The result is NOT a straight line (except in the case where a = a1). The curves, while both catenaries, are not of the same shape.

SOLUTION
Start the poles (height X) a distance Y apart. Bring them together until the rope touches the ground. The poles are now a distance 2s apart.
The equation of the catenary is
y(x)= a cosh(x/a)  Shift the coord system so that the lowest part of the rope is at 0,0. Then
h(x)=a cosh(x/a) – a. We want the value of x that gives X. Thus
X = a cosh(s/a) – a
X+a  = a cosh(s/a)
(X+a)/a =  cosh(s/a)
arccosh ((X+a)/a) = s/a
s = a * arccosh ((X+a)/a)
Now our desired ratio is
R = 2*s/Y
Thus
R =  (a*2/Y) * arccosh ((X+a)/a)
Or if you prefer
R = (2*a/Y)*log [(X+a)/a (+ or –) sqr{{(X+a)/a}^2  - 1}]



Note: The mass of the rope must have an effect on the problem. (It is usually hidden in a in the form of the tension and weight per unit length.) If the mass had no effect, then the rope would not be attracted to the earth and the shape of the rope after the start of this problem would be completely indeterminate. The value of g does not appear because (as long as it is not 0) it affects the weight and tension equally.
Some of my previous comments are at best obscure.

aburr and Gwynforweb:

the only case that the curve shape would change as a result of the wieght of the rope is if the rope would stretch. And as sedgwick has stated the rope does not stretch.

It makes sense.

As long as the rope has SOME mass and the poles can support the full weight of the rope without bending inward etc.
Then: in the case of a thread, as long as the thread was comepletely flexible and there was no wind, etc. It would take the exact same shape as an anchor chain of the same length, poles, distance.

As far as your solutions go:

It seems that you both are very close.

sedgwick?

Aburr,
   Provided the rope has finite mass then the shape is independent of the actual mass.  A general catenery is a.cosh(x/a)+ b, which if it touches at (0,0) is

    y(x)=a.cosh(x/a)-a    (1)

I have a rope of length 3.626860  between  2 poles of height 1.381097 touching at (0,0) then it satisfies the above for some a. I say that a=0.5 and that the poles are 2 part ie

   ie  y(x)=0.5cosh(2)

which if the length is it is calculated it is 3.626860 and at x=1 y(x)=1.381097 ie the height of the poles . This is the equation of the rope between the poles and the poles are 2 units part. I now have the equation of an actual curve between 2 poles touching at (0,0) so I can use it to test my result. The ratios corresspond

  The result for the ratio is independent a.  if we scale the x and y axis both by 1/a ie u=x/a and v=y/a then equation (1) becomes

     v=cosh(u)-1

the ratios are unchanged for the scaled curve and it's  equation is independent of a. Hence the ratio is independent of a. Bascically the a 's cancel out in my analysis since we are taking ratio's. a is a coordinate scaling factor in the catenery equation

  This is a cool problem it is some time since I have played with catenaries. Seeing what elasticity does could be interesting.


   

ie  y(x)=0.5cosh(2) should read y(x)=0.5cosh(2x) - 0.5

GwynforWeb says

Aburr,
Provided the rope has finite mass then the shape is independent of the actual mass. A general catenery is a.cosh(x/a)+ b, which if it touches at (0,0) is
y(x)=a.cosh(x/a)-a      (1)
I have a rope of length 3.626860 between 2 poles of height 1.381097 touching at (0,0) then it satisfies the above for some a. I say that a=0.5 and that the poles are 2 part ie
ie y(x)=0.5cosh(2)
which if the length is it is calculated it is 3.626860 and at x=1 y(x)=1.381097 ie the height of the poles . This is the equation of the rope between the poles and the poles are 2 units part. I now have the equation of an actual curve between 2 poles touching at (0,0) so I can use it to test my result. End quote
-
I would like to think that the shape of the catenary is independent of mass, but I think that the mass is included in a. However that may be, it is clear that the shape of the curve is dependant on a. Just plot ( your equation 1) h(x) = a cosh(x/a) – a for various values of a.
In your example you chose a and said that you had found all curves of the required conditions (0,0 and x,X).
But suppose that you had chosen a = 1 instead of 0.5. Then your answers would be different. (y(1) = 0.543) Your rope length of 3.6 will select out a particular a (in this case 0.5) but you do not, a priori, know the rope length. (It is Y.) Basically IF the mass is such that a = O.5 you have a nice solution. BUT if the mass is such that a does not equal 0.5 you are in trouble.

