Class April 4


Going over Hw 6/7

Hardware –Uses and Application

Hw 8 and Lab Time


Hw 6


Powered rigging

Dead haul – no counterweights/// uses winches to load and unload the weigh

Benefit is you don’t need to unload and load

Brute force – Big motor and safety features


Powered counterweight- counterweight with motor replacing people power

EX: Elevator are counterweight systems


Fixed speed: Only controlled on one speed (cheaper)

Variable speed: multiple speed – Ex show time (match tempo/// more money and programming time)


Money = Material or workmanship quality


Hw 7



Different types of truss




Bowline knot


Over hand

Start with an overhand loop

End of line up through back, behind standing part and behind the loop again

Tail will be in the middle of loop


Ok when loaded and unloaded

Better in more flexible ropes

If you tie it backwards, so the tail is on the outside, you can capsize the know and it will become a slip knot


Bowline is one of the most important knots to know, also clove hitch and figure eight

Should be able to tie any circumstances, rain or shine




Hardware for top to bottom, how to choose


Ex (fig 1.0) start with a batten at the top

Anchor- how are you grabbing the building or scenery

Connectors – shackles, pair rings, (devices that are designed to connect different members)

Tension members- ropes, cables

Adjustor – change the length, making something level


Example Picture 1.1

D ring plate through bottom plate (flat thinks it is sitting on the ground, no added stress and much stronger setup than if bolted to the top)

-Attach rigging at the bottom



Things that attach to scenery:

Standard D ring and plate (very convenient, super cheap, no special tools, common in 20th century when scenery was light) Not rated and should not be used for rigging (Picture 1.2), need to drill out hoes to mount)- very common

Top hanger iron: Run cable so the flat couldn’t move away from wire, only use is for theatrical flats. (Picture 1.3)

Bottom Hanger Iron: (Picture  1.4)


Alternatives to D- Ring and plate: shackle plate, rated. Look at manufacturer spec. Fairly cheap and reusable (Picture 1.5-7)

Eye Bolt: Shouldered forged eyebolt. *** Bent eye bolt is no good.  Eye bolt is designed to be loaded directly in line (Figure 1.9) (Picture 1.8, 2.0)



Galvanized: Protects against rust


Things that attach to I-beam

Wide variety of I- beam clamp

Read Manufacture specs for size beam, weight, can it be loaded at angle, how to inspect and install.

Spanset, steel cable or piece of chain around beam


Things that attach to pipe:

Chain is handy since it is strong, cheap and flexible. Great as going around round objects. Can be both used as an anchor and adjustor.

Most common chain in theatre is grade 30 proof chain: lots of markings “not for overhead lifting” which has a different meaning for chain: use in hoist.

However most of the time we are using is to connect to a batten which is acceptable.

Not idea for going around corners.

Need to size chain and shackles so they fit together; may be overrated so the shackle will fit.

When attaching shackle to chain, ensure spare chain is free to move around

Another method is a batten clamp. Often you will find at permanent points.



Screw pin anchor shackle, tighten to finger tightness. Should be marked with manufacture, country of origin, size, and WLL. Beware of cheap or knockoff shackles, ensure a manufacture is listed. Only buy rigging hardware from reputable dealers. Shackles can be loaded in line or slightly off axis. Many different shackles, for different uses. Screw pin is most common, since it is easy to install and uninstall. For long term use, must mouse the shackle, so it cannot rattle loose. Not good when loaded sideways, must be loaded in line. Other types included, just a pin, bolt and nut… different types have pros and cons for different uses.


Quick links, side screw: not great, very difficult to unscrew once loaded. John just doesn’t like them. Lower profile, could be useful, but rare.


Pair ring: many different shaped of welded or forged rings. Useful in gathering many different points. (Picture 2.1)

“Acrobat Accident Story”: use correct hardware

Failure isn’t always immediate, just because it worked, doesn’t mean it always will.


Carabiner: only one load, in line



Turnbuckle- when turned they expand or retract; many different ends

Very handy, easily adjustable. If inline, adjustor could go anywhere, mechanically or strength wise. Placed so it can be adjusted easily. Mouse turnbuckles so they don’t move. Always have the same side up, so all turnbuckles are adjusted the same way


Look out: solid style, can’t see how much is left. Also avoid anything bent, must be forged. Very careful to not side load.


