Transcript Rigging PPT - Physics of Theatre Home

Physics of Theatre
- Rigging
You can measure
the tension in the room.
Who are we?
Verda Beth Martell
Opera TD
Krannert Center for the Performing Arts
Assistant Professor of Theatre
University of Illinois @ Urbana-Champaign
Dr. Eric Martell
Chair of Physics and Astronomy
Assistant Professor
Millikin University - Decatur, IL
http://www2.kcpa.uiuc.edu/kcpatd/physics/index.htm
(Google “Physics of Theatre”)
Physics of Theatre: Rigging
What are we talking about?
• Measuring the forces in a system.
• Finding the forces acting on the components
in your system.
• Explaining why design factors exist and why
many experts disagree about what the design
factor should be.
• What effects are incorporated into the design
factor and which really shouldn’t be.
Physics of Theatre: Rigging
What are we NOT talking about?
• We are not talking about changing the
rigging process.
– The rigging process grew out of the experience
of sailors and theatrical riggers over literally
hundreds of years of experience.
Much of the Physics of Theatre Project
is about building up intuition.
Physics of Theatre: Rigging
Our approach:
•Identify the forces acting on the objects in
the system.
•Apply Newton’s Laws of Motion.
•Solve for the forces of interest (say, tension
in a cable, force on a sheave).
•Use results to build our intuition.
Physics of Theatre: Rigging
Force = Mass * Acceleration
English Units: lbs = slugs * ft/s2
Metric Units: N = kg * m/s2
Forces have Magnitude & Direction
300 lbs Horizontal & to the Right
172 kN 45 deg off the
horizontal Down and
to the Left
15 kN Vertical Up
77 lbs Vertical Down
Physics of Theatre: Rigging
Velocity
The rate of movement – how fast is it going?
Units: m/s, ft/s, MPH, cubits/fortnight….
Acceleration
The rate of change in velocity
Units: m/s2, ft/sec2
Velocity and Acceleration, like force, are vectors –
They have both magnitude and direction.
Direction - Sense
Direction is arbitrary in any given problem, but it
must be consistent throughout the whole problem.
+
-
+
-
+
+
+
-
OR
-
-
+
OR
-
When you enter a force into an equation, you must
indicate both the magnitude and direction
Vertical Suspension
What are the forces acting on the ball?
PIPE
Tension (Ft)
Vertical Up
ROPE
Bowling
Ball
Gravity (Fg)
Vertical Down
Force of Gravity
Mass * Acceleration of gravity (g) = Fg or “weight”
The bowling ball is not accelerating vertically.
Ft = Fg
Vertical Movement
Why is it important to know Ft?
It’s a matter of tension.
The lifting force has to get to the object
through cable, shackles, turnbuckles, chain…
All of these components must be able to withstand the
tension * a design factor.
Vertical Movement
What if the ball was accelerating?
PIPE
Tension (Ft)
Vertical Up
ROPE
Bowling
Ball
Gravity (Fg)
Vertical Down
Force of Gravity
Mass * Acceleration (g) = Fg or “weight”
Fnet (Net Force) = Fg + Ft = ma
Vertical Movement
Let’s work one out……
Acceleration = 4 ft/s2
Weight = 10 lbs
Fnet = Mass * Acceleration
Mass of Bowling ball =
10 lbs
32 ft/s2
= .3125 slugs
Fnet = .3125 slugs * 4 ft/s2
Fnet = 1.25 lbs
Fnet (Net Force) = Fg + Ft = ma
Vertical Movement
Let’s work one out……
Acceleration = 4 ft/s2
Weight = 10 lbs
Fg = 10 lbs
Fnet = 1.25 lbs
These are the magnitudes of the forces.
Fnet (Net Force) = Fg + Ft = ma
Vertical Movement
Let’s work one out……
Acceleration = 4 ft/s2
Weight = 10 lbs
Fg = -10 lbs
Fnet = 1.25 lbs
1.25 lbs = -10 lbs + Ft
Ft = 11.25 lbs
+
-
+
-
Fnet (Net Force) = Fg + Ft = ma
Vertical Movement
What if we were accelerating downward?
Acceleration = -4 ft/s2
Weight = 10 lbs
Fg = -10 lbs
Fnet = .3125 slugs * -4 ft/s2
Fnet = -1.25 lbs
+
-
+
-
-1.25 lbs = -10 lbs + Ft
Ft = 8.75 lbs
Fnet (Net Force) = Fg + Ft = ma
Vertical Movement
What if we were decelerating?
Acceleration = 4 ft/s2
Weight = 10 lbs
Fg = -10 lbs
Fnet = .3125 slugs * 4 ft/s2
Note that this is the same
force as accelerating
upwards - the directions
of the forces are what
matter.
Fnet = 1.25 lbs
1.25 lbs = -10 lbs + Ft
Ft = 11.25 lbs
Fnet (Net Force) = Fg + Ft = ma
Demo Break
Vertical Movement
What if we wanted to land a 200 lb unit?
