) JIll 1111 11 111111 llll II II I i ll IIl1 ilt III i ll lit HI 12 is 1IJlV Id illto translashyd(IJwl ill(ll l I (ll Hxs ) wlllfI d lvid d II I in Fq (2-1) ) W ltall S its ma nitude li S IlIlIows

MMf =1ff r- =3602 in-lb-stc2 I 12592 in2 = 227 Ih-scc2lin

I rhaps Iht ilion familiar form is the effective weight

Wt ff = M e g = 227 Ib-sec2lin x 386 insec2 = 877 Ib

omparing this figure to the weight of a typical passenger car (2500 Ib) W I S that it adds about 35 to the effective weight of the car during t I ral ion in first gear The inertia of the non-driven wheels will add another 27 Ib to 111( effecti ve weight (I )

2) akulat~ the maximum tractive effort and corresponding road speed in fi ls l and fifth gears of the car described above when inertial losses are 11 g l cted

Sulution

Maximum tractive effort will coincide with maximum torque which (I( urs at 4400 rpm So the problem reduces to finding the tractive effort from till lirst term in Eq (2-9) for that value of torque

Fx= Tc NtfTltfr

= 20 J ft-Ib (428 x 292) (0966 x 099)11259 in x 12 inft

= 2290 Ib

TIll road speed is determined by use of the relationships given in Eq X) Although the equation is written in terms of acceleration the same

Illaliollships hold true for speed That is

IJJ = Nf Ww and we = Nt wd = Nt Nf Ww (2-8a)

Ill wheel rotational speed will be

(ow =we(Nt Nf)=4400revmin 21tradrev I minl60sec(428 x 292)

= 3687 radsec

rhe corresponding ground speed will be found by con verti ng the rotational spcd to translational speed at the circumference of the tire

V x =ffiw r=3687 radsecx 1259 in =4642 inlsec=387 ftlsec =264 mph

34

The sam 111 Ihod is liS d to kulal perforrnan(( in high gear as well

Fx = Te NtfTltfr

= 20 I ft-Ib (1 0 x 292) (099 x 097)12 59 in x 12 inft

= 537 Ib

ffiw= we I(Nt Nf) = 4400 revmin 2 1t radrev I minl60 sec I( 10 x 292)

= 1578 radsec

Vx = Ww r= 1578 radlsec x 1259 in= 1987 insec = 165 ftlsec = 1 13 mph

TRACTION-LIMITED ACCELERATION

Presuming there is adequate power from the engine the acceleration may be Iimited by the coefficient of friction between the tire and road In that case Fx is limited by

(2-13)

where

Il = Peak coefficient of friction

W = Weight on drive wheels

The weight on a drive wheel then depends on the static plus the dynamic load due to acceleration and on any transverse shift ofload due to drive torque

Transverse Weight Shift due to Drive Torque

Transverse weight shift occurs on all solid drive axles whether on the front or rear of the vehicle The basic reactions on a rear axle are shown in Figure 210 The drives haft into the differential imposes a torque T d on the axle As will be seen the chassis may roll compressing and extending springs on opposite sides of the vehicle such that a torque due to suspension roll stiffness T s is produced Any difference between these two must be absorbed as a difference in weight on the two wheels If the axle is of the non-locking type then the torque delivered to both wheels will be limited by the traction limit on the most lightly loaded wheel

W -L + W

y2

Fig 2 10 jretgt-body diagram oja solid drive axle

Writ ing NSL for rotation of the axle about its centerpoint allows the 1t 1I middottiolls to be related When the axle is in equilibrium

L To = (W12 + W Y - W12 + W y) tl2 + Ts - T d = 0 (2-14)

I II W Y =(Td - T s) t

III th above equation Td can be related to the drive forces because

(2-15)

whr

Fx =Total drive force from the two rear wheels

r = Tire radius

Nr =Final drive ratio

11owcver it is necessary to determine the roll torque produced by the IISP nsion which requires an analysis of the whole vehicle because the Hmiddotlt tion of the drive torque on the chassis attempts to roll the chassis on both Iht front and rear suspensions The entire system of interest is illustrated in 1 i1l1 21 I for the case of a rear-wheel-drive car

Th drive torque react ion at the enginetransmission is transferred to the IraiIK and distrihut (I bctwe n the front and rem suspensions It is generally asHlIlIlld that til roll torqu produ t hy a SIISP I1SiOIl is proportional (() roll all k Ilollkl S I aw) III tht middothassis Till-II

