Progettazione Funzionale di Sistemi Meccanici e Meccatronicimech.unibg.it/~righettini/downloads/corsi/pmf2010/slide_ml.pdf · Progettazione Funzionale di Sistemi Meccanici e Meccatronici

Progettazione Funzionale di SistemiMeccanici e Meccatronici

Motion Laws

prof. Paolo Righettini

[email protected]

Dipartimento di Progettazione e Tecnologie

Mechatronics and Mechanical Dynamics Laboratories

prog. funz. sis. Meccanici e Meccatronici - prof. Paolo Righettini – p. 1/34

Payload position

Linear or rotary movements

position pay-load y as a function as time t

y = y(t)

external forces

dynamic actions due to pay-load acceleration


Periodic motion

the problem is how to describe the pay-load motionrequest

rotary motion may hasn’t limits, while linear motion has

a typical representation of a linear movement request maybe


Periodic motion

T total motion time or period

h motion rise (stroke)

four main parts

I rise

II dwell

III return

IV dwell


Periodic motion

rise part, for a total pay-load displacement h

dwell part, holding the position y = h

return part, return of the pay-load to the initial positiony = 0

dwell part, holding the position y = 0


Rise part

analysing the rise motion

the pay-load reaches the final position y = h at the time td

at the starting point A we have null velocity → y(0) = 0

at the ending point B we have null velocity → y(td) = 0


Rise part

four constraints, two for t = 0, two for t = td

t = 0 → y(0) = 0 , y(0) = 0

t = td → y(td) = h , y(td) = 0


Rise part

null velocities at the starting and ending motion points (A,B) are required to avoid shock

the movement profile y(t) must be continuous for thederivatives of order 0 and 1 (c1 function)


Motion Law - displacement

the conditions at the motion starting and ending points,lead to an ∞ of solutions of the displacement problem

which is the best way to choose a motion profile that respects t heboundary conditions?


Motion Law

kinematics: any motion displacement request leads toacceleration request

the acceleration of the mechanical part is the cause ofarising of inertial actions, moreover acceleration in themotion profile shows a positive and a negative part

inertial forces may cause vibration or action overload ofthe mechanical parts

therefore we study the motion law at the accelerationlevel, to control the acceleration amplitude and so theinertial forces

how to configure an acceleration profile y(t) that leads to therequired motion (stroke h in the time td)?


Adimensional approach

given a movement as a function y(t), stroke h and drivetime td

the acceleration profile may be expressed as

y(t) =h

t2d

f(t/td)

dimensional term h

t2d

depends on the two main motionparameters

adimensional function f(x), x = t/td, 0 ≤ x ≤ 1 represents theacceleration shape profile


Velocity profile

starting from the expression

y(t) = h/t2df(t/td)

of the acceleration profile, the velocity profile results

y(t) =

∫y(t)dt

remembering that x = t/td and therefore dt = dxtd we get

y(xtd) =

∫h

t2d

f(x)dxtd =h

td

∫f(x)dx =

h

tdF (x)


Velocity profile

y(t) =h

tdF (t/td)

dimensional term h

tdrepresents the mean velocity of the

motion required

adimensional function F (x) is the primitive of the functionf(x), with x = t/td, 0 ≤ x ≤ 1


Position profile

starting from the expression

y(t) = h/tdF (t/td)

of the velocity profile, the position profile results

y(t) =

∫y(t)dt

remembering that x = t/td and therefore dt = dxtd we get

y(xtd) =

∫h

tdF (x)dxtd = h

∫F (x)dx = hG(x)


Position profile

y(t) = hG(t/td)

dimensional term h represents the stroke of the requiredmotion

adimensional function G(x) is the primitive of the functionF (x), with x = t/td, 0 ≤ x ≤ 1


Motion law description

positiony(t) = hG(t/td)

velocity

y(t) =h

tdF (t/td)

acceleration

y(t) =h

t2d

f(t/td)


