
 George,Y
 Super Member
The Shape of a Suspension Bridge's Main Cable
Aesthetic, light, and strong, suspension bridges can span distances from 2,000 to 7,000 feet  far longer than any other kind of bridge. They also tend to be the most expensive to build. True to its name, a suspension bridge suspends the roadway from huge main cables, which extend from one end of the bridge to the other. These cables rest on top of high towers and are secured at each end by anchorages.
The towers enable the main cables to be draped over long distances. Most of the weight of the bridge is carried by the cables to the anchorages, which are imbedded in either solid rock or massive concrete blocks. Inside the anchorages, the cables are spread over a large area to evenly distribute the load and to prevent the cables from breaking free.
Suspension bridge Golden Gate Bridge, San Francisco, CA See More at This Page
Now my question is, what's the shape of a suspension bridge's main cable in between the two towers? I got an answer from my rough model. And one of the two alternatives is my real answer. A a PARABOLA B an ARC
Of course you may have your own answer. Anyway, say your answer and demonstrate it in this topic
Last edited by George,Y (20060404 22:57:25)
X'(yXβ)=0
 George,Y
 Super Member
Re: The Shape of a Suspension Bridge's Main Cable
Actually, my friend disagrees with my answer, thus A and B are our answers.
And the shape should be like a parabola or an arc, not exactly for sure.
Different people, different models, so Welcome Your Model
X'(yXβ)=0
Re: The Shape of a Suspension Bridge's Main Cable
I would have assumed the shape would be a catenary. But it appears I am wrong:
See here
Last edited by jchristophm (20060403 03:58:57)
 George,Y
 Super Member
Re: The Shape of a Suspension Bridge's Main Cable
Thank you for your enthusiasm, jchristophm. It's really natural to think it as a catenary. (actually it was also my instinct) Myorigenal model (edited): My model is to cut the whole by the same horizontal distance, Δx, and assume that cablepieces are straight, and that the concret road pieces are don't support each other. It finally derives the tangents of the cable pieces develop by 0, k, 3k, 5k, 7k...to each side, where k is an unknown constant related to cable flexibilty. And that's similar to the motion that an arrow does without air resistance.
The solution from your link is more advanced.
Still, to be exact, allow me to say that it's only near parabola for incontinousity.
Any way, next time we see a SB, we will enjoy its beauty and know its cables rest almost parabolas.
Last edited by George,Y (20060404 23:00:30)
X'(yXβ)=0
 George,Y
 Super Member
Re: The Shape of a Suspension Bridge's Main Cable
Why is my uploaded image so small ??? Anyone help me ...
X'(yXβ)=0
Re: The Shape of a Suspension Bridge's Main Cable
George,Y wrote:Still, to be exact, allow me to say that it's only near parabola for incontinousity.
I don't quite understand why you would say "near parabola". With the assumption that each "piece" of cable supports the weight of the bridge directly beneath it you get a function x^2. And this assumption is born out of engineering reality.
Re: The Shape of a Suspension Bridge's Main Cable
George said "Why is my uploaded image so small ??? Anyone help me ...". Answer: Click on the tiny picture and it gets bigger. This confused me once too.
I think George said nearparabola because the weight of the cables throws it off slightly, but probably only millimeters.
igloo myrtilles fourmis
 George,Y
 Super Member
Re: The Shape of a Suspension Bridge's Main Cable
The discontinuity:
n,m,g,W,+ if we name the mid verticle cable the 0th verticle, we can also name all five verticle cables 2nd verticle, 1st verticle, 0th verticle, 1st verticle and 2nd verticle. and then we can define on each Δx is distributed with same mass m. g represents acceleration due to earth gravity.
1/2mg=def=W hence mg=2W
at the upper end of nth verticle(the triknit), forces are always 3, one downward(gravitional force), one leftward, and one rightward. define horizontally rightward and verticly upward +,and use cartesian expression of vectors. Using Newton's 1st and 2nd Law,we will easily get the table below.
nth 0 1 2 3 (not displayed in graph) leftwardforce (h*,W) (h,W)# (h,3W) (h,5W) *h is unkown rightwardforce(h,W)# (h,3W) (h,5W) (h,7W) LF tangent  W/h 3W/h 5W/h heightincreament (W/h)Δx 3(W/h)Δx 5(W/h)Δx height 0(set) (W/h)Δx 4(W/h)Δx 9(W/h)Δx distance 0(set) Δx 2Δx 3Δx
# (h,W) and (h,W) are anti each other (Newton's 2nd Law) at nth knit, LF+RF=2W, always(from 1st Law) hence height= C distance², ignoring measure
but if the main cable behave more like a soft rope rather than a solid dome, it will bend more at the knits, thus not perfect parabola.
