Rope Sag - Catenary

[DDK]

A recent discussion on the relationship between tensions and angles for a fixed load on a rope brought up the question of the weight of the rope (at least it did for me). The fixed load mathematics are fairly straightforward geometry, trigonometry and algebra.

The shape of a rope as it symmetrically sags under its own weight between two points of attachment is described by a hyperbolic cosine function [ cosh(x) ] known as a Catenary.  For the sake of simplicity and brevity in this discussion, I will assume the two points of attachment are at the same height and forego the derivation of these formulas.  The uses of the results of this derivation, however, are not too difficult and provide interesting information.

The form of the function describing the sag of the rope is y = a*cosh(x/a) where x and y respectively are the horizontal and vertical paired position coordinates of each point (x,y) on the curve (rope), the parameter "a" which is equal to [ T₀ / (W/l) ], has the units of distance and influences the degree of "sag" in the curve and the symbol "*" refers to multiplication. The cosh(x) function is symmetric about and at a minimum for x = 0, so, if we describe a 20 meter distance to be spanned by the rope, we will describe it as spanning from x = -10 meters to x = +10 meters.

T₀ refers to the minimum tension in the rope and is found at the lowest point in the rope ( x = 0, y = a ; cosh(0) = 1). As one might guess, the largest tension in the rope is at the upper points of attachment as that part of the rope must lift up on the most rope below. (W/l) refers to the weight per length of the rope.  As one might also guess, if T₀ is large compared to (W/l), the parameter "a" will be larger and there will be less relative sag.

A very useful result is that the tension in the rope is proportional to the height, y, at that part of the rope, that is, tension, T = (W/l)*y. Also of interest is that the length of the rope, s, from the lowest point on the rope, that is at x = 0, to a point at position (x,y) can be found from the equation y² = a² + s².

Example

Choose the minimum tension,
T₀ = 5 Newtons   (approx. equivalent to 0.51 kilograms)

Choose the weight of the rope per length,
(W/l) = 0.8 Newtons/meter   (approx. equivalent to 0.082 kgs/meter or 2.45 kgs/30 meters)

Then, the parameter "a" = [ T₀ / (W/l) ] = [ 5 N / (0.8 N/m) ] = 6.25 meters

Assuming we would like to span 20 meters, select values for x from -10 to 10 and then find the (heights, y), (tensions, T), and (rope length, s) corresponding to those values of x.

We use the formulas: y = a*cosh(x/a), T = (W/l)*y, and s = SquareRoot(y² - a²).

       x         Height, y   Tension, T    Rope Length, s
(meters)     (meters)    (Newtons)        (meters)
   -10             16.11          12.89   
     -9             13.93          11.14   
     -8             12.11            9.69   
     -7             10.60            8.48   
     -6               9.36            7.49   
     -5               8.36            6.69   
     -4               7.57            6.06   
     -3               6.98            5.59   
     -2               6.57            5.26   
     -1               6.33            5.06   
      0               6.25            5.00                0.00
      1               6.33            5.06                1.00
      2               6.57            5.26                2.03
      3               6.98            5.59                3.12
      4               7.57            6.06                4.28
      5               8.36            6.69                5.55
      6               9.36            7.49                6.97
      7             10.60            8.48                8.56
      8             12.11            9.69              10.37
      9             13.93          11.14              12.45
    10             16.11          12.89              14.85

It can be seen that the tension varies from a minimum of 5 Newtons to 12.89 Newtons and that the total length of the rope is twice the 14.85 m calculated for one side, that is, 29.7 meters is needed to span the 20 meter distance between the two points of attachment. The (x,y) data points are plotted below.  Of special note is that the height of zero is not arbitrary and is at a distance of "a" below the minimum in the catenary. Only if the zero height position is correctly placed can the formulas for the tension, T, and the rope length, s, be used.

Also plotted are catenaries with different values of the parameter "a".  Since the maximum height for each curve varies, the curves have been offset in height for the purposes of comparison.  Parameter "a" values less than about 4 have sags too great to display with a height change limit of 20 used.

