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Vector Forces

VECTOR FORCES

Vector forces become apparent whenever there is an internal angle greater than 0° between two or more rigging components or anchorage points.

For ease of explanation, a vector force is typically trying to pull horizontally as well as vertically. This has a multiplying effect on the loads that are felt at the anchor points and likewise the tension exerted within the rigging equipment, be it ropes, slings, strops or chains.

Force is an influence that has both magnitude and direction, it is usually given in the dynamic unit of Newtons (N). For ease of explanation we have used kilograms on this page.

The Basics

To start with the basics, if we imagine a load of 100kg suspended equally from two slings then each sling would equally share half of the loads weight.

In the situation illustrated to the right, the weight of the load = 100kg. The load is supported by two slings of equal configuration with no internal angle, so 100kg / 2 = 50kg. This means that each sling and anchor point is being subject to 50kg or 50% of the loads weight.

Diagram showing equal forces with a zero internal angle

The Ideal Angle

As the internal angle between the rigging slings increases then additional forces (vector forces) begin to be applied to each sling / anchor point.

When rigging ropes the ‘ideal angle’ is approximately 45°, at the ideal angle there would be 54% of the loads weight being distributed to each anchor device. Although this is over half of the original weight of the load we have still gained an advantage by sharing it between the two anchor points.

The ‘OK’ Angle

An internal angle of 90° between ropes and rigging components is sometimes referred to as the ‘OK’ angle. At this angle 71% of the loads weight will be distributed to each anchor component, so in this example that will be 71kg.

It is often easier to roughly estimate a 90° or right-angle when undertaking rigging tasks. By staying at or below this angle ensures that we don’t load our anchor components with excessive forces.

Diagram showing the 'OK' angle of 90 degrees and the vector force applied to each anchor

The Critical Angle

A basic way to understand the effect of vector forces is to imagine that if a full circle equates to 360° and this was split into three equal parts we would end up with three angles of 120°, as show in the illustration on the right. An internal angle of 120° is also defined as the ‘critical angle’.

Because everything is in equilibrium at the critical angle of 120°, whatever the load weighs is what we have being exerted to each anchor point and each item of rigging equipment. So in this example it is 100kg or 100% of the loads weight.

The Calculations

Vector forces can be calculated using mathematical formula. So far on this page we have used kilograms to represent the loads in the illustrations. As a kilogram is a measurement of mass, this should be converted to weight (Newtons) to calculate the resultant force correctly.

Providing that the rigging components are sharing the weight of the load equally, such as in a ‘Y’ hang then the following equation can be used:

Using formula to calculate vector forces in rigging

Where:

F is the resultant force exerted to each anchorage.
W is the weight of the load.
α is the internal angle between the two slings.

Force is an influence that has both magnitude and direction, it is usually given in the dynamic unit of Newtons (N). For simplicity we have used kilograms for the examples on this page.

Calculating vector forces in rigging using mathematical equation

Vector Force Chart

This chart displays the resultant force applied to each anchor point / rigging component when the load is equally shared in a Y-hang rigging configuration. The ratio is also given in percentages as this is often an easier way to calculate forces relevant to the specific weight of the load.

Diagram showing vector force chart and anchor loadings for several angles in rigging

Notice that when the critical angle of 120° is exceeded, then these forces increase dramatically. If an angle of 175° could be achieved (although this would be extremely difficult) then with a 100kg load there would nearly be 1150kg being felt by each anchor component. Something worth keeping in mind when working with Tensioned Lines, Cross Hauls and Tyrolean’s!

Using the Percentage Factor

It is not always the case that the load will weigh 100kg, it is far easier to calculate the relevant vector forces from a percentage ratio. This can be achieved by using the formula:

calculating vector forces in rigging using percentage ratio

For example if we had a load weighing 76kg suspended from a Y-hang rigging configuration with an internal angle of 75° then:

Angle% factor
50.0
50.1
10°50.2
15°50.4
20°50.8
25°51.2
30°51.8
35°52.4
40°53.2
45°54.1
50°55.2
55°56.4
60°57.7
Angle% factor
65°59.3
70°61.0
75°63.0
80°65.3
85°67.8
90°70.7
95°74.0
100°77.8
105°82.1
110°87.2
115°93.1
120°100.0
125°108.9
Angle% factor
130°118.3
135°130.6
140°146.2
145°166.8
150°193.2
155°231.0
160°287.9
165°383.1
170°573.7
175°1146.9
180°

Vector Force Graph

The graph below displays the relationship between the internal angle and the percentage ratio. At 180° this line would continue to rise vertically, meaning that from a mathematical point of view the force exerted on each anchor point would be infinite.

