Chapter 2: Vehicle Motion: Kinematics & Kinetics

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Nov 14, 2013 (3 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




-

curves of road





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



-

Important in roadway design


-

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


ask why?




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






Important in designing curved roads.












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.





f is about 0.16





3

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








Dimension of vehicle plays a role.


o

Suburban overturns often.