performer flying. dr. eric martell associate professor of physics millikin university verda beth...

PHYSICS OF THEATRE PROJECTPerformer Flying

WHO WE ARE

Dr. Eric MartellAssociate Professor of Physics Millikin University

Verda Beth MartellChair of Scenic TechnologyUniversity of Illinois at Urbana-ChampaignTechnical DirectorKrannert Center for the Performing ArtsETCP Certified Rigger – TheatreETCP Recognized Trainer

ETCP CREDITS

There is a sheet at the front of the room.

WHAT WE PLAN TO TALK ABOUT

Identifying the forces in a performer flying system.

Determining the tensions on various components in the system.

Determining the forces on the performer. Applying design factors. What effects are incorporated into the

design factor and which really shouldn’t be.

WHAT WE ARE 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.

Much of the Physics of Theatre Project is about building intuition.

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.

WHERE TO START

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

Metric Units: N = kg * m/s2

Forces 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

DIRECTION AND SENSE

+

-

+-

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

+

-

+-

OR

+

-

+-

OR

VERTICAL SUSPENSION

Tension (Ft)Vertical Up

Gravity (Fg)Vertical Down

SF=maFt - Fg = ma = 0

The person is not accelerating vertically.Ft = Fg

VERTICAL SUSPENSION

SF=maFt - Fg = ma = 0

The performer is not accelerating vertically.Ft = Fg = 160 lb

160 lb

160 lb 160 lb

160 lb

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

pins.

160 lb person

VERTICAL SUSPENSION

SF=maFt - Fg = ma = 0

The performer is not accelerating vertically.Ft = Fg = 160 lb

160 lb

160 lb 160 lb

0 lb

This is the same as the previous slide.

And no one is going anywhere.

In order to work, this system must be person heavy.

160 lb person

160 lb

ACCELERATION

160 lb

160 lb 160 lb

160 lb person

160 lb person

Holding themselves up on the line.

This is the same as the previous slide...

although the potential formotion exists.

Let’s look at the effect of large and small counterweights in these systems.

ACCELERATION (GOING OUT)

SF=maFg1 – Fg2 = (m1+m2)a

240 lb – 160 lb = (400 lb/32.2 ft/s2)* aa = 6.44 ft/s2

Ft – Fg1 = (m1)aFt = 192 lb

How fast is a person going after a 12’ ladder jump?v1

2 – v02 = 2ad

V1 = SQRT (2 * 6.44 ft/s2 * 12 ft) = 12.43 ft/s

192 lb

192 lb

192 lb

160 lb person

160 lb person

80 lb

ACCELERATION (COMING IN)

SF=maFg1 – Fg2 = (m1+m2)a

80 lb – 160 lb = (240 lb/32.2 ft/s2)* aa = -10.73 ft/s2

(Until the stagehand grips the line to slow the fall.)

Fg1 – Ft = (m1)aFt = 107 lb

107 lb

107 lb 107 lb

160 lb person

160 lb person

Lets go of line.

80 lb

ACCELERATION (COMING IN)

SF=maFt - Fg = ma

240 lb – 160 lb = (240 lb/32.2 ft/s2)* aamax = 10.73 ft/s2

Fg1 – Ft = (m1)aFt = 213 lb

213 lb

213 lb 213 lb

160 lb person

160 lb person

Grabbing rope, applying maximum 160 lb force.

80 lb

How long will it take to get from 25’ to the deckif accelerating the whole time? (Note: OW!)

d = v02t + 1/2at2

t = SQRT(d/.5a) = SQRT(-25 ft/(.5*-10.73 ft/s2)) = 2.15 s

v1 = v0 + atv1 = 0 + 10.73 ft/s2 * 1.52 sv1 = -16.3 ft/s

MORE REALISTICALLY…

a = -10.73 ft/s2 a = 10.73 ft/s2

-VELO

CIT

Y

TIME

d = -12.5 ft d = -12.5 ft

d = v0t + ½ at12

-12.5 ft = 0 + ½ * -10.73 ft/s2*t12

t1 = sqrt (-12.5 ft / (½ * -10.73 ft/s2))

t1 = 1.52 s

d = v1t + ½ at22

-12.5 ft = -16.3 ft/s * t2 + ½ * 10.73 ft/s2*t22

0 = ½ * 10.73 ft/s2 * t22 + -16.3 ft/s *t2 + -12.5 ft

FINISH Quadratict2 = 1.52 s

ttot = 3.04 s

LET’S INCREASE THE COUNTERWEIGHT (GOING OUT)

SF=maFt - Fg = ma

310 lb – 160 lb = (470 lb/32.2 ft/s2)* aa = 10.3 ft/s2

Ft – Fg1 = (m1)aFt = 211 lb

So increasing the counterweight increases the acceleration going out from 6.4 ft/s2 to 10.3 ft/s2.

