Modeling and Analysis of Dynamic Systems by Dr. Guillaume Ducard Fall 2016 Institute for Dynamic Systems and Control ETH Zurich, Switzerland 1/21
Outline 1 Lecture 4: Modeling Tools for Mechanical Systems 2 2/21
Outline 1 Lecture 4: Modeling Tools for Mechanical Systems 2 3/21
Lagrange Equations for Constrained Systems Remarks: { } d L L dt q k q k ν µ j α j,k = Q k, k = 1,...,n, (1) j=1 the constraints are included using Lagrange multipliers : µ j j = 1...ν Number of constraints: ν with (ν < n) n may be seen as the number of DOF In the end, we obtain: n+ν coupled equations to be solved for q k and µ j (usually requires computing the time derivative of the constraints, i.e., µ). 4/21
Outline 1 Lecture 4: Modeling Tools for Mechanical Systems 2 5/21
ϕ(t) y(t) r,m,ϑ mg ψ(t) χ(t) u(t) R,Θ wheel: mass moment of inertia around c.o.g. is Θ, radius R, ball: mass m, mass moment of inertia around c.o.g. ϑ, radius r 6/21
Control objective The ball must be kept on top of the wheel. Model objectives Build a model for: system analysis (stability, observability, controllability) control design Assumptions No-slip no-slip equation 7/21
Modeling the ball on the wheel Step 1: Inputs and outputs Input: u(t) torque to the wheel Output: y(t) = (R+r)sinχ horizontal distance of the center of the ball w.r.t. the vertical axis of the wheel u( t) y( t) Torque to the wheel Model of the ball on the wheel Horizontal distance to the center of the ball 8/21
Step 2: n generalized coordinates (n DOF) and energies The system has 3 DOF: rotation of 1 the wheel around its center: angle ψ(t) 2 the ball around the center of the wheel: angle χ(t) 3 the ball around its own center: angle ϕ(t) ϕ(t) y(t) r,m,ϑ mg ψ(t) χ(t) u(t) R,Θ 9/21
Modeling the ball on the wheel Step 3: Lagrange function L(t) = T(t) U(t) Step 4: Differential equations including constraints Lagrange equations nonholonomic case (n = 3, ν = 1, q 1 = ψ, q 2 = χ, q 3 = ϕ) ( ) d L L µα k = Q k dt q k q k with Q 1 = u(t), Q 2 = Q 3 = 0 and α 1 q 1 +α 2 q 2 +α 3 q 3 = 0 with α 1 = R, α 2 = (R+r) and α 3 = r. Finally, we get a set of four equations define the four unknown variables { ψ, χ, ϕ,µ}, where { ϕ,µ} are easy to eliminate. 10/21
Modeling the ball on the wheel Final results Θ+ϑR 2 r 2 ϑ R(R+r) r 2 ϑ R(R+r) r 2 m(r+r) 2 +ϑ (R+r)2 r 2 [ ] [ ψ = χ Mass matrix M is positive definite. Therefore u mg(r+r)sin(χ) ] ψ(t) = [ (mr 2 +ϑ)u(t)+mgrϑsin(χ(t)) ] /Γ χ(t) = [ ϑru(t)+(θr 2 +ϑr 2 )mgsin(χ(t)) ] /[Γ(r+R)] where the scalar Γ is given by Γ = Θϑ +m(ϑr 2 +Θr 2 ) 11/21
Outline 1 Lecture 4: Modeling Tools for Mechanical Systems 2 12/21
Introduction In general they are described by Navier-Stokes equations. For control purposes simpler formulations are necessary, to build networks with building blocks: ducts, compressible nodes, valves, etc. 13/21
Water duct in gravitational field v(t) h p 1 A l p 2 v(t) Objective d dt v(t) = f (p 1(t),p 2 (t),v(t),h,ρ,a,l) 14/21
Change in momentum: Newton s law d p dt = md v dt = Fpressure + F gravity + F friction = [P 1 A P 2 A] x+ g dm+ F friction The mass m of the fluid in the element of tube of length l is given by m = ρ A l dm = ρ A dl tube 15/21
Angle of the duct sinα = dh dl Gravity force g dm = g ( cosα y +sinα x)ρ A dl tube tube [ h = ρ g A 0 cosα sinα dh y + h 0 ] sinα sinα dh x = ρ g A(tanα) 1 h y +ρgah x 16/21
Water duct in gravitational field Dynamics along the x axis (because v = v x) with ρ A l dv(t) dt = A (P 1 P 2 )+ρ g A h F friction,x (t) F friction,x (t) = 1 2 ρv2 (t)sign[v(t)] λ(v(t)) A l d Remark: shape factor: l d 17/21
Outline 1 Lecture 4: Modeling Tools for Mechanical Systems 2 18/21
V(t) V = 0 V in (t) p(t) k = 1/(σ 0 V 0 ) V out (t) Definitions of compressability Property of a body (solid, liquid, gas, etc.) to deform (to change its volume) under the effect of applied pressure. Defined as: σ 0 = 1 V 0 dv dp V 0 : nominal volume [m 3 ], P : pressure [Pa] σ 0 : compressibility [Pa 1 ] k 0 = 1 σ 0 is called elasticity constant [ Pa m 3]. 19/21
Compressibility effects V(t) V = 0 V in (t) p(t) k = 1/(σ 0 V 0 ) V out (t) Modeling d dt V(t) = V in(t) V out(t) = A in v in (t) A out v out (t) P(t) = k V(t) = 1 σ 0 V 0 V(t) V(t) = V(t) V 0 20/21
Next lecture + Upcoming Exercise Next lecture Pelton Turbine Electromagnetic systems Next exercise: Online next Friday Hydro-electric Power plant, Part I 21/21