Performer Flying

Dr. Eric Martell Associate Professor of Physics Millikin University Verda Beth Martell Chair of Scenic Technology University of Illinois at Urbana-Champaign Technical Director Krannert Center for the Performing Arts ETCP Certified Rigger – Theatre ETCP Recognized Trainer

There is a sheet at the front of the room.

 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.

 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.

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 * Acceleration English Units: lbs = slugs * ft/s 2 Metric Units: N = kg * m/s 2 Forces have Magnitude & Direction 300 lbs Horizontal & to the Right 15 kN Vertical Up 172 kN 45 deg off the horizontal Down and to the Left 77 lbs Vertical Down

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

Tension (F t ) Vertical Up Gravity (F g ) Vertical Down  F=ma F t - F g = ma = 0 The person is not accelerating vertically. F t = F g

 F=ma F t - F g = ma = 0 The performer is not accelerating vertically. F t = F g = 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

 F=ma F t - F g = ma = 0 The performer is not accelerating vertically. F t = F g = 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

160 lb person Holding themselves up on the line. This is the same as the previous slide... although the potential for motion exists. Let’s look at the effect of large and small counterweights in these systems.

 F=ma F g1 – F g2 = (m 1 +m 2 )a 240 lb – 160 lb = (400 lb/32.2 ft/s 2 )* a a = 6.44 ft/s 2 F t – F g1 = (m 1 )a F t = 192 lb How fast is a person going after a 12’ ladder jump? v 1 2 – v 0 2 = 2ad V1 = SQRT (2 * 6.44 ft/s 2 * 12 ft) = ft/s 192 lb 160 lb person 80 lb

 F=ma F g1 – F g2 = (m 1 +m 2 )a 80 lb – 160 lb = (240 lb/32.2 ft/s 2 )* a a = ft/s 2 (Until the stagehand grips the line to slow the fall.) F g1 – F t = (m 1 )a F t = 107 lb 107 lb 160 lb person Lets go of line. 80 lb

 F=ma F t - F g = ma 240 lb – 160 lb = (240 lb/32.2 ft/s 2 )* a a max = ft/s 2 F g1 – F t = (m 1 )a F t = 213 lb 213 lb 160 lb person Grabbing rope, applying maximum 160 lb force. 80 lb How long will it take to get from 25’ to the deck if accelerating the whole time? (Note: OW!) d = v 0 2 t + 1/2at 2 t = SQRT(d/.5a) = SQRT(-25 ft/(.5* ft/s 2 )) = 2.15 s

v 1 = v 0 + at v 1 = ft/s 2 * 1.52 s v 1 = ft/s a = ft/s 2 a = ft/s 2 -VELOCITY TIME d = ft d = v 0 t + ½ at ft = 0 + ½ * ft/s 2 *t 1 2 t 1 = sqrt (-12.5 ft / (½ * ft/s 2 )) t 1 = 1.52 s d = v 1 t + ½ at ft = ft/s * t 2 + ½ * ft/s 2 *t = ½ * ft/s 2 * t ft/s *t ft FINISH Quadratic t 2 = 1.52 s t tot = 3.04 s

 F=ma F t - F g = ma 310 lb – 160 lb = (470 lb/32.2 ft/s 2 )* a a = 10.3 ft/s 2 F t – F g1 = (m 1 )a F t = 211 lb So increasing the counterweight increases the acceleration going out from 6.4 ft/s 2 to 10.3 ft/s lb 160 lb person Jumping off a ladder. 150 lb

 F=ma F t - F g = ma 150 lb – 160 lb = (310 lb/32.2 ft/s 2 ) * a a = ft/s 2 (Until the stagehand grips the line to slow the fall.) lb 160 lb person Lets go of line. 150 lb F g1 – F t = (m 1 )a F t = lb

 F=ma F t - F g = ma 310 lb – 160 lb = (310 lb/32.2 ft/s 2 ) * a a max = 15.6 ft/s 2 F g1 – F t = (m 1 )a F t = 238 lb 238 lb 160 lb person Grabbing rope, applying maximum 160 lb force. 150 lb How long will it take to get from 25’ to the deck if accelerating the whole time? (Note: OW?) d = v 0 2 t + 1/2at 2 t = SQRT(d/.5a) = SQRT(-25 ft/(.5*-1.04 ft/s 2 )) = 6.93 s

v 1 = v 0 + a 1 t v 1 = ft/s 2 * t 1 a 1 = 1.04 ft/s 2 a 2 = ft/s 2 VELOCITY TIME d 1 + d 2 = 25 ft d 1 = v 0 t 1 + ½ a 1 t 1 2 = 0 + ½ * 1.04 ft/s 2 *t 1 2 =117ft/s 2 *t 2 2 d 2 = v 1 t 2 + ½ a 2 t 2 2 = 1.04 ft/s 2 * t 1 *t 2 + ½ * ft/s 2 *t 2 2 =7.81 ft/s 2 *t ft = 117ft/s 2 *t ft/s 2 *t ft = ft/s 2 *t 2 2 t 2 = 0.45 s, and t 1 = 6.75 s t tot = 7.2 s v f = v 1 + a 2 t 0 = v ft/s 2 * t ft/s 2 * t 2 = 1.04 ft/s 2 * t 1 t 1 =15 t 2 v max = 7.02 ft/s

With 5% friction opposite rotation at every sheave, the effect is even more pronounced. Load needs to be (1.05) 2 * CW, or lb in order to accelerate. to accelerate. 160 lb person Lets go of line. 150 lbs This is a static system. rotation

With 5% friction opposite rotation at every sheave, the effect is even more pronounced lb lb lb 160 lb person The performer will accelerate upward at 7.2 ft/s 2. The stagehand will accelerate down at 14.4 ft/s 2. rotation 95.5 lb 40 lbs 160 lb person Jumping off a ladder.

