# Calculate the force required to pull a bungee cord taut, creating a straight line

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fascinated
Summary:: How would I calculate the necessary force?

Let's take an example where the bungee is 1000 feet long and weighs 9 pounds.

How much force would be needed to pull it taught across a horizontal plane?

Thanks for any help.

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Lnewqban

## Answers and Replies

Staff Emeritus
If you are talking about an ideal cord, if there is any tension at all, it will be approximately taut.

If it is a physical cord, this is a problem that requires a lot of information about the cord.

If you want an exactly taut cord, the force is infinite.

scottdave, Murray_P, hutchphd and 1 other person
Mentor
The cord assumes a shape known as a catenary. Search the term to learn more and find the equation.

Murray_P, wrobel and vanhees71
fascinated
If you are talking about an ideal cord, if there is any tension at all, it will be approximately taut.

If it is a physical cord, this is a problem that requires a lot of information about the cord.

If you want an exactly taut cord, the force is infinite.
Hi Van,
Thanks for that and I get that.

I'm not sure what could be said about the cord, other than the manufacturers claim that it stretches to 100% of its original length.
The description is here:

• 1/8" diameter shock stretch bungee cord...
• Made in the USA with a Rubber core with 100% elongation (+/- 10%)
• Can hold up in harsh conditions and environmental factors

My problem here is that we actually just stretched the thing for just over a 1/2 mile, or 2640 feet, and there was still a seemingly tremendous amount of sag.

I wanted to keep going, but the wife got scared. We were next to a highway.

I am now wondering if it is even possible. I am hypothesizing that the cord snaps prior or just after being pulled *exactly* taught. Does this sound right?

The cord assumes a shape known as a catenary. Search the term to learn more and find the equation.
This is pretty much what I am talking about, but is there no force great enough to pull a 9 pound cord taught?

Am I missing something?

Gold Member
2022 Award
This is pretty much what I am talking about, but is there no force great enough to pull a 9 pound cord taught?
By "taught (sic)" do you mean "straight"? If so, then @Vanadium 50 has already answered your question.

Dr.D
Am I missing something?
I think you are missing gravity operating on the cord itself.

Gold Member
2022 Award
Am I missing something?
I think you are missing a force diagram that would show you what's going on.

Mentor
There would be a tremendous amount of potential energy in a taut cord or even one with a catenary droop. There's been some spectacular accidents when a cord breaks and I wouldn't want to be around when it did.

Think about garage door springs when they break they can easily take out your car window. That's why now the safety standard into always run a cable through the spring to prevent that kind of accident.

You could do your experiments on a smaller piece of cordage using weights to stretch it downward.

Here's video on bungee cord testing that can give you some idea what happens when it breaks:

or this one using expendable tv hosts as human subjects:

and some pointers should you decide to be the test subject:

Mentor
I think a few of the prior posters have missed the OP's specification that the cord is being "pulled across a horizontal plane". In other words, it's being pulled across the ground.

I'll repeat @Vanadium 50 's answer: what happens here depends on the very specific specifications of the cord.
My problem here is that we actually just stretched the thing for just over a 1/2 mile, or 2640 feet, and there was still a seemingly tremendous amount of sag.
Meaning it is dragging on the ground? How much it sags has very little to do with how far you stretch it when it is dragging on the ground. Please explain why this is an issue.
...the manufacturers claim that it stretches to 100% of its original length...
Ok. I know you said "my problem is..." but really you haven't said what the actual problem is you are trying to solve. So far you have proven only that the manufacturer of this particular cord has included a safety factor in its advertised specifications. Who cares and what does that have to do with why you are stretching it so far?

If your goal here is simply and only to identify exactly what the difference between the advertised and actual performance is, well, I don't see a lot of value in that. But I do see a lot of risk. I would think an elastic cord stretched for a half mile would be storing plenty of energy to kill someone. So please provide more information about what you are doing and why.

phinds
Gold Member
2022 Award
I think a few of the prior posters have missed the OP's specification that the cord is being "pulled across a horizontal plane". In other words, it's being pulled across the ground.

fascinated
Thank-you everyone who popped in. I don't think I was clear. Let me try again.

What I don't understand is this.

This is clearly possible:

When I see that, I see a straight line. I don't see the catenary. Is it still there, but so small that it can not be seen or is that a straight line?

And if that is a straight line, then theoretically is it possible to do the same thing over longer distances.

And if it is theoretically possible to do that over a long distance, is there a way to demonstrate this in reality?

And if the bungee cord does not have the tensile strength to mass ratio to achieve this, would anyone know of a material strong enough and light enough to accomplish my goal of creating a straight line over distance?

