Mechanical System Modeling

Transcription

1 Mechanical Engineer Modeling & Simulation Electro- Mechanics Electrical- Electronics Engineer Sensors Actuators Computer Systems Engineer Embedded Control Controls Engineer Mechatronic System Design K. Craig 1

2 References for Mechanical Systems System Dynamics, E. Doebelin, Marcel Dekker, (This is the finest reference on system dynamics available; many figures in these notes are taken from this reference.) Modeling, Analysis, and Control of Dynamic Systems, W. Palm, 2 nd Edition, Wiley, Vector Mechanics for Engineers: Dynamics, 7 th Edition, F. Beer, E.R. Johnston, and W. Clausen, McGraw Hill, K. Craig 2

3 Mechanical System Elements Three basic mechanical elements: Spring (elastic) element Damper (frictional) element Mass (inertia) element Translational and Rotational versions These are passive (non-energy producing) devices Driving Inputs force and motion sources which cause elements to respond K. Craig 3

4 Each of the elements has one of two possible energy behaviors: stores all the energy supplied to it dissipates all energy into heat by some kind of frictional effect Spring stores energy as potential energy Mass stores energy as kinetic energy Damper dissipates energy into heat Dynamic Response of each element is important step response frequency response K. Craig 4

5 Spring Element Real-world design situations Real-world spring is neither pure nor ideal Real-world spring has inertia and friction Pure spring has only elasticity - it is a mathematical model, not a real device Some dynamic operation requires that spring inertia and/or damping not be neglected Ideal spring: linear Nonlinear behavior may often be preferable and give significant performance advantages K. Craig 5

6 Device can be pure without being ideal (e.g., nonlinear spring with no inertia or damping) Device can be ideal without being pure (e.g., device which exhibits both linear springiness and linear damping) Pure and ideal spring element: f K x x K x s 1 2 s T K K s 1 2 s K s = spring stiffness (N/m or N-m/rad) 1/K s = C s = compliance (softness parameter) x Csf CT s K s x f f x C s K. Craig 6

7 Energy stored in a spring E s Csf K x s Dynamic Response: Zero-Order Dynamic System Model Step Response Frequency Response Real springs will not behave exactly like the pure/ideal element. One of the best ways to measure this deviation is through frequency response. K. Craig 7

8 Spring Element Differential Work Done f dx Total Work Done x 0 Ksx dx 0 K x dx s K x C f s 0 s 0 K. Craig 8

9 Frequency Response Of Spring Elements 0 s 0 f f sin t x C f sin t K. Craig 9

10 Zero-Order Dynamic System Model Step Response Frequency Response K. Craig 10

11 More Realistic Lumped-Parameter Model for a Spring K s K s f, x M B B K. Craig 11

12 Linearization for a Nonlinear Spring 2 2 df d f x x dx xx dx 2! xx y f (x ) x x df y y x x 0 0 dx xx df y y x x yˆ Kxˆ 0 0 dx xx 0 K. Craig 12

13 Real Springs nonlinearity of the force/deflection curve noncoincidence of the loading and unloading curves (The 2 nd Law of Thermodynamics guarantees that the area under the loading f vs. x curve must be greater than that under the unloading f vs. x curve. It is impossible to recover 100% of the energy put into any system.) K. Craig 13

14 Several Types of Practical Springs: coil spring hydraulic (oil) spring cantilever beam spring pneumatic (air) spring clamped-end beam spring ring spring rubber spring (shock mount) tension rod spring torsion bar spring K. Craig 14

15 Spring-like Effects in Unfamiliar Forms aerodynamic spring gravity spring (pendulum) gravity spring (liquid column) buoyancy spring magnetic spring electrostatic spring centrifugal spring K. Craig 15

16 Damper Element A pure damper dissipates all the energy supplied to it, i.e., converts the mechanical energy to thermal energy. Various physical mechanisms, usually associated with some form of friction, can provide this dissipative action, e.g., Coulomb (dry friction) damping Material (solid) damping Viscous damping K. Craig 16

17 Pure / ideal damper element provides viscous friction. All mechanical elements are defined in terms of their force/motion relation. (Electrical elements are defined in terms of their voltage/current relations.) Pure / Ideal Damper Damper force or torque is directly proportional to the relative velocity of its two ends. dx dx dx dt dt dt 1 2 f B B d d d dt dt dt 1 2 T B B K. Craig 17