Batten Splices

Very often riggers will be responsible for replacing battens,

Schedule 40 pipe.


Sleeve to join pieces together. Plumbing fittings are not for rigging. Held together by hope. Different types of sleeves to connect pipe. Find something designed to be used for rigging


Wooden Batten clamp: Out of style


Choose tension member and calculating load. Must also choose all the other hardware. Start in Backstage handbook. Look up shackle WLL. Determine load on hanging point.


Rigging system is only as strong as weakest link.


Crosby Design factor 6:1


Tension Member might be weakest link.




Passed Out Hw 8 and Lab time to work on steel cable

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3/28/17 – Class Notes

Hello Everyone,

Topic: Problem Solving, Span Set, and Cable Termination

Stage Rigging Handbook pg. 189 – 228

Before I start, McCullough announced:

Lab Day

WhenSaturday,  April 22nd

Time: 10AM – 4pm 

What we are doing: Setting up hardware and truss


In the beginning of class, we discussed about our midterms:

Based on the exam, McCullough provided a data analysis of what questions we did/didn’t understand, i.e., question #2, which we discussed in class about finding the reactions for the free body diagram.

In the Stage Rigging Handbook, page 11 – 12, provides examples of free body diagrams, similar to the first page of the midterm.

In the Data Analysis PowerPoint, McCullough asked us how we studied for the midterm, whether we studied alone, or in groups, and we chose to study alone by reviewing our homework and our notes. But for the final, we can try to study in groups by doing practice questions and comparing our notes, helping each other out.

Midterm Questions, based on Fiber Rope, you have to:

  1. Determine the Load
  2. Choose the DF (Design Factor)
  3. Find the Rope
    1. WLL x Efficiency  > Load

1) Problem Solving

Strategy to Problem Solving:

  1. Define the problem
  2. How are you going to solve it
    1. Method for solving
  3. Determine Known Relevant Info
  4. Determine Unknown Info
  5. Find Unknown Info

When problem solving, McCullough made a template of what he knows and don’t know. We should try that method to solve the problem as well.


2) Span Set

Knot of the Week: 3 Round Sling Knots

  1. The Choker Hitch
    1. Same as the Lark’s Head except the ends are connected to each other
    2. Capacity of 2100lbs, which is less than the vertical hitch capacity, roughly 80% less because they wrap around sharp ends
    3. Put on the bottom of the truss, which are smooth and round
    4. Advantage:
      1. It doesn’t move because as you apply the load, it tightens the load so it doesn’t slip and slide, which is called Good Load Handling
      2. It requires less hardware than the vertical hitch
    5. Disadvantage:
      1. Protect the edges of the sling from coming into contact with sharp corner
      2. If the object lifting is fragile, choker hitch is not recommended because it will crush
  2. Vertical Hitch
    1. Uses the full strength of the sling
    2. Capactiy of 6500lbs
    3. You want to make sure they are not twisted
    4. Flat Nylon Sling has a dashed line so that you can indicate that it is not twisted.
    5. The simpliest hitch
    6. Advantage:
      1. Very efficient – you get to use the entire strength of sling
    7. Disadvantage:
      1. It needs 1 or 2 pieces of hardware to connect to the sling, i.e., shackle, hook
  3. Basket Hitch
    1. Capacity of 10,000lbs, which is double the weight of the vertical hitch
    2. Sling goes underneath the object, like a basket
    3. Advantage:
      1. It has great strength
    4. Disadvantage:
      1. It’s going to move around
      2. You need pieces of hardware
    5. Need to be careful with what object you are going to lift with the basket hitch

These three sling knots are called hitches because they attach to a pipe and hitches are the type of knot that attaches the rope to other objects.

Note: Attaching two slings together is not a good idea unless you have a shackle or some piece of hardware in between the two slings.

You need to have equal tension and equal balance when lifting an object with the slings.