-
Scaling
Quote
The result for the ratio is independent a. if we scale the x and y axis both by 1/a ie u=x/a and v=y/a then equation (1) becomes
v=cosh(u)-1
the ratios are unchanged for the scaled curve and it's equation is independent of a. Hence the ratio is independent of a.
End quote
-
I am not as familiar with scaling as I should be, but it seems to me that, if you scale both axes by the same factor, the resulting equation (1) should be
v(ua) = cosh(u) – 1, so the a remains.

Quote
This is a cool problem it is some time since I have played with catenaries. Seeing what elasticity does could be interesting.
End Quote
I agree but considering elasticity would be too much for me. I am already spending too much time on this problem.

I got tired of waving my hand and saying that a depended on the mass. I lost the derivation that I had so turnes the the URL that had so kindly been provided by GwynforWeb

http://pascal7.freeyellow.com/Catenary/Hello%20everyone.htm

There he uses h for our a. His h depends on H and w (my notation) where H is the tension at the botton of the catenary and w is the weight per unit length for the rope. He explicitly states that H does not depend on the mass of the rope but only on our Y and the position of the end points. That leaves the weight per unit length in the demoninator of a. Thus a is dependant on the weight per unit length of the rope. Therefor the exact shape (while always a catenary) will depend on the composition of the rope. Hence our ratio will also depend on the composition of the rope.

As stated previously, because the initial condition is unphysical, there is no independant way of determining a. Hence the problem as originally proposed is insoulable.
However my solution above gives the answer as a function of a.

   Shape is not dependent on mass. In the derivation  he says H is indepenent of M, M is not the mass of the rope it is a position in his diagram. H is linearly dependent on the mass. ie double ropes weight and H must double. This gives that h is independent of ropes mass. ie a is independent of mass

   Given y(x)=a.cosh(x/a)-a and then making the transform u=x/a and v=y/a you do in fact obtain  v(u)=cosh(u)-1

again looking at the derivation we see that h (our 'a') is the ratio of a force over a force ie is dimensionless. 'a'is dependent only on gravity. This interesting in that ropes (and presumably other things?!) hang diffrently on the moon.

sedgwick,
   I am not arguing this any more, the shape is independent of the weight and hence 'a' is independent of the weight. Similary the Ratio is independent of 'a' and is given by
   
    Ratio = k*k/{(1-k*k)*arccosh((1+k*k)/(1-k*k))}

          where k= 2X/Y

I have given an example illustrating that this formula is correct and can give you a couple more if you wish.

GwynforWeb

Ok shape is independant of weight. Ask a physics PhD.

Starting with standard arc length integral.

By using the Fundamental Theorom of Calculus we
derive the formula for length of curve
b/c this derivation uses limits from 0 to x we must double it so:

Length of curve = 2(a*sinh(x/a)

Distance between = 2*X

So then the ratio is: a*sinh(x/a) / x

note ( I found derivation for length in Calculus book) can make more explicit if neccessary

Yes but now you must calculate 'a' in terms of X and Y which if you do you end up with my formula  

Isn't "a" is a constant that raises the curve up or down on the y axis? If so, the ratio shouldn't change for a given "a" and can be assumed 1. So the ratio is sinh(x):x.

a is a constant in each instance, but it varies from instance to instance and cannot assumed to be 1 for all cases.

