Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2021-05-06 20:00:36

vellawitting
Member
Registered: 2021-05-06
Posts: 1

Catenary

Hi,

I work on a tugboat and we tow a barge from Jacksonville, FL to San Juan PR.  I am trying to find a formula to solve for the depth of the tow wire below the surface of the water.

Given the total length of the towing wire and the distance between the back of the tug boat and the front of the barge, how deep below the surface of the water is the lowest point of the tow wire?

Assume length of wire is 2100' and distance between tug and barge is 1650'

Thanks

Offline

#2 2021-05-07 06:02:27

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Catenary

hi vellawitting

Welcome to the forum.

I see you think the wire will follow a catenary curve.  If you used a rope then the weight of water in the rope would probably throw the calculations.  As it's a wire, maybe we're ok.

The catenary is usually used for a suspension bridge between two fixed points. Because the barge is under tow that may upset the symmetry of the curve.  Also the formula that is usually used gives the 'sag' using the distance between the fixed points.  You've given the length of the wire.

So I've got some homework to do looking up the underlying theory that is used to develop the equation.

I'll come back to this when I've made some progress.

Can I assume that the height of the fixed points is close enough to the water level so that we can disregard it?  If not I'll need more measurements.

Are you concerned that the wire will scrape the bottom in shallow water?

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

Offline

#3 2021-05-07 11:19:09

mathland
Member
Registered: 2021-03-25
Posts: 444

Re: Catenary

Bob wrote:

hi vellawitting

Welcome to the forum.

I see you think the wire will follow a catenary curve.  If you used a rope then the weight of water in the rope would probably throw the calculations.  As it's a wire, maybe we're ok.

The catenary is usually used for a suspension bridge between two fixed points. Because the barge is under tow that may upset the symmetry of the curve.  Also the formula that is usually used gives the 'sag' using the distance between the fixed points.  You've given the length of the wire.

So I've got some homework to do looking up the underlying theory that is used to develop the equation.

I'll come back to this when I've made some progress.

Can I assume that the height of the fixed points is close enough to the water level so that we can disregard it?  If not I'll need more measurements.

Are you concerned that the wire will scrape the bottom in shallow water?

Bob

I look forward to your math work here. I also would like to know how this is done.

Offline

#4 2021-05-08 01:06:48

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Catenary

Further analysis:

It looks like the water makes no difference.  see https://thenavalarch.com/safe-towing-ca … y-and-sag/

I've played about with those numbers but no solution has appeared so far.  I'm trying to take the riverbed as the x axis and then find the lowest point on the curve ( made symmetrical so x = 0), which is the number that mathematicians usually call a in the equation.

But I also need to know the height above the bed of the wire fixing points.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

Offline

#5 2021-05-08 03:43:33

mathland
Member
Registered: 2021-03-25
Posts: 444

Re: Catenary

Bob wrote:

Further analysis:

It looks like the water makes no difference.  see https://thenavalarch.com/safe-towing-ca … y-and-sag/

I've played about with those numbers but no solution has appeared so far.  I'm trying to take the riverbed as the x axis and then find the lowest point on the curve ( made symmetrical so x = 0), which is the number that mathematicians usually call a in the equation.

But I also need to know the height above the bed of the wire fixing points.

Bob

This is a tough a problem.

Offline

#6 2021-05-09 19:19:22

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Catenary

The original poster has gone quiet on this.  I'm ready to post the derivation of the equation but it'll take a while as there's lots to show.  Next post hopefully.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

Offline

#7 2021-05-10 04:01:40

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Catenary

Ok. Here we go. Before I get on to the catenary itself, there's a couple of topics I need as preliminaries.

Arc Length Formula.

NLPKqX6.gif

This works for any curve, not just a catenary.  Δs is the length along the curve.

The diagram shows that Δx, Δy and Δs almost make a right angled triangle.  Not quite, because the 'hypotenuse' isn't a straight line.  However, as Δx tends to zero, the curve gets ever closer to straight and so in the limit we can assume pythagoras theorem.

Divide by Δx^2

Let Δx tend to zero:

Hyperbolic Trig Functions.

Sinh is pronounced sinch

I leave the following as exercises for the interested reader:

The Catenary

By observation of hanging ropes etc. it is clear that there is a minimum point.  It is usual to put the y axis through that point.
The curve looks a bit like a parabola, but it isn't quite that. Here's a rough diagram of the situation.  If you have done any mechanics questions with 'light inextensible strings' you may be used to the tension being constant along the string.  The catenary calculation does not have weight free strings.  The assumption is that the string has a constant weight along its length.  This means that, if the weight is w per unit length (let's say feet) the 1 foot weighs w, 2 weighs 2w, and s feet weighs ws.