18 Forces or torques on the two ends of the damper are exactly equal and opposite at all times (just like a spring); pure springs and dampers have no mass or inertia. This is NOT true for real springs and dampers. Units for B to preserve physical meaning: N/(m/sec) (N-m)/(rad/sec) Transfer Function Differential Operator Notation dx dt 2 Dx D x x D (x)dt x D 2 2 d x dt 2 x dtdt K. Craig 18

19 Operational Transfer Functions f T BDx BD f T D BD D BD x x 1 1 D D f BD T BD We assume the initial conditions are zero. Damper element dissipates into heat all mechanical energy supplied to it. dx dx Power force velocity f B dt dt Force applied to damper causes a velocity in same direction. 2 K. Craig 19

20 Power input to the device is positive since the force and velocity have the same sign. It is impossible for the applied force and resulting velocity to have opposite signs. Thus, a damper can never supply power to another device; Power is always positive. A spring absorbs power and stores energy as a force is applied to it, but if the force is gradually relaxed back to zero, the external force and the velocity now have opposite signs, showing that the spring is delivering power. Total Energy Dissipated 2 dx dx Pdt B dt B dx f dx dt dt K. Craig 20

21 Damper Element Step Input Force causes instantly (a pure damper has no inertia) a Step of dx/dt and a Ramp of x K. Craig 21

22 Frequency Response of Damper Elements B 0 t f f sin t dx B dt 1 x x f sin t dt f B 0 1 cos t f0 Ax B 1 A f B f 0 K. Craig 22

23 Sinusoidal Transfer Function x D 1 D i f BD i x 1 M f ib M is the amplitude ratio of output over input φ is the phase shift of the output sine wave with respect to the input sine wave (positive if the output leads the input, negative if the output lags the input) x i 1 M 1 90 f ib B K. Craig 23

24 Real Dampers A damper element is used to model a device designed into a system (e.g., automotive shock absorbers) or for unavoidable parasitic effects (e.g., air drag). To be an energy-dissipating effect, a device must exert a force opposite to the velocity; power is always negative when the force and velocity have opposite directions. Let s consider examples of real intentional dampers. K. Craig 24

25 Viscous (Piston/Cylinder) Damper A relative velocity between the cylinder and piston forces the viscous oil through the clearance space h, shearing the fluid and creating a damping force. = fluid viscosity L h 2 R 2 R 1 B R 3 2 R1 h h 2 h R 2 2 K. Craig 25

26 Simple Shear Damper and Viscosity Definition 2A F V t F 2A B V t fluid viscosity shearing stress velocity gradient F / A V / t K. Craig 26

27 Examples of Rotary Dampers B 4 D 0 16t B 3 DL 4t K. Craig 27

28 Commercial Air Damper (Data taken with valve shut) laminar flow linear damping Air Damper much lower viscosity less temperature dependent no leakage or sealing problem turbulent flow nonlinear damping K. Craig 28

29 Motion of the conducting cup in the magnetic field generates a voltage in the cup. A current is generated in the cup s circular path. A current-carrying conductor in a magnetic field experiences a force proportional to the current. The result is a force proportional to and opposing the velocity. The dissipated energy shows up as I 2 R heating of the cup. Eddy-Current Damper K. Craig 29

30 Temperature Sensitivity of Damping Methods K. Craig 30

31 Other Examples of Damper Forms K. Craig 31

32 The damper element can also be used to represent unavoidable parasitic energy dissipation effects in mechanical systems. Frictional effects in moving parts of machines Fluid drag on vehicles (cars, ships, aircraft, etc.) Windage losses of rotors in machines Hysteresis losses associated with cyclic stresses in materials Structural damping due to riveted joints, welds, etc. Air damping of vibrating structural shapes K. Craig 32

33 Hydraulic Motor Friction and its Components K. Craig 33

34 Coulomb Friction: Modeling and Simulation In most control systems, Coulomb friction is a nuisance. Coulomb friction is difficult to model and troublesome to deal with in control system design. It is a nonlinear phenomenon in which a force is produced that tends to oppose the motion of bodies in contact in a mechanical system. Undesirable effects: hangoff and limit cycling K. Craig 34