In the general form of the catenary it is the distance the global minima of the curve is above the x axis but the shape of the curve will change depending on how luch "sag" there is. IE from straight line to the curve we are trying to describe.

baqi solved this way back and I think Sedgwick should award him the points.  I demurred at first but then saw the error of my ways!

Lengore,
   Given an X and Y explain to me how Bagi's solution gives the ratio?

CalvinDay, You are correct at last some one has seen the point! (But X is the height of the poles not he distance between them so you formula sinh(X):X is not right)

sedgwick
   All hanging ropes are a section of the curve

             y(x)=cosh(x)-1

scaled and translated to fit ie of the form

       Y(X)=a.cosh((X-b)/a)-a-c
 
       (ie Y(X)=A.cosh((X-B)/A) + C)

were a is the scale factor and b is the translation in the x and c in the y, ie the transformation

       X=ax+b

       Y=aY+c

all ratios are clearly unaffected by such a transformation. With

        Ratio = k*k/{(1-k*k)*arccosh((1+k*k)/(1-k*k))}

         where k= 2X/Y

a fact that I have demonstrated with and without an 'a' in the equation

should read
Y(X)=a.cosh((X-b)/a)-a+c   and Y=ay+c

I said I was not going to argue  any more but I feel srongly about this

GwynforWeb,

You ask how baqi solution gives the ratio (I hope baqi approves of the following):

The equation of the catenary which just touches the ground is:

y = a.cosh(x/a) – a

The differential is dy/dx = sinh(x/a)

Length of the catenary S = integral of sqrt(1 + (dy/dx)^2)
S = integral of sqrt(1 + sinh(x/a)^2)
S = integral of cosh(x/a)

Assume range is x = -1 to x = 1 so the final distance apart of the poles is 2.
Then S = sinh(1/a) – sinh(-1/a) = original distance apart

Thus the ratio is (sinh(1/a) – sinh(-1/a)) / 2

The range of –1 to +1 is valid because if, for example, the final distance between the poles was 10 feet we can simply say let x’ = x / 5 and substitute x’ for x in the above equations and still use the range –1 to +1. The length of the chain would then be 5 * S

  I do not see how given say  X/Y= .25 you are obtaining the ratio of the to the distance apart. Demonstrate with adding additional analysis that has not been already  posted how the ratio of height to distance apart is calculated.

  There is also a difference between a relationship and a procedure for calculating a value  
 

Ratio = k*k/{(1-k*k)*arccosh((1+k*k)/(1-k*k))}

        where k= 2X/Y

seems to give it immediately

"ratio of the to the distance apart." should read
"ratio of the height to the distance apart."
 

sedgwick,
   We are all going around in circles and I doubt that we will ever agree,  be merciful on us and close this one off

The original question states "what is the ratio of the two DISTANCES".  I cannot interpret the height of a pole as a distance.

Lengore
    ?
GwynforWeb

E=MC^2 ??? <grins>

 |\ Y  /|
X | \  / | X
  |  \/  |
    ?  ?

  |\
X | \y/2
  |  \
  ----
    ?

X^2 + (Y/2)^2 = ?^2

?= sqrt(X^2 + (Y/2)^2)

???
im no expert but this is what i come up

|\    /|
 | \  / |
 |  \/  |
  ?     ?
X is the height and Y is the lenght of the rope
|\
| \y/2
|  \
----
   ?

X^2 + (Y/2)^2 = ?^2

?= sqrt(X^2 + (Y/2)^2)

???
im no expert but this is what i come up

GwynforWeb: the answer should give the ration between the distances (the initial one and the final one) not the poles height.
since i don't know the answer nor do i know how to decide which one of the answeres is correct, i can't close this thread.
i've decided to increase points and try 2 advice my math lecturer from uni, maybe he'll light my eyes.
i'll make sure to post any updates to this thread and give points to the right answer.