3) Cable Termination

Backstage Handbook pg 103 – 104

There are 2 Primary Methods of Terminating Steel Cable

  1. Swage Sleeve or NicoPress Sleeve:

Related image

  • Efficiency: 95% to 100%, Derate to 95%
  • A lot stronger than wire rope clips
  • Prefer Swages over Wire Rope Clips because they are:
    • Smaller
    • Stronger
    • Take up less space on the cable

2. Wire Rope Clips or Crosby Clips:

Image result for wire rope clips

  • “Never Saddle a Dead Horse” – Saddle should be on the loaded side and the U-bolt should be on the non loaded side
  • Efficiency: 80%
  • Bigger than swage sleeve
  • Cheaper than swage sleeve
  • Common
  • Torque Wrench Needed: Tighten at 15 Ft-lbs
    • You know it’s fully tightened when you hear a click at the end.
  • When using wire rope clips, you need to load the cable and then tighten it again.

Before creating a wire rope with swage sleeves ourselves, McCullough demonstrated how to assemble a wire rope clips on a wire rope with 1/4″ cable

  • Needs minimum of 2 wire rope clips
  • 4-3/4″ Turn back from Thimble (BH pg. 103)


Lab: Creating a 1′-6″ Eye to Eye, Swage Sleeve Wire Rope

Materials used for this lab:                        Tool:

  • 1/8″ Wire Rope                                         1. Cable Cutter
  • 1/8″ Thimble                                                    A. C9 (9mm)
  • 1/8″ NicoPress

2 Tools when Crimping the NicoPress

Image result for bench swaging tool

  1. Swaging Tool
    1. Once crimped, you can’t crimp again
    2. Gauge after you make another crimp
      1. If it fits, you are good to go, but if it doesn’t fit you need to make a new one

Image result for hand swage tool

2. Hand Swage Tool (Compression Tool, BH pg. 104)                                                                            A.  After crimped once, may need to crimp 2 or 3 more times.

Note: Backstage Handbook pg 104 has a crimping order of 2, 3, 1 (CROSS THAT OUT), it’s an incorrect crimping order.

The correct crimping order should be  1, 2, 3.                                                                                   1 is next to the thimble. There is a picture below showing the correct crimping order.

Displaying FullSizeRender.jpg

These are all three topics that was covered on March 28, 2017







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Syllabus Update

Everyone, the new syllabus has been posted. This accounts for the changes we’ve had to make due to the snow day. The rest of the semester will be pretty tight for us to fit in all the information we need to cover, but I think we can do it.

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Class Notes from February 21st

Hello Everyone,

You all know that English is not my strength, so I apologize in advance if my notes are a little bit confusing. However, I will do my best though, do not worry.

Let’s start by the new knot we learned.

The Lark’s Heat or the Lanyard Hitch:

Note:  I used an app called 3D Knot and it is pretty useful, check it out.

This hitch is commonly used in theater to connect a short section tie line to the ring of a curtain.

Pros:   The two ends of the tie line will be the same length.

Cons:  The hitch can loosen itself to easily

Solution to cons:  loosen the hitch/twist it 180º/  grad the ends/ go through again.  The hitch will be locked in place.  You can even drop a little bit of white glue to avoid the tie line robbers to steal them.

Lesson of the day is about HEMP RIGGING

This is a sketch of a hemp rigging system

The hemp rigging system is a system that which uses rope for the lift lines to the pipe batten.  It is the bas of all the other rigging system.  When using natural fiber, manila is the most common, but synthetic rope are usually used in modern theater.

The batten we used are 1.5″ Schedule 40 pipe (but round shape are bad are carrying load).

Loft Blocks can be underhung or overhung.




Head Blocks can have more than one rope going through.






With this system, stagehands carry all the load, but it can be balanced with sandbags up to 50 or 100 pound under the load.

The person who operates the system was call the Flyman.

The allowable batten loads is the amount of load allowed on a batten (logic, but important to mention).

Below is a helpful chart for allowable batten loads:

To calculate the total load, here is the formula:



Design factor

When working with fiber ropes, you want to use a 10:1 design factor.

Reasons Why using 10:1 DF:

  • More susceptible to damage
  • Strength is variable  all the way through

When a knot is made, it reduces the efficiency of the rope by 50%,  so anytime you work with a rope and a knot, a design factor of 20:1 has to be applied.

