lec 3_vehicle motion

8/6/2019 Lec 3_Vehicle Motion

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Vehicle motion


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TransportationEngineering

Dr. Lina Shbeeb2

Definitions

Kinematic is the study of motion irrespective ofthe forces that cause it

Kinetic is the study of motion that accounts the

forces that cause it. The motion of a body can be linear or curvilinear

It can be investigated in relation to a fixedcoordinate system (absolute motion) or in

relation to a moving coordinate system (relativemotion)

Vehicle motion can be described based on kinematic and kinetic equations


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Dr. Lina Shbeeb3

Equation of motion/ Rectilinear Motion

The rectilinear position of x is measured from areference point and has unit of length

The displacement is the difference in its position

between two instants.

Velocity v is the displacement of the particle divided bytime over which the displacement occurs. It is given by

the derivative of the displacement with respect of time

Speed is a scalar quantity and it is equal to the

magnitude of the velocity, which is a vector

dt

dxv

=


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Dr. Lina Shbeeb4


Acceleration a is the rate of change

of velocity with respect to time.

It can be positive, zero or negative.

Negative acceleration or what iscommon known as deceleration is

often denoted as dand its

magnitude is given in the positive

(dof 16 ft/s2 equals the same as an

acceleration of - ft/s2)adxvdv

toleadswhich

vdx

dva

dt

dx

dx

dva

dt

dva

=

=

=

=

Equation derivation


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Dr. Lina Shbeeb5


The simplest case of rectilinear motion is the case of constant acceleration where

( )

oo

oo

o

t

o

v

v

xtvatx

Thus

xxavv

leadwhichadxvdv

inegratingbycedisoffunctionaasressedbecanvelocityThe

vatv

dtadv

givesttotittheoveregratingby

adtdv

tconsadt

dv

o

++=

=

=

+=

=

==

==

2

22

2

1

)(2

1

,

tanexp

0limint

tan


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Dr. Lina Shbeeb6


The acceleration of a vehicle from an initial speed vois

given by the relationship

Acceleration as a function of velocity

)1()1(

)(

,

)1(

)ln(1

tan

2

BtoBt

Bt

o

Bt

o

Bt

v

v

t

o

v

v

eB

ve

B

At

B

Ax

eBvAa

equalsaBvAainsubstituteisvif

eve

B

Av

leadwhich

tBvAB

dt

BvA

dv

consareBandA

BvAdt

dva

o

o

+

=

==

+=

=

=

==


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Dr. Lina Shbeeb

Travel Speed

12

12

tt

xxv

=

Time

Distance

t2t1

x1

x2


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Dr. Lina Shbeeb

Spot Speed

dtdxv =

Time

Distance

t1

x1

V


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Dr. Lina Shbeeb

Spot Speed Measurements

t1t2t3Time

x3

x2

x1

Dista

nce

45.0

40.0

30.0

Distance

x

(ft)

4.0

3.0

2.0

Time

t

(s)

)40-30)/(3-2(

=10.0

---

Speed 1

v

(ft/s)

---

)45-30)/(4-2(

=7.5

---

Speed 2

v

(ft/s)

)45-40)/(3-2(

=5.0


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Dr. Lina Shbeeb

Spot Speed Measurements

Time(s)

Distance(ft)

Speed(ft)

0.0 0.0 -

0.1 2.13 21.50.2 4.30 21.9

0.3 6.51 22.4

0.4 8.78 22.4

0.5 10.99 21.3

0.6 13.04 -


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Dr. Lina Shbeeb

Average Acceleration Rate

12

12

tt

vva

=

Time

Speed

t2t1

v1

v2


12/27

Dr. Lina Shbeeb

Spot Acceleration Rate

dtdva =

Time

speed

t1

v1

a


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Dr. Lina Shbeeb

Measuring Acceleration

Rates

Time(s)

Distance(ft)

Speed(ft/s)

Acceleration(ft/s2)

0.0 0.0 - -0.1 2.13 21.5 -

0.2 4.30 21.9 4.5

0.3 6.51 22.4 2.5

0.4 8.78 22.4 -5.5

0.5 10.99 21.3 -

0.6 13.04 - -


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Dr. Lina Shbeeb

Constant Acceleration Motion

constadt

dv ==

=

tv

v

adtdv00

0vatv +=

avdx

dv=

=xv

v adxvdv 00

a

vvx

2

20

2 =

dtvatvdtdx )( 0+==

+=

x tdtvatdx

0 00 )(

tvatx 02

2

1+=

Remark: The equation used for design is , where the

deceleration rate has a positive value.

a

vvx

2

220 =


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Dr. Lina Shbeeb

Exercise

From the following data,

calculate the acceleration

rate at the distance of 2

feet from the referencepoint.