Last edited by George,Y (20060405 23:07:43)
X'(yXβ)=0
 George,Y
 Super Member
Re: The Shape of a Suspension Bridge's Main Cable
cannot enlarge them in posts don't know how others could do it
X'(yXβ)=0
Re: The Shape of a Suspension Bridge's Main Cable
Alternatively, you can upload an image to imageshack or another free hosting site, and then put it here with the [img] tag.
Why did the vector cross the road? It wanted to be normal.
 George,Y
 Super Member
Re: The Shape of a Suspension Bridge's Main Cable
Many thanks, mathsyperson
X'(yXβ)=0
 George,Y
 Super Member
Re: The Shape of a Suspension Bridge's Main Cable
an alternative model has been developed
A lot of films have such a horrible scene: heros dashing on a broken rope bridge, and luckily reaching the end.
The rope , similar to the cable of a suspension bridge, should break near one end, rather than in the mid point you may verify this.
Last edited by George,Y (20060405 22:30:13)
X'(yXβ)=0
 George,Y
 Super Member
Re: The Shape of a Suspension Bridge's Main Cable
NOTICE the KNOTS (corrected), they are streched downward.
Franklin's guess is right, too. Even the knits doesn't form a parabola but a parabolacatenary mix.
For Post 12, you cannot get a parabola knits trajectory. the tangent (slope) grows by 0, 2, 4, 6 height grows by 0, 2, 6, 10 =n(n+1)
length l= (n1/2)Δx set Δx=1 n=l+1/2 n(n+1)= (l+1/2)(l+3/2) the function of right knits isn't symmetric about the axis in the graph, thus left knits have another function (l1/2)(l3/2) Two parabolas
Last edited by George,Y (20060407 11:22:53)
X'(yXβ)=0
Re: The Shape of a Suspension Bridge's Main Cable
Ok.
But if we only plotted the points were hanger meets main cable, then we have a parabola.
But if we had a bridge that had only one hanger that was planar across the entire main cable, then it would be parabolic.
Re: The Shape of a Suspension Bridge's Main Cable
jchristophm wrote:But if we had a bridge that had only one hanger that was planar across the entire main cable, then it would be parabolic.
Of course, then the massive weight of this planar hanger is going to have to be considered.
 krassi_holmz
 Real Member
Re: The Shape of a Suspension Bridge's Main Cable
There's an interesting math function, called chain fnction. It's exactly the hyperbolic cosinus.
IPBLE: Increasing Performance By Lowering Expectations.
 krassi_holmz
 Real Member
Re: The Shape of a Suspension Bridge's Main Cable
A plot:
Last edited by krassi_holmz (20060407 04:02:24)
IPBLE: Increasing Performance By Lowering Expectations.
 George,Y
 Super Member
Re: The Shape of a Suspension Bridge's Main Cable
But if we had a bridge that had only one hanger that was planar across the entire main cable, then it would be parabolic.
Yes, you could say so. but to be exact, near. because matter is made of sufficient smalls instead of continuous infinite smalls. My point is, most of caculus's success is built on descrete approximation.
Last edited by George,Y (20060407 11:30:13)
X'(yXβ)=0
 mikau
 Super Member
Re: The Shape of a Suspension Bridge's Main Cable
This is an awsome problem! :D
A logarithm is just a misspelled algorithm.
 George,Y
 Super Member
Re: The Shape of a Suspension Bridge's Main Cable
Thank you for your appreciation. Thank you krassi_holmz for your reference.
Let's take a break and appreciate some history.
The first modern suspension bridge ever built is Brooklyn Bridge, which connects Manhatten and Brooklyn, New York. The idea came from a German immgrant engineer, John Augustus Roebling . Avant as he was, he prepared 2 years from 1867 for every detail before start, and worked for 14 more years before he died of an infection.
His son Washington, also an engineer and bridge builder, took over his father's dream. He and his wife cosi[ervised the construction and completed the bridge.
The Brooklyne Bridge
Total Span: (Measures the distance between the two anchorages.) 3,455 feet Main Span: (Measures the distance between the two towers.) 1,595 feet Height of the Towers: 276 feet Engineer(s): John Roebling, Washington A. Roebling Cost: $15 million
Washington and his wife
sourced from this page
More details in suspension bridge development
Last edited by George,Y (20060410 11:21:10)
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