# Rigging Question (!)



## derekleffew (Apr 10, 2010)

A truss forty feet long is suspended from five equidistant points. Hanging from the truss is a 40'W x 24'H velour drape that weighs ten pounds per linear foot. The truss weighs five pounds per foot. Ignoring the weight of suspension hardware, what is the load on each point?

Point A (SR) _____
Point B (SRC) _____
Point C (CTR) _____
Point D (SLC) _____
Point E (SL) _____


As always, professionals are requested to refrain from answering for one week, or until the correct answer is posted. (But are always allowed to kibbitz.)


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## photoatdv (Apr 10, 2010)

So the points are the two ends and every 10' in the middle, right? And assuming we are allowed to answer seeing as this was posted by one of our Senior Team...

I'm pretty sure A and E are 75#/ point and BCD are 150#/ point if it is perfectly balanced.

Now for the extent of rigging I do (checking the riggers math when I do lighting)... I'd calculate it as the weight on both sides of each point as a worst case scenario, and as long as that is within the SWL then I'd be good with it.


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## jonliles (Apr 10, 2010)

let me preface with "I am not a rigger, nor will I attempt to do such."

Ignoring Rigger Math, look at it from purely a vector mechanics standpoint, assuming everything is in equilibrium (static situation) you would have 600 # total hanging (200 for the truss, 400 for the drape). Divide that into 5 equal parts, you would have 120# at each load point...just a physics standpoint...


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## Dionysus (Apr 10, 2010)

On a physics standpoint the load is not 'completely divided mathematically among the points', for two points the load is 50/50 (or .5/.5) for three points its .1875/.625/.1825. I am not a professional rigger, however I do know my rigging, and I aced my rigging class in college. Thus I will refrain from answering the question myself. Notice I did not give figures for 4 or 5 points. You'll have to figure those out kiddies.

I like the fact that the question DOES NOT ask for what you think anything should be rated, or involve adding in any additional information thus helping to keep it out of 'dangerous territory', among other things.

Rigging is a lot more complicated and touchy than many people give it credit for, there is a REASON why it should be left to the professionals.


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## photoatdv (Apr 10, 2010)

Or, you could just use 2 ton rated components and not care exactly what is on each point . Seeing as 2000/600 is a 3.3 : 1 safety margin, and 600# out of 600# on a point means you have bigger problems than maintaining a 3:1 (5:1, 8:1, 10:1, whatever you want) safety margin.

[/end sarcasm]


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## DuckJordan (Apr 11, 2010)

Do i have time to consult my building engineer, my lawyer, my insurance company, my TD, my building owner, My public educator, and my rigging company? if not then i will kindly refuse to answer such question.


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## Footer (Apr 11, 2010)

DuckJordan said:


> Do i have time to consult my building engineer, my lawyer, my insurance company, my TD, my building owner, My public educator, and my rigging company? if not then i will kindly refuse to answer such question.



You guys can answer. You are not specifying the rigging package that is holding it up, just what each point will be holding. This is not against the TOS. 

Lets think of it another way if you don't feel comfortable discussing rigging...

You have the same truss. You have the same curtain attached to the truss. 

5 men are standing on a balcony and holding the truss with the soft good tied to it. All the of weight of the truss and the good attached is being supported by the men. The men are evenly spaced out at 10' intervals on the truss. How much weight will each man be holding?


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## MarshallPope (Apr 11, 2010)

My instincts would tell me that it would be 75# for the outside points, and 150# for the inner points, agreeing with photoatdv. I have never done anything of this sort before, though, so I could be forgetting something major. I divided the truss into 8 5' sections, each having a weight of 75#. I would assume that each lift point would support the sections on either side of it, with the outer sections only having weight on one side. As such:

|---+---+---+---|

supported as:

|-/-+-/-+-/-+-/-|
(the red slashes mark where I see the weight shifting from one lift line to the next.)