Jerry_Pang: the rope's shape is an arc!!!

guys,

please refer to http://mathforum.org/dr.math/faq/faq.circle.segment.html
advice me reagrding case #3, is is our case?

sedgwick
    I see, the question is a little ambiguous but is now clear. Since pole height to final distance is

height/(final distance)  = k*k/{(1-k*k)*arccosh((1+k*k)/(1-k*k))}

       where k= 2X/Y

were X/Y = height/(initial distance) =k/2

then
(intial distance)/(final distance)

     ={ height/(final distance) } / {height/(initial distance) }

     =(2/k)*{k*k/{(1-k*k)*arccosh((1+k*k)/(1-k*k))}  

     =2k/{(1-k*k)*arccosh((1+k*k)/(1-k*k))}  


GwynforWeb

sedgwick, No case 3 has nothing to do with this problem, ropes do not hang as circles or ellipses, the maths is different here.

Here is a numerical example

If the poles are 2 units apart of height X=1.3810978 with the rope toching at (0,0) then y(x) is given by

     y(x)=0.5cosh(2x)-0.5


then checking this
  if final distance apart=2
  X=0.5cosh(2)-0.5=1.381097   (pole height)

so all is correct and
 
Y= sinh(2)=3.626860  (rope length,ie initial distance apart)
 
X/Y=0.380797

     Ratio=(Rope length)/(final distance apart)

          =1.81343

now using the formula then

  k=2X/Y=0.761594 and

so

  2k/(1-k*k)=3.62686
  arccosh((1+k*k)/(1-k*k))=2
 
  Ratio ={k*k/{(1-k*k)}/{arccosh((1+k*k)/(1-k*k))}

        =1.81343

       

GwynforWeb,

I don't understand why you still refer to the height of the sticks as Sedgwick has clarified his question.  Out of interest I computed some values using:

               y = a.cosh(x/a) - a

Value of a      Catenary Length        Height of Sticks
   0.2              29.681                  14.642
   0.3               8.399                   3.910
   0.5               3.627                   1.381
   1.0               2.350                   0.543

In each case the final distance apart of the sticks is 2.

As you can see, the ratio of catenary length to 2 is quite different from the height/distance ratio and the relationship between the ratios varies with a.

Lengore

  If  you read my post carefully you will see that I end up saying

  (intial distance)/(final distance)
 
  = 2k/{(1-k*k)*arccosh((1+k*k)/(1-k*k))}

and I have given a numerical example.  

  Below I give the solution for the now unambigous question.

Lengore In am not argueing the "a" issue any more, you either see it or you do not. In your data the X/Y ratios are different

sedgwick,Here is the complete proof.

    Consider 2 poles of height a.cosh(A/a) - a  a distance 2A apart with rope touching the ground at (0,0). Assuming a caternary then the ropes curve  y(x) is given by ( a being some constant)

   y(x)=a.cosh(x/a)-a


the length of the rope Y is given by

integral(-A to A ) sqrt(1+(y'(x))^2) dx = 2a.sinh(A/a)

now choose A st the ratio of the pole height to rope length is X/Y ie choose A s.t

   (a.cosh(A/a) - a)/(2a.sinh(A/a)) = X/Y

ie  (cosh(A/a) - 1)/(sinh(A/a)) = 2X/Y     (1)

the ratio of  rope length to  distance apart to  is

      Ratio={ 2a.sinh(A/a)}/2A

             =a.sinh(A/a)/A    

where A saisfies equation (1),

Equation (1) can be solved by substituting sinh(A/a)=sqrt(cosh((A/a))^2 - 1), squaring both sides and collecting/cancelling terms to give (taking  the non-degenerate root)

cosh(A/a)  =(1+k*k)/(1-k*k)

where k= 2X/Y

and

sinh(A/a) = 2k/(1-k*k)

so the solution is

 
       Ratio=a.sinh(A/a)/A

              =(a (2*k)/(1-k*k))/ (a*arccosh( (1+k*k)/(1-k*k) )

              ={ 2*k/(1-k*k) } /  arccosh( (1+k*k)/(1-k*k) )
   
             
 

Using the equations for a catenary that touches the ground (y=a*cosh(x/2a-1))and its length (s=2a*sinh(x/2a)), we can find that for a given 'stick height' (X), the ratio of 'rope length' (Y) to 'stick distance when the rope touches the ground' (Y1) is dependent on 'a' as:

Y/Y1=2a(sinh(arccosh(X/a+1)))

'a' is a physical property of the rope that controls how much sag there is for a given distance of traverse and rope length.  In order to define Y/Y1, you need to define 'a' and pole height.