10:1 (using rope) + Knot (-50%) = 20:1 DF

Warning: Always find the rope specs before working in a new venue.

Advantage and disadvantage of three-Strand twisted rope


  • Shock absorption
  • Cheap


  • Length changes with load
  • Any damages affect the entire structure of the rope

Advantage and disadvantage of parallel-core rope


  • Very Strong


  • The inside can slip out


For a heavenly spaced line, the load on the ropes are different on different points.


Formula for choosing the adequate rope :

UBS = Load x DF


  • DF of 20:1
  • Load 425 lbs

UBS =  425# x 20 = 8500 lbs

You could use a 1″ Manila twisted.

When choose a rope, go with Manila, Nylon, or Polyester (Best).

Note: Polypropylene floats.


Anytime a rope is bent, you have a resultant load.  The factor depend on the angle (1.41 for 90º)

Ex:  425 lbs x 1.41 (for 90º) = Resultant 600 lbs.

Sand Bags

When balancing the hemp system with sand bags, attach them progressively to not have to carry the entire load at once.

Pin Rail

Pin rails need to be strong enough to stay on the floor.

Manage your ropes

  • Inspect periodically your rope for any sign of damage.
  • When using your system, look, smell, listen for anything different than usual
  • Every year, check all components of your rigging system (anything found, take the part away and replace it)
  • create a log
Notes:  the best height for a rigging system is 2.5 time the proscenium height.




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Mid-term Review Information

Here are the topics, and point values, for the mid-term. You can also download the test outline here: Mid-term Outline

Calculations (60 points)

  • Calculate stress and strain
  • Draw free body diagrams
  • Calculate reactions
  • Find lengths and tensions for bridles
  • Determine loads for battens and lift lines
  • Find UBS, DF, WLL
  • Convert DF
  • Calculate actual and theoretical mechanical advantage
  • Calculate resultant loads

Terminology (30 points)

  • Identify the parts of common rigging systems (draw pictures with labels, label diagrams, identify physical samples)

Best Practices (60 Points)

  • Choose appropriate design factors
  • Choose appropriate ropes, cables and terminations
  • Tie knots
  • Know the 4 K’s
  • Know safe working procedures (study this in the Stage Rigging Handbook)
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Hoist and Truss Homework

Alright class, at this point we have covered many of the mathematical concepts and procedures we need to solve rigging problems. For the rest of the semester, we are going to put those concepts into practice by solving real-world problems. A big part of being a rigger is reading and understanding technical documentation produced by manufacturers, trade groups, and standards organizations. For the truss lesson, we are going to practice reading some manufacturer’s documentation.

For the hoist and truss homework, you need to:

  1. watch this video
  2. read the CityTech Truss Manual (handed out in class)
  3. read the Lodestar NH Manual
  4. fill in the blanks on the Chain Hoist manual cover page (handed out in class)
    1. You will need to do some investigation to find your answers. It may involve asking Rudy, inspecting the hoist itself, or going through our equipment records.
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Third Class Notes – Feb. 14th 2017

The topics discussed in our third class meeting included:

  1. Knot of the day
  2. Homework #1 notes/common mistakes/questions
  3. Homework #2 questions
  4. Mechanical advantage
  5. Block and fold system lab
  6. Block and fold system lab calculations

1. First off, knot of the day!

Figure 8 knot.

This knot is perfect for preventing the end of a rope from slipping through a hole or pulley.


Link to figure 8 knot how to:

Link to figure 8 knot followthrough how to:


2. Homework #1 notes/common mistakes/questions

Notes from McCollough concerning HW #1

-Units! Pay attention to units and make sure to label all units.

-Check your work.

-Don’t ignore distributed loads (References: Stage Rigging Handbook pg.5-12 and pg.10)


As for questions concerning HW #1 we reviewed/discussed distributed loads. Below is a picture of the example used in class:

3. Homework #2 questions:

-Pertaining to question #7 on HW#2 we went over allowable tensile load

P all= Ft*A

-Pertaining to question #6 on HW#2 we went over area of a circle (in regards to the material used) 

Area of a circle = πr²


4. Mechanical Advantage (a.k.a the ability to lift heavy objects without exerting all the force on your own!)

“Mechanical advantage is the ratio, or comparison, between the force required to move a load and the force of the load.” (Stage Rigging Handbook 3rd edition pp. 49)


Mechanical Advantage incorporates the following elements:

-ratio of effort put in vs. effort put out

-the load

-friction force – proportional to what the load is



Question: Does a single pulley create mechanical advantage?