35 Hangoff (or dc limit cycle) prevents the steadystate error from becoming zero with a step command input. Limit Cycling is behavior in which the steady-state error oscillates or hunts about zero. What Should the Control Engineer Do? Minimize friction as much as possible in the design Appraise the effect of friction in a proposed control system design by simulation If simulation predicts that the effect of friction is unacceptable, you must do something about it! K. Craig 35

36 Remedies can include simply modifying the design parameters (gains), using integral control action, or using more complex measures such as estimating the friction and canceling its effect. Modeling and simulation of friction should contribute significantly to improving the performance of motion control systems. K. Craig 36

37 Modeling Coulomb Friction F f F stick F slip V "Stiction" Coulomb Friction Model K. Craig 37

38 Case Study to Evaluate Friction Model V 0 V m = 0.1 kg k = 100 N/m F stick = 0.25 N F slip = 0.20 N (assumed independent of velocity) V 0 = step of m/sec at t = 0 sec k m F f K. Craig 38

39 Friction Model in Simulink 2 From Reset Integrator 1 Resultant Force 3 Velocity Resultant Force Friction Force From Reset Integrator To Reset Integrator Velocity Friction Model 1 Friction Force 2 To Reset Integrator K. Craig 39

40 Simulink Block Diagram 1 Resultant Force u Abs 0.25 F_stick min Min Product Friction Model Sign hit 2 Hit Crossing 1 From Reset Integrator Hit Crossing Switch Friction Force 3 Velocity 0.2 Sign 2 To Reset Integrator F_slip Product K. Craig 40

41 Example with Friction Model Coulomb Friction Example Ff Position Ramp Resultant Force Friction Force Friction Force Sum 100 k Sum 1/0.1 1/m 1 s Integrator 1 s Integrator From Reset Integrator To Reset Integrator Velocity x position vel velocity Friction Model K. Craig 41

42 2*position, velocity, 0.1*Friction Force Position, Velocity, Friction Force vs. Time time (sec) K. Craig 42

43 Inertia Element A designer rarely inserts a component for the purpose of adding inertia; the mass or inertia element often represents an undesirable effect which is unavoidable since all materials have mass. There are some applications in which mass itself serves a useful function, e.g., accelerometers and flywheels. K. Craig 43

44 Accelerometer Useful Applications of Inertia Flywheels are used as energy-storage devices or as a means of smoothing out speed fluctuations in engines or other machines. K. Craig 44

45 Newton s Law defines the behavior of mass elements and refers basically to an idealized point mass : forces mass acceleration The concept of rigid body is introduced to deal with practical situations. For pure translatory motion, every point in a rigid body has identical motion. Real physical bodies never display ideal rigid behavior when being accelerated. The pure / ideal inertia element is a model, not a real object. K. Craig 45

46 Rigid and Flexible Bodies: Definitions and Behavior K. Craig 46

47 Newton s Law in rotational form for bodies undergoing pure rotational motion about a single fixed axis: torques moment of inertia angular acceleration The concept of moment of inertia J also considers the rotating body to be perfectly rigid. Note that to completely describe the inertial properties of any rigid body requires the specification of: Its total mass Location of the center of mass 3 moments of inertia and 3 products of inertia K. Craig 47

48 Rotational Inertia J (kg-m 2 ) tangential force massacceleration 2 rl drr R R MR 0 total torque 2Lr dr R L J 2 2 K. Craig 48

49 Moments of Inertia for Some Common Shapes K. Craig 49

50 How do we determine J for complex shapes with possibly different materials involved? In the design stage, where the actual part exists only on paper, estimate as well as possible! Once a part has been constructed, use experimental methods for measuring inertial properties. How? K. Craig 50

51 Experimental Measurement Of Moment of Inertia 2 d torques J J dt 2 K 2 d J dt s 2 2 d 2 dt K J s 0 cos t ( 0) 0 n 0 n 2 K J s rad/sec n f n cycles/sec J K s 2 2 n 4 f K. Craig 51

52 Actually the oscillation will gradually die out due to the bearing friction not being zero. If bearing friction were pure Coulomb friction, it can be shown that the decay envelope of the oscillations is a straight line and that friction has no effect on the frequency. If the friction is purely viscous, then the decay envelope is an exponential curve, and the frequency of oscillation does depend on the friction but the dependence is usually negligible for the low values of friction in typical apparatus. K. Craig 52