Or is that simply a mathematical impossibility regardless of the mass to tensile strength ratio?

Or am I looking at this all wrong still?

Thanks again for any help, folks.

Gold Member
2022 Award
OK, now you're OFF of the "across a horizontal plane" and into the open air. That question has already been answered. Several times.

russ_watters
Mentor
It’s not possible unless it’s done in space where there’s no gravity at all And even there one could argue that there is still some curvature however slight due gravity being present everywhere.

Shane Kennedy
Staff Emeritus
So far you have proven only that the manufacturer of this particular cord has included a safety factor in its advertised specifications.
Is it safety?

It's on a bridge 2100 feet in the air. A coyote ties it on himself figuring he has 100 foot of margin. But the cord has an extra 10% too:

As an academic question, yes, you never get completely horizontal. As a practical question, seeing how far it will stretch while you ask your wife to "hold my beer" does not seem very safe to me.

hutchphd, jedishrfu and russ_watters
Homework Helper
Or is that simply a mathematical impossibility regardless of the mass to tensile strength ratio?
Yes. If the line were exactly horizontal, the force from tension would be purely horizontal and there would be no vertical force to support the cord in the middle of the span.

Assume that 100% elongation is achieved at the nominal 100 pound working strength of the 1/8 inch bungee cord and that 1000 feet (unstretched) bungee weighs 9 pounds. Then each end of the span needs to support half the weight of the cord -- 4.5 pounds each. The sag angle at both ends must then be ##\sin^{-1} \frac{4.5}{100}##. That is about 2.5 degrees. But we do not need the angle. We'll be taking its sine in a moment anyway. So we just need the sine: ##\frac{4.5}{100}=0.045##

If we imagine this cord extended to a stretched length of 2000 feet in a perfect V then the sag at the bottom of the V would be 1000 * 0.045 = 45 feet. But since the actual span will be a curve, we can guesstimate a true sag of about half that. Call it twenty or twenty-five feet.

fascinated
Yes. If the line were exactly horizontal, the force from tension would be purely horizontal and there would be no vertical force to support the cord in the middle of the span.

Assume that 100% elongation is achieved at the nominal 100 pound working strength of the 1/8 inch bungee cord and that 1000 feet (unstretched) bungee weighs 9 pounds. Then each end of the span needs to support half the weight of the cord -- 4.5 pounds each. The sag angle at both ends must then be ##\sin^{-1} \frac{4.5}{100}##. That is about 2.5 degrees. But we do not need the angle. We'll be taking its sine in a moment anyway. So we just need the sine: ##\frac{4.5}{100}=0.045##

If we imagine this cord extended to a stretched length of 2000 feet in a perfect V then the sag at the bottom of the V would be 1000 * 0.045 = 45 feet. But since the actual span will be a curve, we can guesstimate a true sag of about half that. Call it twenty or twenty-five feet.
Thank-you, I'm getting it, and bungee is clearly not going to work.

I am wondering if we attempted the same thing with fishing braid that does not stretch (or just barely); how much we could reduce the catenary sag.

Could we calculate the amount of sag if we used this for 5000 feet?
https://www.amazon.com/dp/B0015NA36C/?tag=pfamazon01-20

Am I correct to assume that the sag would be far less?

I am also seeing that the idea of "bar tight" comes up, which seems to occur when the line is straight or almost straight, and then fails/breaks. Does that sound right? Does it become straight for even a moment before breaking?

Thanks again for your help.

Homework Helper
Does it become straight for even a moment before breaking?
No. No amount of force can flatten the line. It is impossible.

If you want to convert this from a mathematical impossibility into an engineering challenge, you will have to reveal what you are actually trying to do and how close to "flat" might achieve the goals that you have not shared with us.

russ_watters and Bystander
fascinated
No. No amount of force can flatten the line. It is impossible.

If you want to convert this from a mathematical impossibility into an engineering challenge, you will have to reveal what you are actually trying to do and how close to "flat" might achieve the goals that you have not shared with us.
It is needed for part of a massive, outdoor art installation.

And the engineering challenge would be to get as close to a horizontal line as possible.

Homework Helper
the engineering challenge would be to get as close to a horizontal line as possible.
No. That is not a viable challenge. You want to get as close to horizontal as is practical while achieving some other, still unstated, aesthetic goals. How much load do you need the "horizontal" line to support?

Edit: if the goal is aesthetic and the viewing distances are large then a light, fat, highly visible cord could be supported in a quite flat configuration by a thinner and less visible cord with a sag.

