Physics of Theatre - Rigging

1. Physics of Theatre- Rigging You can measure the tension in the room.

2. 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”)

3. 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.

4. 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.

5. 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.

6. 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

7. 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.

8. Direction - Sense Direction is arbitrary in any given problem, but it must be consistent throughout the whole problem. + + + - + - + - + - OR OR - - When you enter aforceinto an equation, you must indicate both themagnitude and direction

9. Vertical Suspension What are the forces acting on the ball? Tension (Ft) Vertical Up PIPE 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

10. 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.

11. Vertical Movement What if the ball was accelerating? Tension (Ft) Vertical Up PIPE ROPE Bowling Ball Gravity (Fg) Vertical Down Force of Gravity Mass * Acceleration (g) = Fg or “weight” Fnet (Net Force) = Fg + Ft = ma

12. Vertical Movement Let’s work one out…… Acceleration = 4 ft/s2 Weight = 10 lbs Fnet = Mass * Acceleration 10 lbs 32 ft/s2 Mass of Bowling ball = = .3125 slugs Fnet = .3125 slugs * 4 ft/s2 Fnet = 1.25 lbs Fnet (Net Force) = Fg + Ft = ma

13. 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

14. 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

15. 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

16. Vertical Movement What if we were decelerating? Acceleration = 4 ft/s2 Weight = 10 lbs Note that this is the same force as accelerating upwards - the directions of the forces are what matter. Fg = -10 lbs Fnet = .3125 slugs * 4 ft/s2 Fnet = 1.25 lbs 1.25 lbs = -10 lbs + Ft Ft =11.25 lbs Fnet (Net Force) = Fg + Ft = ma

17. Demo Break

18. 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 32 ft/s2 Mass of Unit = = 6.25 slugs 200 lbs

19. Vertical Movement What if we wanted to land an 200 lb unit? Ft = 425 lbs Acceleration = 36 ft/s2 Fg = -200 lbs Fnet = 6.25 slugs * 36 ft/s2 Fnet = 225 lbs 225 lbs = -200 lbs + Ft 200 lbs Ft =425 lbs Fg = -200 lbs Fnet (Net Force) = Fg + Ft = ma

20. 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

21. Immediate Stop Loading What if that same unit stopped suddenly? Ft = 1325 lbs Acceleration = 180 ft/s2 Fg = -200 lbs Fnet = 6.25 slugs * 180 ft/s2 Fnet = 1125 lbs 1125 lbs = -200 lbs + Ft 200 lbs Ft =1325 lbs Fg = -200 lbs Fnet (Net Force) = Fg + Ft = ma

22. 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

23. Shock Loading What if you need to stop this object from freefall? Ft = 2212.5 lbs Acceleration = 322 ft/s2 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

24. DesignFactor The same load can cause multiple tensions. Ft = ? lbs Manual Rigging: Ft = 425 lbs 2.125x the load Motorized Rigging: Ft = 1325 lbs 6.625x the load Shock Loading: Ft =2212.5 lbs More than 11x the load 200 lbs Fg = -200 lbs

25. 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. • Design factors will not help with shock loading. • Shock loaded equipment should be replaced. 200 lbs

26. 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? • Minor load changes/estimation issues • Human error • Equipment wear • Human acceleration 200 lbs

27. 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, ...

28. 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: Two-point bridle Safety Note: Each cable should be able to support the entire load.

29. Two Cables Cables attach to the object at two different points F2 F1 Question: What is the tension in each cable? 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.

30. 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 x1 F2 F1 Fg Since tnet = 0, t1 = t2, or x1.F1 = x2.F2.

31. Two Cables Cables attach to the object at two different points Newton’s 2nd Law: F1 + F2 = Fg. x2 x1 F2 F1 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) Fg

32. 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.

33. Two Cables Object suspended from a two-point bridle f q F2 F1 Newton’s 2nd Law says F1 + F2 + Fg = 0. Fg 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(q)/tan(f) + cos(q) sin(f)/tan(q) + cos(f)

34. Two Cables Object suspended from a two-point bridle f q What on earth does that mean? F2 F1 When f = q, F1 = F2, as we’d expect. Fg When f = q = 60o, 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).

35. 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.

36. Demo

37. Counterweight systems What is the tension? 283 lbs 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

38. Counterweight Systems What happens if the system is out of weight? Fnet = Fg + Ft 150 lbs Fnet = 200 lbs - 150 lbs Fnet = 50 lbs 200 lbs

39. 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 150 lbs 50 lbs = 10.87 slugs * a a = 4.6 ft/sec2 200 lbs

40. Counterweight Systems What are the tensions in the cables? Fnet = ma Fnet = Fg - Ft or Fg - Ft = ma Ft = -ma + Fg Ft = -6.21 slugs*4.6 ft/s2 + 200 lbs = 171.4 lbs Tension (Ft) Vertical Up 171.4 lbs Gravity (Fg) Vertical Down 200 lbs

41. Counterweight Systems What are the tensions in the cables? Fnet = ma Fnet = -Fg + Ft or -Fg + Ft = ma Ft = ma + Fg Ft = 4.66 slugs*4.6 ft/s2 + 150 lbs = 171.4 lbs Tension (Ft) Vertical Up 171.4 lbs 150 lbs Gravity (Fg) Vertical Down

42. Sheave forces What force is acting on the sheave? At a right angle… 171.4 lbs c a2 + b2 = c2 171.42 lbs +171.42 lbs = c2 c = 242.4 lbs Also 171.4 * SQRT(2) a b 171.4 lbs

43. Sheave forces What force is acting on the sheave? At a non-right angle… 191.7 lbs 171.4 lbs f a Law of Cosines c q a2 + b2 -2ab*cos(q) = c2 171.42 + 171.42 -2*171.4*171.4cos(112) = c2 b 36745.6 lbs2 = c2 171.4 lbs 191.7 lbs = c f = 68 q = 180 - f

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

45. Single Purchase Systems 400 lbs 200 lbs 200 lbs 200 lbs 200 lbs

46. Double Purchase 300 lbs 600 lbs 200 lbs 100 lbs 200 lbs 200 lbs 400 lbs 100 lbs 100 lbs 200 lbs 200 lbs Disadvantage Advantage

47. Double Purchase 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 100 lbs 100 lbs 100 lbs 200 lbs

48. Double Purchase 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 200 lbs 400 lbs 200 lbs 200 lbs

49. Block and Fall 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 250 lbs 50 lbs each 50 lbs 200 lbs