13 replies to this topic

[j.ermengard]

Im designing a new cable system which will extend almost 1.24mi long. terminals have an altitude of 157 ft and the other end is 60 ft. The system will join the mainland and an island.

I need also the assessment regarding the cost of the project.

Anybody want to help me. This is for my school work. any help is great.

Thanks

This post has been edited by j.ermengard: 11 January 2008 - 11:08 PM

[Lift Dinosaur]

How about a ferry boat? A 6500 foot span with an elevation difference of 97 feet ain't gonna cut it, IMHO.

Dino

[j.ermengard]

Lift Dinosaur, on Jan 12 2008, 04:23 PM, said:

How about a ferry boat? A 6500 foot span with an elevation difference of 97 feet ain't gonna cut it, IMHO.

Dino

its just a study of building new way of transportation in the area.

[Callao]

Yes, the catenary curve:

y = a(cosh(x/a)) = (e^(ax) + e^(-ax))/2a

where a = (T/P), and
T is the horizontal component of the tension (a constant) and
P is the weight per length unit.

The equation I found that relates tensile force to sag and rope weight is this:

T = (mL^2)/8S

Where T is tension,
m is the weight of the rope per unit length,
L is the horizontal span between the supports, and
S is the vertical sag.

http://answers.googl...dview?id=594421

Just a note, however, that this equation assumes an equal force at any point on the rope, which is not the case is your rope weighs something. Also, it doesn't account for the fact that one end of your rope is higher than the other. I couldn't find an equation for that. But this should give you a good ball-park estimate about how much tension you will need, to avoid taking your passengers swimming.

[Callao]

Say for instance that your rope is only about 4600 feet long (so that both ends are 60 feet above the water), and that you use a 5/8" diameter, stainless steel cable that weighs .715 lbs per linear foot, and that you will allow a 50-foot sag, putting the rope's lowest point 10 feet above the water.
Then we have:

T = ((.715 lbs)(4600 ft)^2)/(8)(50 ft) = (15,129,400)/(400) = 37,824 lbs.

Too bad; the breaking strength of a 5/8 stainless steel cable is 35,000 lbs, making this particular project unfeasible.
However, if you allow a 55-foot sag, putting your rope 5 feet above the water, your rope won't snap--unless you put somebody on it.
But hey, thanks for the problem, even though it doesn't seem that aerial tramways will work here. We enjoy your questions in the forum.

[AlphaBet]

You might be able to put a tower or two in the water, much like how bridge piers are built in the water.

[Callao]

AlphaBet, on Jan 15 2008, 11:27 AM, said:

You might be able to put a tower or two in the water, much like how bridge piers are built in the water.

That actually would be a good idea, considering that the length between supports is the killing factor (it's squared) when it comes to required tension. Two supports separating the rope into three segments would decrease the required tension by 3^2, or 9 times.
Did I figure that right?

If your transport is informal enough, you could put your two supports on two wobbly floaties.

[j.ermengard]

Actually this is great challenge.. can you suggest the possible tower altitude for the lift system? Small boats and yachts.

What about wind loads/disturbances?
Thanks for the help!!!

I have attach a map of the area

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•  Picture1.jpg (114.54K)
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[Callao]

My one course in Statics never made me much of a useful engineer, so I can't say I'm much of an expert on tower design. As for your question about wind disturbances? That would be solved with a little math and some knowledge on material strengths. The more interesting problem posed here might be with your submerged towers, of which I am no help. I was never much of a water guy anyway.

[zeedotcom]

Would something like the Peak to Peak Gondola at Whistler work? Put a couple of big/tall towers on each end and then let it sag a bit in between or possibly build it more like a funitel so that there are more wires to hang it from?

[Callao]

zeedotcom, on Jan 16 2008, 04:57 PM, said:

Would something like the Peak to Peak Gondola at Whistler work? Put a couple of big/tall towers on each end and then let it sag a bit in between or possibly build it more like a funitel so that there are more wires to hang it from?

Putting in two tall towers and allowing more sag will work. But for the original anti-sag model, it doesn't matter how many cables you use: if steel is steel and is only as strong as steel, more steel cable will not hold up any more of itself than a little steel cable. Its weight-to-strength ratio will be the same.

[j.ermengard]

the water level is very deep, how could this be?

[Peter]

This project could be of interst to you.

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•  gondola.JPG (91.12K)
  Number of downloads: 53

[j.ermengard]

here is the topography of the location..

check the water depth

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•  Picture2.JPG (101.34K)
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