Answer: No, due to friction. A single pulley can only change the direction of a rope.



When referring to Table 2.1  (Stage Rigging Handbook 3rd edition pg. 52) we can see that when adding 1 to 10 lines/rope, the mechanical advantage increases as the load remains the governing factor.

When adding anywhere between 10 to 15 lines/rope mechanical advantage decreases as friction becomes the governing factor.


TMATheoretical Mechanical Advantage

AMAActual Mechanical Advantage

Below is some work that we did in class pertaining to TMA and AMA.



Bronze has a low coefficient of friction. Materials with low coefficient of friction are more expensive!


We can calculate:

LLP (Lead Line Pull) – Required to lift a load

Lines required to lift a load with LLP

MAX load for given system


We do all the above calculations in order to: 

-Know what you need for your mechanical advantage system

Know if we have the man/woman power needed to supply the required force

Figure out the type and  strength of rope needed for your system


Knowing the LLP limit supplies us with: 

the amount of people needed and WLL (Working Load Limit) of rope

Knowing the MAX load supplies us with:

– the fixed LLP (Lead Line Pull)/ the x amount of pounds that can be lifted



LLP= LOAD/R (ratio)

-Ratio in the above and below formulas pertain to the AMA found on the appropriate table (2.1-2.4) in pages 52-54 in the Stage Rigging Handbook 3rd edition.

Below is a sample problem completed in class.


When trying to decide how many people you need to lift a load you must consider the following:

-situation, use judgement

-is lifting the load the first or last thing scheduled in a 16 hour day?

-fatigue, etc.

-average is 50lb per person


Lines required to lift a load w/LLP 


ex: 1,000lbs/250lbs=4

Note: LLP can be the WLL of the rope or force you apply


Maximum load, given system 



5. Block and fold system lab

For lab this week we headed downstairs into the scene shop were we worked with the I-beam, rope, 2 pulleys, a green cart, a genie, 170 pounds of sandbags, crocheting, and a lot of hands.

First we calculated the TMA (Theoretical Mechanical Advantage) of our system. System uses plain bearings with 10% friction.

Drawing of pulley system used for lab

TMA= 4:1

We then attached a green cart to the block closest to the floor (lower block), we quickly noticed that there was a lot more pulling of rope involved than there was increased height of the load.

Next, we retrieved 170 pounds of sandbags and attached them all to the block on the floor. The line pull (rope) measured 5/8″ diameter so it was easy for the volunteers to grab onto and pull comfortably.

After completing the lifting of the load demonstration we proceeded to learn the proper handling and care that must be taken when striking a block and fold system.

When striking a block and fold system you must:

  1. You need 2 people. Each person grabs a block (person who grabs the floor/lower block must hold on to the end of all the extra rope to prevent an unwanted knot to occur) and pull away from each other.
  2. Continue to pull away from each other until all rope is straight and all rope has been feed through the blocks.
  3. Lay it down on the floor, and crochet! Making a loop closes to the lower block to start and feeding the rope in the same direction (to prevent any unwanted knots) until you reach the other block.
  4. Properly store the system.

Note: I know this may be hard to understand, I was unable to get a picture of the final product but it lives right my McCullough’s office if you want to take a closer look.

After the lab, we went back upstairs into the classroom and performed some calculations.


6. Block and fold system lab calculations

Loads on each pulley, where:

F=friction, P=total load,  N=number of ropes, S₁=furthest one away from lead line, and S=sections/parts of rope, LLP=Lead Line Pull, TDL=Total Distributed Load


S₂=S₁* (1+F)

S₃=S₂* (1+F)

S₄=S₃* (1+F)

LLP= S₄* (1+F)


Note: You can continue to add more sections, as long as you follow the above samples ex: S₅=S₄* (1+F) , S₆=S₅* (1+F), etc. Follow through with TDL formula.