Graph showing the relationship between the internal angle and the forces exerted to anchor points

Problem One-Eighty

What would happen if we could actually achieve an internal angle of 180°? Although this would be physically impossible as the rope or slings would even deflect under their own weight, a perfect internal angle of 180° would produce an infinite amount of force exerted to the rigging components and anchor points.

Diagram showing infinite vector forces applied with an internal angle of 180 degrees

The reason for this is that in trying to calculate the forces using the equation would result in a number being divided by 0. It is impossible to divide a number by zero, try this on a calculator and an E or error message will be displayed.

Where else can Vector Forces be found?

These additional vector forces will also occur wherever internal angles between rigging equipment and loads are apparent. In addition to rigging Y-hangs, this will also include;

  • Cross Hauling & Rope to Rope Transfers
  • Tensioned Lines & Ariel Tramways
  • Tyrolean’s & Zip Wires / Lines
  • Included angles in knot and slinging configurations

Angular Vector Forces

Angular vector forces occur when ropes are passed through a deviation or a directional pulley.

Dependent on the angle created, this can have a multiplying effect on the forces that are felt at the deviation / directional anchor point.


Measuring from the angle of deflection or the included angle

Hauling Systems

HAULING SYSTEMS

Hauling systems have many uses, from the basic sack hauling set-ups commonly used during big wall climbs to the more complex systems that are required for advanced rope rescue techniques.

Many different types of hauling systems are practiced around the world and hauling methods and equipment will vary in different disciplines, applications and countries. This collection of pages explains how a variety of effective hauling systems can be created and are intended purely for example purposes only.

Professional industrial and rope-rescue teams train for many years to master the skills and techniques that are required to undertake complicated rigging and hauling exercises. In no way do these pages constitute for the proper training, experience, knowledge and supervision that is required to safely construct and operate hauling systems.

Before reading any further, an appreciation of how conventional pulley systems work would provide a significant advantage in understanding hauling systems. For basic information about pulley systems and how they work, please see the pulley systems page.

Disadvantages of Conventional Pulley Systems

This conventional 3:1 pulley system has been assembled to raise a load weighing 90kg which needs to be lifted over a height of 50m.

Due to the way that the system has been created we would need at least 150m of rope just to form the 3:1 mechanical advantage. In this situation the operator is standing at ground level so an additional 50m of rope will also be needed to redirect the rope back down to where the effort will be applied. All in all this system will require at least 200m of rope.

Also this conventional system has been constructed from standard pulleys and no self-breaking devices have been installed. If the operative was to raise the load but then accidentally let go of the control end of the rope the load would free-fall into a very rapid and uncontrolled descent!

It is possible to redesign the pulley system making it more compact and manageable while using less rope and equipment. Before we look at building a basic hauling system, we need to ensure that we have a haul or lift plan in place. Planning is the key point in any hauling or lifting procedure!

Diagram showing a conventional pulley system with a 3:1 ratio lifting a load over 50 metres

Planning the Haul

The proper planning of hauling tasks is a key feature that should not be overlooked. As with most things, poor preparation can easily result in further complications, potentially causing unnecessary and timely delays to the hauling operation.

A principle focus of any lifting or hauling task should be to fully assess the weight, dimensions, characteristics and security of the load or object that is to be moved, whether it is a 30kg sack of gear or a rope-rescue scenario where the load of two people may need to be hauled simultaneously.

The loads centre of gravity, its correct orientation and its path during the haul should be determined. If the ‘load’ comprises of a person or casualty then thought should be given to the proper positioning and management of the subject before, during and after the lift.

