We are privileged here at RigWorld to have Delbert Hall amongst our membership. I’ve created this page for Delbert as he has graciously agreed to share some of his knowledge here with us.
Delbert has recently published another new book with Brian Sickels
With nearly 400 pages of text and over 550 illustrations, The Theatre Riggers’ Handbook is the most comprehensive book on theatre rigging available. Written by experienced riggers Hall and Sickels, authors of The Arena Riggers’ Handbook, this book clearly describes all aspects of theatre rigging, including hardware, rigging math and techniques, installations, fire curtains, concert shells, hemp and manual counterweight rigging, automated systems, aerial rigging, rigging safety and inspections, and more. Whether you are a beginning rigger or a seasoned professional, The Theatre Riggers’ Handbook is a “must have” for your professional library.
Delbert has recently published a new book with Brian Sickels
Whether you are a student technician or a union rigger, The Arena Riggers’ Handbook is a “must have” book for your library. Written by experienced and certified riggers, this book clearly describes all aspects of arena rigging, including: hardware, rigging techniques, electricity, rigging math, safety and more. It even includes an arena rigging quiz to help you access your preparedness for taking an arena rigging certification exam.
Delbert Hall, Ph.D., is head of the design and technical theatre program in the Division of Theatre and Dance at East Tennessee State University, where he has taught since 1986. Dr. Hall received his Ph.D. from the University of Florida, his MFA from the University of North Carolina at Greensboro, and his B.S.Ed. from Western Carolina University. He is the U.S. Institute for Theatre Technology -Southeast Region’s Outstanding Educator in Theatrical Design and Technology for 2000-2001, and received an Outstanding Achievement Award for special effects rigging from the Kennedy Center for the Arts in 2000. Delbert is the president of D2 Flying Effects, LLC, and has been rigging performer flying effects and aerial dance apparatus for over 30 years. He is an ETCP certified rigger (Theatre) and an ETCP Recognized Trainer. Delbert regularly teaches workshops on rigging and rigging math and is the originator of RigCalc. Delbert also does destructive testing of hardware used in the entertainment industry and fabrics used in aerial dance in order to better understand how these items break and at what loads.
This month we are lucky enough to have a preview lesson from the second edition of Rigging Math Made Simple. I had the pleasure of being one of the Technical editors of this book and it is a valuable tool to have in your library, it will be available in mid January and can be purchased through SpringKnollPress.com or Amazon.com. If you have any questions, please post in the comments section at the bottom of the page and Delbert will be happy to respond.
Delbert has released several great Rigging Tools with instructions in the new edition of the book. Make sure to download Delbert’s Excel workbooks here: Rigging Math Downloads
Two-Point Bridles in an Arena
In Lesson 5 we discussed how to calculate the length of a bridle based on knowing the Horizontal and Vertical distance the Bridle Point would be from the Hanging points. In other words, the head rigger needs to determine the exact position of the Bridle Point and then calculate the lengths of the bridle legs that would create this position. While this will work, it is not always the way it is done.
For example, the head rigger will need the Bridle Point at a specific location in plan view, but instead of needing to put the Bridle Point at an exact height, the head rigger may want it at a minimum height above the deck (high enough to get the truss at the need height). There is also a maximum height above the deck for the Bridle Point, which is determined by the point where the bridle angle reaches 120 degrees (90 degrees is preferred). As long as the Bridle Point is between the minimum and maximum heights, all is well because the chain hoist will control the height of the truss. To understand why the head rigger does not just choose an arbitrary point between the minimum and maximum heights, we need to understand the resources for making bridle leg.
As discussed in Lesson 18, common lengths of steel slings that rigging companies have include: 2’, 2.5’, 5’, 10’, 20,’ 30’, and 50’ lengths. Baskets and bridle legs will be made from a combination of these standard length slings. When a bridle leg is an “odd length,” one that cannot just be made from the standard lengths listed above, links of Deck Chain are added to make the desired length bridle leg.