Setting up the problem….
v = -18 ft/sec
We’re going to make it stop in the last 1/2 s of travel.
a = 36 ft/sec2
200
lbs
Mass of Unit = 200 lbs2
32 ft/s
= 6.25 slugs
Vertical Movement
What if we wanted to land an 200 lb unit?
Acceleration = 36
ft/s2
Ft = 425 lbs
Fg = -200 lbs
Fnet = 6.25 slugs * 36 ft/s2
Fnet = 225 lbs
200
lbs
225 lbs = -200 lbs + Ft
Ft = 425 lbs
Fg = -200 lbs
Fnet (Net Force) = Fg + Ft = ma
Immediate Stop loading
What if that same unit stopped suddenly?
Setting up the problem….
v = -18 ft/sec
We’re going to make it stop in the last .1 s of travel.
af = 180 ft/sec2
200
lbs
Immediate Stop Loading
What if that same unit stopped suddenly?
Acceleration = 180 ft/s2
Ft = 1325 lbs
Fg = -200 lbs
Fnet = 6.25 slugs * 180 ft/s2
Fnet = 1125 lbs
200
lbs
1125 lbs = -200 lbs + Ft
Ft = 1325 lbs
Fg = -200 lbs
Fnet (Net Force) = Fg + Ft = ma
Shock loading
What if you need to stop this object from freefall?
The object fell for 1 sec.
(say it was caught on something and then came free)
v = -32.2 ft/sec
There’s always some stretch in cable,
so let’s say that it took .1 sec to arrest the fall.
af = 322 ft/sec2
200
lbs
Shock Loading
What if you need to stop this object from freefall?
Acceleration = 322 ft/s2
Ft = 2212.5 lbs
Fg = -200 lbs
Fnet = 6.25 slugs * 322 ft/s2
Fnet = 2012.5 lbs
200
lbs
2012.5 lbs = -200 lbs + Ft
Ft = 2212.5 lbs
Fg = -200 lbs
Fnet (Net Force) = Fg + Ft = ma
Design Factor
The same load can cause multiple tensions.
Manual Rigging:
Ft = 425 lbs
2.125x the load
Ft = ? lbs
Motorized Rigging:
Ft = 1325 lbs
6.625x the load
200
lbs
Shock Loading:
Ft = 2212.5 lbs
More than 11x the load
Fg = -200 lbs
Design Factor
The same load can cause multiple tensions.
 Many theatre technicians have been told to use a design factor
of 5:1.
 Many experts now feel that design factors of 7:1, 8:1 and 10:1
are more appropriate.
 What does that mean for you?
Not all suppliers use the same design factor
It may be best to convert everything to UBS and use the
design factor you find most appropriate.
200
lbs
 Design factors will not help with shock loading.
Shock loaded equipment should be replaced.
Design Factor
What’s not included in the design factor?
Dynamic or Shock Loading
Motor Driven Loads
Efficiency
So if you’re going to calculate everything,
why do you still need a design factor?
200
lbs
Minor load changes/estimation issues
Human error
Equipment wear
Human acceleration
Intermission
•When trying to move an object, you are most concerned with
finding the lift force.
•The lift force creates a tension in many of the system
components.
•The lift force created to accelerate an object vertically up is
greater than the weight of the object.
•The greater the acceleration, the more tension is put into the
system components.
•In a vertical movement system, the main forces are the weight
of the load and the tension in a rope or cable. However, other
forces can act on the system – friction (in the pulleys or
tracking), the weight of the cable, ...
Two Cables
What’s different when we suspend the object
from more than one cable?
• Cables attach to the object at
two different points
• Cables come together to suspend
the object from one point: Twopoint bridle
Safety Note: Each cable should be
able to support the entire load.
Two Cables
Cables attach to the object at
two different points
Question: What is the tension in
each cable?
F2
F1
Fg
Assuming it’s dead hung, Fnet=0, which gives F1 + F2 = Fg.
This isn’t enough information to solve for F1 and F2 – we need
to know something else about the system.
Not only is the object not accelerating, it’s also not rotating,
so we can also use the rotational analogue of Newton’s 2nd
Law, tnet = Ia = 0.
Two Cables
Cables attach to the object at two different points
Torque – the action of a force around an axis
Depends on the size and direction of the force and the
distance from the axis of rotation for the system
The tension in each cable applies a torque around the center
of gravity: t1= x1.F1, t2 = x2.F2, where x1 and x2 are the
distances from each cable to a vertical line through the center
of gravity.
x2
F2
F1 x1
Fg
Since tnet = 0, t1 = t2, or x1.F1 = x2.F2.
Two Cables
Cables attach to the object at two different points
Newton’s 2nd Law: F1 + F2 = Fg.
Newton’s 2nd Law (Rotational): x1.F1 = x2.F2.
Putting these together, we get:
F1 = Fg.x2/(x1 + x2)
and
F2 = Fg.x1/(x1 + x2)
F1
x1
Fg
x2
F2
Two Cables
Cables attach to the object at two different points
What does this mean?
If the cables are placed equidistant from the center of gravity,
F1 = F2 = ½ * Fg (as we’d expect).