211 lb

211 lb 211 lb

160 lb person

160 lb person

Jumping off a

ladder.

150 lb

ACCELERATION (COMING IN)

SF=maFt - Fg = ma

150 lb – 160 lb = (310 lb/32.2 ft/s2) * aa = -1.04 ft/s2

(Until the stagehand grips the line to slow the fall.)

154.8 lb

154.8 lb 154.8 lb

160 lb person

160 lb person

Lets go of line.

150 lb

Fg1 – Ft = (m1)aFt = 154.8 lb

ACCELERATION (COMING IN)

SF=maFt - Fg = ma

310 lb – 160 lb = (310 lb/32.2 ft/s2) * aamax = 15.6 ft/s2

Fg1 – Ft = (m1)aFt = 238 lb

238 lb

238 lb 238 lb

160 lb person

160 lb person

Grabbing rope, applying maximum 160 lb force.

150 lb

How long will it take to get from 25’ to the deckif accelerating the whole time? (Note: OW?)

d = v02t + 1/2at2

t = SQRT(d/.5a) = SQRT(-25 ft/(.5*-1.04 ft/s2)) = 6.93 s

v1 = v0 + a1tv1 = 0 + 1.04 ft/s2 * t1

HOW FAST?

a1 = 1.04 ft/s2a2 = -15.58 ft/s2

VELO

CIT

Y

TIME

d1 + d2 = 25 ft

d1 = v0t1 + ½ a1t12 = 0 + ½ * 1.04 ft/s2*t1

2=117ft/s2*t22

d2 = v1t2 + ½ a2t22 = 1.04 ft/s2 * t1*t2 + ½ * -15.58 ft/s2*t2

2=7.81 ft/s2*t2

2

25 ft = 117ft/s2*t22+7.81 ft/s2*t2

2

25 ft = 124.81 ft/s2*t22

t2 = 0.45 s, and t1= 6.75 s ttot = 7.2 s

vf = v1 + a2t0 = v1 + -15.58 ft/s2 * t2

15.58 ft/s2 * t2= 1.04 ft/s2 * t1

t1=15 t2

vmax = 7.02 ft/s

FRICTION REDUCTION

With 5% friction opposite rotationat every sheave,

the effect is even more pronounced.Load needs to be (1.05)2* CW,

or 165.4 lb in order to accelerate.

160 lb person

160 lb person

Lets go of line.

150 lbs

This is a static system.

rotation

2:1 ADVANTAGE (GOING OUT)

With 5% friction opposite rotationat every sheave,

the effect is even more pronounced.

100.3 lb

105.3 lb 110.6 lb

160 lb person

The performer will accelerate upward

at 7.2 ft/s2. The stagehand will

accelerate down at 14.4 ft/s2.

rotation

95.5 lb40 lbs

160 lb person

Jumping off a

ladder.

2:1 ADVANTAGE (COMING IN)

With 5% friction opposite rotationat every sheave,

the effect is even more pronounced.

62.1 lb

59.2 lb 56.3 lb

160 lb person

160 lb person

Lets go of line.

Performer will accelerate downward

at 13.2 ft/s2.

rotation

65.2 lb40 lbs

2:1 ADVANTAGE (COMING IN)

With 5% friction opposite rotationat every sheave,

the effect is even more pronounced.

144.9 lb

138 lb 131.5 lb

160 lb person

Performer will accelerate upward

at 55.2 ft/s2.

rotation

152.2 lb

40 lb

160 lb person

Grabbing rope, applying maximum 160 lb force.

v1 = v0 + a1tv1 = 0 + 13.2 ft/s2 * t1

HOW FAST?

a1 = 13.2 ft/s2a2 = -55.2 ft/s2

VELO

CIT

Y

TIME

d1 + d2 = 25 ft

d1 = v0t1 + ½ a1t12 = 0 + ½ * 13.2 ft/s2*t1

2=116.4ft/s2*t22

d2 = v1t2 + ½ a2t22 = 13.2 ft/s2 * t1*t2 + ½ * -55.2 ft/s2*t2

2=27.8 ft/s2*t2

2

25 ft = 116.4 ft/s2*t22+27.8 ft/s2*t2

2

25 ft = 144.2 ft/s2*t22

t2 = 0.42 s, and t1= 1.74 s ttot = 2.16 s

vf = v1 + a2t0 = v1 + -55.2 ft/s2 * t2

55.2 ft/s2 * t2= 13.2 ft/s2 * t1

t1=4.2 t2

vmax = 23 ft/s

2:1 ADVANTAGE SYSTEM

What makes this better? What are the downsides?

PENDULA - PERIOD

160 lb person

This length of pendulum will swing over and back in 4.28 s.

As the oscillation dies down and the angle decreases, the period of the swing remains the same as long as the length of the cable between the performer and sheave stays the same.

