# Rigging angled trusses



## derekleffew

STUDENTS ONLY FOR ONE WEEK PLEASE. 06/12/11: Question now open to all.

So my Lackadaisical Designer (henceforth known as LD) just sent me the FINAL (yeah, right) light plot for our upcoming tour. Included are two angled sidelight trusses. Here's the SR one (reverse&repeat for SL).



Truss will be two sections of 10' Tomcat MD2020, weight: 85 lbs. per section.
Fixtures are VL3000Spot, weight: 91 lbs. each, but I'm calling it 100 lbs/ea to account for clamps, cable, etc.

Now the kicker: LD wants the truss to hang at ~45°, with the DS point at +30' and the US point at +16'. 
*
How do I calculate the weight on each point (excluding suspension hardware)?* If the truss were level (horizontal), it would be (2x85)+(6x100)=770; /2=385 lbs on each point. If the truss were vertical, the top point would take all the weight, 770 lbs. But what if it's halfway? 

Venues are going to want the exact load per point. The house riggers will handle any bridles that need to be done, to accommodate house structural steel/points.


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## mstaylor

Are the 30ft and 16 ft mean height from floor? I would be more concerned with where on the truss you plan to pick it.


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## derekleffew

The truss will be picked at the ends, with SpanSet s+steel back-up or GAC Flex.

Funny, the designer didn't specify whether the dimensions were from the deck or the arena floor. I'm also assuming the trim is to the bottom truss chord. I don't think any of this has any bearing on the original question though, or does it?


Sent from my iPhone using Tapatalk


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## mstaylor

Only to show the angle. That was the reason for the question, I wanted to take that confusion out of the equation. A fourteen foot difference states the angle and knowing whether you are end picking or in by the first V makes a difference in the math. Either way, a 14 ft difference on a 20ft truss doesn't make a 45 degree angle.


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## LavaASU

mstaylor said:


> Only to show the angle. That was the reason for the question, I wanted to take that confusion out of the equation. A fourteen foot difference states the angle and knowing whether you are end picking or in by the first V makes a difference in the math. Either way, a 14 ft difference on a 20ft truss doesn't make a 45 degree angle.



14 ft on a 20 ft truss would roughly make a 45 degree angle. The 20' truss is the hypotenuse of an equilateral triangle. Therefore with the vertical distance as A, the horizontal distance as B and the truss (angled length) as C:
A^2 + B^2 = C^2
14^2+14^2 = 20^2
196 + 196 = 400
392 = 400

Thats pretty close on a truss someone's trying to rig at an angle by sight.


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## mstaylor

That's why I asked if it was height differential or pick points. If not the numbers were wrong because of the 30. I will step out of the math now and let the students have at it. Just remember if you properly pick it with spansets it will be picked top and bottom so the low end actually won't be at the end. The bottom pick will be but the top will be further up the truss. How far will depend on the size, 12, 15 or 20,5 inch truss.


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## derekleffew

LavaASU said:


> ...The 20' truss is the hypotenuse of an equilateral triangle. ...


Err, isosceles right triangle.


Isosceles Right Triangle -- from Wolfram MathWorld


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## LavaASU

derekleffew said:


> Err, isosceles right triangle.
> 
> 
> Isosceles Right Triangle -- from Wolfram MathWorld



Opps... yes that's what I meant. Thats what I get for doing math in the wee hours of the morning.


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## shiben

I cant figure out how to do it even after reading The Stage Rigging Handbook... I might have missed it tho... Its kinda late.


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## BrianWolfe

Don't we need the spacing on the lights to do this calculation accurately?


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## sk8rsdad

It may be fair to assume uniform distribution since Derek assumed so in the original question. If not, then summing the weights at various points can be determined from the manufacturers information for the various components in conjunction with the LD's diagrams. 

If the load on the pick point comes down to the decimal point then the truss should not be lifted. Some rounding is inevitable since the square root of 2 forms part of the trigonometry of this question. Bringing in the dynamic loading while the piece is being lifted and any additional dynamic loading while the VL3000s are in motion would really complicate things.


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## erosing

770/2= 385 lbs
385*1.414= 544.39 lbs
385* (1 or .6760)= 385 lbs or 260.26 lbs

Those are my quick guesses, but I'm really working off the assumption that I can find the answer by calculating the LAF and that I did so correctly, which I'm fairly sure I didn't since I couldn't reverse engineer the opposing angle with it.

So my silly answer for the day is this, is the LAF 1.0, meaning that the weight on each point would still be 385 lbs? If the LAF is (770/2)x(S/H) where S is sling length and H is height, in this instance, S and H are the same and must equal 1.


