Rigging angled trusses
- Thread starterderekleffew
- Start date
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.
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.
BrianWolfe
Active Member
kylebjordahl
Member
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.
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.
JChenault
Well-Known Member
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.
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.
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.
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.
JonCarter
Well-Known Member
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.)
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.
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.
