Chapter 2: Vehicle Motion: Kinematics & Kinetics

Mechanics

Nov 14, 2013 (4 years and 8 months ago)

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1

Chapter 2: Vehicle Motion: Kinematics & Kinetics

Introduction

* Why we care about vehicle motion?

Kinematics:

1. Need to know
s, v, a

to allow calculation of,
e.g.

-

determining stopping sight distance

-

brake distance

-

passing distance

etc.

Kinetics:

2. Need to know
forces

so to design

-

guard rail

-

attenuation devices

-

safe design of vehicles

-

etc.

2

* Eq
uation
s. of Motion

2.2.1
Rectilinear motion

o

If position (x) is a function of time,

dt
dx
v

x
d
a
v
d
v

dt
dv
a

o

If
a

is constant

t
a
v
v
o

a
2
v
v
x
x
2
o
2
o

2
2
1
t
a
t
v
x
x
o
o

3

Example 2.1
Constant Acceleration

G: Vehicle appro
aches an intersection @ a speed = 30 mph

= 44 f/s

o

At t = 0, it started to decelerate @ a = 16 f/s
2

o

Vehicle was 55 ft away from stop line

R: 1) Did it stop legally?

2) Plot a

t, v

t & x

t graphs

1)

a
2
v
v
x
x
2
o
2
o

)
16
(
2
44
0
0
x
2

x = 60.5 ft > 55

Therefore, it did not stop legally….

2)

a

t

graph

a is constant

v

t

v = v
o

+ at

v = 44 + (
-
16) t

= 44

16 t

x

t

Other

x

x
o

= v
o

t
+
2
1

at
2

x = 44t +
2
1
(

16) t
2

x = 44t

8 t
2

or

dt
dx
v

= 44t

8 t
2

t
o
2
t
8
t
44
dt
)
t
16
44
(
dt
v
dx
x

55

x

a

-
16

t

v

44

t

t

x

4

Questions & Notes on Example:

1.

Why we have (. . . . ) pa
rt in graphs?

look @ v

t, final v is o. Driver will not reverse.

2.

Once you plot x

t, you can get the other graphs. Why?

dt
dx
v

we did this

dt
dv
a

3.

In reality, it is often that
a

is not
constant. But if assumed as in this case

-

it is reasonably OK.

5

Example: Variable acceleration:

G: a = (18 t + 2)m/s
2

for a vehicle.

vehicle @ rest at t = 0

R: 1) What are

x at t = 3

v at t = 3

1)

2
t
18
dt
dv
a

t
o
2
t
2
t
9
dt
a
v

s
/
m
87
|
v
3
t

3
o
3
2
3
|
t
t
3
dt
v
x

= 90

Recommended HW

Problem 1, p.94

(see the HW sheets @ the end of Notes of this chapter)

6

2.2.2 Braking Distance

-

-

Kinematic equa
tions in example 2.1 deals with a =

16 f/s
2

“a” there represents
“net effect” of forces according to Newton’s 2
nd

Law.

-

More practical, we consider the following:

A car is
decelerating

on an up
-
hill with incline of

º. Surface
-
tire friction is f.

“car is
sliding

up

when braking. So

F is down”.

+

F
y

= ma
y

= 0

N

w cos

= 0

N = w cos

+

F
x

= ma
x

w sin

-

F = ma
x

w sin

f w cos

=
a
g
w

a
=

g sin

f g cos

=

g cos

(tan

+ f)

7

In

a
2
v
v
x
x
2
o
2
o

)
f
(tan
cos
g
2
v
v
x
x
2
o
2
o

for small angles, cos

1
~

Also, tan

is G = grade in decimal = slope.

e.g.

075
.
40
3

not 7.5%

Notes:

0. For braking , v
2

= 0

1. For downhill, x

x
o

=
)
G
f
(
g
2
v
v
2
2
o

2. f is a variable. It varies with:

-

speed

-

condition of surface : wet , dry

-

the surface : concrete , asphalt

3.

Wet f is taken for design

4. On asphalt:

f
wet

= 0.3

f
dry

= 0.6

)
G
f
(
g
2
v
v
x
x
2
2
o
o

8

5. x

x
o

=
)
G
f
(
g
2
v
v
2
2
o

; also

x

x
o

=
a
2
v
v
2
o
2

limit of comfortable a (8 to 10 f/s
2
)

Recommended HW

Probl
ems 2, 5, 6

a = g (f ± G)

9

2.2.3 Curvilinear Motion

v is tangent

a is to the concave side

a

can be resolved into a
n

& a
t

a
n

points to center at the

instant

a
t

is tangent to path.

dt
dv
a
t

&

2
n
v
a

where

= radius of curve @ the instant

For a vehicle

FBD

KD

OR

F
t

=
dt
dv

F
n

=

2
v
m

F
z

= 0

v

a

a

a
n

a
t

Σ F
n

Σ F
t

ma
n

ma
t

10

Notes:

o

If v is constant, then a
t

= 0

o

In a circle,

= R = constant

Consider vehicle on a flat surface turning to left

tendency to overturn

Provide “
banking

superelevation

+

F
y

=
ma
y

:

N

W cos

= ma
n

sin

= 0

1

+

F
x

= ma
x

:

W sin

+ F = ma
n

cos

2

f = f N

3

1 & 3 in 2

W sin

+ f (W cos

) = ma
n

cos

=

cos
R
v
g
w
2

Dividing by W cos

gives

R

F

N

W

R

F

N

W

β

β

n

11

tan

+ f =
R
g
v
2

tan

= e = superelevation rate

e + f =
R
g
v
2

Notes
:

1

This equation provides combination of the parameters (e, f, v & R) so to

avoid
slipping
.

2

f is the

side friction coefficient developed. It is different from f in SSD.

3

Tipping is about to occur when N is @ the outer wheel.

Dimension of vehicle plays a role.

o

Suburban overturns often.