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## coldnorth57 (Apr 11, 2010)

Hmmmmmmm very intresting ....I found a little program that helped me 
"Rig them all..."
Kool


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## DuckJordan (Apr 11, 2010)

rounding up it would be about /--60#--172#--140#--172#--60#--\


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## JChenault (Apr 11, 2010)

Not a rigger - but I did once take a course in 'Structural engineering for Architects'. It seems to me the answer lies in what the OP means by '5 equidistant points'. If he means that each pick point is 8 feet apart, and the end pick points are 4 feet from the end of the truss, I believe the weight would be evenly distributed on each pick point ( IE 120# ). 


Now if you have a pick point at each end of the truss, and are spaced 10 feet apart, I agree with marshalPopes analysis ( assuming that the truss is rigid and that bending effects can be ignored).

Given no restrictions on pick points - what would be the correct locations to choose? It seems to me that one would logically place the pick points at 8 foot intervals. If I had five guys ( very strong guys) carrying a 40 foot beam that weighed 600 lb's, I would expect that they would not put a guy at each end, and space the guys at 10 feet, but instead would space themselves at 8 feet.

But I will be interested in hearing what the guys who do this for real say.

Good discussion.


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## gafftaper (Apr 12, 2010)

photoatdv said:


> And assuming we are allowed to answer seeing as this was posted by one of our Senior Team...



 It's a trap. Don't answer.

Actually the answers submitted have already shown exactly why rigging is such a touchy subject. You can't just assume that the logical answer to you, is the correct one.


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## What Rigger? (Apr 12, 2010)

gafftaper said:


> It's a trap. Don't answer.



Ah yes, the immortal words of Admiral Ackbar. 

Actually, isn't this a scenario that illustrates why you should just pull out the ol' iPhone and use the iRigging app? (I am JOKING here, folks. It's so seldom that humor arises in rigging...sigh)


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## TechSooth (Apr 12, 2010)

Footer said:


> You guys can answer. You are not specifying the rigging package that is holding it up, just what each point will be holding. This is not against the TOS.
> 
> Lets think of it another way if you don't feel comfortable discussing rigging...
> 
> ...




If the five guys are :

Point A (SR) ______Footer (the confident) __________ 75#
Point B (SRC) _____gafftaper (sweating profusely) ___ 46#
Point C (CTR) _____Dionysus (WTF???) ____________ 358#
Point D (SLC) _____derekleffew (keepin it taught!) ____ 1#
Point E (SL) ______jonliles (physics rules!) _________ 120#

Or... if the same truss and good are supported on five ladders, what is the vertical load applied to each ladder (not a rigging question 'cause what riggers own ladders)?

sum of the forces vertical = +15#/' x 40' + sum of the reactions vertical (-600#)= 0

if the five reactions are spaced 8' apart and the load is centered on C, thereby subdividing the load into 5 equal loads:

Point A (SR) ______120#
Point B (SRC) _____120#
Point C (CTR) _____120#
Point D (SLC) _____120#
Point E (SL) ______120#


if the five reactions are spaced 10' apart and the load is centered on C, thereby subdividing the load into 4 equal loads: 

Point A (SR) _______75#
Point B (SRC) _____150#
Point C (CTR) _____150#
Point D (SLC) _____150#
Point E (SL) _______75#


the real question is -- What is the load applied to the reactions when lifting the mass at 32'/sec, or when the SL wagon hooks the chain pocket as it travels SR during the 17 second change between II-2 and II-3?


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## derekleffew (Apr 12, 2010)

DuckJordan, can you tell us how you arrived at your answer? 

(CB never ceases to amaze me. Less than four hours after I posted the question, Arez submitted the answer I was looking for. In order to promote discussion, I asked him to delete it.)
http://www.controlbooth.com/forums/members/duckjordan.html


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## KeepOnTruckin (Apr 12, 2010)

Since I don't know much about the mathematics of rigging, I would say that whatever you use needs to be rated for 10 times the load it might have to carry. 200 lbs for the truss and 400 lbs for the curtain = 600 lbs, so just use hardware that is all rated for 6 tons and as long as it is hung and connected properly by a trained person, you will be ok.