Below are my calculations for the lab pulley system used.


Hope this brings a little more clarity to you if you were having some problems.

Feel free to shoot me a question through here or in person.


– Yarie

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Notes on second class 02/07/17

Hello classmates, here are the notes from the second class. Material is from lecture, questions, lab presentation and the board.

The knot of the week was the clove hitch. A hitch is a knot that ties to something fixed. We practiced with our fingers or arms. To make the clove hitch wrap one of the rope completely around the finger/arm and over itself. Wrap a second time and have the rope go under the second wrap. Pull tightly.

Dead hang: Something hung under a support and stays still

Figure 2-1

A rigger must notify the engineer for a building for all the bridles that will be hung in a space/arena. Certain beams and support are used for the building/space exclusively.

Can you  put a motor on a bridle? No, a motor works for linear support and doesn’t do well in angle loads Correction, a motor can be on a bridle but not as one of the bridles.

What’s the difference with forces between a dead hang and a bridle? Dead Hang has only vertical forces while bridle has vertical and horizontal forces.


It’s arbitrary which axis is used to measure the angle but for class purposes we’ll use horizontal axis.

When vectors are added we end up with a new vector.
For each vector we have a vertical component & horizontal component.

Figure 2-2

Trigonometry: “If you get a cold sweat from hearing it , then don’t.”
An easy way of remembering trigonometric functions to solve for theta (angle) is to use:
Sine(θ) = Opposite/Hypotenuse
                            Cosine(θ) = Adjacent/Hypotenuse
                            Tangent(θ) = Opposite/Adjacent

Note: Symbol for theta is θ. Theta is the angle.

Magnitude = 13.7 lbs
θ = 30°

Sin(30°) = Opp/137 lbs -> Sin(30°) = 0.5
0.5 = Opp/137 lbs -> 0.5 * 137lbs = Opp = 68.5 lbs

Note: Make sure your calculator is set to degree and not radiant.

Figure 2-3

We must use pythagorean theorem to solve this problem.

Figure 2-4
Displaying Office Lens 20170213-213114.jpg

T is the tension force countering the pull by the rope in the bridle.

T₁ = (L₁ * H₂ * P (P stands for Load))/ (V₁ * H₂ + V₂ * H₁)
T₂ = (L₂ * H₁ * P)/(V₁ * H₁ + V₂ * H₁)
Vf₁ = (V₁ * H₂ * P)/(V₁ * H₂ + V₂ * H₁)
Vf₂ = (V₂ * H₁ * P)/(V₁ * H₂ + V₂ * H₁)

To check if correct:
Vf₂ + Vf₁ + P = 0

Pythagorean Triplets are the proportions that always match on a right triangle. The lowest of these are 3-4-5, with 5 always the hypotenuse.

Lab time!

New piece of hardware introduced: Beam Clamp
It has a plate that contains Max load and badge # in case of product failure and
manufacturer can then recall.

The beam clamp used had 1 ton max load. We scaled a sand bag at 49 lbs.
Max put a clove hitch with a rope and onto the beam clamp. Professor McCullough put a half clove hitch on a carabiner to hang the scale and sand bag. We created first a dead hang. It was a dead hang because all our components can withstand a 49 lbs. fource.

We took a measuring tape and with a second scale we pulled the sand bag 2 feet. Pulling the sand bag was over 30 lbs. force! This shows that pulling a 300 lbs. curtain can easily take over 100 lbs.

Figure 2-4

Figure 2-5
Displaying 20170207_191255.jpg

A pear ring can withstand forces from various angles.
We also used a munter hitch on one of our hangs.

Now scaling two sand bags on a bridle. Stage right had 40 lbs. Stage left had just under 35. We made an experimentation to shorten the stage left rope of the bridle. Only a little change increased on each rope. We then spread the beam clamps further apart and then each rope had 55 lbs. in tension. This showed that we can create higher loads than what we hung.

For the next lab exercise, we hung a 10 lbs. sand bag. Two people held the rope on each side of the rope respectively. If they both held the rope  vertically over the sand bag, each person would have half the load of the sand bag. But by moving away from each other, the load kept increasing for each person. The vertical load is not changing. The θ (theta) is changing and the smaller the θ the greater the force. But it could never be straight, it would become an infinite force. It’s impossible to hang curtain over a perfectly straight cable. Even a beam sags a little bit but the sag is unnoticeable.