[DDK]

Another Approach to Rope Sag - Rope Sections and Numerical Calculation

Consider a rope spanning a gap of known width, X, and attached at the same height on either side.  The rope can be treated as divided into equal sections with the weight of each section acting at its center of mass.  A rope of known total length, L, and total weight, W, if divided into, for example, five sections, then each section would be of length, L/5, and weight, W/5.  If the rope were to be laid flat and the value x = 0 assigned to one end, the centers of mass of these equal sections would be at positions x = L/10, 3L/10, 5L/10, 7L/10 and 9L/10.  A weight of W/5 will be placed at each of these positions on this otherwise "weightless" rope.  See the first attached picture.

This configuration will be approached as a typical force statics problem summing the vertical and horizontal forces at each point to zero.  The origin will be taken as the upper left-hand point of rope attachment.  The x-axis will be horizontal and positive to the right. The y-axis will be vertical and positive downward.

There are some simplifications due to symmetry. For 5 sections, positions #5 and #1 are equivalent as are positions #4 and #2 and provide no additional information or constraints.  Odd numbers and even numbers produce slightly different sets of equations that must be solved.  The examples below will be for an odd number of sections which also happen to have a center of mass point hanging at the midpoint of the span where x = X/2.  Also used is the fact that the distances betwen adjacent points are known. If (x₁,y₁) corresponds to point #1 and (x₂,y₂) corresponds to point #2, then the distance between them which is equal to L/5 is equivalent to SquareRoot[ (x₂ - x₁)² + (y₂ - y₁)² ].

The problem will be solved by guessing the x-value of the first point, x₁, and calculating the remaining quantities.  The x-value of the point in the middle of the span (point #3 for the 5-point configuration), that is, x₃, will be compared to the value X/2 and the process repeated until they are essentially equal.  The Excel spreadsheet program has an Add-in software called "Analysis Toolpak" and "Solver" which does this nicely.  In Excel, this is loaded from the FILE > OPTIONS > ADD-INS screen and afterwards is accessed from the Excel "Data" tab.

Sum of Forces

In reference to the first attached picture, T represents the tensions in the rope, θ, the angles, and as before, W, the total weight of the rope.  The sum of horizontal forces at point #1 gives T₁*sin(θ₁) = T₂*sin(θ₂). Similarly, the sum of horizontal forces at point #2 gives T₂*sin(θ₂) = T₃*sin(θ₃).

In general, T₁*sin(θ₁) = T₂*sin(θ₂) = T₃*sin(θ₃) = . . . etc. independent of the number of points.

For the summing of vertical forces, it is best to start at the midpoint, that is, point #3 for the 5-point configuration. Here the sum of forces gives T₃*cos(θ₃) = (W/5)/2 = W/10.  We divide the load by two because there is an equal contribution from T4 which by symmetry, exactly equals T3.  The summing of vertical forces at point #2 gives T₂*cos(θ₂) = T₃*cos(θ₃) + W/5, but, T₃*cos(θ₃) was just determined. Therefore, T₂*cos(θ₂) = 3W/10. Likewise, T₁*cos(θ₁) = 5W/10.

In summary for the vertical components of force for the 5-point configuration,
T₁*cos(θ₁) = 5W/10
T₂*cos(θ₂) = 3W/10
T₃*cos(θ₃) = 1W/10

In general, for any odd number of points = Npts, T_i*cos(θ_i) = [Npts - 2*(i - 1)]*W/(2*Npts) for point #1 to the midpoint.

Example: 21-Point Configuration

The 21-point configuration requires consideration of 11 points.  The 11 columns of Excel data fit into a full screen without scrolling.  Since the formulas repeat for column 3 and upward, this process should be relatively straightforward to extend to a higher number of points using Excel's auto-fill feature.

First the sheet viewing the formulas will be given followed by the sheet viewing the calculated values. The variable names were defined using the FORMULAS Tab > CREATE FROM SELECTION feature of Excel.  The columns data was highlighted before the "CREATE FROM NAME" feature was used so that the header variables to the left would be set up as arrays.

First the formulas.  The starting parameters were chosen to equal the parameters in the original post.

