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Q: How much force to keep a cable horizontal and taut at a given length? ( Answered ,   6 Comments )
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 Subject: How much force to keep a cable horizontal and taut at a given length? Category: Science Asked by: rleitzell-ga List Price: \$20.00 Posted: 17 Nov 2005 17:28 PST Expires: 17 Dec 2005 17:28 PST Question ID: 594421
 ```I'm an artist working on an installation of photos. For the installation I need to stretch two cables horizontally for a great distance and then clip photos snugly between the two. For aesthetic reasons it is very important that these cables can maintain a true (or very close to true) horizontal level throughout their length. The total length is approximately 1,000 feet, and I would like to know how often I will need to add new vertical brace units to keep the cable vertical and taut, and what the formula would be to calculate (or approximate) that number. Thanks!```
 Subject: Re: How much force to keep a cable horizontal and taut at a given length? Answered By: redhoss-ga on 17 Nov 2005 19:14 PST Rated:
 ```Hello rleitzell, thanks to our favorite search engine's new book scanning project I was able to find your answer. http://books.google.com/books?hl=en&id=qF3AxLlv1o4C&pg=PA196&lpg=PA196&dq=tension+in+horizontal+cable&prev=http://books.google.com/books%3Fq%3Dtension%2Bin%2Bhorizontal%2Bcable%26lr%3D&sig=2tJuXZa1c9TW__H34ISCfCs0X_0 The formula you need is: f = (w x L^2) / (8 x H) Where: f is the vertical sag w is the weight of the cable per unit horizontal length L is the horizontal span between supports H is the horizontal component of the cable tension You don't say what you plan to use for cable. I will choose 3/8 IWRC from this table: http://www.wrca.com/wr_619_636.html Looking in the table 3/8 IWRC weighs .26 #/ft. So that we will end up with sag in inches we need to use weight in #/in. or .26/12 = .022 #/in. Also, from the table the minimum breaking strength is 6.56 tons x 2,000 #/ton = 13,120# So, if we tension the 3/8 cable to 13,120# we can calculate the expected sag in a 100 foot = 1,200 inch span. f = .022 x 1,200^2 / 8 x 13,120 = 0.3 inches Notice that the sag increases as the square of the span. So, increasing the span by a factor of two you will get 4 times the sag. If you want to choose a cable size and type, ask for a clarification and I will calculate the sag for you. Or, you may want to do the calculation yourself. Either way I will be glad to answer any question you might have. Good luck with your art project, Redhoss```
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 Subject: Re: How much force to keep a cable horizontal and taut at a given length? From: qed100-ga on 17 Nov 2005 17:50 PST
 `Do you mean 1,000 non-stop feet of horizontal cable?`
 Subject: Re: How much force to keep a cable horizontal and taut at a given length? From: myoarin-ga on 17 Nov 2005 18:08 PST
 ```I expect that you will have to use trial and error to solve this. You will definitely need many vertical supports if you want the photos to hang without noticable gaps at the bottom. Sounds like the project would be very sensitive to wind.```
 Subject: Re: How much force to keep a cable horizontal and taut at a given length? From: mechestang-ga on 17 Nov 2005 21:07 PST
 ```3 Things: 1: Don't forget to include the weight of the pictures into the equation above for the top cable (if they are hung) or half the weight of the pictures into each cable if they are attached to both cables (ie the weight of the pictures is evenly distributed into each cable). 2: Allow for adjustment, the cables WILL stretch. 3: You may also want to consider a substantial safty factor (sometimes called design factor). It is not wise to load the cable to its yield strength (breaking strength) with people around. A safty factor of 2 means that you would pretend the yield strength is half of the book value (6.56 tons for 3/8 cable according to redhoss). So your maximum tension allowable would become 3.28 tons with a SF of 2 and 2.18 tons with a SF of 3. If this is displayed outside, exposed to the elements, you must also worry about corrosion, increase your safty factor accordingly.```
 Subject: Re: How much force to keep a cable horizontal and taut at a given length? From: xplusak-ga on 21 Nov 2005 09:58 PST
 ```For aesthetic reasons the total sag at center should be a FRACTION of vertical dimensions of the pictures. Therefore in my opinion it is an impossible task with a set of two wires over 1000 ft. A set of three wire scenario will give a perfectly horizontal layout of pictures. To visualise this just imagine 1. any suspension bridge most notably Golden Gate Bridge and its twin decks http://www.goldengatebridge.org/ 2. Overhead wires on Railway lines - catenary http://en.wikipedia.org/wiki/Image:Tile_Hill_train_550.jpg The catenary wire size reduces drastically if catenary profile is allowed to be deep. To add aesthetics to the scene, the drop wires can be transparent and very thin as in fishing lines.```
 Subject: Re: How much force to keep a cable horizontal and taut at a given length? From: myoarin-ga on 21 Nov 2005 14:18 PST
 ```Xplusak-ga, That certainly sounds like a better solution that vertical supports, especially when one considers that the underlying ground will almost surely not be perfectly level, so that adjusting their heights to provide support for a "perfectly" straight cable would be extremely difficult. Of course, adjusting the droplines would also be a tedious project, and the "pylons" for the catenary cable and their support will also be a problem. 1000 meters - we are talking about a major suspension bridge span. And you have 1000 meters of photos ...? (I am gaining respect for Christo!) If the project does materialize, I hope the construction reaps appropriate recognition.```
 Subject: Re: How much force to keep a cable horizontal and taut at a given length? From: xplusak-ga on 23 Nov 2005 01:21 PST
 ```First, It is 1,000 ft ... not 1,000 meters. This simplifies the problem by a factor 3 raised to power ... something ... > 2; I mean substantially. That difference keeps the probelm into light engineering domain as against heavy engineering project. Second, Drop lines height adjustment is to be done on lower end of drop lines. Their initial length (estimate) can be calculated once the profile of catenary is fixed (designed). So adjustment happens at the same level where you hang the pictures, I mean no issues about accessibility. Fine tuning of drop lines heights to get a horizontal layout of pictures will be necessary otherwise it gets complex. Third, Even after considering the catenary type design, the cable for hanging the picture need be under good tension, if number of drop lines are to be limited, else this cable can be same as fishing line used for drop lines. Fourth all aspects discussed above can be put under calculations. Finally, one aspect missing from original question - it is outdoors? You have to consider winds and it get murky there.```