Vehicle motion

Transportation

Engineering

Dr. Lina Shbeeb 2

Definitions

• Kinematic is the study of motion irrespective of the 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 fixed coordinate system (absolute motion) or in relation to a moving coordinate system (relative motion)

Vehicle motion can be described based on kinematic and kinetic equations

Transportation

Engineering

Dr. Lina Shbeeb 3

Equation of motion/ Rectilinear Motion

• The rectilinear position of x is measured from a

reference 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 by

time 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

Transportation

Engineering

Dr. Lina Shbeeb 4

Equation of motion/ Rectilinear Motion

• Acceleration a is the rate of change

of velocity with respect to time.

• It can be positive, zero or negative.

Negative acceleration or what is

common known as deceleration is

often denoted as d and its

magnitude is given in the positive

(d of 16 ft/s2 equals the same as an

acceleration of - ft/s2) adxvdv

toleadswhich

vdx

dva

dt

dx

dx

dva

dt

dva

Equation derivation

Transportation

Engineering

Dr. Lina Shbeeb 5

Equation of motion/ Rectilinear Motion • 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

Transportation

Engineering

Dr. Lina Shbeeb 6

…Equation of motion/ Rectilinear Motion

• The acceleration of a vehicle from an initial speed vo is

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

eveB

Av

leadwhich

tBvAB

dtBvA

dv

consareBandA

BvAdt

dva

o

o

Dr. Lina Shbeeb

Travel Speed

12

12

tt

xxv

Time

Distance

t2

t1

x1

x2

Dr. Lina Shbeeb

Spot Speed

dt

dxv

Time

Distance

t1

x1

V

Dr. Lina Shbeeb

Spot Speed Measurements

t1 t2 t3 Time

x3

x2

x1

Dis

tance

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

Dr. Lina Shbeeb

Spot Speed Measurements

Time

(s)

Distance

(ft)

Speed

(ft)

0.0 0.0 -

0.1 2.13 21.5

0.2 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 -

Dr. Lina Shbeeb

Average Acceleration Rate

12

12

tt

vva

Time

Speed

t2

t1

v1

v2

Dr. Lina Shbeeb

Spot Acceleration Rate

dt

dva

Time

speed

t1

v1

a

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 - -

Dr. Lina Shbeeb

Constant Acceleration Motion

constadt

dv

tv

vadtdv

00

0vatv

avdx

dv

xv

vadxvdv

00

a

vvx

2

20

2

dtvatvdtdx )( 0

x t

dtvatdx0 0 0 )(

tvatx 0

2

2

1

Remark: The equation used for design is , where the

deceleration rate has a positive value.

a

vvx

2

220

Dr. Lina Shbeeb

Exercise

•From the following data,

calculate the acceleration

rate at the distance of 2

feet from the reference

point.

Distance

(ft) Speed

(ft/s)

0 19.4

1 19.6

2 20.0

3 20.8

4 21.3

a=5.91ft/s2???

Transportation

Engineering

Dr. Lina Shbeeb 16

Constant Acceleration Motion

constadt

dv

tv

vadtdv

00

0vatv

avdx

dv

xv

vadxvdv

00

a

vvx

2

20

2

dtvatvdtdx )( 0

x t

dtvatdx0 0 0 )(

tvatx 0

2

2

1

Remark: The equation used for design is , where the

deceleration rate has a positive value.

a

vvx

2

220

Transportation

Engineering

Dr. Lina Shbeeb 17

Braking Distance

ag

w

w

sinw

u

coswf

Db

G

1.0

Distance to stop vehicle

Transportation

Engineering

Dr. Lina Shbeeb 18

Braking on Grades

sincos WWfag

W

a

vvx

2

220

x

Db

cos2

cos22

0

a

vvxDb

bDvva

2

cos)( 22

0

cos

sincos2

cos)(

1 220

f

Dvv

g b

cos

sin

2

1)(

1 220

f

Dvv

g b

G

tan

cos

sin)(2

220

Gfg

vvDb

Transportation

Engineering

Dr. Lina Shbeeb 19

Braking distance

• Braking Distance (Db)

• Db = distance from brakes enact to final speed

• Db = f(velocity, grade, friction)

• Db = (V02 – V2)/[30(f +/- G)]

• or

• Db = (V02 – 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

Transportation

Engineering

Dr. Lina Shbeeb 20

Braking Distance

Db = braking distance

u = initial velocity when brakes are

applied

a = 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 function

of the roadway, tires, etc

• Can use when deceleration is known (usually not) or

use previous equation with friction

Db = _____u2_____

30({a/g} ± G)

Transportation

Engineering

Dr. Lina Shbeeb 21

Vehicle Braking Distance

• Factors

• Braking System

• Tire Condition

• Roadway Surface

• Initial Speed

• Grade

• Braking Distance Equation

• db = (V2 - U2) / 30( f + g )

Transportation

Engineering

Dr. Lina Shbeeb 22

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

Transportation

Engineering

Dr. Lina Shbeeb 23

Motion on Circular Curves

dt

dvat

R

van

2

Transportation

Engineering

Dr. Lina Shbeeb 24

coscossin ns amWfW

coscos)(cossin

2

WR

v

g

WWfW s

e

tan

cos

sin

gR

vfe s

2

Motion on

Circular

Curves

Transportation

Engineering

Dr. Lina Shbeeb 25

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

Transportation

Engineering

Dr. Lina Shbeeb 26

Relative Motion • It is common to examine the motion of one

object 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)

Transportation

Engineering

Dr. Lina Shbeeb 27

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

system, RA and RB and the position vector rB/A with respect to the moving object A is

Y

Z

y

X

x

z

RA

RB

RA/B

ABAB

ABAB

ABAB

aaa

and

vvv

givestimetorespectwithatingDifferenti

rrr

/

/

/