Ideally the hauling system should be kept as simple as possible by selecting the correct equipment relevant to the task and assembling it to provide an adequate mechanical advantage. Appropriate safety or back-up systems may also need to be incorporated to protect against the possibility of equipment failure within the primary system.

Additionally, effective team management is another crucial aspect which will require attention. A clear and methodical haul or lift plan should be devised, how would this plan be affected if a member of the team became incapacitated or a component failed?

All team members should be competent and fully understand their individual roles and responsibilities in the operation. Adequate communication between the members of the team will also be a necessity.

Before we dive into building a basic haul system, let’s familiarize ourselves with some common types of hauling equipment and how they can be used to construct effective hauling systems.

Fall Factors

FALL FACTORS

A fall factor is a simple representation of the severity of a fall. It can be used to evaluate the potential loadings exerted on the climber, the equipment and the anchor components.

Calculating a fall factor is achieved by simply taking the distance of the fall and then dividing it by the length of rope or lanyard available to absorb the energy;

Equation for calculating fall factors. Distance of fall divided by the length of rope available to absorb energy

While a fall factor does identify the possible loadings that will be exerted on both the climber and the equipment, even a low fall factor could have fatal consequences. Big falls can still result in low fall factors where attention should be made towards the amount of distance fallen and the necessary clearance required.

This page uses diagrams and references to testing standards to assist in the understanding of fall factors, these have been included purely for illustrative purposes only! In no way do they suggest a recommendation or represent how the equipment should or could be used.

Fall Factor Chart

The chart below displays several examples of different fall factors. All the lanyards are made from a two metre length of dynamic rope which has been clipped to a fixed anchor point.

Chart showing fall factors with fall factor 2, fall factor 1 and fall factor 0. The diagram displays the relationship between a climbers position prior to a fall in relation to a fixed anchor point with a set length of dynamic rope

Fall Factor Two – FF2

In this situation the climber has a lanyard made from a section of dynamic rope that is two metres in length which is connected to a fixed anchor point. The climber has climbed two metres past the anchor point and can no longer climb any further.

If the climber were to fall from this point then he or she would fall two metres (2m) to the anchor point and then an additional two metres (2m) past the anchor point (the length of the rope or lanyard). Overall the climber would fall a total distance of four metres (4m) until the fall is arrested by the rope. Using the equation above to work out the fall factor;

Equation for calculating fall factors resulting in a fall factor of two FF2

Normally we should not be able to achieve a fall factor greater than two (FF2). A fall factor two or FF2 could potentially exert a significant amount of force both on the climber and the equipment in the system, this will also include the anchor point! Dynamic climbing ropes certified to the European EN 892 standard are tested to withstand only a certain number of fall factor 1.77 falls prior to failure occurring.

It is bad practice to accept a fall factor two in any situation. Fall factors should be kept as minimal as possible to reduce the likelihood of excessive forces being applied to the climber and the equipment.

Fall Factor One – FF1

In this diagram the climber has a two metre dynamic rope lanyard that is connected to an anchor point. The anchor point is at the same height in relation to the harness attachment point, meaning that the climber could fall a distance of two metres. So;

This would result in a fall factor one (FF1). Low-stretch or semi-static ropes conforming to the European EN 1891 standard are tested to hold a specific number of fall factor one (FF1) falls before failure occurs. Although using semi-static rope to arrest falls should always be avoided as significant and possibly fatal forces can be exerted on the climber, the anchor point and the equipment.

Fall Factor Zero – FF0

A fall factor zero (FF0) would be achieved if the climber could not fall any distance, such as when the rope or lanyard is anchored directly above them. This provides a good level of protection for the climber and the equipment, top-rope and bottom-rope climbing methods can provide minimal fall factors.

Fall Factors in Lead Climbing

The same principle is applied when lead climbing. The climber in the illustration has climbed 30m above the main belay and has placed running belays for protection during the ascent. The last piece of equipment that the climber placed was 15m above the main belay, the climber then climbs a further 15m but then falls off the route.

The climber falls a distance of 30m before the rope and the equipment are able to begin arresting the fall. Due to the overhanging section of the route there is 40m of rope between the climber and the belay device which results in a fall factor of 0.75.

