Rigging Math ?

chawalang

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Hi everyone,

Hopefully someone can point me in the right direction. I’m trying to find the correct formula for my question.

If I have a load that is suspended between two line sets and have the DS higher than US, what is the forNyla to figure out how much weight is on each line set.

I figured it’s similar to when you carry a couch up stairs, the person on the bottom hsd most of the weight.

I’m trying to figure it out for installing a sound reflector that will be at a 45 degree between two line sets.
 

derekleffew

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If I have a load that is suspended between two line sets and have the DS higher than US, what is the forNyla to figure out how much weight is on each line set.

I figured it’s similar to when you carry a couch up stairs, the person on the bottom hsd most of the weight.
Working in only one axis instead of two (hoists instead of linesets)...
https://www.controlbooth.com/threads/rigging-angled-trusses.24756/
I don't think we ever got a definitive answer.
@Ted jones , @Delbert , @egilson1 et al ?
 
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Ted jones

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chawalang

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The piece will be a 4'x48' sound reflector, weighing 360lbs. hanger pacing is at the DS and US edge of the 4' width. Line sets will be straight up above each pint at 90 degrees
 

Delbert

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Do you mean how much the arbor can hold?
This is a center of gravity of a three-dimentional load problem. I would start by finding the COG of the truss and use that to calculate the load on each of the two suspension points. Then do the same for the lighting instruments. The higher lineset/point will be holding the most weight because the COG of the truss will be shifted in that direction. Of course, this calculated load does not take into account the weight of the cable. We would have to know the weight per foot of the cable and how the cable is run across and hangs off of the truss to add this into our load equations.
 

chawalang

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This is a center of gravity of a three-dimentional load problem. I would start by finding the COG of the truss and use that to calculate the load on each of the two suspension points. Then do the same for the lighting instruments. The higher lineset/point will be holding the most weight because the COG of the truss will be shifted in that direction. Of course, this calculated load does not take into account the weight of thecable. We would have to know the weight per foot of the cable and how the cable is run across and hangs off of the truss to add this into our load equations.

is there supposed to be an image in your post?
 

egilson1

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It’s all about Center of Gravity. If the CG is half way between the two linesets then each will take 50% of the load. If the CG moves closer to one suspension then the load on that suspension increases. The trick is knowing where the CG is on any given object. Don’t think I’m stating anything people here don’t already know. I’m a little slammed at the moment but I can go into more detail next week.

Ethan
 
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Ted jones

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Personally, I would make a bracket that will allow you to hang the piece from one set and adjust the angle. But, I have a metal shop at my disposal.

So, to follow your question about hanging on two sets to make an adjustable reflector, start by hanging it flat. You will need to acquire/find/make brackets to support it 4' OC along both edges. These brackets will have to allow the pivot of the edges as you change angles. Load up your arbors equally until they balance the load. Raise both sets about a foot and then lock the one that will be low. Then secure it with either a stick twisted into the control lines or a chain on the arbor. Raise the one set that will be high to the angle you like. This will also draw the battens toward each other.

This is where it gets fun. Marry the arbors together so they operate as one. Not seeing your stage, I cannot help you with design, but two 3 x 3 x 1/4 angles with two 1" U-bolts at each end around the arbor rods, assuming they are still overlapping each other, is the simplest way to connect the arbors. So we are clear, use a total of eight U-bolts and two pieces of angle. With two U-bolts at each end about 2" apart, the angles will not be able to pivot and will keep the arbors solidly attached to each other.

At this point, you will have one lock open during the shows and operate the "set" with the other line and lock. Lock both between shows and tag/label them as one set and caution people to not touch them if they are not the trained operator.

To answer your original question, would one set carry more than the other. Maybe. But the amount will be negligible at such a low weight. If the shell were 3,600 lb, there would be some noticeable weight shift, still slight. Depending on the angle. The upper one may end up lifting more of the weight if the angle is steep enough.

T
 

BillConnerFASTC

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Ted explains a good method to do what you think you want to do but since one set will balane the load, consider hanging with lines - I'd use chain but wire rope is fine - attached along up and down stage edges and over batten. It will be much simpler and safer as it will always be in balance. Some effort to install but thats on the ground and not overhead. You loose instant adjustability of tilt but safer.
 

bobgaggle

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To answer the OP, I'm not sure of my logic here, but have a look at the pic.
image1.jpeg

Makes sense to me. I figured 2lbs/degree gets you to 180lbs at 90 degrees (hanging horizontally). So 45 degrees would be a 75%/25% split. I have a nagging suspicion that its not that simple. Maybe an quadratic or log curve. I would also like to know how to figure this out properly. Anyone have the formula for this? Maybe its in one of the many rigging books on my shelf that hasn't seen the light of day since college.

edit: This assumes a uniform load across the span
 
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Ted jones

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Ted explains a good method to do what you think you want to do but since one set will balane the load, consider hanging with lines - I'd use chain but wire rope is fine - attached along up and down stage edges and over batten. It will be much simpler and safer as it will always be in balance. Some effort to install but thats on the ground and not overhead. You loose instant adjustability of tilt but safer.
Bill is right. Hence my opening comment, hang it on one set. I like his method with chain. Cheap and simple. It only works to a certain angle where the weight of the shell ceiling overcomes the friction of the chain. Wrap the chain around the pipe once before making your second connection to add friction.
 

Ted jones

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To answer the OP, I'm not sure of my logic here, but have a look at the pic.
View attachment 17091

Makes sense to me. I figured 2lbs/degree gets you to 180lbs at 90 degrees (hanging horizontally). So 45 degrees would be a 75%/25% split. I have a nagging suspicion that its not that simple. Maybe an quadratic or log curve. I would also like to know how to figure this out properly. Anyone have the formula for this? Maybe its in one of the many rigging books on my shelf that hasn't seen the light of day since college.

edit: This assumes a uniform load across the span
Your load will probably be 50/50, or close to it until your angle becomes so high that the top point starts to lift the shell ceiling. Probably 45ish degrees. Take a 2' long bar and hang it from two scales and start changing the angle and see when the weight shifts.
 

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