It

Tsf

Tsr

Klt1gt

where

Tsf

T sr

Klt1gtf

Klt1gtr

Klt1gt

(2-16a)=Klt1gtf lt1gt

(2-16b)= Klt1gtr lt1gt

(2-16c)= Klt1gtf+ Klt1gtr

= Roll torque on the front suspension

= Roll torque on the rear suspension

= Front suspension roll stiffness

= Rear suspension roll stiffness

= Total roll stiffness

~ y

Fig 2 J J Diagram ojdrive torque reactions on the chassis

7

Now I~ r ( III Ill 111 11 1 1 tn III IIdl 1111 I IllJ 11111111 all I all h I Ial -II 1111111 driw llIqllt t o foll()ws Ill n 11 all I i s sinlply Ihl driv lorll - divided I III IlIld roll slilTII ss

(I) =Id IKltgt =1lt1 I(K(prf- Kltgtr) (2-17)

lit 11 111 11 ~llhslillling in E l) (2- 16b)

lsr = K(pr T d I(K(pf + Kltgtr)

~ h i s ill llIllI ~ltIn be substituted into Eq (2-14) along with the expression 11 11 ohl1I11 J from Ell (2-15)

F r Kltjgtr (2- 18a) Wy = x I 1- - 1 Nfl Kltjgtr+-Kltjgtf

Ill( I nil in the brackets collapses to yield

W _ Fx r K~)r (2-18b) y - Nrl Kltjgt

Thistqllalioll g i ves the magnitude of the lateral load transfer as a function II I IIw Ir ~1 I iv for(e and a number of vehicle parameters such as the final drive 11 II Ir(ad uflk axle lire radius and suspension roll stiffnesses The net load 11 III I r uaxl during acceleration will be its static plus its dynamic composhy111 111 (s Joq ( 1-7raquo For a rear axle

(2-19)

Nl I - ling the rolling resistance and aerodynamic drag forces the accelshy( 1 111111 IS simply (he tractive force divided by the vehicle mass

(2-20)

lIIlIIII1 w -ighl on the rig ht rear wheel WIT will be W2 - W or y

(2-21 ) F x h 2 I

II III I

(2-22)II ( hI WI I

IH

Traction Limits

Solving or Fx gives the final expression for the maximum tractive force that can be developed by a solid rear axle with a non-locking differential

fl Wb F - L (2-23)

xmax - h 2 fl r Kltjgtfl--fl+---shy

L Nft Kqgt

For a solid rear axle with a locking differential additional tractive force can be obtained from the other wheel up to its traction limits such that the last term in the denominator of the above equation drops out This would also be true in the case of an independent rear suspension because the driveline torque reaction is picked up by the chassis-mounted differential In both ofthese cases the expression for the maximum tractive force is

fl Wb F - L (2-24)

xmax - h I --fl

L

FinaJJy in the case of a front axle the foreaft load transfer is opposite from the rear axle case Since the load transfer is reflected in the second term of the denominator the opposite direction yields a sign change Also the term W blL arose in the earlier equations to represent the static load on the rear drive axle For a front-wheel-drive vehicle the term becomes W clL For the solid front drive axle with non-locking differential

fl We F xmax =_ ___L=-____ (2-25)

h 2 fl r KqgtrI +-fl+-- --shy

L Nft Kltjgt

And for the solid front drive axle with locking differential or the indepenshydent front drive axle as typical of most front-wheel-drive cars today

(2-26)xmax =

11

H INIlIMI NJ I ( II VI IIII I 1 1 I tN Mil

EXAMPI E PROBLEMS

I) JoiIlU Ih Ira middottion-lillliltU C1cccl l ratiol fOf (he flur-drivc passlnger car wi lllll llU wilhoul a 10 kin J differential on a surface of rnoueralt friction level (it ill flllllIalioll thai will he needed is as follows

Wlil hl~ Front shy 21()) Ib Rear - 1850 Ib Total shy 3950 Ib

G h i middothl 210 in Wheelbase - 108 in

(I ffic i nl or liiction 062 Tread shy 590 in

1I llal drive ratio 290 Tire size - 130 in

Roll sliffnesscs Fronl- 1150 ft-Ibdeg Rear shy 280 ft-Ibldeg

SlIllIUon

Ill equation for the maximum tractive force of a solid axle rear-drive ( 1 willI t non locking differential was given in Eg (2-23)

(2-23)

