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Puzzle for you - Zaphod - 08-16-2018 09:45 PM
![[Image: BF033_A29-64_DC-4_D33-_AF81-_D2_A79046_A0_E5.jpg]](https://s8.postimg.cc/6pg4yi36d/BF033_A29-64_DC-4_D33-_AF81-_D2_A79046_A0_E5.jpg)
If you’d like, Calculate the distance between the poles
RE: Puzzle for you - Albert Chan - 08-16-2018 10:19 PM
The rope is too short !
RE: Puzzle for you - HP-Collection - 08-16-2018 11:29 PM
Distance must be zero as the rope half lengh (3 feet) is the difference between the full height (5 feet) and the minimal height over ground (2 feet). The rope must hang straight down and up (sorry about my english)
RE: Puzzle for you - ttw - 08-17-2018 03:13 AM
I assume it's a weightless rope hanging by the force of gravity.
RE: Puzzle for you - Thomas Klemm - 08-17-2018 03:59 AM
We can model the rope as a catenary:
\(y=a\cosh(\frac{x}{a})\)
where \(a\) is a scaling factor.
Then with \(h\) as the difference of the heights and \(s\) as half the length of the rope we get:
\(h=a(\cosh(\frac{x}{a})-1)\)
\(s=a\sinh(\frac{x}{a})\)
With:
\(u=\frac{x}{a}\)
and
\(v=\frac{h}{s}=\frac{a(\cosh(u)-1)}{a\sinh(u)}=\tanh(\frac{u}{2})\)
Thus
\(u=2\tanh^{-1}(v)=2\tanh^{-1}(\frac{h}{s})\)
This allows us to calculate:
\(a=\frac{s}{\sinh(u)}=\frac{s}{\sinh(2\tanh^{-1}(\frac{h}{s}))}=\frac{s^2-h^2}{2h}\)
Now we plug both \(a\) and \(u\) in to calculate:
\(\begin{align*}
x=a\cdot u &=\frac{s^2-h^2}{2h}\cdot2\tanh^{-1}(\frac{h}{s}) \\
&=\frac{s^2-h^2}{h}\cdot\tanh^{-1}(\frac{h}{s}) \\
&=s\cdot(\frac{s}{h}-\frac{h}{s})\cdot\tanh^{-1}(\frac{h}{s}) \\
&=s\cdot(\frac{1}{v}-v)\cdot\tanh^{-1}(v)
\end{align*}\)
Example:
\(h=3\)
\(s=4\)
\(x=4\cdot(\frac{4}{3}-\frac{3}{4})\cdot\tanh^{-1}(\frac{3}{4})\approx 2.27022850723\)
Limit \(h\to s\)
For \(v\to1\) the value \(\tanh^{-1}(v)\to\infty\) but at the same time \(\frac{1}{v}-v\to0\).
Thus we can apply L'Hôpital's rule to find:
\(\begin{align*}
\lim_{v\to1}(\frac{1}{v}-v)\cdot\tanh^{-1}(v) &= \lim_{v\to1}\frac{1-v^2}{v}\cdot\tanh^{-1}(v) \\
&= \lim_{v\to1}\frac{\tanh^{-1}(v)}{\frac{v}{1-v^2}} \\
&= \lim_{v\to1}\frac{\frac{1}{1-v^2}}{\frac{1+v^2}{(1-v^2)^2}} \\
&= \lim_{v\to1}\frac{1-v^2}{1+v^2}=\frac{1-1}{1+1}=0
\end{align*}\)
And thus:
\(x=s\cdot0=0\)
Which we already knew.
Cheers
Thomas
RE: Puzzle for you - Albert Chan - 08-17-2018 12:02 PM
(08-17-2018 03:59 AM)Thomas Klemm Wrote: We can model the rope as a catenary:
\(y=a\cosh(\frac{x}{a})\)
...
Example:
\(h=3\)
\(s=4\)
\(x=4\cdot(\frac{4}{3}-\frac{3}{4})\cdot\tanh^{-1}(\frac{3}{4})\approx 2.27022850723\)
Out of curiosity, had the curve a parabola, with same h and x, s ~ 3.409 (less rope)
With a fixed h and x, can I assume catenary always curvier than a parabola ?
Regarding this puzzle, my original guess were also 0 distance between poles.
But, the rope cannot even fit between the poles ...