NPts		21		# of rope sections
WtPerLength		0.80		wt. per length of the rope
LRope		29.69460		total length of the rope
XWidth		20.0		distance for rope to span
Weight		=WtPerLength*LRope/NPts		rope weight per section
T1SinAlpha		=TensionSinAngle		tension in first rope section times sin(alpha)

Index

0

1

2

3

4

5

6

7

8

9

10

Xpos

0.2745

=DelX+XposPrev

*

*

*

*

*

*

*

*

*

Ypos

=SQRT(Length^2-Xpos^2)

=DelY+YposPrev

*

*

*

*

*

*

*

*

*

XposPrev

0.0000

=INDEX(Xpos,1,Index)

*

*

*

*

*

*

*

*

*

YposPrev

0.0000

=INDEX(Ypos,1,Index)

*

*

*

*

*

*

*

*

*

DelX

=Xpos-XposPrev

=SQRT(Length^2/(1+(1/TanAngle^2)))

*

*

*

*

*

*

*

*

*

DelY

=Ypos-YposPrev

=SQRT(Length^2-DelX^2)

S

S

S

S

S

S

S

S

S

Length

=LRope/(2*NPts)

=LRope/NPts

A

A

A

A

A

A

A

A

A

Angle

=ATAN(TanAngle)

=ATAN(TanAngle)

M

M

M

M

M

M

M

M

M

SinAngle

=SIN(Angle)

=SIN(Angle)

E

E

E

E

E

E

E

E

E

CosAngle

=COS(Angle)

=COS(Angle)

*

*

*

*

*

*

*

*

*

TanAngle

=Xpos/Ypos

=T1SinAlpha/(WeightCos/2)

*

*

*

*

*

*

*

*

*

WeightCos

=(NPts-2*Index)*Weight

=(NPts-2*Index)*Weight

*

*

*

*

*

*

*

*

*

Tension

=(WeightCos/2)/CosAngle

=(WeightCos/2)/CosAngle

*

*

*

*

*

*

*

*

*

TensionSinAngle

=Tension*SinAngle

=Tension*SinAngle

*

*

*

*

*

*

*

*

*

2*TensionCosAngle

=2*Tension*CosAngle

=2*Tension*CosAngle

*

*

*

*

*

*

*

*

*

Below are the calculations.

NPts		21		# of rope sections
WtPerLength		0.80		wt. per length of the rope
LRope		29.69460		total length of the rope
XWidth		20.0		distance for rope to span
Weight		1.1312		rope weight per section
T1SinAlpha		5.0044		tension in first rope section times sin(alpha)

Index		0		1		2		3		4		5		6		7		8		9		10
Xpos		0.2745		0.8714		1.5242		2.2426		3.0382		3.9245		4.9158		6.0247		7.2558		8.5949		10.000
Ypos		0.6515		1.9334		3.1877		4.4057		5.5746		6.6765		7.6848		8.5622		9.2579		9.7119		9.8708
XposPrev		0.0000		0.2745		0.8714		1.5242		2.2426		3.0382		3.9245		4.9158		6.0247		7.2558		8.5949
YposPrev		0.0000		0.6515		1.9334		3.1877		4.4057		5.5746		6.6765		7.6848		8.5622		9.2579		9.7119
DelX		0.2745		0.5969		0.6528		0.7184		0.7956		0.8862		0.9913		1.1089		1.2311		1.3391		1.4051
DelY		0.6515		1.2819		1.2543		1.2179		1.1690		1.1018		1.0084		0.8773		0.6957		0.4541		0.1588
Length		0.7070		1.4140		1.4140		1.4140		1.4140		1.4140		1.4140		1.4140		1.4140		1.4140		1.4140
Angle		0.3987		0.4358		0.4799		0.5329		0.5976		0.6774		0.7769		0.9015		1.0564		1.2439		1.4583
SinAngle		0.3883		0.4221		0.4617		0.5081		0.5626		0.6268		0.7010		0.7842		0.8706		0.9470		0.9937
CosAngle		0.9215		0.9065		0.8871		0.8613		0.8267		0.7792		0.7131		0.6205		0.4920		0.3211		0.1123
TanAngle		0.4213		0.4657		0.5205		0.5898		0.6806		0.8043		0.9831		1.2640		1.7695		2.9492		8.8477
WeightCos		23.756		21.493		19.231		16.968		14.706		12.443		10.181		7.9186		5.6561		3.3937		1.1312
Tension		12.889		11.855		10.840		9.8501		8.8944		7.9846		7.1384		6.3812		5.7482		5.2842		5.0362
TensionSinAngle		5.0044		5.0044		5.0044		5.0044		5.0044		5.0044		5.0044		5.0044		5.0044		5.0044		5.0044
2*TensionCosAngle		23.756		21.493		19.231		16.968		14.706		12.443		10.181		7.9186		5.6561		3.3937		1.1312

The second picture shows a comparison of the 21-point configuration fit to a catenary.  The parameter "a" was 6.25 in the original post.

