Translational and Rotational Dynamics!

Transcription

1 Translational and Rotational Dynamics Robert Stengel Robotics and Intelligent Systems MAE 345, Princeton University, 217 Copyright 217 by Robert Stengel. All rights reserved. For educational use only. 1 Reference Frame Newtonian Inertial) Frame of Reference Unaccelerated Cartesian frame Origin referenced to inertial non-moving) frame Right-hand rule Origin can translate at constant linear velocity Frame cannot rotate with respect to inertial origin Position: 3 dimensions What is a non-moving frame? r = x y z Translation = Linear motion 2

2 Velocity and Momentum of a Particle Velocity of a particle v = dr = r = x y z = v x v y v z Linear momentum of a particle p = mv = m v x v y v z 3 Newton s Laws of Motion: Dynamics of a Particle First Law. If no force acts on a particle, it remains at rest or continues to move in straight line at constant velocity,. Inertial reference frame. Momentum is conserved d mv) = ; mv t1 = mv t2 4

3 d Newton s Laws of Motion: Dynamics of a Particle Second Law Particle acted upon by force Acceleration proportional to and in direction of force Inertial reference frame Ratio of force to acceleration is particle mass mv) = m dv Force = f x f y f z = ma = Force = force vector dv = 1 m Force = 1 m I 3 Force = 1/ m 1/ m 1/ m f x f y f z 5 Newton s Laws of Motion: Dynamics of a Particle Third Law For every action, there is an equal and opposite reaction Force on rocket motor = Force on exhaust gas F R = F E 6

4 One-Degree-of-Freedom Example of Newton s Second Law ) dx 1 t dx 2 t ) 2 nd -order, linear, time-invariant ordinary differential equation d 2 xt) 2 xt) = v x t) = f xt) m Corresponding set of 1 st -order equations State-Space Model) x 1 t) x 2 t) v x t) x 1 t) = x 2 t) = v x t) = f xt) m x 1 t) xt), ) x 2 t) dx t Defined as Displacement, Rate 7 State-Space Model is a Set of 1 st - Order Ordinary Differential Equations State, control, and output vectors for the example x 1 t) xt) = x 2 t) ; ut) = u t ) = f t); yt) = x 1 t) x Stability and control-effect matrices x 2 t) F= 1 ; G= Dynamic equation xt) = Fxt) + Gut) 1 / m 8

5 State-Space Model of the 1-DOF Example Output equation yt) = H x xt) + H u ut) Output coefficient matrices H x = 1 1 ; H u = 9 Dynamic System Dynamic Process: Current state may depend on prior state x : state dim = n x 1) u : input dim = m x 1) w : disturbance dim = s x 1) p : parameter dim = l x 1) t : time independent variable, 1 x 1) Observation Process: Measurement may contain error or be incomplete y : output error-free) dim = r x 1) n : measurement error dim = r x 1) z : measurement dim = r x 1) 1

6 State-Space Model of Three- Degree-of-Freedom Dynamics x t) rt) vt) = x y z v x v y v z = v x v y v z f x / m f y / m f z / m = Fx t) + Gu t) x t) rt) vt) = x y z v x v y v z = x y z v x v y v z + 1/ m 1/ m 1/ m f x f y f z 11 What Use are the Dynamic Equations? Compute time response Compute trajectories Determine stability Identify modes of motion 12

7 Forces 13 External Forces: Aerodynamic/Hydrodynamic f aero = X Y Z = air density, function of height = sealevel e h V = airspeed = v 2 x + v 2 2 y + v z 1/2 = v T 1/2 v A = referencearea = C X C Y C Z C X C Y C Z = 1 2 V 2 A dimensionless aerodynamic coefficients Inertial frame or body frame? 14

8 External Forces: Friction 15 External Forces: Gravity Flat-earth approximation g is gravitational acceleration mg is gravitational force Independent of position z measured up mg flat = m g o ; g o = 9.87 m / s 2 16

9 External Forces: Gravity r = x 2 + y 2 + z 2 ) 1/2 L = Latitude = Longitude Spherical earth, inertial frame Inverse-square gravitation Non-linear function of position µ = x 1 14 m/s Spherical earth, rotating frame Inverse-square gravitation Centripetal acceleration = 7.29 x 1 5 rad/s g x g round = g y = g gravity g z = µ x y r 3 = µ x / r r 2 y / r = µ cos L cos r 2 cos Lsin z z / r sin L g r = g gravity + g rotation = µ x y r 3 z x + 2 y 17 State-Space Model with Round-Earth Gravity Model Non-Rotating Frame) x y z v x v y v z = x y z v x v y v z µ / r 3 µ / r 3 µ / r 3 x y z = µ / r 3 µ / r 3 µ / r 3 Inverse-square gravity model introduces nonlinearity x y z v x v y v z 18

