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