Distance(ft)

Speed(ft/s)

0 19.41 19.6

2 20.0

3 20.8

4 21.3

a=5.91ft/s2???


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Dr. Lina Shbeeb16

Constant Acceleration Motion

constadt

dv ==

=

tv

v

adtdv00

0vatv +=

avdx

dv=

=xv

v adxvdv 00

a

vvx

2

20

2 =

dtvatvdtdx )( 0+==

+=

x tdtvatdx

0 00 )(

tvatx 02

2

1+=

Remark: The equation used for design is , where the

deceleration rate has a positive value.

a

vvx

2

220 =


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Dr. Lina Shbeeb17

Braking Distance

ag

w

w

sinw

u

coswf

Db

G

1.0

Distance to stop vehicle


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Dr. Lina Shbeeb18

Braking on Grades

sincos WWfa

g

W=

a

vvx

2

220 =

x

Db

cos2

cos22

0

a

vvxDb

==

bDvva

2

cos)( 220

=

cossincos

2cos)(1

220 =

f

Dvv

g b

cos

sin

2

1)(

1 220 =

f

Dvv

g bG==

tan

cos

sin)(2

220

Gfg

vvDb

=


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Dr. Lina Shbeeb19

Braking distance

Braking Distance (Db) Db = distance from brakes enact to final speed Db = f(velocity, grade, friction)

Db = (V0

2 V2)/[30(f +/- G)]

or Db = (V

0

2 V2)/[254(f +/- G)] metric Db = braking distance (feet or meters)

V0= initial velocity (mph or kph)

V = final velocity (mph or kph) f = coefficient of friction G = Grade (decimal)

30 or 254 = conversion coefficient


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Dr. Lina Shbeeb20

Braking Distance

Db = braking distance

u = initial velocity when brakes are

applieda = vehicle acceleration

g = acceleration of gravity (32.2 ft/sec2)

G = grade (decimal), level roads G=zero

AASHTO represents friction as a/g which is a functionof the roadway, tires, etc Can use when deceleration is known (usually not) or

use previous equation with friction

Db = _____u2_____

30({a/g} G)


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Dr. Lina Shbeeb21

Vehicle Braking Distance

Factors

Braking System

Tire Condition

Roadway Surface Initial Speed

Grade

Braking Distance Equation

db= (V2- U2) / 30( f+ g )


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Dr. Lina Shbeeb22

Coefficient of friction

Pavement condition Maximum Slide

Good, dry 1.00 0.80

Good, wet 0.90 0.60

Poor, dry 0.80 0.55

Poor, wet 0.60 0.30

Packed snow and Ice 0.25 0.10


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Dr. Lina Shbeeb23

Motion on Circular Curves

dt

dvat =

R

van

2

=


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Dr. Lina Shbeeb24

coscossin ns amWfW =+

coscos)(cossin

2

WR

v

g

W

WfW s =+

e==

tan

cos

sin

gR

vfe s

2

=+

Motion on

CircularCurves


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Dr. Lina Shbeeb25

Minimum Radius of a Circular Curve

where u = vehicle velocity (mph)

e = tan (rate of superelevation)

fs= coefficient of side friction (depends on design speed)

Example

design speed = 65 mph

rate of superelevation = 0.05

coefficient of side friction = 0.11

Solution

minimum radius

R = (65)2/[15(0.05+0.11)] = 1760 ft

)(15

2

sfe

uR

+=


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TransportationEngineeringDr. Lina Shbeeb26

Relative Motion

It is common to examine the motion of oneobject in relation to another, for example the

motion of vehicles on a highway may be studies

from the point of view of the driver of a moving

vehicle. The simplest case of relative motion involves the

motion of one object B relative to a coordinate

system (x, y, z) that is translating but not rotating

with respect to a fixed coordinate system (X, Y,

Z)


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TransportationEngineeringDr. Lina Shbeeb27

Relative Motion The relationship between the position vectors of the two objects in relation to the fixed

system, RAand R

Band the position vector r

B/Awith respect to the moving object A is

Y

Z

y

X

x

z

RA

RB

RA/B

ABAB

ABAB

ABAB

aaa

and

vvvgivestimetorespectwithatingDifferenti

rrr

/

/

/

+=

+=

+=

lec 3_vehicle motion

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