The weight of the person has a negligible affect on the period.

Let’s look at this…

Assuming that the mass of the cable is very small compared tothe performer.

The period (swing and back)

t = 2p* sqrt(L/g)t = 2p * sqrt(15 ft/32.2 ft/s2)t = 4.28 s

15 ft between sheave and performer

PENDULA - TENSION

Ft = 138 lbs

Fg = mg = 160 lb

Assuming that the mass of the cable is negligible.Ft = mg*cosq + (mv2)/L

At top of swing, velocity passes through zero.

Ft = mg*cos(q) = 138 lbs

30 deg

PENDULA - TENSION

Ft = 203 lbs

Fg = mg = 160 lb

At bottom of swing q=0o.

Ft = mg*cosq + (mv2)/LWhere v = sqrt(2gh)v = 11.34 ft/s

Ft = mg*cos (0)+(mv2)/LFt = 160 lb * 1 + (5 slugs*(11.34 ft/s)2)/15 ftFt = 202.6 lb

PENDULA - TENSION

As the swing dampens, the height of the drop reduces.

If v = sqrt(2gh), then smaller h = smaller v

If Ft = mg*cosq + (mv2)/L, smaller v = smaller Ft

Let’s look at this…

CONE OF INTERFERENCE The higher the point, the smaller the angle, The smaller the angle, the shorter the height of the

fall, The shorter the height of the fall, the smaller the

velocity, The smaller the velocity, the lower the tension.

The higher the point, the slower the oscillation. But you can run into stuff. Remember to think in 3 dimensions.

Grid Height

Electric and lower

PENDULUM ON A TRACK (SWING)

SF=max: Ftx = ma

y: Fty – mg = 0Ftsin(q) = maFtcos(q) = mgtan(q) = (a/g)

160 lb person

a

q

MOTORS

What is different? Might eliminate counterweight and thus mass

of the counterweight. In many cases, we set velocity and

acceleration in the control system. If there is no counterweight, we have the whole range of speed.

What tensions can the motor impart to the cable?

What effect does that have on the performer?

HIGH SPEED (GOING OUT)

SF=maFt – Fg = maFt = ma + Fg

Ft = (160 lb/32.2 ft/s2)*16 ft/s2 + 160 lbFt = 240 lbs

Let’s say that we want a person to reach a velocity similar to a ladder

jump, but we want to get there in 1s.

a = Dv/t = (16 ft/s - 0 ft/s)/1 s = 16 ft/s2

HIGH SPEED – SETTING THE BRAKE

If the motor stops abruptly – 0 s ramp.

First, they go up…v1

2 - v02 = 2ad

(0 ft/s)2 – (16 ft/s)2 = 2 * 32.2 ft/s2 * dd = 4 ft

Then, they come down.They will travel downward the same distance at the same rate, ending at

the same speed in the opposite direction. When the cable goes taut:

v = -16 ft/s

HIGH SPEED – SETTING THE BRAKE

When the cable snaps taut:Between cable and harness stretch

d = 3 in

v12 - v0

2 = 2ad(16 ft/s)2 – (0 ft/s)2 = 2 * a * .25 ft

a = 512 ft/s2

Ft – Fg = maFt = ma + Fg

Ft = (160 lb/32.2 ft/s2)*512 ft/s2 +160 lbFt = 2704 lb

HIGH SPEED – SETTING THE BRAKE

Ft = 2704 lb

Compare to Fall Arrest:

OSHA 1915.159(b)(6)(i)Limit the maximum arresting force on a falling employee to 900 pounds (4 Kn) when used

with a body belt

OSHA 1915.159(b)(6)(ii)Limit the maximum arresting force on a falling employee to 1,800 pounds (8 Kn) when used

with a body harness

HIGH SPEED – SLIP

Assume that the brake allows for slip.d = ?

Ft – Fg = maa = (Ft – Fg )/m

a = (1800 lb – 160 lb)/(160 lb/32.2 ft/s2)

a = 330.5 ft/s2

v12 - v0

2 = 2add = ((16 ft/s)2 – (0 ft/s)2) / (2 * 330.5

ft/s2)d = .39 ft or 4.6 in

WHAT IS REASONABLE?

Use OSHA guidelines as an absolute max. Force.

Calculate a max arresting acceleration based on this force.

Controlling this acceleration and force

TERMINAL VELOCITY

What happens if you try to drive a motor downward faster than gravity will allow?

How does air resistance come into play? As object falls faster, drag force increases until it

equals weight. vt=SQRT(2mg/rACd) For falling person (belly flop), vt~54 mph. 50% of vt is reached after only ~ 3 seconds,

while it takes 8 seconds to reach 90%, 15 seconds to reach 99%, etc.

QUESTIONS?

Thank you for coming!