physics of theatre - rigging you can measure the tension in the room

Physics of Theatre- Rigging

You can measure the tension in the room.

Who are we?

Verda Beth MartellVerda Beth MartellOpera TDOpera TDKrannert Center for the Performing ArtsKrannert Center for the Performing ArtsAssistant Professor of TheatreAssistant Professor of TheatreUniversity of Illinois @ Urbana-ChampaignUniversity of Illinois @ Urbana-Champaign

Dr. Eric MartellDr. Eric MartellChair of Physics and AstronomyChair of Physics and AstronomyAssistant ProfessorAssistant ProfessorMillikin University - Decatur, ILMillikin University - Decatur, IL

http://www2.kcpa.uiuc.edu/kcpatd/physics/index.htm(Google “Physics of Theatre”)

Physics of Theatre: Rigging

• 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.

What are we talking about?What are we 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.

What are we NOT talking about?What are we NOT talking about?

Much of the Physics of Theatre Much of the Physics of Theatre Project Project

is about building up intuition.is about building up intuition.


Our approach: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.

Force = Mass * AccelerationForce = Mass * AccelerationEnglish Units: lbs = slugs * ft/s2

Metric Units: N = kg * m/s2

Forces have Magnitude & DirectionForces have Magnitude & Direction300 lbs Horizontal & to the Right

15 kN Vertical Up

172 kN 45 deg off the horizontal Down and to the Left77 lbs Vertical Down


VelocityVelocityThe rate of movement – how fast is it going? Units: m/s, ft/s, MPH,

cubits/fortnight….

AccelerationAccelerationThe rate of change in

velocityUnits: m/s2, ft/sec2

Velocity and Acceleration, like force, are Velocity and Acceleration, like force, are vectors – vectors –

They have both magnitude and direction.They have both magnitude and direction.


Direction is arbitrary in any given problem, Direction is arbitrary in any given problem, but it must be consistent throughout the but it must be consistent throughout the

whole problem.whole problem.++

--

++--

When you enter a forceforce into an equation, you must indicate both the magnitude and magnitude and

directiondirection

+

-

+-

OR

+

-

++--

OR

Direction - Sense

BowlingBall

PIPE

ROPE

Tension (FTension (Ftt))Vertical UpVertical Up

Gravity (FGravity (Fgg))Vertical DownVertical Down

What are the forces acting on the ball?What are the forces acting on the ball?

Force of GravityMass * Acceleration of gravity (g) = Fg or “weight”

The bowling ball is not accelerating vertically.Ft = Fg

Vertical Suspension

Why is it important to know FWhy is it important to know Ftt??

It’s a matter of tension.

The lifting force has to get to the objectthrough cable, shackles, turnbuckles, chain…

All of these components must be able to withstand the

tension * a design factor.

Vertical Movement

BowlingBall

PIPE

ROPE

What if the ball was accelerating?What if the ball was accelerating?

Force of GravityMass * Acceleration (g) = Fg or “weight”



Vertical Movement

FFnet net (Net Force)(Net Force) = Fg + Ft = ma

Let’s work one out……Let’s work one out……

FFnetnet = Mass * Acceleration = Mass * AccelerationMass of Bowling ball = = .3125 slugs

10 lbs32 ft/s2

Fnet = .3125 slugs * 4 ft/s2

Acceleration = 4 ft/s2

Weight = 10 lbs

FFnetnet = 1.25 lbs = 1.25 lbs

Vertical Movement


Let’s work one out……Let’s work one out……

Fg = 10 lbs


Weight = 10 lbs

Fnet = 1.25 lbs

These are the magnitudes of the forces.These are the magnitudes of the forces.

Vertical Movement


Let’s work one out……

Fg = -10 lbs


Weight = 10 lbs

Fnet = 1.25 lbs

1.25 lbs = -10 lbs + Ft

Ft = 11.25 lbs

+

-

+-

Vertical Movement


What if we were accelerating downward?What if we were accelerating downward?

Fg = -10 lbs

Acceleration = -4 ft/s2

Weight = 10 lbs

Fnet = -1.25 lbs

-1.25 lbs = -10 lbs + Ft

FFt t == 8.75 lbs8.75 lbs

+

-

+-Fnet = .3125 slugs * -4 ft/s2

Vertical Movement


What if we were decelerating?What if we were decelerating?


Weight = 10 lbs

1.25 lbs = -10 lbs + Ft

FFt t == 11.25 lbs11.25 lbs

Fnet = .3125 slugs * 4 ft/s2

Fnet = 1.25 lbs

Fg = -10 lbs

Vertical Movement

Note that this is the same force as

accelerating upwards - the directions of

the forces are what matter.


Demo Break

What if we wanted to land a 200 lb unit?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/seca = 36 ft/sec22

Mass of Unit = = 6.25 slugs6.25 slugs200 lbs32 ft/s2

200 lbs

Vertical Movement

What if we wanted to land an 200 lb unit?What if we wanted to land an 200 lb unit?