Here are some values (X=1):
a     Y/Y1
-------------
0.5     2.83
1     3.46
1.5     4.00
2     4.47
2.5     4.90
3     5.29
3.5     5.66
4     6.00
4.5     6.32
5     6.63
5.5     6.93
6     7.21

WHOA!  Left something out of my equations . . . hold on, I'll repost

1. Equation for catenary that touches the ground:

   y=a*(cosh(x/a)-1)


2. Using this, pole height (Y) is:

   Y=a*(cosh(X/2a)-1)

   where X is the total distance between the sticks.


3. This can be rearranged into an expression for X:

   X=2a*arccosh(Y/a + 1)


4. The length of the rope, as GwynforWeb has shown, is:

   S=2a*sinh(X/2a)


5. Then S/X, substituting #3 for X, is:

   S/X = (2a*sinh(arccosh(X/a + 1)))/(2a*arccosh(x/a + 1))

       = sinh(arccosh(X/a + 1)))/(arccosh(x/a + 1))


6. Take the arcsinh of both sides yieds:

   arcsinh(S/X)= (arccosh(X/a + 1)))/(arccosh(x/a + 1))
 
               = 1

7. Therefore:

   S/X = arcsinh(1) = 1.1752



I've rethought this through and it makes sense (as GwynforWeb has been saying) that it is not dependent on 'a . . . assuming that the rope is not elastic/plastic.

Holy Crap!  I've done it again!  This is getting ridiculous . . . I'll try again

Steps 6 & 7 have been removed . . . I confused my parentheses . . . .

--------------------------------------------

1. Equation for catenary that touches the ground:

  y=a*(cosh(x/a)-1)


2. Using this, pole height (Y) is:

  Y=a*(cosh(X/2a)-1)

  where X is the total distance between the sticks.


3. This can be rearranged into an expression for X:

  X=2a*arccosh(Y/a + 1)


4. The length of the rope, as GwynforWeb has shown, is:

  S=2a*sinh(X/2a)


5. Then S/X, substituting #3 for X, is:

  S/X = (2a*sinh(arccosh(X/a + 1)))/(2a*arccosh(x/a + 1))

      = sinh(arccosh(X/a + 1)))/(arccosh(x/a + 1))


So let me re-reevaluate.  Now it seems S/X IS dependent on a, and alas, I am now thoroughly confused.  Here are some numbers, then I swear I won't post another solution:

a           Y/Y1
0.01        19.03
0.05         5.61
0.1             3.55
0.5         1.60
1         1.32
5         1.07
10         1.03
15         1.02
20         1.02
50         1.01
100         1.00
200         1.00

I've rechecked several times . . . my above post is correct.  

Aburr, can you provide numerical solutions to your equation and compare to mine?  I used a pole height of 1.

oic. <grins>

sedgwick -My solution can be written equivalently as below which looks prettier

Ratio = { 2*k/(1-k*k) } /  arcsinh( (2*k/(1-k*k) )

   where k=2X/Y

Just out of curiosity, how the equation would be affected if the poles were from different heights?

The equation is as below and is a bit messy

    y(x)= a.cosh((x-b)/a) - a.cosh((-b)/a)

however the rope shape would still be a section of the basic catenary cosh(x) - 1 scaled to size and translated into place. The analysis to calculate initial to final distance would be the same but very messy and have the extra ratio of the pole height.