For safety you want low stretch. Otherwise, that factor is irrelevant. All you care about is tensile strength versus density per unit distance. You can get up to 7000 MPa for carbon fiber. See the table here which also gives numbers for density. You probably do not want to spring for the next step up -- carbon fiber nanotubes.

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russ_watters
Gold Member
A fishing line, being both light and strong, might get you pretty close to horizontal, but over the kind of distance you're talking about, there would still be measurable sag. Piano wire might be even better as it will have even less stretch than fishing line.

Just taking a shot in the dark here, but would this have anything to do with directly measuring the curvature of the Earth?

Gold Member
2022 Award
Could we calculate the amount of sag if we used this for 5000 feet?
Do you seriously want to stretch a cord ALMOST A MILE, and up in the air? What kind of "art installation" are we talking about here? This is all beginning to sound ridiculous. If you are serious and actually want help, tell us EXACTLY what it is you are planning, including what kind of stuff is going to be hung from the cord.

jbriggs444
Homework Helper
Could we calculate the amount of sag if we used this for 5000 feet?
This 5000 feet contemplated braided fishing line rated at 150 pounds. The quoted ad indicated 10 pounds weight per 500 yards. We are talking about 5000 feet so that's about 33 pounds. Not a very good ratio of tensile strength to line weight (the longer you make the span, the worse that ratio gets).

The catenary formulas I've seen scare me. So I'll just use the back of an envelope again. The sine of the angle at each end will be ##\frac{16.5}{150}= 0.11##. (6 degrees below the horizontal). Multiplying by 2500 feet, that is a sag of about 275 feet for a Vee shape. Divide by two for a near-parabolic catenary and that's a sag of about 140 or 150 feet for a mile of braided fishing line.

russ_watters and Bystander
Homework Helper
Gold Member
Divide by two for a near-parabolic catenary and that's a sag of about 140 or 150 feet for a mile of braided fishing line.
Starting to see how it's done?

fascinated
No. That is not a viable challenge. You want to get as close to horizontal as is practical while achieving some other, still unstated, aesthetic goals. How much load do you need the "horizontal" line to support?
Yes, practical, sorry and the idea was to hang light, but visible ribbons, but based on your edit, that might not matter.
Edit: if the goal is aesthetic and the viewing distances are large then a light, fat, highly visible cord could be supported in a quite flat configuration by a thinner and less visible cord with a sag.
Yes! Can you elaborate here? How can a highly visible cord could be supported in a quite flat configuration and how close to horizontal would the end result be?

For safety you want low stretch. Otherwise, that factor is irrelevant. All you care about is tensile strength versus density per unit distance. You can get up to 7000 MPa for carbon fiber. See the table here which also gives numbers for density. You probably do not want to spring for the next step up -- carbon fiber nanotubes.
Any idea if 5000 feet of this can be purchased? I checked and couldn't find more than 100 meters and it seemed very expensive.

And if we went this route, how much would we reduce the sag?

Thanks again

Homework Helper
Yes! Can you elaborate here? How can a highly visible cord could be supported in a quite flat configuration and how close to horizontal would the end result be?
Work it like a suspension bridge. You have an sagging structural cable and run vertical support wires down to a level deck. Paint the structural cable and the support wires blue to reduce their visibility in day. Or black to reduce their visibility at night.

You may have problems with wind load. But this design is worth what you paid for it.

fascinated
Work it like a suspension bridge. You have an sagging structural cable and run vertical support wires down to a level deck. Paint the structural cable and the support wires blue to reduce their visibility in day. Or black to reduce their visibility at night.

You may have problems with wind load. But this design is worth what you paid for it.
I assumed this is where you were going and if I understood your previous points, this would mean that the support line would have to be somewhere above 150 feet in the air?

A fishing line, being both light and strong, might get you pretty close to horizontal, but over the kind of distance you're talking about, there would still be measurable sag. Piano wire might be even better as it will have even less stretch than fishing line.
Any idea for how much sag with piano wire and if these lengths are accessible?

Just taking a shot in the dark here, but would this have anything to do with directly measuring the curvature of the Earth?
Lol, can you do that?

Do you seriously want to stretch a cord ALMOST A MILE, and up in the air?
Ideally, yes.

What kind of "art installation" are we talking about here? This is all beginning to sound ridiculous. If you are serious and actually want help, tell us EXACTLY what it is you are planning, including what kind of stuff is going to be hung from the cord.
It is a bit ridiculous, lol. But, nothing needs to hang except the for the possibility of bright ribbons to make the line more apparent. The horizontal line is for reference, to 'frame the scene,' if you will. Individual sculptures will be positioned relative to that line after considering multiple viewing positions.

Starting to see how it's done?
Indeed, y'all are great.