The following applies to ropes, strings, chains and wires.  As the original post was for a wire I shall use that from now on.

The tension varies from point to point.  O is the lowest point and P some point elsewhere along the wire.  There are three forces acting on OP.  The horizontal tension T0 at O; the tension T at P, acting at angle alpha, and the weight of OP, ws, acting vertically downwards.


5AXJcsX.gif

The rope is in equilibrium, meaning the forces are balanced so that it isn't moved left or right, up or down.  So the vertical forces must balance and the horizontal forces must separately balance.

The tension, T, varies, but can be eliminated from the equations by dividing the first by the second.  And tan alpha is the gradient of the line and so may be written dy/dx.

T0 and w are both constants so it makes sense to simplify by replacing with a single constant. For reasons which will become clear later, w/T0 = 1/a rather than the other way up.  Don't get hung up on this; it's just a constant.

So we have a differential equation:

Now you might want to jump in with an easy integral, s/a, but wait!  The variable is 's' and the variable we are wanting to integrate with respect to is x.  Can't be done!  You have to have both variables the same.

That's why we need the arc length formula. We can change the 'dx' into a 'ds' using this. 

We have

and

If either is solved alone I'll have a equation not y in terms of x, but y (or x) in terms of s.  Not what we want.  So both need to be solved and then I have to see if s can be eliminated.  Don't worry; I've sneaked a look ahead and it can smile

The y one is directly integrable:

If we choose that minimum point to be (0,a) then C = 0.

The other is harder.  It can be done using substitution.  This technique sometimes works if you can find a good substitution.  Experience helps.  In this case we've got a ready sub that we can try.  You have to change all instants of the old variable to the new one.  At the end you have to convert back to the old variable.

Try s = a sinh(u/a) where u is the new variable.  ds/du = cosh(u/a). 

Therefore

This constant is also zero as sinh(0) = 0

So x = a arc sinh (s/a)   =>    s = a . sinh(x/a)

I can sub that into the y equation thus:

I'll take a break at this point and come back to the actual question later.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

Offline

#8 2021-05-10 11:08:27

mathland
Member
Registered: 2021-03-25
Posts: 444

Re: Catenary

Bob wrote:

The original poster has gone quiet on this.  I'm ready to post the derivation of the equation but it'll take a while as there's lots to show.  Next post hopefully.

Bob

You are right but I'm still here waiting to see how this is done. Don't you just hate that? Someone decides to post a thread and then the person disappears.

Offline

#9 2021-05-10 11:10:46

mathland
Member
Registered: 2021-03-25
Posts: 444

Re: Catenary

Bob wrote:

Ok. Here we go. Before I get on to the catenary itself, there's a couple of topics I need as preliminaries.

Arc Length Formula.

https://i.imgur.com/NLPKqX6.gif

This works for any curve, not just a catenary.  Δs is the length along the curve.

The diagram shows that Δx, Δy and Δs almost make a right angled triangle.  Not quite, because the 'hypotenuse' isn't a straight line.  However, as Δx tends to zero, the curve gets ever closer to straight and so in the limit we can assume pythagoras theorem.

Divide by Δx^2

Let Δx tend to zero:

Hyperbolic Trig Functions.

Sinh is pronounced sinch

I leave the following as exercises for the interested reader:

The Catenary

By observation of hanging ropes etc. it is clear that there is a minimum point.  It is usual to put the y axis through that point.
The curve looks a bit like a parabola, but it isn't quite that. Here's a rough diagram of the situation.  If you have done any mechanics questions with 'light inextensible strings' you may be used to the tension being constant along the string.  The catenary calculation does not have weight free strings.  The assumption is that the string has a constant weight along its length.  This means that, if the weight is w per unit length (let's say feet) the 1 foot weighs w, 2 weighs 2w, and s feet weighs ws.

The following applies to ropes, strings, chains and wires.  As the original post was for a wire I shall use that from now on.

The tension varies from point to point.  O is the lowest point and P some point elsewhere along the wire.  There are three forces acting on OP.  The horizontal tension T0 at O; the tension T at P, acting at angle alpha, and the weight of OP, ws, acting vertically downwards.


https://i.imgur.com/5AXJcsX.gif

The rope is in equilibrium, meaning the forces are balanced so that it isn't moved left or right, up or down.  So the vertical forces must balance and the horizontal forces must separately balance.

The tension, T, varies, but can be eliminated from the equations by dividing the first by the second.  And tan alpha is the gradient of the line and so may be written dy/dx.

