Poll

 I know! (explain) 5 votes (26.31%) I'm on it Wiz. 2 votes (10.52%) I'm going to cheat. 1 vote (5.26%) I just don't know, but I should. 3 votes (15.78%) I just don't know, and it doesn't bother me. 7 votes (36.84%) My integral calculus is obviously a bit rusty. 4 votes (21.05%) I'm tempted to get a penny farthing. 2 votes (10.52%) When is the 2019 WoV spring fling? 5 votes (26.31%) Congratulations SOOPOO on winning the dead pool. 2 votes (10.52%) My gender is x. 4 votes (21.05%)

19 members have voted

Wizard Joined: Oct 14, 2009
• Posts: 19912
January 8th, 2019 at 3:04:16 PM permalink
I 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)
It's not whether you win or lose; it's whether or not you had a good bet.
unJon Joined: Jul 1, 2018
• Posts: 931
January 8th, 2019 at 4:10:02 PM permalink
Having now refreshed on the Wikipedia page, I�m 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?
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Wizard Joined: Oct 14, 2009
• Posts: 19912
January 8th, 2019 at 4:40:04 PM permalink
Quote: unJon

Having now refreshed on the Wikipedia page, I�m 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.
It's not whether you win or lose; it's whether or not you had a good bet.
ChesterDog Joined: Jul 26, 2010
• Posts: 736
January 8th, 2019 at 4:44:57 PM permalink
Quote: Wizard

I 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.
Wizard Joined: Oct 14, 2009
• Posts: 19912
January 8th, 2019 at 4:55:49 PM permalink
Quote: ChesterDog

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.

Thanks, that makes sense. I should have figured that out myself, but I was overthinking it.
It's not whether you win or lose; it's whether or not you had a good bet.
ChesterDog Joined: Jul 26, 2010
• Posts: 736
January 8th, 2019 at 5:56:40 PM permalink
Quote: CrystalMath

Quote: Wizard

Quote: ChesterDog

Thanks!

I did not know that sinh-1=ln[x+(1+x2)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 ) ]
Last edited by: ChesterDog on Jan 9, 2019
AxelWolf Joined: Oct 10, 2012
• Posts: 15962
January 8th, 2019 at 6:03:24 PM permalink
Does 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?
♪♪Now you swear and kick and beg us That you're not a gamblin' man Then you find you're back in Vegas With a handle in your hand♪♪ Your black cards can make you money So you hide them when you're able In the land of casinos and money You must put them on the table♪♪ You go back Jack do it again roulette wheels turinin' 'round and 'round♪♪ You go back Jack do it again♪♪
ChesterDog Joined: Jul 26, 2010
• Posts: 736
January 8th, 2019 at 6:27:35 PM permalink
Quote: AxelWolf

Does 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.
Wizard Joined: Oct 14, 2009
• Posts: 19912
January 8th, 2019 at 6:27:52 PM permalink
Quote: AxelWolf

Does 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).
It's not whether you win or lose; it's whether or not you had a good bet.
unJon Joined: Jul 1, 2018 