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## sk8rsdad

Arez said:


> 770/2= 385 lbs
> 385*1.414= 544.39 lbs
> 385* (1 or .6760)= 385 lbs or 260.26 lbs



544.39lb + 260.26lb = 804.65 > 770lb. Does angling the truss actually make it heavier, or is there something funny in the calculation?


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## erosing

sk8rsdad said:


> 544.39lb + 260.26lb = 804.65 > 770lb. Does angling the truss actually make it heavier, or is there something funny in the calculation?


 
Not more weight, just more tension. 

Also, looking back, I made the possible grave choice of assuming that the truss would be suspended from 2 points at 90 degree angles so that they are directly above.


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## shiben

sk8rsdad said:


> 544.39lb + 260.26lb = 804.65 > 770lb. Does angling the truss actually make it heavier, or is there something funny in the calculation?


 
I would imagine that the lower pick would take more of the mass of the truss, but Im not a professional rigger so I dont know for sure.


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## sk8rsdad

shiben said:


> I would imagine that the lower pick would take more of the mass of the truss, but Im not a professional rigger so I dont know for sure.


 
Certainly the pick supporting the low end of the truss takes more of the weight. The QOTD is interested in how much weight is transferred from the pick supporting upper end of the truss to the pick supporting the lower end.

However, the proposed solution has the downward force vector somehow increasing. Newtonian physics say that in a system in static equilibrium all the forces acting on the system are equal. Rotating the truss does not change its mass, nor does it change the way gravity acts upon it. Since force = mass x acceleration, and both mass and acceleration (gravity) are unchanged by the orientation of the truss, there can be no increase in the downward force vector. 

Arez is on the right track by thinking about the angles. It helps to analyze the force vectors at each bridle in order to determine the magnitude of the downward force vector at each pick point.


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## DuckJordan

sk8rsdad said:


> Certainly the pick supporting the low end of the truss takes more of the weight. The QOTD is interested in how much weight is transferred from the pick supporting upper end of the truss to the pick supporting the lower end.
> 
> However, the proposed solution has the downward force vector somehow increasing. Newtonian physics say that in a system in static equilibrium all the forces acting on the system are equal. Rotating the truss does not change its mass, nor does it change the way gravity acts upon it. Since force = mass x acceleration, and both mass and acceleration (gravity) are unchanged by the orientation of the truss, there can be no increase in the downward force vector.
> 
> Arez is on the right track by thinking about the angles. It helps to analyze the force vectors at each bridle in order to determine the magnitude of the downward force vector at each pick point.


 

Ah but the newton physics assume a single load not different points. The Physics from newton would have to be applied to each individual pick point. Which is where the issue is.


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## shiben

DuckJordan said:


> Ah but the newton physics assume a single load not different points. The Physics from newton would have to be applied to each individual pick point. Which is where the issue is.


 
Well... You could do a summation of forces, I just have zero idea how to figure out the fraction. Actually... might there be a force multiplier? I remember something about this from physics...


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## Sigilion

If both of the lines holding up the points are vertical, the Newtonian rotational physics indicates that both points must have the same weight, so that the torques sum to zero.


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## erosing

Okay, so I took some liberties in an attempt to get an answer, but that's easy to fix later, mostly by floor to grid height, and assuming that I can't calculate exactly where the pick will end up being on the truss, so I don't know the exact lengths.





So I did a small experiment and I found that as close as I could get the item to 45º it was close to half the total weight, but I couldn't get it to register at 45º true. Given the margin of error it should be pretty close to an even distribution at 45º, if not exactly. but anything less than 45º and it the load on the bottom point will increase a lot.


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## mstaylor

However big the truss is, 20.5", 12" or 20" will be how far from the end the bottom pick will be so the width of the pick has to be adjusted accordingly.


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## SanTai

I don't know if I have missed some information, since I do not am fluent in English rigging terms.

If I am not misstaken doesn't Newton tell us that if assuming uniform mass distrubution and that the wires run straight up, we get even even weight distribution on the load points.(No resulting forces or momentum on the system) So unless there is some other kind of hanging point distribution than the ideal one, it will be half on each point.

If I have not missed some info(which is very possible since I am new to theatre and rigging and not nativity English speaker), I would say we need to know how long the space between the load points in ceiling they have to calculate the load on each of the points.


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## BrianWolfe

Agreed. 385 lbs on each point.


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## kylebjordahl

As long as the picks are both vertical (from the truss to the bridle/pick on the ceiling) it doesn't matter how the truss is angled. The compound forces within the bridles (if needed) are just a breakdown of half the total weight (per bridle).