Characteristics of f (x)

the function f(x) has some useful characteristics

∫ td

0

y(t)dt = [y(t)]td0= y(td)− y(0) = 0

the velocity at the starting and ending points is null toavoid shock

taking into account the variable x, it results

∫ td

0

y(t)dt =

∫ 1

0

h

t2d

f(x)dx = 0 ⇒

∫ 1

0

f(x)dx = 0



therefore the acceleration profile y(t) and theadimensional function f(x) have a null mean value

∫ 1

0

f(x)dx = 0

the two area A and B must have the same surface



it is necessary to calculate the scale factor of the functionf(x)

we must relate the amplitude of the function f(x) to therise h ∫ ∫

td

0

y(t)dt = h

remembering that

y(t) = hG(t/td) and y(td) = h

it resultsy(td) = hG(td/td) = h ⇒ G(1) = 1



the twice integration of the function f(x) from 0 to 1 leadsto 1

the function f(x) is an adimensional acceleration profile,defined in the interval 0− 1, that leads to unitary stroke



the scale factor of the function f(x) may be more easilydetermined by means of the momentum of the functionf(x)

∫ 1

0

f(x)xdx



∫ 1

0

f(x)xdx = [F (x)x]10−

∫ 1

0

F (x)dx

but F (0) = 0 and F (1) = 0 and∫ 1

0

F (x)dx = [G(x)]10= G(1)−G(0) = 1

unless the minus sign the integration results∫ 1

0

f(x)xdx = 1



the resultants of the areas, positive and negative, havethe same modulus (f(x) has a null mean value)

R = R1 = R2

the momentum of f(x) may be calculated as∫ 1

0

f(x)xdx = Rδ = 1 ⇒ Rδ = 1



the momentum of f(x) may be calculated as∫ 1

0

f(x)xdx = Rδ = 1 ⇒ Rδ = 1

this relation allows an easy evaluation of the scale factor


Constant acceleration motion law

symmetric law

two zones at constant acceleration, one positive and onenegative

due to symmetric profile, the positive and negativeaccelerations have the same modulus (A)

one parameter xv allows shape changing of the profile



scale factor calculation

remembering that must results Rδ = 1

R = Axv , δ = 1− xv

results

Axv(1− xv) = 1 ⇒ A =1

xv(1− xv)



the amplitude of the constant piece function f(x) dependson the parameter xv

A =1

xv(1− xv)

each value of the parameter xv leads to a differentposition profile but with the same boundary condition


Acceleration coefficientCa

it is useful a motion low index that gives indications on themaximum value of the acceleration

y(t)max =h

t2d

f(t/td)max

we introduce a kinematic coefficient that represents themaximum value of the function f(x)

Ca = max (|f(x)|)

we also introduce others two acceleration coefficientsregarding the positive and the negative part of thefunction f(x)

C+a = max(f(x)) , C−

a = |min(f(x))|


Acceleration coefficientCa

Ca = max(C+a , C

−

a )

maximum value of the acceleration

y(t)max =h

t2d

Ca

moreover these coefficients make more easy to comparetwo acceleration profile expressed as f1(x) and f2(x).



for symmetric constant acceleration law, it results

Ca = C+a = C−

a =1

xv(1− xv)



the minimum value of Ca occur for xv = 1/2



this configuration leads to the minimum value of maximumacceleration

for a half of the time drive the pay-load has a positiveacceleration, for the other half the pay-load has a negativeacceleration



remembering that

y(t) =h

t2d

f(t/td)

it results

y(t)max =h

t2d

max (f(t/td)) =h

t2d

Ca

it is impossible to realize a movement with stroke h in thetime interval td with a maximum acceleration lower thanh

t2d

4

y(t)max ≥h24



it is a good practice to limit the acceleration coefficient to 8

twice the value of the minimum value of Ca

what occurs to the velocity profile while xv changes from 1/2 to0?


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Progettazione Funzionale di Sistemi Meccanici e Meccatronicimech.unibg.it/~righettini/downloads/corsi/pmf2010/slide_ml.pdf · Progettazione Funzionale di Sistemi Meccanici e Meccatronici