The third picture shows a comparison between 21-point and 5-point configurations.

[agent_smith]

Thanks DDK,

All very interesting.
I install test and commission zip lines and have to do calculations in the field as well as verification testing.

Zip lines and high lines are within the scope of your post.

Getting the zip line tension exactly 'right' is tricky - slight changes can have dramatic results.
It gets tricky where there is a steep decline leading out from the launch point - and then a gradual (or partial) incline toward the end termination.

Participants must be sufficiently decelerated before they reach the end termination!
In the field, I normally do the following:
1. Rough adjustment by eye
2. Find the lowest point in the span and suspend 100kg mass - and bounce that mass to confirm sufficient ground clearance (ground strikes are very 'bad').
3. Install digital load cell in the zip line and confirm loading in real time (with 100k mass at lowest point in span).
4. Release a dummy mass (100kg) and observe and measure results.
5. Make adjustments in accordance with collected data - including adjustments to primary and secondary deceleration system.
6. Real live human runs to verify and commission (usually me or my assistant - if I die, at least I died doing what I loved!).

Here is an interesting paper that mostly sums up how I approach the calculations in the field.
Link: http://www.ropelab.com.au/files/physics.pdf

All in all, installing zip lines is 80% sweat and 20% art.

[DDK]

agent_smith - yes, thank you as well.  I enjoyed the comments and reference you shared.

The difference between the theoretical and the practical is not wasted on me.  One of my favorite, self-denigrating jokes includes the punch line "And the Physicist replied, "First, assume the horse is a sphere .".".

What types of primary and secondary deceleration systems do you favor or tend to use? 

[agent_smith]

What types of primary and secondary deceleration systems do you favor or tend to use? 

Commercial-in-confidence (trade secrets) spring to mind

However, given that many people are aware of my systems due to the existing installed user base - such 'secrets' are already out in the wild (the genie has left the bottle long ago).

My go to favourite primary systems will be either:
1. Counter-weights
2. Bungy cord (14mm diameter)

The real trick is how you install and configure it.

The system has to be almost maintenance free - and the risk of failure must adhere to the ALARP principle (as low as reasonably possible).
The 2 best arrester systems are:

1. Counter-weight system consisting of dual 12-13mm (half inch) chain.
When the participant contacts with a 'sliding-impactor' - which is fitted to the steel wire rope zip line - forward momentum is dissipated as the chains are lifted off the ground (the are 2 chains - one either side of the zip line). This system scales perfectly with different body masses. Heavier adults obviously will lift more chain off the ground. Younger children lift less. The amount of chain lifted is proportional to participant momentum (mass x velocity).

The sliding impactor has a rubber interface and is very light-weight - so the participant feels no jolt on contact.
It is a relatively gentle deceleration experience (ie not a sudden jerk and whiplash).

2. Bungy cord (minimum 14mm diameter)
This also works well but needs to be properly configured.
The bungy cords are attached to the sliding impactor - one either side.

...

There are proprietary deceleration systems available - eg check out the company 'Headrush Technologies'.
Pricing can be sky high - so customers need to have plenty of cash... there is also the ongoing cost of ownership (eg cost of mandatory annual servicing).
Link: https://headrushtech.com/zip-stop-zip-line-brake

EDIT:
Secondary deceleration system:
Secondary deceleration is no secret either.
3 or 4 deformable tyres are fitted to the steel wire rope zip line.
(cant use steel belted tyres as they are too rigid and wont deform).
If the primary system fails, the participant crashes into the deformable tyres.

The tyres basically slide and deform on impact.
Biggest issues are:
1. Need drain holes for rain water that collects inside
2. Wasps like to build their nests inside the tyres - so they need to be sprayed and monitored.

[DDK]

. . . such 'secrets' are already out in the wild (the genie has left the bottle long ago) . . .

Thank you for sharing, nonetheless.

. . . The real trick is how you install and configure it . . .

I just watched a zipline installation video - they saddled a dead horse.  Apparently, a few tricks shy of what one might hope.