Equation for calculating fall factors in lead climbing

Initially the climber fell 30m, also the rope stretches to absorb the energy and there maybe slippage through the belay device. Although the fall factor is quite low, the actual distance that the climber has fallen is quite significant.

Pulley Systems

PULLEY SYSTEMS

Pulley systems are used to provide us with a mechanical advantage, where the amount of input effort is multiplied to exert greater forces on a load.

They are typically used for hauling and lifting loads but can also be used to apply tension within a system such as in a Tensioned Line or Tyrolean. This page explains the basic principles of pulley systems and how they work, for information on how to use them in hauling see the hauling systems page.

Force is an influence that has both magnitude and direction, it is usually given in the dynamic unit of Newtons (N). For ease of explanation we have used kilograms on this page. Additionally, the examples on this page do not take into account the effects of angular vector forces or the coefficients of friction.

Disadvantages of Conventional Pulley Systems

This conventional 3:1 pulley system has been assembled to raise a load weighing 90kg which needs to be lifted over a height of 50m.

Due to the way that the system has been created we would need at least 150m just to form the 3:1 mechanical advantage, plus an additional 50m to redirect the rope down towards the user who will apply the effort at ground level so all in all this system will require at least 200m of rope!

It is possible to redesign the pulley system making it more compact and manageable while using less rope. To see how this can be done then have a look at the hauling systems page.

Diagram showing a conventional pulley system with a 3:1 ratio lifting a load over 50 metres

Angular Vector Forces

ANGULAR VECTOR FORCES

Angular vector forces occur when ropes are passed through a deviation or a directional pulley.

Dependent on the angle created, this can have a multiplying effect on the forces that are felt at the deviation or directional pulley components and the associated anchor points.

Force is an influence that has both magnitude and direction, it is usually given in the dynamic unit of Newtons (N). For ease of explanation we have used kilograms on this page.

The Basics

Here we have a load weighing 100kg. A rope has been attached to the load and then passed through a directional pulley which returns the rope back down to the user at ground level. The two sections of rope are in parallel with each other so there is no internal or included angle.

In this situation the directional pulley and its anchor point are actually being loaded with 200kg, as there is the weight of the load on one side and a effort of 100kg is being applied by the user on the other to hold the load in place.

Diagram showing the angular vector forces applied to a directional pulley or deviation with an internal angle of 0 degrees

Deflection Angle of 20°

Here the rope has been deviated away from its original line (in this illustration the vertical) by 20°. While all of the weight of the load is being exerted to the main anchor point, a percentage of the loads weight is also being applied to the deviation / directional pulley components.

With a deviation of 20°, there will be 34% of the loads weight being applied to the deviation / directional pulley anchor point. In this situation, where we have a 100kg load this would produce a force of 34kg on the deviation / directional pulley.

Deflection Angle of 45°

As the angle of deflection increases then the force exerted on the deviation or directional pulley anchor also increases. An angle of deflection measuring 45° would produce a force on the deviation anchor point equivalent to 76% of the loads weight, so in this example where the load weighs 100kg that would be 76kg.

Diagram showing a deviation with a 45 degree angle of deflection

Deflection Angle of 60°

With an angle of deflection measuring 60° 100% of the loads weight will be applied to the deviation or directional pulley anchor point. This can be compared to the critical angle found in vector forces when rigging y hangs, as all three angles are equal.

Deflection Angle of 90°

A 90° angle of deflection results in a force equal to 141% of the loads weight being applied to the deviation or directional pulley anchor point. Note that this would also be the same loading for an included angle of 90° as they are both the same.

Diagram showing a deviation with a 90 degree angle of deflection

Deflection Angle: The Calculations

Angular Vector forces can be calculated using mathematical formula. Here is a formula that we can use to calculate the amount of force that is being applied to the deviation or directional pulley anchor point when measuring from the angle of deflection.

So far on this page we have used kilograms to represent the loads in the illustrations. As a kilogram is a measurement of mass, this should be converted to weight (Newtons) to calculate the resultant force correctly.

Where:

F is the resultant force exerted to the deviation or directional pulley anchor point.
W is the weight of the load.
α is the angle that the rope has been deflected away from its original line.