Thanks again

Gold Member
Lol, can you do that?
No. Because you'd need a straight cable with no sag and you won't be able to do that.
(The better approach would be to use a leveled laser.)

Homework Helper
Any idea for how much sag with piano wire and if these lengths are accessible?
At 3 millimeters, steel wire has a tensile strength around 1500 MPa according to this manufacturer. Steel has a specific gravity of about 8 (8 kg/liter or 8000 kg/m3).

We should be able to use these figures to determine both the tensile strength of a 3 millimeter steel wire and the linear density of the same.

1500 MPa is 1500 million Newtons per square meter. We are talking about a circular cross-section with a radius of 1.5 mm. ##a = \pi r^2## so that is 0.000007 square meters (seven millionths of a square meter). Multiplied by 1500 so we have about 10,000 Newtons.

But you are more comfortable with imperial units. So call that one ton. 2000 pounds, more or less.

Meanwhile, a cross section of seven millionths of a square meter multiplied by about 1.5 kilometers should give us one one hundredth of a cubic meter. Or about 80 kg. Call it 160 pounds mass. Which under one gee is also 160 pounds force. (Yay, imperial units can be handy -- once in a blue moon).

So now we have the sine of our sag angle: 160/2000. Multiply by 2500 and we have a sag of 200 feet fopr a Vee shape. Divide by 2 for the catenary curve and we have a sag of 100 feet, give or take.

Really, you should start doing these calculations yourself. You learn more by working problems than by asking for finished answers.

I am a bit surprised that steel does this poorly. Googling at random, I found some 1/8 inch kevlar cord with a polyester jacket. 825 pound breaking strength. I did not see density figures. But 3600 MPa with a specific gravity of 1.4 is a heck of a lot better than steel (1500 and 8 as indicated above).

Gold Member
It is a bit ridiculous, lol. But, nothing needs to hang except the for the possibility of bright ribbons to make the line more apparent. The horizontal line is for reference, to 'frame the scene,' if you will. Individual sculptures will be positioned relative to that line after considering multiple viewing positions.
How high off the ground will this cord be strung?

Homework Helper
Measuring curvature of the earth.
Lol, can you do that?
Over a one nautical mile span, the Earth curves by one minute of arc (1/60th of one degree). The idea would be to stretch a cord straight enough that we could compare the curvature of the Earth against the straightness of the cord.

So far we've been talking about sag angles in the area of one to ten degrees (60 to 600 minutes of arc) for the two ends of a one mile span of cord. One would have to step this game up by two or three orders of magnitude just to get close.

Gold Member
Using the original numbers of 9 lbs. per 1000 ft, converting to metric units, I get that a one mile cord (1.6km) sags 12m with the tension equal to 825 lbs or 3670N. To get the sag down to 0.12m or about 5 inches the tension would rise 100 fold to about 37 metric tonnes. Of course it would break long before that. Including the 100% stretch might halve that but I think it's still way beyond the bungee cord specs.

My conclusion is that you will have to find another way to create the illusion of framing these sculptures. I like @DaveC426913's idea of using laser light. It reminds me of this artist using laser beams to make 'intergalactic sculpture'. You could literally use smoke and mirrors.

http://maarav.org.il/english/2018/04/17/intergalactic-sculpture-1986-2001/

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Staff Emeritus
It would really help if you were to stop messing with us and tell us what the problem you are trying to solve is and not make us guess.

You should also take a look at telephone or electrical wires. Ever see them strung between poles a mile apart? There's a reason for that.

phinds
fascinated
You guys are great. I have been googling away, learning about lasers a bit, and don't understand how the beam would be visible over that distance? Wouldn't it just project a dot on the other side, wherever it hits?

And a video of this guy shooting a laser over a lake popped up. He claims the laser was four feet above the water and his cell phone picked it up from just inches above the water on the other side.

Measuring curvature of the earth.
Over a one nautical mile span, the Earth curves by one minute of arc (1/60th of one degree).
What would it be over that distance? He claims 16.4 miles.

Is this a trick?

It would really help if you were to stop messing with us and tell us what the problem you are trying to solve is and not make us guess.

You should also take a look at telephone or electrical wires. Ever see them strung between poles a mile apart? There's a reason for that.
I'm not messing with anyone. I'm just curious and not super smart, and I appreciate everyone who takes the time to explain things that must seem so obvious to them.

And no, I have never seen that, but when I googled it, I found this:
The Ameralik Span is the longest span of an electrical overhead powerline in the world. It is situated near Nuuk on Greenland and crosses Ameralik fjord with a span width of 5.37 km (3.34 miles).