Now, if you were going to pick it from a non-vertical position, that would be a different story. Don't know if that's a good idea or not, so no comment.


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## JChenault

derekleffew said:


> If the truss were level (horizontal), it would be (2x85)+(6x100)=770; /2=385 lbs on each point. If the truss were vertical, the top point would take all the weight, 770 lbs. But what if it's halfway?



I'm not site I agree with your statement re the weight if vertical. If the truss were vertical and there were two lines, it seems to me that I could support the truss with each line taking a strain of 385. I could also use unequal strains if I desired. Now since both lines are at the same pick point, that point needs to support 770 but that is because you have both lines to the same point, not be because the truss is at an angle to the floor.

So I am on the side of those who say that the angle makes no never-mind.


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## sk8rsdad

I'm on the side of the angle makes no never-mind as long as you can neglect the center of mass of the truss. In this configuration the center of mass of the truss will most likely shift resulting in a non-zero increase on the lower end of the truss. In the world of zero depth beams, frictionless pulleys and massless strings, the angle doesn't matter.


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## JChenault

sk8rsdad said:


> I'm on the side of the angle makes no never-mind as long as you can neglect the center of mass of the truss. In this configuration the center of mass of the truss will most likely shift resulting in a non-zero increase on the lower end of the truss. In the world of zero depth beams, frictionless pulleys and massless strings, the angle doesn't matter.



It's not clear to me that the load on an unevenly loaded truss that is parallel to the floor would be different from one at 45 degrees. Am I missing something?


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## rochem

This was actually a subject that I was really interested in back a few months ago, and I spoke to a number of rigging professionals to try to figure out how one would actually calculate the loads and forces involved. Interestingly, the responses I received from some of the most respected riggers in our industry varied widely, from "it's identical to a truss parallel to the floor" all the way to "it requires all these calculations with center of gravity" and so on and so on. All the responses, however, stated that they just calculated the full load of the truss on each individual point whenever hanging angled truss in order to prevent against possible overloading. So it seems like, at least in one faction of the industry, this isn't an important calculation.

As I understand it (with the help of an old physics professor), the change in loads is caused by the changing of the positioning of the truss itself in relation to the points. As you lift one of the points of the truss, the center of gravity of the truss starts to shift, causing it to move horizontally. Assuming the points were originally directly above their attachment points on the truss, each pick point would no longer be completely vertical, thus adding a horizontal force component and changing the weight distribution on each point. As it was explained to me, if the supports were something rigid (steel beams or ground supported truss) the forces would remain the same, but since the aircraft cable/chain holding up the truss has the ability to move off vertical, this changes the equation. Again, I don't know if any of this is right, but it's what seemed the most logical to me when I was trying to figure this out, and perhaps it will provoke new discussion.

Out of curiosity - does anyone actually know the answer to this (and how exactly to get it), or is everyone actually stumped. You don't need to step in and answer it just yet, I'm just wondering because it's been open to all responses for almost two weeks, but no one's given the answer.


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## JonCarter

I worked for a very experienced, very good T.D. many years ago who used to say, "You're building a SHOW, not the Empire State Building. Don't 'mother' it!" So, let's see now, 770 lbs. total. From the Cordage Institute standards: 3/4" manila rope breaking strength = 4860#. For 5:1 safety factor, load no higher than 972# "Give me 4, 3/4" manila spot lines, load on each NTX 800#, at the locations shown on the plans." Problem solved. (This from an old timer who hung one H. of a lot of stuff over the years and never dropped anything.)


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## shiben

JonCarter said:


> I worked for a very experienced, very good T.D. many years ago who used to say, "You're building a SHOW, not the Empire State Building. Don't 'mother' it!" So, let's see now, 770 lbs. total. From the Cordage Institute standards: 3/4" manila rope breaking strength = 4860#. For 5:1 safety factor, load no higher than 972# "Give me 4, 3/4" manila spot lines, load on each NTX 800#, at the locations shown on the plans." Problem solved. (This from an old timer who hung one H. of a lot of stuff over the years and never dropped anything.)


 
Well with that Max load you just need 2 of them. But my gut would say tell the house riggers to plan for 770lbs per point, just in case the designer decides that he wants a pile more MLs on the truss.


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## mstaylor

When the ttruss is flat or the truss is at rest in the 45 degree position the numbers will be equal between motors. The problem comes when moving from flat to 45, the lead motor will take somewhat of a dynamic load but even out as it gets to trim. 
Either way, the loads are so light you aren't going to exceed any part of the system except possibly the building steel. Even if you are using 1/2 ton motors everything else is going to far exceed the parts of the equation.


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