Calculating vector forces in rigging using mathematical equation

Deflection Angle: Using the Percentage Factor

The percentage factor can be used to calculate the amount of force that is being applied to the deviation or directional pulley anchor point. This can be achieved by using the following formula:

calculating vector forces in rigging using percentage ratio

Angle% factor
0.0
8.7
10°17.4
15°26.1
20°34.7
25°43.3
30°51.8
35°60.1
40°68.4
45°76.5
50°84.5
55°92.3
60°100.0
Angle% factor
65°107.5
70°114.7
75°121.7
80°128.6
85°135.1
90°141.4
95°147.4
100°153.2
105°158.7
110°163.8
115°168.8
120°173.2
125°177.4
Angle% factor
130°181.3
135°184.8
140°187.9
145°190.7
150°193.2
155°195.3
160°197.0
165°198.3
170°199.2
175°199.8
180°200.0

Looking at it from a different angle

We can measure the angle of a deviation or directional pulley in two different ways, which ever is easiest and more relevant to the specific set-up. Here we have detailed both methods; measuring from the angle of deflection and measuring from the included angle.

An angle of deflection is a measurement of how many degrees the rope has been deviated or diverted away from its original line. These angles will be easier to estimate when using small deviations.

The included angle is measured between the section of rope that passes through the deviation or directional pulley. It is usually easier to calculate this type of angular vector force when using pulley systems and high re-directs.

When measuring either deviations or directional pulleys, both the included angle and the angle of deflection should always result in 180° when added together.

Measuring from the angle of deflection or the included angle

Included Angle of 75°

In this situation we have measured the included angle between the section of rope which passes around the pulley.

The included angle is 75°, this would produce a force on the directional pulley or deviation equivalent to 159% of the loads weight. So in this example that is 159kg!

Diagram showing a deviation with a 75 degree included angle

Included Angle of 60°

Again we have measured the included angle formed where the rope passes through the directional pulley or deviation.

With an included angle of 60°, a force equivalent to 173% of the loads weight will be applied to the directional pulley or deviation.

Included Angle of 45°

In this illustration the included angle measures 45°. As the included angle created by the rope passing through the pulley or deviation decreases towards 0° the force applied to the directional pulley or deviation increases.

In this example the force applied will be roughly equal 185% of the loads weight.

Diagram showing a deviation with a 45 degree included angle

Included Angle: The Calculations

Angular Vector forces can be calculated using mathematical formula. Here is a formula that we can use to calculate the amount of force that is being applied to the deviation or directional pulley anchor point when measuring from the included angle.

So far on this page we have used kilograms to represent the loads in the illustrations. As a kilogram is a measurement of mass, this should be converted to weight (Newtons) to calculate the resultant force correctly.

Using formula to calculate the angular vector forces on deviation and directional pulley anchor points

Where:

F is the resultant force exerted to the deviation or directional pulley anchor point.
W is the weight of the load.
α is the angle that the rope has been deflected away from its original line.

Calculating vector forces in rigging using mathematical equation

Included Angle: Using the Percentage Factor

The percentage factor can be used to calculate the amount of force that is being applied to the deviation or directional pulley anchor point. This can be achieved by using the following formula:

calculating vector forces in rigging using percentage ratio

Angle% factor
200.0
199.8
10°199.2
15°198.3
20°197.0
25°195.3
30°193.2
35°190.7
40°187.9
45°184.8
50°181.3
55°177.4
60°173.2
Angle% factor
65°168.8
70°163.8
75°158.7
80°153.2
85°147.4
90°141.4
95°135.1
100°128.6
105°121.7
110°114.7
115°107.5
120°100.0
125°92.3
Angle% factor
130°84.5
135°76.5
140°68.4
145°60.1
150°51.8
155°43.3
160°34.7
165°26.1
170°17.4
175°8.7
180°0.0

Angular Vector Force Graph

The graph below displays the force applied to the deviation or directional pulley anchor point in comparison to the relationship between the deflection angle and the included angle.

Graph showing the relationship between the percentage ratio and the deflection and included angle applied to deviations and redirect pulleys