Download Two-axle vehicle, on incline, traction-limited model (steady-state solution, etc.)

Longitudinal dynamics and performance
Traction-Limited Drive
Prof. R.G. Longoria
Spring 2016
ME 360/390 – Prof. R.G. Longoria
Vehicle System Dynamics and Control
Department of Mechanical Engineering
The University of Texas at Austin
Overview
These slides review the performance model in steady-state
longitudinal motion.
This gives a basis for understanding the system (plant) that needs to
be controlled either in open loop or closed loop.
These slides review traction-limited performance and demonstrate
how performance can be over-predicted when only looking at ideal
traction, motivating need to include power limited source models.
ME 360/390 – Prof. R.G. Longoria
Vehicle System Dynamics and Control
Department of Mechanical Engineering
The University of Texas at Austin
Longitudinal dynamics
Recall the model for longitudinal dynamics
of a vehicle,
•
•
•
•
Fg = grade
pɺ x = mvɺx = ∑ Fx
Fr = rollingresistance
∑F
Fd = drawbar
x
•
Ftx = traction/braking
Fa = aerodynamic
= Ftx − Fg − Fr − Fa − Fd
Maintaining a set speed requires adjusting the traction force to counteract
external forces (road loads) and/or changes in mass.
If you want to apply basic linear control design concepts, then a linear
model is required.
However, the forces depend on the vehicle velocity in a nonlinear
manner.
So, this equation would need to be linearized with respect to the
forward velocity if you wanted to used linear control methods.
Let’s first review the open-loop model.
ME 360/390 – Prof. R.G. Longoria
Vehicle System Dynamics and Control
Department of Mechanical Engineering
The University of Texas at Austin
Two-axle vehicle on an incline (ideal traction forces)
Fa
Along the longitudinal (x) axis:
pɺ x = m
dvx
= m ⋅ ax = ∑ Fx
dt
cf. Wong, Chapter 3, Fig. 3.1
Frf
vɺx
Fg
Rolling resistance forces placed
at wheel centers. Note, RR is
often ignored in braking since the
effect in that case may not be
significant.
ha
r
A
h
W
Ftf
B Frr
Fd
hd
θs
Frf
l1
L
l2
∑F
x
ME 360/390 – Prof. R.G. Longoria
Vehicle System Dynamics and Control
Note: For passenger cars you
can assume that: h ≈ h ≈ h
a
d
= Tractive force − Road Loads
Department of Mechanical Engineering
The University of Texas at Austin
Longitudinal tractive force (effort)
In the longitudinal (x) direction, need to estimate forces,
∑F
x
= Ftf + Ftr − Fa − Frf − Frr − Fd − Fg
Rolling resistance forces
These are the tractive forces (effort).
∑F
x
The tractive forces should be modeled to
estimate the force propelling a vehicle for
given road/terrain conditions.
It is necessary to make assumptions or
gather information about the tire and the
road/terrain conditions.
= Ftf + Ftr − Fa − Frf − Frr − Fd − Fg
Ftf ,r = tractive effort on front and rear
Fa = aerodynamic resistance force
Frf ,rr = rolling resistance on front and rear
Fd = drawbar load
Fg = grade resistance = W sin θ s
NOTE: See Wong Chapter 3 and/or the appended slides for some details on modeling these forces.
ME 360/390 – Prof. R.G. Longoria
Vehicle System Dynamics and Control
Department of Mechanical Engineering
The University of Texas at Austin
Dynamic model
For vehicle on incline, general relations – for 2D motion:
pɺ x = mvɺx = ∑ Fx
NOTE:
hr = h − r
l1
y
l2
l1 + l2 = L
l1 − hr f r
 1

 − f r
pɺ z = mvɺz = ∑ Fz
hɺ = I ωɺ = M
−l2 − hr f r
1
− fr
ME 360/390 – Prof. R.G. Longoria
Vehicle System Dynamics and Control
y
∑
y
One way to solve for the unknowns is
to formulate as 3 simultaneous
equations:

0  W f   −hFt





0  Wr  =  W cos θ 
−m   vɺx   Fg + Fa − Ft 
Department of Mechanical Engineering
The University of Texas at Austin
Normal forces and acceleration
Solution of the three unknowns yields the two normal loads and an
expression for the vehicle acceleration in terms of ‘known’
quantities,
(l2 + hr f r )
h