Whichever cable is closest to the center of gravity bears more
weight.
If the object is accelerated up or down, the tensions in the
cables change just like they did for the single cable example.
For a symmetric object, if one side is pulled up while the
other stays in place (tilting the object), the tension in the cable
that’s moving goes up while it’s moving, but once it stops and
the object is stationary, the tensions return to their original
values.
Two Cables
Object suspended from a
two-point bridle
Newton’s 2nd Law says F1 + F2 + Fg = 0.
F1
Since the vectors don’t just point horizontally
or vertically, we must break them into
components using trigonometry before
adding (we’ll skip that part in this talk).
Results:
F1 =
Fg
sin(q)/tan(f) + cos(q)
F2 =
Fg
sin(f)/tan(q) + cos(f)
f q
Fg
F2
Two Cables
Object suspended from a two-point bridle
What on earth does that mean?
When f = q, F1 = F2, as we’d expect.
When f = q =
60o,
F1 = F2 = Fg.
F1
f q
F2
Fg
It is impossible to pull the cables with enough tension to make
both completely horizontal.
If the object accelerates up or down, the vertical components
of F1 and F2 change, but the horizontal components remain
the same (the numerators of each formula change from Fg to
Fg + ma).
Three (or more) Cables
Object suspended from a three-point bridle
Using Newton’s 2nd Law, we can calculate the tension in each
cable, but it’s messy (see website after conference).
Object suspended from a three or more points
Need to use engineering techniques to find tensions. See
tables in The Stage Rigging Handbook, Glerum, for example.
Demo
Counterweight systems
What is the tension?
More on the sheave tension later.
200
lbs
200 lbs
Notice that the tension in the cable is the same all
the way from the object to the counterweight.
200 lbs
200
lbs
0 lbs
200
lbs
Counterweight Systems
What happens if the system is out of weight?
Fnet = Fg + Ft
Fnet = 200 lbs - 150 lbs
200
lbs
Fnet = 50 lbs
150
lbs
Counterweight Systems
What happens if the system is out of weight?
How fast will the system accelerate?
Fnet = m * a
m = 200 lbs + 150 lbs / 32 ft/sec2
m = 10.87 slugs
50 lbs = 10.87 slugs * a
200
lbs
a = 4.6 ft/sec2
150
lbs
Counterweight Systems
What are the tensions in the cables?
171.4 lbs
200
lbs
Fnet = ma
Fnet = Fg - Ft
or
Fg - Ft = ma
Ft = -ma + Fg
Tension (Ft)
Vertical Up
Gravity (Fg)
Vertical Down
Ft = -6.21 slugs*4.6 ft/s2 + 200 lbs
= 171.4 lbs
Counterweight Systems
What are the tensions in the cables?
Tension (Ft)
Vertical Up
Gravity (Fg)
Vertical Down
Fnet = ma
Fnet = -Fg + Ft
or
-Fg + Ft = ma
Ft = ma + Fg
171.4 lbs
150
lbs
Ft = 4.66 slugs*4.6 ft/s2 + 150 lbs
= 171.4 lbs
Sheave forces
What force is acting on the sheave?
At a right angle…
c
a
b
171.4 lbs
a2 + b 2 = c 2
171.42 lbs +171.42 lbs = c2
c = 242.4 lbs
171.4 lbs
Also 171.4 * SQRT(2)
Sheave forces
What force is acting on the sheave?
191.7
At a non-right angle…
lbs
171.4
f
a
lbs
c
q
b
Law of Cosines
a2 + b2 -2ab*cos(q) = c2
171.42 + 171.42 -2*171.4*171.4cos(112) = c2
36745.6 lbs2 = c2
171.4 lbs
f = 68
q = 180 - f
191.7 lbs = c
Physics of Theatre: Rigging
Questions?
http://www2.kcpa.uiuc.edu/kcpatd/physics/index.htm
(Google “Physics of Theatre”)
Single Purchase Systems
400 lbs
200 lbs
200
200 lbs
lbs
200
lbs
Double Purchase
300 lbs
600 lbs
200 lbs
100 lbs
200 lbs
200 lbs
400 lbs
100 lbs
100 lbs
200
200
lbs
lbs
Advantage
Disadvantage
Double Purchase
Double Purchase Advantage
2:1 ratio
100 lbs
100
lbs
100 lbs
100 lbs
200
lbs
1/2 the force to lift the object
1/2 the speed – operator pulls
1’, object travels 1/2’
1/2 the total object travel –
counterweight travels 2x object
travel
Double Purchase
Double Purchase Disadvantage
1:2 ratio
200 lbs
200 lbs
2x the force to lift the object
2x the speed – operator pulls
1’, object moves 2’
400
lbs
200 lbs
200
lbs
2x the total object travel counterweight travels 1/2 of
object travel
Block and Fall
250 lbs
Block and Fall Advantages
Various ratios - 4:1, 6:1
To allow relatively weak
machines to move heavy
objects
50 lbs
each
200
lbs
50 lbs
- Lifting sandbags to
counterweight hemp lines
-Storage pipes on crossover
- Cranes