Stress = Actual force/Area

f₁ = Pressure/Area
F₁ = Maximum Allowable Stress
f₁ < F₁ is OK while f₁ > F₁ is NOT OK!

Strain = Deformation
ΔL/L₀ or Change in length/Original length

Types of forces:

Elastic rouge of material deforms and returns to original shape.

Ultimate Breaking Strength or UBS sometimes referred as Ultimate Strength or Breaking Strength is the capacity of the material to withstand a load.

Design Factor (DF), or Safety factor but riggers have strayed from using it because people thought it was a perfect safe system.
WLL = Working Load = UBS/DF

Riggers mostly use a design factor of 5 Throughout the semester we’ll talk about which design factors to use. Design factor should correspond to how much possible negative consequences and account possible unkoown.

New WLL = (Old WLL * Old DF)/New DF

End of notes.


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Notes for First Class- 1/31/17

Hey Everyone,

Here is a briefing on what we learned and did during our first class of rigging.

We started off learning the Square Knot, also known as the Reef or Thief Knot. To practice with your tieline you can take the two ends, place the right side over the left, loop it, place the left side over the right, loop it, and secure the knot tightly. Make sure you do not loop the tieline left over left –> left over left , or vise versa. This can actually create the Granny Knot and the two extensions would be perpendicular rather than going along the line. The Square Knot is best used to hang objects that aren’t going to be moved a lot. It can create on and off tension on the line and loosen or cap-size the knot, which is a bad thing. It needs to constantly be monitored for safety precaution.  In addition, we learned that rope becomes a “line” when you cut it for a specific job or purpose.

Here’s a link to the animated knots website for the Square Knot. McCullough suggested we practice all the knots we know 10 times a day until we can do with our eyes closed. He also explained when you practice you “do it, evaluate it, and modify it.”

We moved along to  “What is Rigging?”

From different answers given in class, we summed it up to “any materials or persons in the air that are being flown in and out of the stage.”

Next we were introduced to the 4 K’s of Rigging

  1. Know the System. Research and check out the system you’ll be working with.
  2. Keep it in good working order. It needs to have regular inspections and/or replacement. Also, know when it’s damaged or worn out.
  3. Know how to use it. Know what it’s meant to be used for and  what’s it not meant to be used for.
  4. Keep your concentration. “When rigging, think about rigging.”- John McCullough ; You need to concentrate on what you’re doing in the moment, no distractions. The seriousness of the subject should be equivalent to the seriousness of  consequences.

Following the 4 K’s of Rigging, McCullough told us another focus is on Rigging Safety.

As a class, we made a list on what to focus on concerning safety.

  • Codes and requirements.
  • Capacity of equipment.
  • Space, size, and position of the objects.
  • Type of equipment.
  • Load- we need to know how much it weighs and where is/isn’t it safe to attach it.

So the focus for safety is on equilibrium- all of the forces on a load balance each other out. With every force bouncing out, we want to make sure our system is “strong enough” to support it.

We also made a list to know how to find the weight of an object

  • Specification Sheets
  • Fish scale or home scales for small objects
  • Measure, calculate, or look it up

This lead us to the math part. We reviewed Newton’s Laws of Motion:

  1. If an object is at rest, it stays at rest. If an object is in motion, it tends to stay in motion.
  2. Force=mass*acceleration (F=m*a)
  3. Every actions causes an immediate reaction.

Then, we participated in a class lab where students had to measure a sandbag and the weight of the sandbag unto a pipe


We concluded that the location of the sandbag affects the scale on whichever side it’s on. If it’s 3/4 to the right then the right scale has about 3/4 more weight added.

Then, we learned how to draw a free body diagram. A Free Body Diagram uses symbols to show how our forces and loads relate to each other. Remember that any distributed force (can be cables, lights, speakers, curtains, etc.) is drawn like a brick bar.

We continued onto the sum of vertical forces, sum of horizontal forces, sum of moments (twisting forces). The formula for moment is Moment=F*d(perpendicular)  The d represents distance.