T0 and w are both constants so it makes sense to simplify by replacing with a single constant. For reasons which will become clear later, w/T0 = 1/a rather than the other way up.  Don't get hung up on this; it's just a constant.

So we have a differential equation:

Now you might want to jump in with an easy integral, s/a, but wait!  The variable is 's' and the variable we are wanting to integrate with respect to is x.  Can't be done!  You have to have both variables the same.

That's why we need the arc length formula. We can change the 'dx' into a 'ds' using this. 

We have

and

If either is solved alone I'll have a equation not y in terms of x, but y (or x) in terms of s.  Not what we want.  So both need to be solved and then I have to see if s can be eliminated.  Don't worry; I've sneaked a look ahead and it can smile

The y one is directly integrable:

If we choose that minimum point to be (0,a) then C = 0.

The other is harder.  It can be done using substitution.  This technique sometimes works if you can find a good substitution.  Experience helps.  In this case we've got a ready sub that we can try.  You have to change all instants of the old variable to the new one.  At the end you have to convert back to the old variable.

Try s = a sinh(u/a) where u is the new variable.  ds/du = cosh(u/a). 

Therefore

This constant is also zero as sinh(0) = 0

So x = a arc sinh (s/a)   =>    s = a . sinh(x/a)

I can sub that into the y equation thus:

I'll take a break at this point and come back to the actual question later.

Bob

Wow! I didn't know that you were such a mathematician. This is very impressive work. I now know you are the right person to help me on my calculus journey and my occasional revisit of courses before calculus.

Offline

#10 2021-05-15 21:17:12

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Catenary

hi vellawitting

My apologies for the long wait for a final reply on this.  I have been distracted by other forum business.

To sum up we have:



I do not know how deep the river is.  The x axis does not represent the river bed.

Using your figures I decided to assume the tug fixing point and the barge fixing point to be at the same height above the x axis and used the symbol h for this.  The coordinates of one of the fixing points is then (825,h)  That's half the distance between tug and barge. I also used s = 1050 (half the wire length) at the fixing point.

I then used Wolfram Alpha (a maths software engine) to solve the third equation (using s = 1050 and x = 825).

WA gave a = 669.814.  Let's say 670.

Then using the second equation I can calculate h.  I got a value of 1245.452.

These two answers seem very large to me and not anywhere near the solution you are looking for.  I decided to get a 'second opinion' by pretending that the wire goes in a straight line from the tug to the lowest point and then in a second straight line back up to the barge. That creates an isosceles triangle with slope length 1050 and 'base' = 1650.  By splitting it down the middle I can use Pythagoras' theorem to compute the height of the triangle.  I get 649.5191 for this.  That together with the catenary calculations is a consistent result.  ie. It's a good approximation of the catenary figures.  So I think they are probably correct.

So, what's going on here?  I need you to check your original figures.  Is it possible that different units have crept in? yards maybe or metres?

Because, as it stands, the wire sags roughly 650 feet down in the middle and I doubt there's a river in the world that deep. ** see last paragraph!!!

Here's a scaled graph showing the curve and the wire sag.

ruqGFIO.gif

Later edit: up until now I've just treated this as an interesting maths problem but now I've started to think about the actual practicalities.  No one would tow a barge with a line that is about 1/3 mile long.  It would be a major shipping hazard.  If a ship is going the other way, port to port, it would steer to the right to safely pass the tug.  But there's nothing to tell the captain not to re-occupy the centre of the waterway once passed.  The wire would be out of sight, below the water.  A steel wire towline would rip a glass-fibre hull to shreds.  And around a bend I expect the barge has personnel on board with independent steerage so the barge can follow the bend.  But the wire wouldn't.  It would straight line across the bank, slicing off branches and innocent walkers' legs.

So there's definitely something wrong with the measurements.

Assume length of wire is 2100' and distance between tug and barge is 1650'

  Perhaps that is Assume length of wire is 2100'' and distance between tug and barge is 1650'', in other words inches.  I haven't had to introduce units into the calculations so you can just substitute inches for feet and we have a more sensible value of about 670 inches = 56 feet.  Still too big but you'd be ok down the centre of  Loch Ness now. smile

even later edit:

I should probably looked this up at the start.  Now I've used Google Earth to locate these places, I can see this is not a river journey at all. Whoops!   This is across the ocean from Florida to Puerto Rica.  Silly me.  All comments about damage to innocent by-standers are therefore null and void.  I expect the ocean is deep enough once you're underway.  In which case I consider this problem solved.


Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

Offline

Board footer

Powered by FluxBB