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## photoatdv (Apr 12, 2010)

You mean 3 tons .


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## DuckJordan (Apr 12, 2010)

derekleffew said:


> DuckJordan, can you tell us how you arrived at your answer?
> 
> (CB never ceases to amaze me. Less than four hours after I posted the question, Arez submitted the answer I was looking for. In order to promote discussion, I asked him to delete it.)




i cheated and looked online for some rigging formulas i wouldn't trust them but it didn't seem too off.


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## mstaylor (Apr 13, 2010)

Where you pick it makes a huge difference on the outside motor points. I would generally pick it about two feet in from the ends. Others have said to pick it at four feet. The further in you pick it, the more it adds to the end motors because of the canterlever effect. The difference between two ft and four ft isn't going to be big difference but should be remembered. I have seen guys pick rag trusses in ten or twelve ft thinking it is no big deal but it can lead to huge problems.


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## philhaney (Apr 13, 2010)

I'm an aspiring rigger, so I went to Jay O. Glerum's book, _Stage Rigging Handbook_, section 3.04.B.2 Batten Rigging. He says to calculate the load on a beam supported by more than two points we can use the three-moment theorem which is beyond the scope of the book. However, he does provide a handy diagram from Peter Albrecht Corp. that shows the reactions of an evenly distributed load on loft lines and loft blocks as percentages of total load on the batten. The numbers crunch out as follows:

Truss: 40 ft @ 5 lb/ft = 200 lbs
Drape: 40 ft @ 10 lb/ft = 400 lbs
Total Load (not counting the rigging hardware): 600 lbs

Point A (SR) 9.8% of total load = 600 lbs x .098 = 58.8 lbs
Point B (SRC) 28.6% of total load = 600 lbs x .286 = 171.6 lbs 
Point C (CTR) 23.2% of total load = 600 lbs x .232 = 139.2 lbs
Point D (SLC) 28.6% of total load = 600 lbs x .286 = 171.6 lbs 
Point E (SL) 9.8% of total load = 600 lbs x .098 = 58.8 lbs

Which, if you round up, are the numbers DuckJordan came up with.


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## Dionysus (Apr 13, 2010)

Remember that the percentages in that diagram assume some things:

1) The End lift lines are at the END of the batten/truss
2) All lines are perfectly evenly spaced
3) The load is symmetrically distributed onto the batten/truss
4) If in the air it is level (ie not pulled up on one end or slack on one point, causing some points to take more load)

Thanks philhaney! Ding!


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## erosing (Apr 13, 2010)

Derekleffew asked me to repost my original answer and how I came to it. 

Original Post: 
Assuming I'm correct in what a linear foot is, a 600 pound load on a batten with 5 points would be:

Point A (SR) 58.8 lbs.
Point B (SRC) 171.6 lbs.
Point C (CTR) 139.2 lbs.
Point D (SLC) 171.6 lbs.
Point E (SL) 58.8 lbs. 

How I came to my answer: 
I was taught that for a 5 point load (our most common usage, conveniently) to calculate it as follows: 10%, 30%, 25%, 30%, 10%, if you add that all together you get 105% of your total load. The reasoning I was taught for this is that it gives you a little bit of play (which could be argued as not the best thing to say to a student) if your load total ends up being slightly off or if your points are not spaced out exactly as the intended formula (see below). However, I have an unhealthy attraction to hemp rigging so I have read a number of books on the subject, in this case I knew that Jay Glerum's book, Stage Rigging Handbook contained the exact percentages for finding the point loads on an evenly distributed batten under the hemp rigging section (for a 5 point batten: .098, .286, .232, .286, .098). Since Derekleffew's wording was, "from five equidistant points," I grabbed my book and used the exact percentages.


