Modeling and Analysis of Dynamic SystemsOutline 1 Lecture 4: Modeling Tools for Mechanical Systems Lagrange Formalism Lagrange Method with Kinematic Constraints Pendulum on a Cart

Modeling and Analysis of Dynamic Systems

Dr. Guillaume Ducard

Fall 2017

Institute for Dynamic Systems and Control

ETH Zurich, Switzerland

G. Ducard c© 1 / 21

Outline

1 Lecture 4: Modeling Tools for Mechanical SystemsLagrange FormalismLagrange Method with Kinematic ConstraintsPendulum on a Cart

2 Lecture 4: Hydraulic SystemsWater DuctCompressible Duct Element


Lecture 4: Modeling Tools for Mechanical SystemsLecture 4: Hydraulic Systems

Lagrange FormalismLagrange Method with Kinematic ConstraintsPendulum on a Cart

Outline






Lagrange: 1736 -1813




Lagrange Formalism: Recipe

1 Define inputs and outputs2 Define the generalized coordinates:

q(t) = [q1(t), q2(t), . . . , qn(t)] andq(t) = [q1(t), q2(t), . . . , qn(t)]

3 Build the Lagrange function:

L(q, q) = T (q, q)− U(q)

4 System dynamics equations:

d

dt

{

∂L

∂qk

}

−

∂L

∂qk= Qk, k = 1, . . . , n

Notes:

Qk represents the k-th “generalized force or torque” acting onthe k−th generalized coordinate variable qkn: number of degrees of freedom in the systemalways n generalized variables G. Ducard c© 5 / 21



Outline






Lagrange Equations for Constrained Systems

d

dt

{

∂L

∂qk

}

−

∂L

∂qk−

ν∑

j=1

µjαj,k = Qk, k = 1, . . . , n, (1)

Remarks:

the constraints are included using “Lagrange multipliers”: µj

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 solvedfor qk and µj (usually requires computing the time derivativeof the constraints, i.e., µ).




Outline






Nonlinear Pendulum on a Cart

y(t)

u(t)M = 1kg

ϕ(t)

l = 1m

mg

m = 1kg

Figure: Pendulum on a cart, u(t) is the force acting on the cart(“input”), y(t) the distance of the cart to an arbitrary but constantorigin, and ϕ(t) the angle of the pendulum.




Step 1: Inputs & Outputs

Input: force acting on the cart: u(t)

Output: angle of the pendulum: ϕ(t)

Step 2: System’s coordinate variables

q1 = y, q1 = y

q2 = ϕ, q2 = ϕ

Step 3: Lagrange functions

L1(t) = T1(t)− U1(t)

L2(t) = T2(t)− U2(t)

L(t) = L1(t) + L2(t)G. Ducard c© 10 / 21



Step 4: System’s dynamics equations

d

dt

{

∂L

∂q1

}

−

∂L

∂q1= Q1

d

dt

{

∂L

∂q2

}

−

∂L

∂q2= Q2

We are looking for dynamic equations of the form:

y(t) = f(ϕ(t), ϕ(t), u(t))

ϕ(t) = g(ϕ(t), ϕ(t), u(t))



Water DuctCompressible Duct Element

Outline






Introduction

In general

they are described by Navier-Stokes equations.

For control purposes

simpler formulations are necessary, to build networks with buildingblocks:

ducts,

compressible nodes,

valves, etc.




Water duct in gravitational field

l

A

v(t)

v(t)

p1

p2

h

Objective

d

dtv(t) = f (p1(t), p2(t), v(t), h, ρ,A, l)




Change in momentum: Newton’s law

d~p

dt= m

d~v

dt= ~Fpressure + ~Fgravity + ~Ffriction

= [P1A− P2A] ~x+

∫

tube

~g · dm+ ~Ffriction

The mass m of the fluid in the element of tube of length l is givenby

m = ρ ·A · l → dm = ρ ·A · dl




Angle of the duct

sinα =dh

dl

Gravity force

∫

tube~g dm = g

∫

tube(− cosα ~y + sinα ~x) ρ · A · dl

= ρ · g ·A[

∫ h

0−

cosαsinα

dh ~y +∫ h

0

sinαsinα

dh ~x]

= −ρ · g · A (tanα)−1 h ~y + ρ g Ah ~x




Water duct in gravitational field

Dynamics along the ~x axis (because ~v = v~x)

ρ A ldv(t)

dt= A (P1 − P2) + ρ g A h− Ffriction,x(t)

with

Ffriction,x(t) =1

2ρ v2(t) sign [v(t)] · λ (v(t)) ·

A l

d

Remark: shape factor: ld




Outline






∗

V in (t)∗

V out (t)p(t)

∆V = 0

∆V (t)

k = 1/(σ0V0)

Definitions of compressability

Property of a body (solid, liquid, gas, etc.) to deform (to changeits volume) under the effect of applied pressure.Defined as:

σ0 =1

V0

dV

dP

V0: nominal volume [m3], P : pressure [Pa]σ0: compressibility [Pa−1]

k0 =1

σ0is called elasticity constant

[

Pa ·m−3]

.G. Ducard c© 19 / 21



Compressibility effects

∗

V in (t)∗

V out (t)p(t)

∆V = 0

∆V (t)

k = 1/(σ0V0)

Modeling

ddtV (t) =

∗

V in(t)−∗

V out(t) = Ainvin(t)−Aoutvout(t)

∆P (t) = k∆V (t) = 1

σ0V0∆V (t)

∆V (t) = V (t)− V0




Next lecture + Upcoming Exercise

Next lecture

Pelton Turbine

Electromagnetic systems

Next exercise: Online next Friday

Hydro-electric Power plant, Part I


Documents

Modeling and Analysis of Dynamic SystemsOutline 1 Lecture 4: Modeling Tools for Mechanical Systems Lagrange Formalism Lagrange Method with Kinematic Constraints Pendulum on a Cart