Here are a couple of graphs using the equations from my post:

Here are different curves using the catenary equation (y=a*(cosh(x/a)-1))with different values of 'a'.  You can see that for a given distance apart, the height of the poles must compensate for different values of 'a' to keep the catenary going through pt(0,0):
http://filebox.vt.edu/users/acardina/catenaryshapeva.bmp

Therefore, if we keep the height of the poles constant and change 'a', the distance apart must compensate to keep the catenary going through pt(0,0).  We can see that in the graph below, which shows the ratio S/X = sinh(arccosh(X/a + 1)))/(arccosh(x/a + 1)) as a function of 'a':
http://filebox.vt.edu/users/acardina/catenaryratiova.bmp

We know that, but now choose 'a' st the ratio of pole height to rope length is X/Y then read off distance apart and take the ratio with rope length

'a' is a scale factor defined by the dimensions of the coordinate system you are using to represent the problem. It has no effect on any other ratios for a given ratio of X to Y.

a=X
b=Y/2
c=sqrt(a^2-b^2)

ratio X(=height of poles) to (distance between poles) is:
X/2c

or:
X/(2*sqrt(X^2-(Y/2^2)))

Stics must form a semi circle so
X = pi*r
r = X/pi
R = 2X/pi
since they are pushed towards distance between them is
Y - (2X1/pi + 2X2/pi)

Y / (Y - (2X1/pi + 2X2/pi))
must be the solution if I understand the problem.

ok
 I did this problem Experimentally using a 2 sticks (X) of length 210 mm and let the rope was 425mm. I know that Y (distance between the rope) was supposed to be 2X  (420 mm)but it was hard to find one fitting the criteria. But the basic principle is the same.
 The distance between the sticks after the rope touched the ground was 485 mm.
  The difference between the two distances was obtained to be 60 mm.

Therefore the ratio between the two didtances is 485/425
which would be equal to 1:1.1412.

so the distance increased by 14%.

djsqueeze. In your example the distance between the sticks is greater than the length of the rope?

sedgwick, Time to close this one off?

GwynforWeb, i had said The difference between the two distances was obtained to be 60 mm. i should have clarified my explanation.

it reads to me that
   rope length=425mm.
   The distance between the sticks after the rope touched  the ground was 485 mm.

(the sticks are to be kept upright)

ohhh and the sticks brought together untill the rope touched the ground right?

if that's the case i have foung my major mistake. I actually slanted the 2 sticks untill the rope touched the ground sorry.

i'll just re-do the experiment

djsqueeze
  I thought thats what you had done. Your results would be interesting to see and will be represented well by the catenary provided your rope is not too stiff.

I did this  re-did problem Experimentally using a 2 sticks (X) of length 8.6 inches(21.59 cm) and let the rope length(Y)(ie. distance between the sticks) be 17.2 inches (43.688 cm).

the distance between the sticks when the rope touched the ground is 4.5 inches(11.43 cm).

therefore the ratio of the distances in inches :-
 
     17.2/4.5    is  1:3.8222

Using the inches:-

43.688/11.43    is 1:1:3.8222


I wonder why i worked out the both of them like i expected them to be different well i just wanted to be sure i guess or just plain idle. :).

I did this  re-did problem Experimentally using a 2 sticks (X) of length 8.6 inches(21.59 cm) and let the rope length(Y)(ie. distance between the sticks) be 17.2 inches (43.688 cm).
note:-
the sticks were kept upright.

   The distance between the sticks when the rope touched the ground is 4.5 inches(11.43 cm).

therefore the ratio of the distances in inches :-
 
     17.2/4.5    is  1:3.8222

Using the inches:-

43.688/11.43    is 1:1:3.8222


I wonder why i worked out the both of them like i expected them to be different well i just wanted to be sure i guess or just plain idle. :).

typo here i'm correcting it
Using the inches:-

43.688/11.43    is 1:3.8222

i hate when my computer re-postsa message twice darn.!.

well i've  posted the correction but i'll just re-post with it to make sure people understand it. Well i doubt  i'll get any points for this,  but I think doing it different from everyone else (ie. not using a bag of formulae) and doing it experimentally allowed be to see things in perspective.