53 Inertia Element Real inertias may be impure (have some springiness and friction) but are very close to ideal. x 1 1 D D f MD T JD 2 2 Inertia Element stores energy as kinetic energy: Mv J or K. Craig 53

54 A step input force applied to a mass initially at rest causes an instantaneous jump in acceleration, a ramp change in velocity, and a parabolic change in position. The frequency response of the inertia element is obtained from the sinusoidal transfer function: x i f Mi 2 M 2 At high frequency, the inertia element becomes very difficult to move. The phase angle shows that the displacement is in a direction opposite to the applied force. K. Craig 54

55 Useful Frequency Range for Rigid Model of a Real Flexible Body A real flexible body approaches the behavior of a rigid body if the forcing frequency is small compared to the body s natural frequency. K. Craig 55

56 Analysis: 2AE x x ALx L 2 L x o x o x i 2E D x i o o 2 2E 1x x 1 x o i n 2 n L i D i i xo D xo 1 i n n n K. Craig 56

57 xo 1 i 1.05 xi max n max n E L cycles/min for a 6-inch steel rod max is the highest frequency for which the real body behaves almost like an ideal rigid body. Frequency response is unmatched as a technique for defining the useful range of application for all kinds of dynamic systems. K. Craig 57

58 Motion Transformers Mechanical systems often include mechanisms such as levers, gears, linkages, cams, chains, and belts. They all serve a common basic function, the transformation of the motion of an input member into the kinematically-related motion of an output member. The actual system may be simplified in many cases to a fictitious but dynamically equivalent one. K. Craig 58

59 This is accomplished by referring all the elements (masses, springs, dampers) and driving inputs to a single location, which could be the input, the output, or some selected interior point of the system. A single equation can then be written for this equivalent system, rather than having to write several equations for the actual system. This process is not necessary, but often speeds the work and reduces errors. K. Craig 59

60 Motion Transformers Gear Train Relations: T m m N 1 m m Tm T m N N N N N 1 N N 2 T m m K. Craig 60

61 Translational Equivalent for A Complex System Refer all elements and inputs to the x 1 location and define a fictitious equivalent system whose motion will be the same as x 1 but will include all the effects in the original system. x 1, x 2, are kinematically related K. Craig 61

62 Define a single equivalent spring element which will have the same effect as the three actual springs. Mentally apply a static force f 1 at location x 1 and write a torque balance equation: L f L K x L x K L f K x 2 1 s 1 1 s s2 2 L1 L1 1 se 1 2 L 1 K K K K 2 se s1 s2 2 s L1 L1 xk K. Craig 62

63 The equivalent spring constant K se refers to a fictitious spring which, if installed at location x 1, would have exactly the same effect as all the springs together in the actual system. To find the equivalent damper, mentally remove the inertias and springs and again apply a force f 1 at x 1 : f L x B L x B L B f L x x B L x B B 1 e L1 L1 B x 2 L 2 1 Be B1 B2 B 2 L1 L 1 K. Craig 63

64 Finally, consider only the inertias present. x x x f L M L M L J L1 L1 L1 f M x 1 e 1 2 L 2 1 Me M1 M2 J 2 L1 L 1 While the definitions of equivalent spring and damping constants are approximate due to the assumption of small motions, the equivalent mass has an additional assumption which may be less accurate; we have treated the masses as point masses, i.e., J = ML 2. K. Craig 64

65 To refer the driving inputs to the x 1 location we note that a torque T is equivalent to a force T/L 1 at the x 1 location, and a force f 2 is equivalent to a force (L 2 /L 1 )f 2. If we set up the differential equation of motion for this system and solve for its unknown x 1, we are guaranteed that this solution will be identical to that for x 1 in the actual system. Once we have x 1, we can get x 2 and/or immediately since they are related to x 1 by simple proportions. K. Craig 65

66 Rules for calculating the equivalent elements without deriving them from scratch: When referring a translational element (spring, damper, mass) from location A to location B, where A s motion is N times B s, multiply the element s value by N 2. This is also true for rotational elements coupled by motion transformers such as gears, belts, and chains. When referring a rotational element to a translational location, multiply the rotational element by 1/R 2, where the relation between translation x and rotation (in radians) is x = R. For the reverse procedure (referring a translational element to a rotational location) multiply the translational element by R 2. K. Craig 66