III this lquation W bL is just the rear axle weight which is known othl dor W do not have to find the value for the parameter b Likewise all

l il t tlth or t rfllS arc known and can be substituted into the equation to obtain

(062) 1850 IbI ~ x ilia x =

1 _ 21 0 62 2 (062) 13 in 1150 108 + 29 59 in 1430

1147lb 1147 - 1 - 0121 + 00758 =0 9548 = 1201 Ib

1201 Iha x I o ()4 1 g s - 97() flIt) (1111

s (c

Wilh a lOCking Jifferential the thinj term in the dlnominator disappears CEq (2 24raquo so that we obta in

F _ (062) 1850 Ib 11471b = 1147 = 1305 lb xmax - 21 1-0121 0879

1 - 108 062

a -~ xmax = 1305 lb =0 330 g s = 10 64 ~ x- Mg 39501b sec 2

Notes

a) For both cases the numerator term is the weight on the drive axle times the coefficient of friction which is equivalent to 1147 Ib of tractive force

b) Similarly the dynamic load transfer onto the rear (drive) axle from acceleration is accounted for by the second term in the denominator which diminishes the magnitude of the denominator by 121 thereby increasing the tractive force by an equivalent percentage

c) The lateral load transfer effect appears in the third term of the denominator increasing its value by approximately 76 which has the effect of decreasing the tractive force by about the same percentage Comparing the two answers the loss from lateral load transfer on the drive axle with a nonshylocking differential is 104 lb On higher friction surfaces a higher loss would be seen

2) Find the traction-limited performance of a front-wheel-drive vehicle under the same road conditions as the problem above The essential data are

Weights Front - 1950 Rear - 1150 Total- 3100

CG Height 190 in Wheelbase - 105 in

Coefficient of friction 062 Tread - 60 inches

Final drive ratio 370 Tire size - 1259 inches

Roll sliffnesses Front - 950 ft-Ibdeg Rear - 620 ft-lbdeg

llJNJ)AMENTALS OF VEHICLE DYNAMICS

Solution

Most front-wheel-drive vehicles have an independent front suspension Ihll s til equation for maximum tractive effort is given by Eq (2-26) and we 111 i Illat all the data required to calculate lateral load transfer on the axle are ( Ill middotJeu The maximum tractive force is calculated by substituting in the j1llalillll as follows

~ We I - L (2-26)xmax shy

l+hJlL

I _ (062) 1950 lb = 1209lb xmax - 19 1 + 01122 = 1087 lb

1 + 105 062

F xmax 1087 Ib ft a x = Mg = 3100Ib = 03506 g s= 1129shy

sec 2

Nj11

I) e ven though the front-wheel-drive vehicle has a much higher percentshyIe of its weight on the drive axle its performance is not proportionately better lb r ~ason is the loss of load on the front (dri ve) axle due to longitudinal weight LI ullsf J during acceleration

I~EFEI~ENCES

I Gillespie TD MethodsofPredicting Truck Speed Loss on Grades Th ~ University of Michigan Transportation Research Institute Reshyport No UM-85-39 November 1986 J69 p

SL John AD and Kobett DR Grade Effects on Traffic Flow Staoility alld aracity Interim Report National Cooperative lIighway RlS anh Pro ram Project 3- 19 December 1972 173 p

I Mnlslla ll 111 M llli llllllll 11111 PrlIhlhl 111 I Jo onolll) of JlItomoshyhi k s S A I ~ 111 WI Nil HOII I I I)HO H p

CHAPTER 2 - ACCELERATION PERFORMANCE

4 Cole D Elementary Vehicle Dynamics course notes in Mechanishycal Engineering The University ofMichigan Ann Arbor Michigan

1972

5 Smith GL Commercial Vehicle Performance and Fuel Economy SAE Paper SP-355 1970 23 p

6 Buck RE A Computer Program (HEVSIM) for Heavy Duty Vehicle Fuel Economy and Performance Simulation US Departshyment of Transportation Research and Special Projects Administrashytion Transportation Systems Center Report No DOT -HS-805-912

September 1981 26 p

Zub RW A Computer Program (VEHSIM) for Vehicle Fuel 7 Economy and Performance Simulation (Automobiles and Light Trucks) US Department ofTransportation Research and Special Projects Administration Transportation Systems Center Report No DOT-HS-806-040 October 198150 p