I did this  re-did problem Experimentally using a 2 sticks (X) of length 8.6 inches(21.59 cm) and let the rope length(Y)(ie. distance between the sticks) be 17.2 inches (43.688 cm).
note:-
the sticks were kept upright.

  The distance between the sticks when the rope touched the ground is 4.5 inches(11.43 cm).

therefore the ratio of the distances in inches :-
 
    17.2/4.5    is  1:3.8222

Using the inches:-

43.688/11.43    is 1:3.8222

Your numbers do not make sense, your rope is exactly twice the height of the sticks, the only way the rope can touch the ground is if the sticks are right next to each other. (ie distance =0). The rope for this problem must be more than twice the height of the sticks.

today must be a really bad day for me  cuz i'm making way too many errors. And if they were big ones it wouldn't be so bad but it's the little ones I  make that piss me off

GwynforWeb do these numbers make any sense now. I changes my whole apparatus for a nother set. Hopefully no errors this time:-

Experimentally using a 2 sticks (X) of length 15.8 cm and let the rope length(Y)(ie. distance between the sticks) be 43cm).
note:-
the sticks were kept upright.

 The distance between the sticks when the rope touched the ground is 28 cm.


therefore ratio is:-
 28/43   is 2.0869:1    or u can say 0.6511 :1.

GwynforWeb do these numbers make any sense now. I changes my whole apparatus for a nother set. Hopefully no errors this time:-

Experimentally using a 2 sticks (X) of length 15.8 cm and let the rope length(Y)(ie. distance between the sticks) be 43cm).
note:-
the sticks were kept upright.

 The distance between the sticks when the rope touched the ground is 28 cm.


therefore ratio is:-
 28/43   is 2.0869:1    or u can say 1: 0.6511.

darn why can't i stop making mistakes

therefore ratio is:-
28/43   is 1.5357:1    or u can say 1: 0.6511.

My formula a gives ratio =1.701:1 for your numbers, not too bad a correspondence.  

thanks GwynforWeb. as you know experimentally I could not obtain an answer aS ACCURATE as urs.

I gather you are being sarcastic?. However the difference is caused by the fact the catenary equation assumes the rope has no stiffness at all which is not true in a real rope, and will only hold for ropes if they are long relative for their thickness. The catenary equation was originally developed for chains which do not display any resitance to being bent.

thanks GwynforWeb  for the info. where is sedgwick
anyway?

thanks GwynforWeb  for the info. where is sedgwick
anyway?

Y is the length of the rope(initial distance between the sticks)
X being the length of the sticks(both of them)

The ratio would be

y : square root(y square - 4*x square)


Maanas

Please no more semi-circle 'solutions'

sedgwick, I have unsubscribed (will pop in occasionally)

its time to put an end to this thread.
without taking under consideration the weight of the rope which has an effect on the rope's curve and without getting into physical equasions it seems that GwynforWeb was giving the first answer.
thank u all for your help/suggestions/replies.

sedgwick,
   Thanks for the points. (if the rope is long enough the ropes weight has essentially no effect, even when it is fairly short it's effects is limited, I assure this is true)

What if the poles are too tall for the length of the rope to ever touch the ground?

my comment from Date: 04/15/2003 04:59PM CST:
assume this: Y > X1+X2
Y = rope length
X1, X2 = sticks hight

Sorry for this comment
 
First the poles are like this  
                                             
-------------------------                                          
|          Y                 |
|                             |
|                             |
|                             |
| X                          |X
|                             |
|                             |
|                             |


When we start to pull from each end  it will come like this only right??


------------------------ ------------------------ ------------------------
       x                              y                                  x


 Ratio is     ( Y ) : (2X+Y)  
i am making guess as our height is negligible. Then this problem will be simple. If i guess like this then this puzzle will look like childish one.  can any of u explain with picture??