# Hanging Cable Problems -- Catenary Math

TR Smith is a product designer and former teacher who uses math in her work every day.

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If a cable, chain, rope, or string is flexible and has a uniform density, then the shape of the curve it makes when you suspend the rope between two points is called a catenary. This name is derived from the Latin catena, meaning chain. Superficially, a catenary looks like a parabola when the sag is relatively small. The true equation for a catenary in Cartesian coordinates is

f(x) = (1/c)cosh(cx)
= [e^(cx) + e^(-cx)]/(2c)

where "c" is some constant that governs the shape of the curve. The function "cosh" is called the hyperbolic cosine. Mathematically, catenaries have many surprising properties. For example, it is one of the few curves you can find the arc length of exactly, since the function sqrt(1 + f'(x)^2) is integrable, something that isn't true in general for most functions f(x). And surprisingly, the weight of the rope makes no difference in the shape. Ten meters of sewing thread hung between two posts 7 meters apart makes the same shape as ten meters of iron chain hung between the same posts!

In any catenary or hanging cable problem, there are three quantities to consider: the length of the cable L, the distance between the two fixed ends D, and the amount of sag S. Using calculus, if you know two of the variables you can compute the third.

For example, you can determine what length of cable, rope, or chain you need to produce a given amount of sag given the distance between the two ends of the cable. Or, given the distance between the ends and the length, you can mathematically compute the sag. Here is a derivation of the equations needed along with some example problems.

## Relating D, L, and S

To find a mathematical relation among D, L, and S, we need to consider the coordinate equation of a catenary with given distance, length, and sag. If a catenary has the equation f(x) = (1/c)cosh(cx), then its derivative is f'(x) = sinh(cx), where sinh is the hyperbolic sine. The arc length integral formula gives us

∫ sqrt[1 + (sinh(cx))^2] dx
= ∫ sqrt[cosh(cx)^2] dx
= ∫ cosh(cx) dx
= (1/c)*sinh(cx)

To find the arc length L between x = -D/2 and x = D/2, we plug the endpoints into the antiderivative and subtract, this gives us

L = (1/c)*sinh(cD/2) - (1/c)*sinh(-cD/2)
= (2/c)*sinh(cD/2)

The sag is the difference in height between the ends and the middle, so we have

S = f(D/2) - f(0)
= (1/c)*cosh(cD/2) - (1/c)cosh(0)
= (1/c)*cosh(cD/2) - 1/c

Now let's simplify the two equations we obtained that relate the variables D, L, S, and c and call them Equations 1 and 2:

Equation 1: cL/2 = sinh(cD/2)
Equation 2: cS = cosh(cD/2) - 1

Even though we ended up with a fourth variable c to account for, we could still use these two equations to solve catenary problems. However, if we do more algebraic manipulations on the equations, we can get more out of them. By using the relation cosh(x)^2 - sinh(x)^2 - 1, we can derive a third equation without hyperbolic functions:

Equation 3: c = 8S/(L^2 - 4S^2)

And finally, plugging this expression for c into the first or second equation and solving for D gives us an equation that just involves the three variables D, L, and S:

Equation 4: D = [(L^2 - 4S^2)/(4S)]*log[(L+2S)/(L-2S)]

where log is the natural logarithm, also written as ln or LN. This last equation gives us an exact value of D when L and S are known, but it cannot be solved algebraically for L or S. We can now use these four equations to solve a variety of hanging rope/cable/chain problems. For catenaries that do not deviate much from a straight line, i.e., the sag is negligible, you can also use simpler approximation formulas to find the measurements of D, L, and S.

## Example Problem 1: Finding S Given D and L

A cable has a length of 10 meters and is hung between two poles set 7 meters apart. How much does the cable sag?

In this problem we have L = 10 and D = 7. The first equation gives us

cL/2 = sinh(cD/2)
5c = sinh(3.5c)

This equation cannot be solved algebraically, but it can be solved with any level of precision using a numerical solving algorithm such as Newton's Method, or with the numerical solving feature on any graphing or advanced scientific calculator. Using a solver gives us c ≈ 0.4327014.