FFnetnet = 225 lbs = 225 lbs

225 lbs = -200 lbs + Ft

FFt t == 425 lbs425 lbs

Fnet = 6.25 slugs * 36 ft/s2

FFgg = -200 lbs = -200 lbs

Fg = -200 lbs

Ft = 425 lbs

200 lbs

Vertical Movement


What if that same unit stopped suddenly?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.aaff = 180 ft/sec = 180 ft/sec22

200 lbs

Immediate Stop loading

What if that same unit stopped suddenly?What if that same unit stopped suddenly?


1125 lbs = -200 lbs + Ft

FFt t == 1325 lbs1325 lbs



FFgg = -200 lbs = -200 lbs

Fg = -200 lbs

Ft = 1325 lbs

200 lbs

Immediate Stop Loading


What if you need to stop this object from freefall?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.

aaff = 322 ft/sec = 322 ft/sec22

200 lbs

Shock loading


2012.5 lbs = -200 lbs + Ft

FFt t == 2212.5 lbs2212.5 lbs


FFnetnet = 2012.5 lbs = 2012.5 lbs

FFgg = -200 lbs = -200 lbs

Fg = -200 lbs

Ft = 2212.5 lbs

200 lbs

Shock LoadingWhat if you need to stop this object from freefall?What if you need to stop this object from freefall?


Manual Rigging:

Shock Loading:

FFt t == 2212.5 lbs2212.5 lbsMore than 11x the loadMore than 11x the load

Motorized Rigging:FFtt = 1325 lbs = 1325 lbs

6.625x the load6.625x the load

FFtt = 425 lbs = 425 lbs2.125x the load2.125x the load

Fg = -200 lbs

Ft = ? lbs

200 lbs

Design FactorThe same load can cause multiple tensions.The same load can cause multiple tensions.

200 lbs

The same load can cause multiple tensions.The same load can cause multiple tensions.Many theatre technicians have been told to use a Many theatre technicians have been told to use a

design factor of 5:1.design factor of 5:1.

Many experts now feel that design factors of 7:1, Many experts now feel that design factors of 7:1, 8:1 and 10:1 are more appropriate.8:1 and 10:1 are more appropriate.

What does that mean for you?What does that mean for you?Not all suppliers use the same design factorNot all suppliers use the same design factorIt may be best to convert everything to UBS It may be best to convert everything to UBS

and use the design factor you find most and use the design factor you find most appropriate.appropriate.

Design factors will not help with shock loading.Design factors will not help with shock loading.Shock loaded equipment should be replaced.Shock loaded equipment should be replaced.

Design Factor

200 lbs

What’s not included in the design factor?What’s not included in the design factor?Dynamic or Shock LoadingDynamic or Shock LoadingMotor Driven LoadsMotor Driven LoadsEfficiencyEfficiency

Design Factor

So if you’re going to calculate everything, So if you’re going to calculate everything, why do you still need a design factor?why do you still need a design factor?

Minor load changes/estimation issuesMinor load changes/estimation issuesHuman errorHuman errorEquipment wearEquipment wearHuman accelerationHuman acceleration

•When trying to move an object, you are most When trying to move an object, you are most concerned with finding the lift force.concerned with finding the lift force.

•The lift force creates a tension in many of the The lift force creates a tension in many of the system components.system components.

•The lift force created to accelerate an object The lift force created to accelerate an object vertically up is greater than the weight of the vertically up is greater than the weight of the

object.object.

•The greater the acceleration, the more tension is The greater the acceleration, the more tension is put into the system components.put into the system components.

•In a vertical movement system, the main forces In a vertical movement system, the main forces are the weight of the load and the tension in a rope are the weight of the load and the tension in a rope

or cable. However, other forces can act on the or cable. However, other forces can act on the system – friction (in the pulleys or tracking), the system – friction (in the pulleys or tracking), the

weight of the cable, ...weight of the cable, ...

Intermission

Two CablesWhat’s different when we suspend the objectWhat’s different when we suspend the objectfrom more than one cable?from more than one cable?

• Cables come together to suspend the object from one point: Two-point bridle

• Cables attach to the object at two different points

Safety Note: Each cable should be able to support the entire load.

Two CablesCables attach to the Cables attach to the object at two different object at two different pointspointsQuestion: What is the tension in each cable?

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, net = I = 0.

F2F1

Fg

Two CablesCables attach to the object at two different pointsCables attach to the object at two different points

The tension in each cable applies a torque around the center of gravity: 1= x1

.F1, 2 = x2.F2, where x1

and x2 are the distances from each cable to a vertical line through the center of gravity.

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

Since net = 0, 1 = 2, or x1.F1 = x2

.F2.

x1 x2 F2F1

Fg


Putting these together, we get:

F1 = Fg.x2/(x1 + x2)

and

F2 = Fg.x1/(x1 + x2)

Newton’s 2nd Law (Rotational): x1.F1 = x2

.F2.

