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 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?

4. Physics of Theatre: Rigging 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? 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. 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 Physics of Theatre: Rigging

7. 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/s 2, ft/sec 2 Velocity and Acceleration, like force, are vectors – They have both magnitude and direction. Physics of Theatre: Rigging

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

9. Bowling Ball PIPE ROPE Tension (F t ) Vertical Up Gravity (F g ) Vertical Down What are the forces acting on the ball? Force of Gravity Mass * Acceleration of gravity (g) = F g or “weight” The bowling ball is not accelerating vertically. F t = F g Vertical Suspension

10. Why is it important to know F t ? 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. Vertical Movement

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

12. Let’s work one out…… F net = Mass * Acceleration Mass of Bowling ball = =.3125 slugs 10 lbs 32 ft/s 2 F net =.3125 slugs * 4 ft/s 2 Acceleration = 4 ft/s 2 Weight = 10 lbs F net = 1.25 lbs Vertical Movement F net (Net Force) F net (Net Force) = F g + F t = ma

13. Let’s work one out…… F g = 10 lbs Acceleration = 4 ft/s 2 Weight = 10 lbs F net = 1.25 lbs These are the magnitudes of the forces. Vertical Movement F net (Net Force) F net (Net Force) = F g + F t = ma

14. Let’s work one out…… F g = -10 lbs Acceleration = 4 ft/s 2 Weight = 10 lbs F net = 1.25 lbs 1.25 lbs = -10 lbs + F t F t = 11.25 lbs + - + - Vertical Movement F net (Net Force) F net (Net Force) = F g + F t = ma

15. What if we were accelerating downward? F g = -10 lbs Acceleration = -4 ft/s 2 Weight = 10 lbs F net = -1.25 lbs -1.25 lbs = -10 lbs + F t F t = 8.75 lbs + - + - F net =.3125 slugs * -4 ft/s 2 Vertical Movement F net (Net Force) F net (Net Force) = F g + F t = ma

16. What if we were decelerating? Acceleration = 4 ft/s 2 Weight = 10 lbs 1.25 lbs = -10 lbs + F t F t = 11.25 lbs F net =.3125 slugs * 4 ft/s 2 F net = 1.25 lbs F g = -10 lbs Vertical Movement Note that this is the same force as accelerating upwards - the directions of the forces are what matter. F net (Net Force) F net (Net Force) = F g + F t = ma

17. Demo Break

18. 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/sec 2 6.25 slugs Mass of Unit = = 6.25 slugs 200 lbs 32 ft/s 2 200 lbs Vertical Movement

19. What if we wanted to land an 200 lb unit? Acceleration = 36 ft/s 2 F net = 225 lbs 225 lbs = -200 lbs + F t F t = 425 lbs F net = 6.25 slugs * 36 ft/s 2 F g = -200 lbs F t = 425 lbs 200 lbs Vertical Movement F net (Net Force) F net (Net Force) = F g + F t = ma

20. 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. a f = 180 ft/sec 2 200 lbs Immediate Stop loading

21. What if that same unit stopped suddenly? Acceleration = 180 ft/s 2 1125 lbs = -200 lbs + F t F t = 1325 lbs F net = 6.25 slugs * 180 ft/s 2 F net = 1125 lbs F g = -200 lbs F t = 1325 lbs 200 lbs Immediate Stop Loading F net (Net Force) F net (Net Force) = F g + F t = ma

22. 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. a f = 322 ft/sec 2 200 lbs Shock loading

23. Acceleration = 322 ft/s 2 2012.5 lbs = -200 lbs + F t F t = 2212.5 lbs F net = 6.25 slugs * 322 ft/s 2 F net = 2012.5 lbs F g = -200 lbs F t = 2212.5 lbs 200 lbs Shock Loading What if you need to stop this object from freefall? F net (Net Force) F net (Net Force) = F g + F t = ma

24. Manual Rigging: Shock Loading: F t = 2212.5 lbs More than 11x the load Motorized Rigging: F t = 1325 lbs 6.625x the load F t = 425 lbs 2.125x the load F g = -200 lbs F t = ? lbs 200 lbs Design Factor The same load can cause multiple tensions.