There are proprietary deceleration systems available - eg check out the company 'Headrush Technologies'.
Pricing can be sky high - so customers need to have plenty of cash... there is also the ongoing cost of ownership (eg cost of mandatory annual servicing).
Link: https://headrushtech.com/zip-stop-zip-line-brake

For those prices, it should include a connection to the grid and being able to sell the energy back to the electric company.

Counter-weight system consisting of dual 12-13mm (half inch) chain.
When the participant contacts with a 'sliding-impactor' - which is fitted to the steel wire rope zip line - forward momentum is dissipated as the chains are lifted off the ground (the are 2 chains - one either side of the zip line). This system scales perfectly with different body masses. Heavier adults obviously will lift more chain off the ground. Younger children lift less. The amount of chain lifted is proportional to participant momentum (mass x velocity).

The sliding impactor has a rubber interface and is very light-weight - so the participant feels no jolt on contact.
It is a relatively gentle deceleration experience (ie not a sudden jerk and whiplash).

Nice.  When the chains fall back to the ground (or when the bungee cords contract), is it the friction in the impactor which prevents or reduces the energy being returned to the rider propelling them back in the opposite direction?

DDK

[DDK]

"Hybrid" Catenary - Numerical Calculation

Consider the weight, W, of a rope/cable divided into sections with the addition of a fixed weight at the midpoint, W_mp.  As before, we assume there is no elastic stretch of the rope/cable.  For the purposes of discussion, I will call this a "hybrid" catenary.  The numerical calculation formulas remain the same except for the the vertical components of force.

The vertical components of force for the 5-point configuration without the weight at the midpoint were,
T₁*cos(θ₁) = 5W/10
T₂*cos(θ₂) = 3W/10
T₃*cos(θ₃) = 1W/10

With an additional weight, W_mp, at the midpoint, these become,
T₁*cos(θ₁) = 5W/10 + W_mp/2
T₂*cos(θ₂) = 3W/10 + W_mp/2
T₃*cos(θ₃) = 1W/10 + W_mp/2

These formulas can be generalized similar to before and the number of sections (one section per Excel column of data) increased.  The data given below will be for catenaries divided into 1001 points.  Due to symmetry, this results in 501 data columns.

Results

For each of the three graphs attached, 200 catenaries were optimized and numerically calculated.  This was done using fairly simple Excel VBA programming which looped through the values of the independent variable of interest for the graph while invoking the Excel Solver routine twice for each value (to improve the optimization).  The calculations required about 65 seconds to optimize the 200 catenaries.  Key quantities were written to a table on the spreadsheet by the program and then graphed.

The first attached graph shows the sag and tension as one applies load at the midpoint.  The weight of the rope itself causes a sag of about 1.5 meters.  This sag quickly increases with load approaching the loaded "weightless" rope value of about 1.734935 meters.  The sag at 100 kg of load is 1.7348 meters. The tension is proportional to the applied load.

The second attached graph again shows the sag and tension as one applies load at the midpoint but for loads only up to 20 kg.

The third attached graph shows the sag and tension as one changes the length of the rope while keeping 100kg at the midpoint.  As one tries to remove that last bit of sag, the tension rises tremendously.

DDK

Edit: The fourth attached graph was added to show the shape of the cable for various loadings.  The "Weightless" Cable curve is the shape one would get for any loading ignoring the weight of the cable. The height of the attachment points is arbitrary.

[DDK]

Elastic Hybrid Catenary - Numerical Calculation

Previous calculations assumed that the rope/cables were "rigid", that is, would not stretch (infinite modulus).  At a finite modulus, the stretch in a cable can be modelled as linear and recoverable with the tension, so called, "linear elastic".  Each section of the cable will experience a different amount of stretch because the tension varies from a maximum at the point of attachment of the catenary to a minimum at the midpoint of the catenary.

The numerical calculations have changed to account for this variation in stretch.  The quantity which is guessed is the tension at the point of attachment (section 1).  This allows the calculation of the increased length of the first section.  Knowing the full weight being supported, the vertical force at the first section allows the calculation of the angle for the first section and then all the remaining quantities for the first section.