Newton’s 2nd Law: F1 + F2 = Fg. x1 x2 F2F1

Fg


Whichever cable is closest to the center of gravity bears more weight.

If the cables are placed equidistant from the center of gravity, F1 = F2 = ½ * Fg (as we’d expect).

What does this mean?

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 CablesObject suspended Object suspended from a two-point from a two-point bridlebridleNewton’s 2nd Law says F1 + F2 + Fg = 0. 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

F2 = Fg

sin()/tan() + cos()

sin()/tan() + cos()

F1 F2

Fg

Two CablesObject suspended from a two-point bridleObject suspended from a two-point bridle

What on earth does that mean?When = , F1 = F2, as we’d expect.

When = , F1 = 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).

F1 F2

Fg

Three (or more) CablesObject suspended from a three-point bridleObject 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 pointsObject 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.

200 lbs

200

lbs200 lbs

200 lbs

283 lbs

200 lbs

0 lbs

Counterweight systemsWhat is the tension?What is the tension?

Notice that the tension in the cable is the Notice that the tension in the cable is the same all the way from the object to the same all the way from the object to the

counterweight.counterweight.

More on the sheave tension later.More on the sheave tension later.

200 lbs

150lbs

FFnetnet = F = Fgg + F + Ftt

Fnet = 200 lbs - 150 lbs


Counterweight SystemsWhat happens if the system is out of weight?What happens if the system is out of weight?

200 lbs

150

lbs

FFnetnet = m * a = m * am = 200 lbs + 150 lbs / 32 ft/sec2

m = 10.87 slugs

50 lbs = 10.87 slugs * a

a = 4.6 ft/seca = 4.6 ft/sec22

Counterweight SystemsWhat happens if the system is out of weight?What happens if the system is out of weight?

How fast will the system accelerate?How fast will the system accelerate?

200 lbs

171.4 lbs

FFnetnet = ma = maFFnetnet = F = Fgg - F - Ftt

ororFFgg - F - Ftt = ma = ma

FFtt = -ma + F = -ma + Fgg

FFtt = -6.21 slugs*4.6 ft/s = -6.21 slugs*4.6 ft/s22 + 200 lbs + 200 lbs= 171.4 lbs= 171.4 lbs

Counterweight SystemsWhat are the tensions in the cables?What are the tensions in the cables?



150lbs

Counterweight SystemsWhat are the tensions in the cables?What are the tensions in the cables?

FFnetnet = ma = maFFnetnet = -F = -Fgg + F + Ftt

oror-F-Fgg + F + Ftt = ma = maFFtt = ma + F = ma + Fgg

FFtt = 4.66 slugs*4.6 ft/s = 4.66 slugs*4.6 ft/s22 + 150 lbs + 150 lbs= 171.4 lbs= 171.4 lbs



171.4 lbs

Sheave forcesWhat force is acting on the sheave?What force is acting on the sheave?

At a right angle…At a right angle…

171.4 lbs

171.4 lbs

aa22 + b + b22 = c = c22

171.4171.42 2 lbs +171.4lbs +171.422 lbs = c lbs = c22

c = 242.4 lbsc = 242.4 lbs

Also 171.4 * SQRT(2)Also 171.4 * SQRT(2)

a

b

c

Sheave forcesWhat force is acting on the sheave?What force is acting on the sheave?

At a non-right angle…At a non-right angle…

a

b

c

171.4

lbs

a2 + b2 -2ab*cos() = c2

Law of CosinesLaw of Cosines

171.42 + 171.42 -2*171.4*171.4cos(112) = c2

36745.6 lbs2 = c2

191.7 lbs = c

= 68 = 180 -

191.7

lbs

171.4 lbs

Questions?


http://www2.kcpa.uiuc.edu/kcpatd/physics/index.htm

(Google “Physics of Theatre”)

200lbs

200lbs

200 lbs

200 lbs

400 lbs

Single Purchase Systems

200lbs

100 lbs

100 lbs

300 lbs

100 lbs

Advantage

200lbs

200 lbs

600 lbs

200 lbs

Disadvantage

200 lbs

400 lbs

Double Purchase

200lbs

100 lbs

Double Purchase Advantage

2:1 ratio

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

100 lbs

100lbs

100 lbs

Double Purchase

200lbs

200 lbs

200 lbs

Double Purchase Disadvantage

1:2 ratio

2x the force to lift the object

2x the speed – operator pulls 1’, object moves 2’

2x the total object travel - counterweight travels

1/2 of object travel

200 lbs

400lbs

Double Purchase

50 lbs each

Block and Fall Advantages

Various ratios - 4:1, 6:1

To allow relatively weak machines to move heavy

objects

- Lifting sandbags to counterweight hemp

lines

-Storage pipes on crossover

- Cranes

50 lbs200lbs

250 lbs

Block and Fall

physics of theatre - rigging you can measure the tension in the room

Documents

force of gravitymass

lbsfnet net force

fg ft

lifting force

physics of theatre project

physics of theatrephysics

lbs ftft

mavertical movementlets