10 Vector-Matrix Form of Round-Earth Dynamic Model r v = I 3 µ r I 3 3 What other forces might be considered, and where would they appear in the model? r v 19 Point-Mass Motions of Spacecraft For short distance and low speed, flat-earth frame of reference and gravity are sufficient For long distance and high speed, round-earth frame and inverse-square gravity are needed 2

11 m = Mass of an Object dm = x, y,z)dx dydz Body x max y max z max x min y min z min x, y,z) = density of the body Density of object may vary with x,y,z) 21 More Forces 22

12 External Forces: Linear Springs Scalar, linear spring f = k x x o ) = kx; k = springconstant Uncoupled, linear vector spring f S = k x k y k z x x o ) y y o ) ) z z o = k x k y k z x y z x o, y o,z o ) : Equilibrium reference) position 23 External Forces: Viscous Dampers f = d v v o Scalar, linear damper ) = dv; d = dampingconstant Uncoupled, linear vector damper f D = d x d y d z v x v xo ) v y v yo ) v z v zo ) d x d y d z v x v y v z v xo,v yo,v zo ) : Equilibrium reference) velocity [usually = ] 24

13 State-Space Model with Linear Spring and Damping x y z v x v y v z = x y z v x v y v z + k x d x m m k y m k z m d y m d z m x x o ) y y o ) ) z z o v x v y v z Spring Effects Damping Effects 25 State-Space Model with Linear Spring and Damping Stability Effects Reference Effects x y z v x v y v z = k x d x m m k y m k z m d y m d z m x y z v + x v y v z k x m d x m k y m d y m k z m d z m x o y o z o 26

14 Vector-Matrix Form r v = I 3 K / m D / m r r o ) v r v = I 3 K / m D / m r v + I 3 K / m D / m r o 27 Rotational Motion 28

15 Assignment 2 Document the physical characteristics and flight behavior of a Syma X11 quadcopter Center of Mass r cm = x cm y cm z cm x max y max z max x min y min z min = 1 m x y z Body r dm x, y,z)dx dydz Reference point for rotational motion 3

16 Angular Momentum of a Particle Particle in Inverse- Square Field Moment of linear momentum of differential particles that make up the body Differential mass of a particle times Component of velocity perpendicular to moment arm from center of rotation to particle dh = r dmv) = r v)dm 31 Cross Product of Two Vectors r v = i j k x y z = yv z zv y )i + zv x xv z ) j + xv y yv x )k v x v y v z i, j, k) : Unit vectors along x, y,z) This is equivalent to = yv z zv y ) zv x xv z ) xv y yv x ) = z y z x y x v x v y v z = rv 32

17 Cross-Product-Equivalent Matrix r v = rv Cross-product equivalent of radius vector r r = z y z x y x Velocity vector v = v x v y v z 33 Angular Momentum of an Object Integrate moment of linear momentum of differential particles over the body h r v)dm = r v)x, y,z)dx dydz Body x max y max z max x min y min z min x max y max z max = rv )x, y,z)dx dydz x min y min z min h x h y h z 34

18 Angular Velocity and Corresponding Velocity Increment Angular velocity of object with respect to inertial frame of reference = x y z I Linear velocity increment at a point, x,y,z), due to angular rotation v x, y,z) = v x v y v z = ) r = * r ) ) 35 Angular Momentum of an Object with Respect to Its Center of Mass Choose center of mass as origin about which angular momentum is calculated = center of rotation) i.e., r is measured from the center of mass h = r v cm + v) dm = [ r v cm ]dm + [ r v]dm Body Body Body = [ rdm] v cm r r ) dm Body Body = r r dm = r r dm) I Body Body ) By symmetry, [ rdm] v cm = r cm v cm = Body I Inertia Matrix 36

19 Angular Momentum h = I Three components of angular momentum h x h y h z = r r dm ) Body I xx I xy I xz I xy I yy I yz I xz I yz I zz ) x ) y ) z 37 Inertia Matrix, I Angular rate has equal effect on all particles I = r r dm = Body Body z y z x y x z y z x y x dm = Body y 2 + z 2 ) xy xz xy x 2 + z 2 ) yz xz yz x 2 + y 2 ) dm 38

20 Inertia Matrix I = y 2 + z 2 ) xy xz xy x 2 + z 2 ) yz dm = xz yz x 2 + y 2 ) Body I xx I xy I xz I xy I yy I yz I xz I yz I zz Moments of inertia on the diagonal Products of inertia off the diagonal If products of inertia are zero, x, y, z) are principal axes, and I = I xx I yy I zz 39 Angular Momentum and Rate are Vectors Can be expressed in either an inertial or a body frame Vectors are transformed by the rotation matrix and its inverse h B = I B B = H I B h I = H I B I I I B = H I B I h I = I I I = H B I h B = H B I I B B I = H B I B 4