Note: Actual deck chain is very seldom used anymore – it has been replaced by Special Theatrical Alloy Chain (sometimes called Special Theatrical Adjusting Chain) or STAC chain. Many, if not most, riggers use the term “Deck Chain” when referring to “STAC chain.” Throughout the rest of this book, I will use the term “Deck Chain” when referring to “STAC chain.”
STAC chain, which can be purchased in 3-foot and 5-foot lengths, is VERY expensive, especially when compared to the price of cable slings, so rigging companies typically carry only a minimum amount of this chain for rigging. The fewer components in the rig, the fewer chances for mistakes. Because of this, it is usually necessary that at least one bridle leg be able to be made by using ONLY the standard length steel slings. Not using any Deck Chain in a bridle is even better.
In the drawing below, we have a bridle where the Bridle Point (also called the “apex”) needs to be 10 feet from one I-beam and 15 feet from the second I-beam. This establishes a vertical line on which the Bridle Point must be placed, but where on that line? We see that the I-beams are 80 feet above the deck, and have determined that the minimum height of the bridle point needs to be 40 feet above the deck, and the maximum height of the bridle point is 65 feet above the deck. So, the Bridle Point can be anywhere along the 25-foot long space between these points.
As you see, I have indicated that the first bridle leg is 25 feet long. You might be asking how did I come to choose this length? First, it is a bridle length that I could make with standard length slings (a 20 footer and a 5 footer); and second, it is a distance that I knew from experience would reach from the I-beam to “somewhere” on the line, between the minimum and maximum height points. Of course, if you are not sure if a certain length will work, make a quick, scaled drawing and measure. I will give you more hints on estimating this length shortly, but first let’s use this length.
It is important to note now that this leg will be longer than 25 feet. Why? Because I have not added-in the Effective Length of Hitch (ELOH) for the basket, and because I have not added-in the additional length of the two shackles in the leg (remember, each piece of steel will require a shackle). Assuming we use a basket make with a 5-foot sling that is going around a 12” x 9” I-Beam (the example in Lesson 18), we get:
Leg 1 = 25 + 1.6093 + 0.3966
Leg 1 = 27.01
Remember, this bridle leg, as calculated, is made without any Deck Chain.
Not only do we know the length of the leg, but also we now have two of the three lengths that make up the bridle triangle. The bridle leg (the hypotenuse) is 27.01 feet long and one leg (the Horizontal distance) is 10 feet long. With this information we can calculate the third leg of the triangle (the Vertical distance). Why this distance is important will become clear soon.
So, using the following variation of the Pythagorean Theorem to calculate this distance using the equation:
V =[math]\sqrt(Leg x Leg) – (H x H)[/math]
So, plugging in the numbers, we get…
V =[math]\sqrt(27.1 x 27.1) – (10 x 10)[/math]
V =[math]\sqrt729.5401 – 100[/math]
V = 25.09 feet
Before we go on the next step, as promised, here are a couple of suggestions that might help you in estimating the length of the first bridle leg, if you choose not to make a drawing and measure.
Option 1: After you estimate the bridle length and calculate the height of the Bridle Point from the beams, subtract that number from the height of the beam and compare it to the minimum and maximum Bridle Point heights. Using the problem above:
Bridle Point Height = 80 – 22.9
Bridle Point Height = 57.1 feet
Since this measure is greater that the minimum Bridle Point height (40 feet) and less than the maximum Bridle Point Height (65 feet), it is “Good.”
Option 2: Take the beam height and subtract the minimum Bridle Point height (80 – 40 = 40 feet). This is going to the somewhere near the maximum length of the bridle. We also know that the bridle length must be greater than the distance from the beam to the maximum Bridle Point height (80-65 = 15 feet). A good compromise length is often somewhere in between, so 40 – 15 = 25 feet.
As we now are beginning to see, several standard bridle lengths would have put the Bridle Point between the minimum and maximum Bridle Point heights.
Option 3: We can see the span of lengths that would work by calculate the bridle lengths if the Bridle Point was at both the minimum and maximum Bridle points heights that falls between these numbers.