Edit: catenary is indeed curvier (flatter bottom, steeper rise): http://mathforum.org/library/drmath/view/65729.html
RE: Puzzle for you - Csaba Tizedes - 08-19-2018 06:57 PM
(08-16-2018 09:45 PM)Zaphod Wrote:
If you’d like, Calculate the distance between the poles
a.) In engineering practice the headroom at least 4.5 meter (approx 15 feet).
b.) 2 feet is less than 15 feet.
c.) Therefore this is not a real life problem.
d.) Engineers do not waste their time to a non real-life fictious problems.
e.) I am an engineer.
f.) Therefore in my sight this problem is useless and only playing with numbers - I do not want to do anything with it.
Maybe if you have data of distance of poles vs headroom below the rope, in this case this will be more interesting.
Csaba
RE: Puzzle for you - Albert Chan - 08-19-2018 07:53 PM
(08-19-2018 06:57 PM)Csaba Tizedes Wrote: a.) In engineering practice the headroom at least 4.5 meter (approx 15 feet).
b.) 2 feet is less than 15 feet.
c.) Therefore this is not a real life problem.
Why not a real problem ?
5 feet pole, 6 feet rope ... it is called a fence
RE: Puzzle for you - Zaphod - 08-24-2018 08:02 PM
(08-16-2018 11:29 PM)HP-Collection Wrote: Distance must be zero as the rope half lengh (3 feet) is the difference between the full height (5 feet) and the minimal height over ground (2 feet). The rope must hang straight down and up (sorry about my english)
(08-17-2018 03:59 AM)Thomas Klemm Wrote: We can model the rope as a catenary:
\(y=a\cosh(\frac{x}{a})\)
where \(a\) is a scaling factor.
Then with \(h\) as the difference of the heights and \(s\) as half the length of the rope we get:
\(h=a(\cosh(\frac{x}{a})-1)\)
\(s=a\sinh(\frac{x}{a})\)
With:
\(u=\frac{x}{a}\)
and
\(v=\frac{h}{s}=\frac{a(\cosh(u)-1)}{a\sinh(u)}=\tanh(\frac{u}{2})\)
Thus
\(u=2\tanh^{-1}(v)=2\tanh^{-1}(\frac{h}{s})\)
This allows us to calculate:
\(a=\frac{s}{\sinh(u)}=\frac{s}{\sinh(2\tanh^{-1}(\frac{h}{s}))}=\frac{s^2-h^2}{2h}\)
Now we plug both \(a\) and \(u\) in to calculate:
\(\begin{align*}
x=a\cdot u &=\frac{s^2-h^2}{2h}\cdot2\tanh^{-1}(\frac{h}{s}) \\
&=\frac{s^2-h^2}{h}\cdot\tanh^{-1}(\frac{h}{s}) \\
&=s\cdot(\frac{s}{h}-\frac{h}{s})\cdot\tanh^{-1}(\frac{h}{s}) \\
&=s\cdot(\frac{1}{v}-v)\cdot\tanh^{-1}(v)
\end{align*}\)
Example:
\(h=3\)
\(s=4\)
\(x=4\cdot(\frac{4}{3}-\frac{3}{4})\cdot\tanh^{-1}(\frac{3}{4})\approx 2.27022850723\)
Limit \(h\to s\)
For \(v\to1\) the value \(\tanh^{-1}(v)\to\infty\) but at the same time \(\frac{1}{v}-v\to0\).
Thus we can apply L'Hôpital's rule to find:
\(\begin{align*}
\lim_{v\to1}(\frac{1}{v}-v)\cdot\tanh^{-1}(v) &= \lim_{v\to1}\frac{1-v^2}{v}\cdot\tanh^{-1}(v) \\
&= \lim_{v\to1}\frac{\tanh^{-1}(v)}{\frac{v}{1-v^2}} \\
&= \lim_{v\to1}\frac{\frac{1}{1-v^2}}{\frac{1+v^2}{(1-v^2)^2}} \\
&= \lim_{v\to1}\frac{1-v^2}{1+v^2}=\frac{1-1}{1+1}=0
\end{align*}\)
And thus:
\(x=s\cdot0=0\)
Which we already knew.
Cheers
Thomas
Both these
RE: Puzzle for you - Thomas Klemm - 12-22-2018 09:22 PM
(08-17-2018 03:59 AM)Thomas Klemm Wrote: Example:
\(h=3\)
\(s=4\)
\(x=4\cdot(\frac{4}{3}-\frac{3}{4})\cdot\tanh^{-1}(\frac{3}{4})\approx 2.27022850723\)
Just stumbled upon this video where the same problem (although scaled by 10) is solved in a slightly different way:
Can You Solve Amazon's Hanging Cable Interview Question?
Cheers
Thomas