With 5% friction opposite rotation at every sheave, the effect is even more pronounced lb 59.2 lb 56.3 lb 160 lb person Lets go of line. Performer will accelerate downward at 13.2 ft/s 2. rotation 65.2 lb 40 lbs

With 5% friction opposite rotation at every sheave, the effect is even more pronounced lb 138 lb lb 160 lb person Performer will accelerate upward at 55.2 ft/s 2. rotation lb 40 lb 160 lb person Grabbing rope, applying maximum 160 lb force.

v 1 = v 0 + a 1 t v 1 = ft/s 2 * t 1 a 1 = 13.2 ft/s 2 a 2 = ft/s 2 VELOCITY TIME d 1 + d 2 = 25 ft d 1 = v 0 t 1 + ½ a 1 t 1 2 = 0 + ½ * 13.2 ft/s 2 *t 1 2 =116.4ft/s 2 *t 2 2 d 2 = v 1 t 2 + ½ a 2 t 2 2 = 13.2 ft/s 2 * t 1 *t 2 + ½ * ft/s 2 *t 2 2 =27.8 ft/s 2 *t ft = ft/s 2 *t ft/s 2 *t ft = ft/s 2 *t 2 2 t 2 = 0.42 s, and t 1 = 1.74 s t tot = 2.16 s v f = v 1 + a 2 t 0 = v ft/s 2 * t ft/s 2 * t 2 = 13.2 ft/s 2 * t 1 t 1 =4.2 t 2 v max = 23 ft/s

 What makes this better?  What are the downsides?

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 to the performer. The period (swing and back) t = 2  * sqrt(L/g) t = 2  * sqrt(15 ft/32.2 ft/s 2 ) t = 4.28 s 15 ft between sheave and performer

F t = 138 lbs Fg = mg = 160 lb Assuming that the mass of the cable is negligible. F t = mg*cos  + (mv 2 )/L At top of swing, velocity passes through zero. F t = mg*cos(  ) = 138 lbs 30 deg

F t = 203 lbs Fg = mg = 160 lb At bottom of swing  =0 o. F t = mg*cos  + (mv 2 )/L Where v = sqrt(2gh) v = ft/s F t = mg*cos (0)+(mv 2 )/L F t = 160 lb * 1 + (5 slugs*(11.34 ft/s) 2 )/15 ft F t = lb

As the swing dampens, the height of the drop reduces. If v = sqrt(2gh), then smaller h = smaller v If F t = mg*cos  + (mv 2 )/L, smaller v = smaller F t Let’s look at this…

 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

 F=ma x: F tx = ma y: F ty – mg = 0 F t sin(  ) = ma F t cos(  ) = mg tan(  ) = (a/g) 160 lb person a 

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?

 F=ma Ft – Fg = ma Ft = ma + Fg Ft = (160 lb/32.2 ft/s 2 )*16 ft/s lb Ft = 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  v/t = (16 ft/s - 0 ft/s)/1 s = 16 ft/s 2

If the motor stops abruptly – 0 s ramp. First, they go up… v v 0 2 = 2ad (0 ft/s) 2 – (16 ft/s) 2 = 2 * 32.2 ft/s 2 * d d = 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

When the cable snaps taut: Between cable and harness stretch d = 3 in v v 0 2 = 2ad (16 ft/s) 2 – (0 ft/s) 2 = 2 * a *.25 ft a = 512 ft/s 2 F t – F g = ma F t = ma + F g F t = (160 lb/32.2 ft/s 2 )*512 ft/s lb F t = 2704 lb

Compare to Fall Arrest: OSHA (b)(6)(i) Limit the maximum arresting force on a falling employee to 900 pounds (4 Kn) when used with a body belt OSHA (b)(6)(ii) Limit the maximum arresting force on a falling employee to 1,800 pounds (8 Kn) when used with a body harness

Assume that the brake allows for slip. d = ? F t – F g = ma a = (F t – F g )/m a = (1800 lb – 160 lb)/(160 lb/32.2 ft/s 2 ) a = (1800 lb – 160 lb)/(160 lb/32.2 ft/s 2 ) a = ft/s 2 v v 0 2 = 2ad d = ((16 ft/s) 2 – (0 ft/s) 2 ) / (2 * ft/s2) d =.39 ft or 4.6 in

 Use OSHA guidelines as an absolute max. Force.  Calculate a max arresting acceleration based on this force.  Controlling this acceleration and force

 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.  v t =SQRT(2mg/  AC d )  For falling person (belly flop), v t ~54 mph.  50% of v t is reached after only ~ 3 seconds, while it takes 8 seconds to reach 90%, 15 seconds to reach 99%, etc.

 Thank you for coming!