Now we can plug this value of c into the second equation to find S. This gives us

cS = cosh(cD/2) - 1
0.4327014*S = cosh(1.514455) - 1
S = [cosh(1.514455) - 1]/0.4327014
S = 3.197205

Therefore, the cable hangs or sags by about 3.2 meters.

## Example Problem 2: Finding D Given L and S

You have 12 meters of rope. If you hang it between two posts, how far apart must they be so that the sag is 4 meters?

Here we have L = 12 and S = 4. Using the fourth equation, we can plug L and S directly into it to compute D without having to compute c.

D=[(L^2-4S^2)/(4S)]*log[(L+2S)/(L-2S)]
= [(144 - 64)/16]*log[(12+8)/(12-8)]
= (80/16)*log(20/4)
= 5*log(5)

The approximate value of 5*log(5) is 8.0472. This means you need to place the ends of the rope about 8.05 meters apart.

## Example Problem 3: Finding L Given D and S

You want to hang a chain between two poles that are 8 meters apart. How long should the chain be so that the vertex of the curve is 0.9 meters lower than the ends?

In this catenary problem we have D = 8 and S = 0.9. Plugging these values into the fourth equation gives us

D = [(L^2 - 4S^2)/(4S)]*log[(L+2S)/(L-2S)]
8 = [(L^2 - 3.24)/3.6]*log[(L+1.8)/(L-1.8)]

This cannot be algebraically solved for L, so using a solver tool we find L = 8.2639. Therefore, the chain needs to be about 8.26 meters long to produce 0.9 meters of sag with the poles are 8 meters apart.

## Upside Down Catenaries and Catenary Arches

Click thumbnail to view full-size
The Gateway Arch in St. Louis, Missouri is an inverted flattened catenary. (Wikimedia)
Catenary Arch at La Pedrera - Casa Mila. Courtesy of Tripadvisor.com | Source
Keleti railway station in Budapest, Hungary. (Wikimedia)
Sheffield Winter Garden in Sheffield, UK. (Wikimedia)

Inverted catenaries are popular shapes for arches because they are stable and strong. The Gateway Arch in St. Louis is not actually a true catenary because its equation is not of the form f(x) = (1/c)cosh(cx). Its form is instead f(x) = b*cosh(cx), where b does not equal 1/c. This can be called a flattened catenary.

## Scalability of Catenaries

One of the most convenient things about catenaries is that they are scalable. Suppose a catenary has a width of D, an arc length of L, and sags by S. If you double (or triple) any two of the three variables, then the third variable is also doubled (or tripled). This means if you know the ratio of L to D, you can find the ratios S to D and S to L even if you don't have the absolute numbers. You can solve such problems simply by setting one variable equal to 1.

The scalability of catenaries makes the following table possible. Assuming the distance D between the ends of the catenary (i.e., the width) is 1 unit, if the length is L times the width, then the sag is S times the width.

## Suppose D = 1 ...

If L = ....
then S = ...
|
If L = ...
then S = ...
1.0
0.000
|
2.5
1.069
1.1
0.201
|
2.6
1.123
1.2
0.292
|
2.7
1.176
1.3
0.369
|
2.8
1.229
1.4
0.438
|
2.9
1.282
1.5
0.503
|
3.0
1.334
1.6
0.565
|
3.1
1.387
1.7
0.624
|
3.2
1.439
1.8
0.683
|
3.3
1.491
1.9
0.740
|
3.4
1.543
2.0
0.796
|
3.5
1.595
2.1
0.852
|
3.6
1.646
2.2
0.907
|
3.7
1.698
2.3
0.962
|
3.8
1.750
2.4
1.016
|
3.9
1.801

Example of How to Use Table: Suppose the width of a hanging rope is 7.8 meters and the length of the rope is 9.2 meters. Since D = 7.8 and L = 9.2, we rescale it so that D = 7.8/7.8 = 1 and L = 9.2/7.8 = 1.179.

In the table, we have entries for L = 1.1 and L = 1.2. Since our value of L = 1.179 is closer to 1.2, we use that entry. According to the table, if a catenary has a width of 1 and a length of 1.2, then the sag is 0.292.