For section two and beyond, the horizontal component of force, T*sin(θ), is constant and known and the vertical component of force, T*cos(θ), is known because the amount of weight being supported at each point is known.  Dividing these components produces a value for tan(θ) and therefore, θ, and the tension for each section.  With this, the stretch and all other quantities can be calculated.  As before, the optimization will iterate through guesses (values of tension) until the the x-value of the midpoint section converges to the value, the span width divided by two.

The parameters to be used include:
Number of Sections = 1001
Span Width = 30 meters
Length of Unstretched Cable = 30.2 meters
Weight per Length of Cable = 1.5 newtons/meter
Weight of Cable = 45.30 newtons (equiv. to 4.619 kg)
O.D. of Cable = 0.25 inches
Elastic Strand Modulus of Cable = 47,500 newtons/(millimeter squared)
Rigid Strand Modulus of Cable = 1 x 10¹⁰ newtons/(millimeter squared)

Results

The first attached graph shows the sag for a rigid and elastic cable in which the load applied at the midpoint is varied.  The large number shown above for the rigid modulus was used to simulate an "infinite" modulus.  At zero additional load, the elastic cable already has an additional sag of 0.0085 meters compared to the rigid cable. This difference in sag increases as more load is applied. 

At 100 kilograms of load, the elastic cable has an additional sag of 0.300 meters and an additional length of 0.075 meters compared to the rigid cable.  Interestingly, although, maybe not surprisingly, the maximum tension for the elastic cable is 635 newtons less than that for the rigid cable.  This lower tension in the elastic cable is equivalent (edit: very similar would be a more accurate choice of words) to the tension of a rigid cable with this same increased length, that is 30.275 meters.

[DDK]

Catenary - "Equivalent Weight"

A cable with a weight fixed at its midpoint can be modelled as a catenary of a longer cable without a weight.  The midpoint weight can be replaced with an additional length of cable of the exact same weight.  The reason for this in part is that the vertical component of force exactly equals the weight it is supporting below.  The added catenary section provides exactly this weight.  In addition, the added catenary section provides the horizontal force (which is a constant) needed.

Example - Rigid Hybrid Catenary

Parameters include:
Number of Sections = 1001
Span Width = 30 meters
Length of Cable = 30.2 meters
Weight per Length of Cable = 1.5 newtons/meter
Weight of Cable = 45.30 newtons (equiv. to 4.619 kg)

Results

The first attached graph shows the catenary for a 30.2 meter (4.62 kg) cable with a 1 kg weight at its midpoint and the catenary for a 36.74 meter (5.62 kg) cable without a weight.  The catenary with a midpoint weight can be considered to be of two sections, each that are 2.31 kg in weight, and a midpoint weight of 1 kg.  The longer catenary without a midpoint weight can be considered to be of three sections: two that are 2.31 kg in weight and one that is 1 kg in weight. 

[GrandpaBig]

...
The shape of a rope as it symmetrically sags under its own weight between two points of attachment is described by a hyperbolic cosine function [ cosh(x) ] known as a Catenary.  ....

Did you know that a catenary curve turned upside down can be called a "Nubian Arch" and creates a stable structure where "mud" can be used to make a stable roof structure?   Dr. Gernot Minke, professor retired of Kassel University in Germany has taught and written books on "mud architecture"   Thought you might be interested?   Blessings.

[KC]

i think we will find this inversion trick to work for many things of arches.
The rope has the grace to lightly and Naturally assume the most efficient arc structure to guide us in making bridges, domes etc.
>>can pick a width and depth of arc or even corbell(wiki) and ask a string how to build it before computers etc..
.
Is a specialty arc not quite parabolic nor fave Brachistochrone (wiki).
Brachistochrone curve(instructables) can win you lots of bets for fastest/furthest marble race vs. other peoples choices of curves...
.
A Round Arc of squared proportions is what we see in knotting
>>but not where the catenary guides us for structure.
>>there are different types of arcs, that have different extreme specialty utility purposes.

[DDK]

@ GrandpaBig

I was aware of the catenary shape being used in arches but was unfamiliar with its use in Nubia or in mud.  Thanks for the reference on mud architecture.

Oddly enough, I have borrowed and made use of the phrase "Aaaah Nubian!" numerous times.  In this case, the reference is to the planet Nubia in the Star Wars film "Phantom Menace".  Thanks for the additional context.

[GrandpaBig]

DAK,

Thanks for noticing it!  Happy New Year.