W cos θ 
 − L Ft +
L
W f  

h
(l1 − hr f r )
W  = 

θ
F
+
W
cos
t
 r 

L
L
 vɺx  

1
 [ Ft − W sin θ − Fa − f rW cos θ ]
 m

The traction force depends on the condition of the front and rear
tires: rolling or slipping?
Ft = Ftf + Ftr
front
ME 360/390 – Prof. R.G. Longoria
Vehicle System Dynamics and Control
rear
Department of Mechanical Engineering
The University of Texas at Austin
Shouldn’t the front to rear axle loads show the effect of
acceleration?
Only if you re-derive the equations, assuming you are given
acceleration (or deceleration). From previous case, rewrite x-direction
Newton’ law to solve for traction forces in terms of known quantities:
Ft = max + f rW f + f rWr + Fg + Fa
Then insert into
the other two
equation to solve
for the axle loads:
l1 −l2  W f   −h ( max + Fg + Fa ) 

 1 1  W  = 
W cos θ

  r  

l2
h
W f = W cos θ −  max + Fg + Fa 
L
L
l1
h
Wr = W cos θ +  max + Fg + Fa 
L
L
ME 360/390 – Prof. R.G. Longoria
Vehicle System Dynamics and Control
Department of Mechanical Engineering
The University of Texas at Austin
Consider this problem from Wong, Chapter 3
This problem seeks a steady-state solution for vehicle speed.
ME 360/390 – Prof. R.G. Longoria
Vehicle System Dynamics and Control
Department of Mechanical Engineering
The University of Texas at Austin
Finding solutions for velocity
1. Steady state: assume maximum traction, e.g., Ft max f = µW f
Solve for the front and rear forces:
Ft max r
 l1 − hr f r (v) 
= µW cos θ 

L
−
h
µ


Ft max f
 l2 + hr f r (v) 
= µW cos θ 

µ
L
+
h


Plot these forces and all loads as functions of velocity.
Look for intersection between traction force and load curve.
2. Dynamic: need to solve for velocity as a function of time by
integrating the 1st order ODE, considering all the loads, which
are also nonlinear functions of velocity. Need a model for the
traction force – for now assume maximum traction as above.
These solutions estimate the ‘traction-limited’ performance of the
vehicle. It is necessary to also look at ‘power-limited’ performance.
ME 360/390 – Prof. R.G. Longoria
Vehicle System Dynamics and Control
Department of Mechanical Engineering
The University of Texas at Austin
Solving source load problems: source and load
characteristics intersect, a steady-state speed is achieved
The source and load lines are just equations of effort (torque or force) versus flow
(angular velocity or velocity), so when you plot them together the intersection
corresponds to a solution of the two simultaneous equations. Basically, you are
solving a set of algebraic equations for a steady-state (operating) condition.
Graphical solution
e
(T or F )
Source
Load
eo
( eo , fo )
P = eo ⋅ f o
fo
Area = power delivered
ME 360/390 – Prof. R.G. Longoria
Vehicle System Dynamics and Control
f
(ω or v)
Next slide shows
how this approach
is used by Wong to
solve Problem 3.1.
Department of Mechanical Engineering
The University of Texas at Austin
Solution to 3.1 from Wong
Make note of how the total road
loads are plotted here on the
same graph as the traction
forces (the source).
ME 360/390 – Prof. R.G. Longoria
Vehicle System Dynamics and Control
Department of Mechanical Engineering
The University of Texas at Austin
My solution to 3.1 from Wong
ME 360/390 – Prof. R.G. Longoria
Vehicle System Dynamics and Control
Department of Mechanical Engineering
The University of Texas at Austin
Summary of vehicle on incline
• Plane motion analysis of rigid vehicle, an assumption valid for
many practical applications.
• Typical of what is required for basic vehicle weight transfer
calculations, or simple force analysis, etc.
• Analysis not useful for simulation of braking or traction, or for
control design
• Good for ‘go/no-go’ type assessment, design analysis, etc.
• This model assumes best case scenario with maximum traction.
• To be more realistic, we need to consider power-limited
performance by the including a model of the ‘source’
ME 360/390 – Prof. R.G. Longoria
Vehicle System Dynamics and Control
Department of Mechanical Engineering
The University of Texas at Austin