25. 200 lbs The same load can cause multiple tensions. Many theatre technicians have been told to use a design factor of 5:1. 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. Many experts now feel that design factors of 7:1, 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 factor 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. 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. Design factors will not help with shock loading. Shock loaded equipment should be replaced. Shock loaded equipment should be replaced. Design Factor

26. 200 lbs What’s not included in the design factor? Dynamic or Shock Loading Dynamic or Shock Loading Motor Driven Loads Motor Driven Loads Efficiency Efficiency Design Factor So if you’re going to calculate everything, why do you still need a design factor? Minor load changes/estimation issues Minor load changes/estimation issues Human error Human error Equipment wear Equipment wear Human acceleration Human acceleration

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

28. Two Cables What’s different when we suspend the object 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.

29. Two Cables Cables attach to the object at two different points Question: What is the tension in each cable? Assuming it’s dead hung, F net =0, which gives F 1 + F 2 = F g. This isn’t enough information to solve for F 1 and F 2 – 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 2 nd Law,  net = I  = 0. F2F2 F1F1 FgFg

30. Two Cables Cables attach to the object at two different points The tension in each cable applies a torque around the center of gravity:  1 = x 1. F 1,  2 = x 2. F 2, where x 1 and x 2 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 x 1. F 1 = x 2. F 2. x1x1 x2x2 F2F2 F1F1 FgFg

31. Two Cables Cables attach to the object at two different points Putting these together, we get: F 1 = F g. x 2 /(x 1 + x 2 ) and F 2 = F g. x 1 /(x 1 + x 2 ) Newton’s 2 nd Law (Rotational): x 1. F 1 = x 2. F 2. Newton’s 2 nd Law: F 1 + F 2 = F g. x1x1 x2x2 F2F2 F1F1 FgFg

32. Two Cables Cables attach to the object at two different points Whichever cable is closest to the center of gravity bears more weight. If the cables are placed equidistant from the center of gravity, F 1 = F 2 = ½ * F g (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.

33. Two Cables Object suspended from a two-point bridle Newton’s 2 nd Law says F 1 + F 2 + F g = 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:F 1 = F g F 2 = F g sin(  )/tan(  ) + cos(  ) sin(  )/tan(  ) + cos(  ) F1F1 F2F2 FgFg  

34. Two Cables Object suspended from a two-point bridle What on earth does that mean? When  = , F 1 = F 2, as we’d expect. When  =  , F 1 = F 2 = F g. 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 F g to F g + ma). F1F1 F2F2 FgFg  

35. Three (or more) Cables Object suspended from a three-point bridle Using Newton’s 2 nd 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. 200 lbs 200 lbs 200 lbs 283 lbs 200 lbs 0 lbs Counterweight systems What is the tension? Notice that the tension in the cable is the same all the way from the object to the counterweight. More on the sheave tension later.

38. 200 lbs 150 lbs F net = F g + F t F net = 200 lbs - 150 lbs F net = 50 lbs Counterweight Systems What happens if the system is out of weight?

39. 200 lbs 150 lbs F net = m * a m = 200 lbs + 150 lbs / 32 ft/sec 2 m = 10.87 slugs 50 lbs = 10.87 slugs * a a = 4.6 ft/sec 2 Counterweight Systems What happens if the system is out of weight? How fast will the system accelerate?

40. 200 lbs 171.4 lbs F net = ma F net = F g - F t or F g - F t = ma F t = -ma + F g F t = -6.21 slugs*4.6 ft/s 2 + 200 lbs = 171.4 lbs Counterweight Systems What are the tensions in the cables? Tension (F t ) Vertical Up Gravity (F g ) Vertical Down

41. 150 lbs Counterweight Systems What are the tensions in the cables? F net = ma F net = -F g + F t or -F g + F t = ma F t = ma + F g F t = 4.66 slugs*4.6 ft/s 2 + 150 lbs = 171.4 lbs Tension (F t ) Vertical Up Gravity (F g ) Vertical Down 171.4 lbs

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

43. Sheave forces What force is acting on the sheave? At a non-right angle… a b c 171.4 lbs a 2 + b 2 -2ab*cos(  ) = c 2 Law of Cosines 171.4 2 + 171.4 2 -2*171.4*171.4cos(112) = c 2 36745.6 lbs 2 = c 2 191.7 lbs = c  = 68  = 180 -    191.7 lbs 171.4 lbs

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

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

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

47. 200 lbs 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 100 lbs 100 lbs Double Purchase

48. 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 400 lbs Double Purchase

49. 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 lbs 200 lbs 250 lbs Block and Fall