8 Phillips AW Assanis DN and Badgley P Development and Use of a Vehicle Powertrain Simulation for Fuel Economy and Performance Studies SAE Paper No 900619 1990 14 p

Now I~ r ( III Ill 111 11 1 1 tn III IIdl 1111 I IllJ 11111111 all I all h I Ial -II 1111111 driw llIqllt t o foll()ws Ill n 11 all I i s sinlply Ihl driv lorll - divided I III IlIld roll slilTII ss

(I) =Id IKltgt =1lt1 I(K(prf- Kltgtr) (2-17)

lit 11 111 11 ~llhslillling in E l) (2- 16b)

lsr = K(pr T d I(K(pf + Kltgtr)

~ h i s ill llIllI ~ltIn be substituted into Eq (2-14) along with the expression 11 11 ohl1I11 J from Ell (2-15)

F r Kltjgtr (2- 18a) Wy = x I 1- - 1 Nfl Kltjgtr+-Kltjgtf

Ill( I nil in the brackets collapses to yield

W _ Fx r K~)r (2-18b) y - Nrl Kltjgt

Thistqllalioll g i ves the magnitude of the lateral load transfer as a function II I IIw Ir ~1 I iv for(e and a number of vehicle parameters such as the final drive 11 II Ir(ad uflk axle lire radius and suspension roll stiffnesses The net load 11 III I r uaxl during acceleration will be its static plus its dynamic composhy111 111 (s Joq ( 1-7raquo For a rear axle

(2-19)

Nl I - ling the rolling resistance and aerodynamic drag forces the accelshy( 1 111111 IS simply (he tractive force divided by the vehicle mass

(2-20)

lIIlIIII1 w -ighl on the rig ht rear wheel WIT will be W2 - W or y

(2-21 ) F x h 2 I

II III I

(2-22)II ( hI WI I

IH

Traction Limits

Solving or Fx gives the final expression for the maximum tractive force that can be developed by a solid rear axle with a non-locking differential

fl Wb F - L (2-23)

xmax - h 2 fl r Kltjgtfl--fl+---shy

L Nft Kqgt

For a solid rear axle with a locking differential additional tractive force can be obtained from the other wheel up to its traction limits such that the last term in the denominator of the above equation drops out This would also be true in the case of an independent rear suspension because the driveline torque reaction is picked up by the chassis-mounted differential In both ofthese cases the expression for the maximum tractive force is

fl Wb F - L (2-24)

xmax - h I --fl

L

FinaJJy in the case of a front axle the foreaft load transfer is opposite from the rear axle case Since the load transfer is reflected in the second term of the denominator the opposite direction yields a sign change Also the term W blL arose in the earlier equations to represent the static load on the rear drive axle For a front-wheel-drive vehicle the term becomes W clL For the solid front drive axle with non-locking differential

fl We F xmax =_ ___L=-____ (2-25)

h 2 fl r KqgtrI +-fl+-- --shy

L Nft Kltjgt

And for the solid front drive axle with locking differential or the indepenshydent front drive axle as typical of most front-wheel-drive cars today

(2-26)xmax =

11

H INIlIMI NJ I ( II VI IIII I 1 1 I tN Mil

EXAMPI E PROBLEMS

I) JoiIlU Ih Ira middottion-lillliltU C1cccl l ratiol fOf (he flur-drivc passlnger car wi lllll llU wilhoul a 10 kin J differential on a surface of rnoueralt friction level (it ill flllllIalioll thai will he needed is as follows

Wlil hl~ Front shy 21()) Ib Rear - 1850 Ib Total shy 3950 Ib

G h i middothl 210 in Wheelbase - 108 in

(I ffic i nl or liiction 062 Tread shy 590 in

1I llal drive ratio 290 Tire size - 130 in

Roll sliffnesscs Fronl- 1150 ft-Ibdeg Rear shy 280 ft-Ibldeg

SlIllIUon

Ill equation for the maximum tractive force of a solid axle rear-drive ( 1 willI t non locking differential was given in Eg (2-23)

(2-23)

III this lquation W bL is just the rear axle weight which is known othl dor W do not have to find the value for the parameter b Likewise all

l il t tlth or t rfllS arc known and can be substituted into the equation to obtain

(062) 1850 IbI ~ x ilia x =

1 _ 21 0 62 2 (062) 13 in 1150 108 + 29 59 in 1430

1147lb 1147 - 1 - 0121 + 00758 =0 9548 = 1201 Ib

1201 Iha x I o ()4 1 g s - 97() flIt) (1111

s (c

Wilh a lOCking Jifferential the thinj term in the dlnominator disappears CEq (2 24raquo so that we obta in