Multiplying by the original width D, we get S = 0.292*7.8 = 2.2776. Rounding down, this means the sag of the rope is about 2.2 meters.

## Approximation Formulas for D, L, and S

Simple hand calculators don't have numerical equation solving capabilities, but for certain cases of catenaries, you can use approximations formulas to compute L or S when you know the values of the other two variables. The formulas provide reasonable estimates provided that one of the following conditions is met:

• S is no more than 22% of D
• D is at least 88% of L

If one of these conditions is met, then you can use these equations

S ≈ sqrt[0.375*D*(L-D)]
L ≈ (8S^2 + 3D^2)/(3D)

We don't need an approximation formula for D since we have its exact equation in terms of L and S.

2

2

2

13

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Submit a Comment

• Yael 3 years ago

I think this is related to the problem I was trying to solve, but I don't really understand the math. I wanted to design some earrings that have a straight rigid bar connected in the center to the earring hook. Connected to the ends of the bar are three loops of chain. I guess they are called catenaries. The bar is 1.5 inches long, and I want the loops to drop down 2 inches, 1.5 inches, and 1 inch. So I need to figure out how long each chain should be.

• Author

TR Smith 3 years ago from Germany

It's solved as in the third example, where D and S are known and L is unknown. The distance between the ends of the chains is D = 1.5 inches, and the sag or drop values are S = 2, 1.5, and 1 inches.

Sparing you the mathematical detail, the values of L work out to be 4.5 inches, 3.56 inches, and 2.66 inches. In total, you need 10.72 inches of chain per earring so that the three loops hang just so.

• Yael 3 years ago

Thank you! I tested it with a piece of string and it works out. I guess could have used trial and error just as well. Thanks again.

• Billy 2 years ago

I need help with a problem similar to the third example. I'm trying to figure out what L would be if two poles are at a distance D of 22 feet and the rope needs to sag 2 inches or .166667 ft.

• Author

TR Smith 2 years ago from Germany

Since S is very very small compared to D, you can use the approximation formula

L ≈ (8S^2 + 3D^2)/(3D)

= (8*(1/6)^2 + 3*22^2)/(3*22)

= 22.0033670034 feet

If you use the exact equation and plug it into a numerical solver, you get L = 22.0033666427317, so not much difference between the simpler approx formula and the exact formula in this case.

• Nate 2 years ago

How do you get Equation 4 from 2 and 3 or 1 and 3?

In my problem one of the links is considerably longer and heavier than the others. To date, I haven't been able to solve this case.

• Jamie 2 months ago

Hi,

Thank you for the article. Can you tell me how exactly you got from Equation 3 from Equation 1 and 2? I have tried to use various method of substituting the rule: cosh^2(x) - sinh^2(x) = 1, but I was unable to make a the subject. This is the equation I stayed stuck on after algebraically manipulation:

(S^2)(c^2+2Sc)=(L^2)/(4c^2 )

• Author

TR Smith 2 months ago from Germany

Hi Badger, thanks for the comments. It's a harder problem to solve if the cable does not have uniform density. If the heaviest link is in the center and it is really heavy, then the shape will approach an inverted triangle.

• Author

TR Smith 2 months ago from Germany

Hi Jamie, thanks for the question. Start by multiplying both sides of Equation 1 by 2, then square both Equations 1 and 2.

(cL)^2 = 4*sinh(cD/2)^2

(cS)^2 = cosh(cD/2)^2 - 2*cosh(cD/2) + 1

Now multiply both sides of the second by 4

(cL)^2 = 4*sinh(cD/2)^2

(2cS)^2 = 4*cosh(cD/2)^2 - 8*cosh(cD/2) + 4

Now subtract the first from the second

(2cS)^2 - (cL)^2 = 4*cosh(cD/2)^2 - 8*cosh(cD/2) + 4 - 4*sinh(cD/2)^2

= -8*cosh(cD/2) + 8

= -8*(cosh(cD/2) - 1)

= -8*cS

So you get (2cS)^2 - (cL)^2 = -8*cS, which you can then solve for c.