Rigging angled trusses

derekleffew

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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).
SdltAngledTruss.jpg

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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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.
 
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?


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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.
 
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.
 
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.
 
I cant figure out how to do it even after reading The Stage Rigging Handbook... I might have missed it tho... Its kinda late.
 
Don't we need the spacing on the lights to do this calculation accurately?
 
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. :)
 
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.
 
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.
 
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.
 
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.
 
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.
 
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...
 
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.
 
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.

option A_lowres.png

option-b_lowres2.jpg

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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