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## derekleffew (Apr 13, 2010)

Props to Arez, DuckJordan, and philhaney. 



JChenault said:


> ... If he means that each pick point is 8 feet apart, and the end pick points are 4 feet from the end of the truss, I believe the weight would be evenly distributed on each pick point ( IE 120# ). ...


Brings up an interesting point: *Trusses are usually rigged from the ends, but battens usually have some cantilever.
*
Anyone know how to (or want to) do the math on a beam/truss/batten supported thusly?

_4'_*A*__8'__*B*__8'__*C*__8'__*D*__8'__*E*_4'_


Terms to Google: statically indeterminate structure, theorem of three moments.


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## mjw56 (Apr 22, 2010)

I know how the 3 members listed above got to their answer, (just did this 2 days ago in mechanics of materials) but i don't agree. the load is evenly distributed and acts on all the reactions equally except for the end 2. just to check i popped the problem into FEA software, and got equal reactions except for the ends which were half. A POINT load with statically indeterminate analysis is the only way to get the load distribution shown in their answers. 
Can anyone explain?

the attached picture shows the shear and moment diagram for the new question( 8' spacings with cantilevers). mathematically the loads are all equal because the tributary area of each reaction is equal 
8'x15#/LF=120# 120#x5=600#
but the FEA anlysis come out a little differently.

the loads tally left to right as: 126.43;111.43;124.28;111.43;126.43
still = 600# but best i can figure is that the "imperfect" shear stresses are related to the nature of the "pin" reaction and deflections.


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## dcollins (Apr 22, 2010)

I wasn't going to respond, then Dave posted a link on the SML and I figured I would...I am not a rigger, but I am an engineering student and figured I would try it out

There are five unknowns, therefore we need five distinct equations. We can assume symmetry, that gives us RA=RE and RB=RD. We have the sum of the forces acting vertically must be zero, or 600=2RA+2RB+RC. We have the three moments equation applied to points ABC and applied to BCD.

For the three moments theorem, we need to do some fancy math. I worked out the shear and moment diagrams on paper. Bending moments are:
MA=0, MB=10RA-750, MC=20RA+10RB-3000, MD=30RA+20RB+10RC-6750, ME=40RA+30RB+20RC+10RD-12000.you simplify that equation
We can check this result by finding that the bending moment at each end is zero. ME=0, and if you simplify that equation it is the same as the first equation for forces.
For the first three moments calculation:
0*10+2*(10RA-750)*20+(20RA+10RB-3000)*10=1/4*15*10^3+1/4*15*10^3
And for the second:
(10RA-750)*10+2*(20RA+10RB-3000)*20+(30RA+20RB+10RC-6750)*10=1/4*15*10^3+1/4*15*10^3
Solving these yields, according to my calculator, RA=91.1, RB=128.6, RC=160.7

Upon realizing that I am the only one with this answer I ran the calculations again. Here are all the equations:
(VAR is the shear force (V) just slightly to the right side (R) or point (A), the bending moments are calculated by (Area of the triangle)-(Width)*(Displacement below 0))
RA unknown
RB unknown
RC unknown
RD=RB
RE=RA
VAR=RA
VBL=VAR-150
VBR=VBL+RB
VCL=VBR-150
VCR=VBR+RC
VDL=VCR-150
VDR=VDL+RD
VEL=VDR-150
MA=0
MB=MA+10*(VAR-VBL)/2+10*VBL
MC=MB+10*(VBR-VCL)/2+10*VCL
MD=MC+10*(VCR-VDL)/2+10*VDL
ME=MD+10*(VDR-VEL)/2+10*VEL
RA+RB+RC+RD+RE=600
MA*10+2*MB*20+MC*10=1/4*15*10^3+1/4*15*10^3
MB*10+2*MC*20+MD*10=1/4*15*10^3+1/4*15*10^3


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