Math Problem 


As you guys can see, I redid the whole problem with color. The math problem was a bit shaky at first for me too, but I hope the colors and my explanation helps as I go along with how I think the problem got solved. If anyone sees any errors, please point it out for the sake of all of us passing this class. On the left photo, you see arrows. Keep in mind throughout this whole process that those arrows represent the direction of the force and that we’re focusing on twisting forces (sum of moment). So that means that the forces are either moving clockwise or counterclockwise and we will have a pivot point to reference from.

So our goal was to find the sum of the vertical forces. On the right side, you see a picture that demonstrates which direction would be considered positive or negative when we solve for these forces. The last one, sum of moment, is what we’ll focus on. Bare with me.

In order to find the sum of vertical forces, we must find the sum of moment for R1 and R2 individually. As you look unto the picture on the left, you can see that we have more information to plug in numbers for R1. Remember the formula for moment is F*d(perpendicular).

I apologize you can’t see the photo as clear as it is on my phone or the colors, but I’ll keep explaining the process. I’ll bring the sheets to class with me next Tuesday, if anyone needs to look over it to understand it more.

R1 becomes our pivot point and starts at 0′-0″. We have two objects: one that is 4′-0″ away from R1, weighing 300 lbs., and another that is 6′-0″ away from R1 weighing 100 lbs. Then we have a pipe of 12′-0″.

So our equation starts off being equaled to 0.

0= (R1*0′) + (4’*300lbs) + (6’*100lbs) – (R2*12′)

What did we do here? We simply followed the formula for moment, multiplying the force with it’s perpendicular distance.

How did we know whether it was positive or negative? Due to the direction (the arrows) the force was going, keeping in mind that R1 is the pivot point and we’re focusing on finding the twisting force.

As we went along with solving the problem. We came to this point:

0= 1800 ft./lbs -(R2*12′)

R2*12′ = 1800 ft./lbs

R2= 150 lbs

We wanted to make sure there weren’t any more negatives around, so we added (R2*12). This canceled out the negative on the right side of the equation and becomes positive on the left side of the equation. Then to solve for R2, we needed to isolate it. Therefore, we divided 12′. On the right side, the feets canceled out each other, leaving us with only pounds (lbs). We concluded that R2= 150 lbs.

You would think we would be finished there but wait, there’s more!

Side note: I realize now that I didn’t have to rewrite the problems with color because I could have just explained all this by typing it out. Since, I already did all that work, I’ll still be showing the photos.

R2 becomes our new pivot point. Our equation becomes:

0=(R2*0′) -(6’*100lbs) -(8’*300lbs) + (R1*12′)

How did we get 6′-0″ and 8′-0″? The distance from R2 (our new pivot point) and the first object is 6′-0″ weighing 100lbs. The distance from R2 and the second object is 8′-0″ weighing 300lbs. The formula for moment is force * perpendicular distance.

We came to this point in the equation:

0= -3,000 ft./lbs + (R1*12′)

3000 ft./lbs = R1 * 12′

250 lbs. = R1

Remember that we want to get all the negativity out. So that’s why we made -3000 ft./lbs positive by adding it on both sides of the equation. Then to find R1, we needed to isolate it.  We divided 12′ on both sides, feet cancelling each other out, resulting in our answer that R1= 250 lbs.

To double check this is correct, we solved for the sum of Vertical Forces. We basically just added R1 and R2 and the answer was 400 lbs. If you scroll up to our first photo, you will see that 400 lbs is exactly the amount of weight that is shown.

Thanks for your patience and baring with me. I hope this really helps a lot of you. Don’t be afraid to ask questions because we may all need some more clarification on certain things.

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Beginning of Spring 2017

Welcome back. I hope everyone had a good break and that everyone is ready to get back into the swing of things. We have a lot of material to cover this semester, and we start covering serious topics from day one. One this openlab site, you can find all the important information for the class.

The current syllabus will always be posted on the syllabus page, so you can stay up to date.

Homework problems and other assignments are posted on the assignments page.

Additional resources, like video tutorials and manufacturers’ websites are posted on the resources page. Be sure to check it out when you are working on HW 1.

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