## Poll

5 votes (26.31%) | |||

2 votes (10.52%) | |||

1 vote (5.26%) | |||

3 votes (15.78%) | |||

7 votes (36.84%) | |||

4 votes (21.05%) | |||

2 votes (10.52%) | |||

5 votes (26.31%) | |||

2 votes (10.52%) | |||

4 votes (21.05%) |

**19 members have voted**

One step I struggle with in this whole problem is scaling a catenary. The general form is

y = a * cosh(x/a).

I can see how this scales the catenary, but would these curves be catenaries:?

y = cosh(x/a)

y = a*cosh(x)

Quote:unJonHaving now refreshed on the Wikipedia page, Im confused. The a in the catenary function appears to equal the Horizontal Force dividend by the mass per unit of length and the distance between poles. Is that correct, and if so, what assumptions about force and mass were used to derive a=11.25?

I looked at it more as a math problem. See my posted solution for how I arrived at the value for a.

Quote:WizardI forgot to congratulate you on looking at this more like a physics problem.

One step I struggle with in this whole problem is scaling a catenary. The general form is

y = a * cosh(x/a).

I can see how this scales the catenary, but would these curves be catenaries:?

y = cosh(x/a)

y = a*cosh(x)

According to http://xahlee.info/SpecialPlaneCurves_dir/Catenary_dir/catenary.html, "It is worthwhile to note that catenary has only one shape (it is not a family of curves)."

So, if the formula for a curve can be transformed into the general form of the catenary, y = a * cosh(x/a), by making these substitutions, Y = ky and X = kx, then it's a catenary, too. Note that the same scaling factor, k, must be used for both X and Y.

For example, is Y = coshX a catenary? Let Y=ky and X=kx. Then ky = cosh(kx). y = (1/k) cosh(kx). Set 1/k = a, then y = a cosh(x/a). So Y=coshX is a catenary.

Regarding Y = cosh(X/a), let Y=ky and X=kx. Then ky = cosh(kx/a). y = (1/k) cosh(kx/a). Let A = 1/k, then y = A cosh(x/(aA)). This is a catenary only if a=1, in which case the formula would be y = A cosh(x/A).

The same goes for y=a*cosh(x). It's a catenary only if a = 1.

Quote:ChesterDogSo, if the formula for a curve can be transformed into the general form of the catenary, y = a * cosh(x/a), by making these substitutions, Y = ky and X = kx, then it's a catenary, too. Note that the same scaling factor, k, must be used for both X and Y.

Thanks, that makes sense. I should have figured that out myself, but I was overthinking it.

Quote:CrystalMathQuote:WizardQuote:ChesterDogThanks!

I did not know that sinh^{-1}=ln[x+(1+x^{2})^{1/2}].

And it's amazing to me that your (45 ln(40/9+(1+(40/9)^{2})^{1/2}))/2 simplifies to 45ln3.

Good catch! I didn't know it simplified to that. Very elegant. For what it's worth, I chose the pole and rope length arbitrarily.

That is crazy. All the stuff inside the ln evaluates to 9, so we have 45*ln(9)/2, and ln(9)/2 = ln(3).

I checked to see if the simplification happens for other choices of rope length and maximum dip of the rope, and it does. If we let r = half of the rope length and h = maximum dip of the rope, then a formula for the distance, D, between the supports is:

D = ( 1 / h ) * ( r + h ) * ( r - h ) * ln[ ( r + h ) / ( r - h ) ]

Quote:AxelWolfDoes it matter how thick the rope is? I have to asuume that will make a difference? I have to assume all calculations are done without factoring in the thickness of the rope?

I'm sure the thickness of the rope would matter in the real world. Also, the elasticity of the rope and the attachments of the rope's ends would matter.

We have only been dealing with the ideal catenary.

Quote:AxelWolfDoes it matter how thick the rope is? I have to asuume that will make a difference? I have to assume all calculations are done without factoring in the thickness of the rope?

It assume thickness and weight of the rope is not an issue. Perhaps I should have said a string or the type of chain they use to secure pens at the bank (which never seem to have ink).

While I too have heard that girth matters, it appears that length trumps all, when it comes to the catenary position.Quote:AxelWolfDoes it matter how thick the rope is? I have to asuume that will make a difference? I have to assume all calculations are done without factoring in the thickness of the rope?