# Catenary **URL:** \ **Category:** The Show\ **Created:** [April 18, 2011, 3:34am UTC](https://community.cartalk.com/t/catenary/63816 "2011-04-18T03:34:09Z")\ **Posts on this page:** 13\ **Page:** 1
**Author:** ![impala61](https://avatars.discourse-cdn.com/v4/letter/i/77aa72/32.png) [@impala61](https://community.cartalk.com/u/impala61)\ **Post date:** [April 18, 2011, 3:34am UTC](https://community.cartalk.com/t/catenary/63816/1 "2011-04-18T03:34:09Z")
Yes, this is another of those _Stream of Consciousness_ posts. I’m thinking 5\*sqrt(2) feet. And I know how to properly spell “catenary”. ---
**Author:** ![impala61](https://avatars.discourse-cdn.com/v4/letter/i/77aa72/32.png) [@impala61](https://community.cartalk.com/u/impala61)\ **Post date:** [April 18, 2011, 3:42am UTC](https://community.cartalk.com/t/catenary/63816/2 "2011-04-18T03:42:58Z")
Sorry, was only thinking of one side of the curve. Should have said 10\*sqrt(2), about 14 feet. ---
**Author:** ![texases](https://sea1.discourse-cdn.com/cartalk/user_avatar/community.cartalk.com/texases/32/862_2.png) [@texases](https://community.cartalk.com/u/texases)\ **Post date:** [April 18, 2011, 3:46am UTC](https://community.cartalk.com/t/catenary/63816/3 "2011-04-18T03:46:33Z")
No math needed to solve it, really. Think about it. ---
**Author:** ![impala61](https://avatars.discourse-cdn.com/v4/letter/i/77aa72/32.png) [@impala61](https://community.cartalk.com/u/impala61)\ **Post date:** [April 18, 2011, 6:35am UTC](https://community.cartalk.com/t/catenary/63816/4 "2011-04-18T06:35:28Z")
Would you like to share with the class? ---
**Author:** ![mattedgar75](https://avatars.discourse-cdn.com/v4/letter/m/ad7895/32.png) [@mattedgar75](https://community.cartalk.com/u/mattedgar75)\ **Post date:** [April 18, 2011, 1:48pm UTC](https://community.cartalk.com/t/catenary/63816/5 "2011-04-18T13:48:14Z")
I think Texases is suggesting that you think about the rope length. The discussion about catenaries seems to be a red herring. I should have applied a little logic before I spent an hour trying to solve the puzzler. You can use the equation for a catenary to get an answer, but use that answer to find the arc length. ---
**Author:** ![texases](https://sea1.discourse-cdn.com/cartalk/user_avatar/community.cartalk.com/texases/32/862_2.png) [@texases](https://community.cartalk.com/u/texases)\ **Post date:** [April 18, 2011, 1:49pm UTC](https://community.cartalk.com/t/catenary/63816/6 "2011-04-18T13:49:02Z")
100’ poles, 150’ chain, 25’ from bottom of ‘catenary’ to ground, isn’t that right? Draw a picture. ---
**Author:** ![impala61](https://avatars.discourse-cdn.com/v4/letter/i/77aa72/32.png) [@impala61](https://community.cartalk.com/u/impala61)\ **Post date:** [April 18, 2011, 5:57pm UTC](https://community.cartalk.com/t/catenary/63816/7 "2011-04-18T17:57:53Z")
I did that first thing. I’m not saying you’re wrong (since I don’t know your answer) just that I couldn’t get anywhere with that approach. I fully expect to feel stupid when I see the answer, that’s no problem. Here’s an example of how I’m overthinking it. A straight line drawn between the vertex of the catenary curve and one of the anchor points at the top of the pole must be less than 75 feet long, since there are 75 feet of rope on each side of the curve. This forms the hypotenuse, c, of a right triangle, and the 75 feet of height left over from the vertex being 25 feet off the ground forms one of the other two sides, a. Ultimately we want to solve for the remaining side, b, and double it to get the distance between the poles. But in a right triangle, no other side can be longer than the hypotenuse, which must be less than 75 feet. Clearly there is no solution or more likely I am totally missing it. Even fumbling around like this is much more fun than last week’s tow-truck Puzzler. ---
**Author:** ![texases](https://sea1.discourse-cdn.com/cartalk/user_avatar/community.cartalk.com/texases/32/862_2.png) [@texases](https://community.cartalk.com/u/texases)\ **Post date:** [April 18, 2011, 7:44pm UTC](https://community.cartalk.com/t/catenary/63816/8 "2011-04-18T19:44:21Z")
Ok, here’s a (big) hint - what’s half of 150? ---
**Author:** ![impala61](https://avatars.discourse-cdn.com/v4/letter/i/77aa72/32.png) [@impala61](https://community.cartalk.com/u/impala61)\ **Post date:** [April 19, 2011, 1:31am UTC](https://community.cartalk.com/t/catenary/63816/9 "2011-04-19T01:31:30Z")
Clearly in my case you’re no Heloise because that wasn’t much of a hint. See the numbers in bold in the post just above. Why don’t you be a dear lamb and just spell it out? ---
**Author:** ![texases](https://sea1.discourse-cdn.com/cartalk/user_avatar/community.cartalk.com/texases/32/862_2.png) [@texases](https://community.cartalk.com/u/texases)\ **Post date:** [April 19, 2011, 1:33am UTC](https://community.cartalk.com/t/catenary/63816/10 "2011-04-19T01:33:48Z")
150/2+25=100, right? So the poles must be 0" apart, there’s only enough chain to reach 25’ above the ground if the chains folded exactly in half. Capiche? ---
**Author:** ![impala61](https://avatars.discourse-cdn.com/v4/letter/i/77aa72/32.png) [@impala61](https://community.cartalk.com/u/impala61)\ **Post date:** [April 19, 2011, 4:15am UTC](https://community.cartalk.com/t/catenary/63816/11 "2011-04-19T04:15:10Z")
Remember what I told you about expecting to feel stupid? I want to say that’s not a catenary, but it’s probably a special case of a catenary. I was on the right track with my Pythagorean objection, but I still feel stupid. Thanks for spelling it out. ---
**Author:** ![Kevingets](https://avatars.discourse-cdn.com/v4/letter/k/e79b87/32.png) [@Kevingets](https://community.cartalk.com/u/Kevingets)\ **Post date:** [April 22, 2011, 9:03pm UTC](https://community.cartalk.com/t/catenary/63816/12 "2011-04-22T21:03:03Z")
uhmmm good hint, I think you are suggesting that the poles must be apart more than 75 feet. ---
**Author:** ![ericosman](https://sea1.discourse-cdn.com/cartalk/user_avatar/community.cartalk.com/ericosman/32/2382_2.png) [@ericosman](https://community.cartalk.com/u/ericosman)\ **Post date:** [April 23, 2011, 1:56am UTC](https://community.cartalk.com/t/catenary/63816/13 "2011-04-23T01:56:40Z")
The poles are 100 ft tall, so if the poles were RIGHT next to each other, then the 150 feet of chain would drop 75 feet from one pole, and then go 75 feet back up to the top of the other one. That means the (very) sharp V of chain is dangling 25 feet above the ground as requested in the puzzle. If you move the poles away from each other any distance what so ever, the bottom of the V starts rising up, so you better leave the poles right next to each other. QED. Do we have a winner ?