Leg Length (Min) =[math]\sqrt(15 x 15) + (10 x 10) [/math]
Leg Length (Min) =[math]\sqrt(225) + (100) [/math]
Leg Length (Min) =[math]\sqrt325 [/math]
Leg Length (Min) = 18 feet
Leg Length (Max) =[math]\sqrt(40 x 40) + (10 x 10) [/math]
Leg Length (Max) =[math]\sqrt(1600) + (100) [/math]
Leg Length (Max) =[math]\sqrt1700 [/math]
Leg Length (Max) = 41.23 feet
So, the following bridle lengths would all have worked: 20’, 25’, 30’, 35’, and 40 feet. Whichever one you choose, you will need to calculate the distance from the beam to the bridle point.
So, why is the 25.09 foot distance so important? It is because we need to know this distance in order to find the length of the other bridle leg. We now know the Horizontal distance (15 feet) and the Vertical distance (25.09 feet), so we can plug them into the Pythagorean Theorem and find the length of the leg. (This assumes both beams are at the same height above the deck).
Leg =[math]\sqrt(15 x 15) + (25.09 x25.09) [/math]
Leg =[math]\sqrt(225) + (629.54) [/math]
Leg = 29.23 feet
Finally, we need to figure out how to make a bridle leg that is 29.23 feet long.
Figuring out how to create this bridle will take a bit of trial and error. However, before you begin you should know that there is no one correct answer, and how you do it will be based on what lengths of Steel and Deck Chain are available to you. But, let’s look at you might approach this problem.
We can get most of the way to 29.23 feet by using one 20-foot and one 5-foot pieces of Steel (and the lengths of two shackles). So,
L2 = 20 + 5 + 0.1983 + 0.1983
L2 = 25.396 feet
We still need 3.8334 feet of leg.
We know we need a basket, so let’s add 1.6093 feet to our leg for a 5’ basket.
L2= 25.3966 + 1.6093
L2 = 27.0059 feet
That still leaves us 2.2241 feet short. The big question now is “how do we create this length?”
The answer depends on what we have to use. Deck (STAC) chain is a common material, but if we have a 2-foot long piece of Steel, that could be used as well. No matter what we use, we need to remember that each new piece of hardware will require an additional shackle (and we can also use shackles to add length). Let’s look at two ways two ways to create that additional length of the leg.
Since we know that we will need a shackle, subtract the length of the shackle (0.1983 feet) from desired length (2.2241 feet). We get 2.0258 feet. If you added a 2-foot Steel here, the remainder would be a mere 0.0258 feet, or 0.3096 inches. That is REALLY close.
If we do not have a 2-foot Steel, we will need to make this length entirely from Deck (STAC) chain. Again, we subtract the length of the shackle (0.1983 feet) from desired length (2.2241 feet to get 2.0258 feet. Now, to determine the number of needed links of chain, we divide this length by the internal measurement of one link of Deck Chain. One common size of Deck Chain had 3.174 inches (or 0.2645 feet) links, so we will use this measurement.
Number of links = 2.0258/0.2645
Number of links = 7.658
Since you cannot add a fraction of a link, you have to choose either 7 or 8 links. Seven links would be a little short of the desired length and eight would be a little longer than the desired. Of these two, eight is the closer, but it is still more than an inch off.
Another option is to add seven links plus an additional shackle. This would create 2.0498 feet of addition length. This is only 0.024 feet (0.288 inches) longer than the desired addition length. Again, REALLY close to our desired length.
If a finer adjustment was needed, you could have added the shackle to the basket. This would have added only half of the length of the shackle to the length of the leg.
As stated earlier, there is more than one possible solution to this problem. Had we selected to use a longer basket, the ELOH would have been greater, which would have resulted in different lengths of Steel and Deck Chain needed. It would be extremely time consuming to manually calculate every single configuration for making up a bridle legs. In Lesson 20, we will look at tools that can help you look at lots of options very quickly.