67 When referring a force at A to get an equivalent force at B, multiply by N (holds for torques). Multiply a torque at by 1/R to refer it to x as a force. A force at x is multiplied by R to refer it as a torque to. These rules apply to any mechanism, no matter what its form, as long as the motions at the two locations are linearly related. K. Craig 67

68 Mechanical Impedance When trying to predict the behavior of an assemblage of subsystems from their calculated or measured individual behavior, impedance methods have advantages. Mechanical impedance is defined as the transfer function (either operational or sinusoidal) in which force is the numerator and velocity the denominator. The inverse of impedance is called mobility. K. Craig 68

69 Mechanical Impedance for the Basic Elements f Ks ZS D D v D f ZB D D B v f ZM D D MD v K. Craig 69

70 Measurement of impedances of subsystems can be used to analytically predict the behavior of the complete system formed when the subsystems are connected. We can thus discover and correct potential design problems before the subsystems are actually connected. Impedance methods also provide shortcut analysis techniques. When two elements carry the same force they are said to be connected in parallel and their combined impedance is the product of the individual impedances over their sum. K. Craig 70

71 For impedances which have the same velocity, we say they are connected in series and their combined impedance is the sum of the individual ones. Consider the following systems: B K B K x 1, v 1 f, v Series Connection f, v Parallel Connection K. Craig 71

72 Parallel Connection K f B KB D D v K B BD K D Series Connection f K BD D B K v D D K. Craig 72

73 Force and Motion Sources The ultimate driving agency of any mechanical system is always a force not a motion; force causes acceleration, acceleration does not cause force. Motion does not occur without a force occurring first. At the input of a system, what is known, force or motion? If motion is known, then this motion was caused by some (perhaps unknown) force and postulating a problem with a motion input is acceptable. K. Craig 73

74 There are only two classes of forces: Forces associated with physical contact between two bodies Action-at-a-distance forces, i.e., gravitational, magnetic, and electrostatic forces. There are no other kinds of forces! (Inertia force is a fictitious force.) The choice of an input form to be applied to a system requires careful consideration, just as the choice of a suitable model to represent a component or system. Here are some examples of force and motion sources. K. Craig 74

75 Force and Motion Inputs acting on a Multistory Building K. Craig 75

76 A Mechanical Vibration Shaker: Rotating Unbalance as a Force Input K. Craig 76

77 Electrodynamic Vibration Shaker as a Force Source K. Craig 77

78 Force Source Constructed from a Motion Source and a Soft Spring K. Craig 78

79 Energy Considerations A system can be caused to respond only by the source supplying some energy to it; an interchange of energy must occur between source and system. If we postulate a force source, there will be an associated motion occurring at the force input point. The instantaneous power being transmitted through this energy port is the product of instantaneous force and velocity. If the force applied by the source and the velocity caused by it are in the same direction, power is supplied by the source to the system. If force and velocity are opposed, the system is returning power to the source. K. Craig 79

80 The concept of mechanical impedance is of some help here. The transfer function relating force and velocity at the input port of a system is called the driving-point impedance Z dp. Z (D) We can write an expression for power: dp f (D) v f Z dp(i ) (i ) v f f P fv f Z Z dp 2 dp K. Craig 80

81 If we apply a force source to a system with a high value of driving-point impedance, not much power will be taken from the source, since the force produces only a small velocity. The extreme case of this would the application of a force to a perfectly rigid wall (drivingpoint impedance is infinite, since no motion is produced no matter how large a force is applied). In this case the source would not supply any energy. The higher the driving-point impedance, the more a real force source behaves like an ideal force source. The lower the driving-point impedance, the more a real motion source behaves like an ideal motion source. K. Craig 81

82 Real sources may be described accurately as combinations of ideal sources and an output impedance characteristic of the physical device. A complete description of the situation thus requires knowledge of two impedances: The output impedance of the real source The driving-point impedance of the driven system K. Craig 82

83 Mechanical System Examples Rack-and-Pinion Gear System Problem Statement Develop the equivalent rotational model of the rack-and-pinion gear system shown. The applied torque T is the input variable, and the angular displacement is the output variable. Neglect any twist in the shaft. Bearings are frictionless. The pinion gear mass moment of inertia about its CG (geometric center) is I p Im Is Ip mrr cr kr T K. Craig 83