21 Similarity Transformations for Matrices Alternative expressions for angular momentum h B = I B B = H I B h I = H I B I I I = H I B I I H B I B Inertia matrices are transformed from one frame to the other by similarity transformations I B = H I B I I H B I I I = H B I I B H I B 41 Newton s 2 nd Law Applied to Rotational Motion in Inertial Frame dh I dh B Rate of change of angular momentum = applied moment or torque), m Inertia Matrix Expressed in Inertial Frame is Not Constant if Body is Rotating ) = d I I I ) = d I B B = di I d I + I I I = di B d B + I B B = m I [moment vector] I In a body reference frame, inertia matrix is constant d = ) B + I B B = I B d B = m B [moment vector] B 42

22 How do We get Rid of di I )/ in the Angular Momentum Rate Equation? Write the dynamic equation in bodyreferenced frame With constant mass, inertial properties are unchanging in body reference frame... but the frame is non-newtonian or non-inertial I 43 dh I Vector Derivative Expressed in a Rotating Frame Chain Rule = H B I dh B + dh I B h B = H B I dh B I I dh + I H B )h B = H B B + I H I B h B ) dh I = H B I dh B + I h I = H B I dh B + I h I Cross-product-equivalent matrix of angular rate: Similarity transformation = z y z x y x I = H B I B H I B 44

23 Rate of Change of Body-Axis Angular Momentum Substitute dh I = H B I dh B + H B I B H I B h I = m I = H B I m B Eliminate H B I dh B + H B I B H I B h I = H B I m B Rearrange and Substitute dh B = m B B h B 45 Rate of Change of Body-Axis Angular Velocity d I B B d B = m B B I B B = I 1 B m B B I B B ) 46

24 Rate of Change of Body-Axis Translational Velocity Inertial Velocity v I = H B I v B dv I = H B I dv B ) + d H I B v B = H B I dv B + I v I Rate of change = H B I dv B + H B I B H I B v I = H B I dv B + H B I B v B = 1 m f = 1 I m H I Bf B Substitution H B I dv B + H I B B v B = 1 m H I Bf B Result dv B = 1 m f B B v B 47 Rate of Change of Translational Position Express derivative in inertial frame r I = v I = H B I v B 48

25 Rate of Change of Angular Orientation Euler Angles) Body-axis angular rate vector components are orthogonal B = x y z B p q r Euler angles form a non-orthogonal vector * * * ) Therefore, Eulerangle rate vector is not orthogonal d = * * + * ), x, y, z * * +, I * )* I 49 Transformation From Euler-Angle Rates to Body-Axis Rates is measured in the Inertial Frame is measured in Intermediate Frame 1 is measured in Intermediate Frame 2 5

26 Sequential Transformations from Euler-Angle Rates to Body-Axis Rates is measured in the Inertial Frame is measured in Intermediate Frame 1 is measured in Intermediate Frame 2 p q r B = I 3 + H 2... which is + H B 2 2 H 1 ) p q r = 1 sin cos) sin) cos sin) cos) cos ) * L B I + 51 Inversion to Transform Body-Axis Rates to Euler-Angle Rates Transformation is not orthonormal 1 sin ) L B I = cos sin cos ) sin cos cos ) Inverse transformation is not the transpose B L I ) 1 I L B ) T L I B B L I ) 1 = 1 sin ) cos sin cos ) sin cos cos ) 1 ) ) = Adj L B I B det L I 52

27 Euler-Angle Rates ) ) = B L I ) 1 = Adj L B I det L I B 1 sin ) cos sin cos ) sin cos cos ) 1 = 1 sin tan cos tan ) cos sin ) sin sec cos sec ) Euler-angle rates from body-axis rates ) ) ) = 1 sin tan cos tan cos *sin sin sec cos sec ) ) ) p q r ) ) = L I B + B ) Can the inversion become singular? What does this mean? 53 Summary of Six-Degree-of-Freedom Rigid Body) Equations of Motion Rigid-body dynamic equations are nonlinear r I = H B I v B v B = 1 m f B B v B = L I B B B = I 1 B m B B I B B ) Translational position and velocity r I = x y z ; v B = u v w Rotational position and velocity p * = *; + B = q * ) r * * * ) 54

28 Next Time: Flying and Swimming Robots 55 )* +),-)* 56

29 Moments and Products of Inertia for Common Constant-Density Objects are Tabulated 57 Moments and Products of Inertia Bedford Fowler) 58

30 I = I xx I xy I xz I xy I yy I yz I xz I yz I zz Construction of Inertia Matrix Build up moments and products of inertia from components using parallel-axis theorem, e.g., I xxairplane = I xxwings + I xx fuselage + I xxhorizontal tail + I xxvertical tail +... or use software, e.g., Autodesk, Creo Parametric, SimMechanics, 59