84 Multi-Gear System Problem Statement A load inertia I 5 is driven through a double-gear pair by a motor with inertia I 4, as shown. The shaft inertias are negligible. The gear inertias are I 1, I 2, and I 3. The speed ratios are 1 / 2 = 2 and 2 / 3 = 5. The motor torque is T 1 and the viscous damping coefficient c = 4 lb-ft-sec/rad. Neglect elasticity in the system, and use the following inertia values (sec 2 -ft-lb/rad): I 1 = 0.1, I 2 = 0.2, I 3 = 0.4, I 4 = 0.3, I 5 = 0.7. Derive the mathematical model for the motor shaft speed 1 with T 1 as the input I I I I I c T K. Craig 84

85 Dynamic Vibration Absorber Physical Model Problem Statement A dynamic vibration absorber consists of a mass and an elastic element that is attached to another mass in order to reduce its vibration. The figure is a representation of a vibration absorber attached to the cantilever support. For a cantilever beam with a force at its end, k = Ewh3/4L3 where L = beam length, w = beam width, and h = beam thickness. (a) Obtain the equation of motion for the system. The force f is a specified force acting on the mass m, and is due to the rotating unbalance of the motor. The displacements x and x2 are measured from the static equilibrium positions when f = 0. (b) Obtain the transfer functions x/f and x 2 /f. Physical System x m D k F mm D m k k mk D kk x k F mm D m k k mk D kk K. Craig 85

86 Rigid Body Dynamics: Kinematics y 1 P Reference Frames R - Ground xyz R 1 - Body x 1 y 1 z 1 y R 1 A x 1 R1 R1 v v r v R P R A R AP P z 1 R1 R1 R1 a a r r R P R A R R AP R AP z O R x a 2 v R P R R R P Note: For any vector q R dq dt R dq dt 1 R R1 q K. Craig 86

87 Rigid-Body Kinematics Example Given: Find: R R R R1 R P a 5i ˆ constant 4k ˆ constant R r = 0.06 m R 1 R 2 O θ = 30º y 2 ˆi ˆ i ˆj ˆ 1 0 cos sin j kˆ 0 sin cos kˆ 1 y 1 x 2 y 1 y Reference Frames: R ground: xyz R 1 shaft: x 1 y 1 z 1 R 2 disk: x 2 y 2 z 2 O x 1 z O z 1 K. Craig 87

88 2 2 R2 R2 R2 a a r r R P R O R R OP R OP R O a 0 R R R R P a 0 P v a 2 v Point O at end of rotating shaft fixed in R Point P fixed in R 2 (disk) R2 R R1 R1 R2 5i R R R2 R R d d 2 5i ˆ 4kˆ dt dt ˆ 4kˆ 1 1 R dk1 R R kˆ 1 dt 4 5i ˆ kˆ 20j ˆ K. Craig R P R R R P 20 ˆjcos ksin ˆ OP r rcos i1 rsin j1 ˆ ˆ

89 After Substitution and Simplification: ˆ ˆ ˆ R P a 16r cos i1 41r sin j1 40r cos k1 Alternate Solution: R R R1 R1 R1 a a r r R P R O R R OP R OP R O a 0 R1 5i ˆ constant R R d dt R 1 R a 2 v R P R R R P OP r rcos i1 rsin j1 ˆ ˆ K. Craig 89

90 a a r r (P is fixed in R 2 ) 1 O a 0 R P R O R R R R OP R R OP R R R R 1 R1 R1 R2 R1 R 1 R P R O R R OP R O v 0 4kˆ d d 4kˆ 0 dt dt v v r After Substitution and Simplification: OP r rcos i1 rsin j1 ˆ ˆ ˆ R P a 16r cos i1 41r sin j1 40r cos k1 ˆ ˆ (same result) K. Craig 90

91 Rigid Body Dynamics: Kinetics Point C: mass center of a rigid body of mass m. R C Linear Momentum L m v y 1 C Angular Momentum about point C R 1 H H ˆ y A x i 1 1 H ˆ y j 1 1 Hz kˆ 1 1 R R1 Hx I 1 x1x I 1 x1y I 1 x1z 1 x 1 R R 1 z H 1 y1 Iy 1x I 1 y1y I 1 y1z1 y 1 R R1 H z I z O R x 1 z1x I 1 z1y I 1 z1z 1 z1 R R C d v Fm Reference Frames dt Equations of Motion R - Ground xyz R dh R 1 - Body x 1 y 1 z M 1 dt y 1 x 1 K. Craig 91

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