F _ (062) 1850 Ib 11471b = 1147 = 1305 lb xmax - 21 1-0121 0879

1 - 108 062

a -~ xmax = 1305 lb =0 330 g s = 10 64 ~ x- Mg 39501b sec 2

Notes

a) For both cases the numerator term is the weight on the drive axle times the coefficient of friction which is equivalent to 1147 Ib of tractive force

b) Similarly the dynamic load transfer onto the rear (drive) axle from acceleration is accounted for by the second term in the denominator which diminishes the magnitude of the denominator by 121 thereby increasing the tractive force by an equivalent percentage

c) The lateral load transfer effect appears in the third term of the denominator increasing its value by approximately 76 which has the effect of decreasing the tractive force by about the same percentage Comparing the two answers the loss from lateral load transfer on the drive axle with a nonshylocking differential is 104 lb On higher friction surfaces a higher loss would be seen

2) Find the traction-limited performance of a front-wheel-drive vehicle under the same road conditions as the problem above The essential data are

Weights Front - 1950 Rear - 1150 Total- 3100

CG Height 190 in Wheelbase - 105 in

Coefficient of friction 062 Tread - 60 inches

Final drive ratio 370 Tire size - 1259 inches

Roll sliffnesses Front - 950 ft-Ibdeg Rear - 620 ft-lbdeg

llJNJ)AMENTALS OF VEHICLE DYNAMICS

Solution

Most front-wheel-drive vehicles have an independent front suspension Ihll s til equation for maximum tractive effort is given by Eq (2-26) and we 111 i Illat all the data required to calculate lateral load transfer on the axle are ( Ill middotJeu The maximum tractive force is calculated by substituting in the j1llalillll as follows

~ We I - L (2-26)xmax shy

l+hJlL

I _ (062) 1950 lb = 1209lb xmax - 19 1 + 01122 = 1087 lb

1 + 105 062

F xmax 1087 Ib ft a x = Mg = 3100Ib = 03506 g s= 1129shy

sec 2

Nj11

I) e ven though the front-wheel-drive vehicle has a much higher percentshyIe of its weight on the drive axle its performance is not proportionately better lb r ~ason is the loss of load on the front (dri ve) axle due to longitudinal weight LI ullsf J during acceleration

I~EFEI~ENCES

I Gillespie TD MethodsofPredicting Truck Speed Loss on Grades Th ~ University of Michigan Transportation Research Institute Reshyport No UM-85-39 November 1986 J69 p

SL John AD and Kobett DR Grade Effects on Traffic Flow Staoility alld aracity Interim Report National Cooperative lIighway RlS anh Pro ram Project 3- 19 December 1972 173 p

I Mnlslla ll 111 M llli llllllll 11111 PrlIhlhl 111 I Jo onolll) of JlItomoshyhi k s S A I ~ 111 WI Nil HOII I I I)HO H p

CHAPTER 2 - ACCELERATION PERFORMANCE

4 Cole D Elementary Vehicle Dynamics course notes in Mechanishycal Engineering The University ofMichigan Ann Arbor Michigan

1972

5 Smith GL Commercial Vehicle Performance and Fuel Economy SAE Paper SP-355 1970 23 p

6 Buck RE A Computer Program (HEVSIM) for Heavy Duty Vehicle Fuel Economy and Performance Simulation US Departshyment of Transportation Research and Special Projects Administrashytion Transportation Systems Center Report No DOT -HS-805-912

September 1981 26 p

Zub RW A Computer Program (VEHSIM) for Vehicle Fuel 7 Economy and Performance Simulation (Automobiles and Light Trucks) US Department ofTransportation Research and Special Projects Administration Transportation Systems Center Report No DOT-HS-806-040 October 198150 p

8 Phillips AW Assanis DN and Badgley P Development and Use of a Vehicle Powertrain Simulation for Fuel Economy and Performance Studies SAE Paper No 900619 1990 14 p

H INIlIMI NJ I ( II VI IIII I 1 1 I tN Mil

EXAMPI E PROBLEMS

I) JoiIlU Ih Ira middottion-lillliltU C1cccl l ratiol fOf (he flur-drivc passlnger car wi lllll llU wilhoul a 10 kin J differential on a surface of rnoueralt friction level (it ill flllllIalioll thai will he needed is as follows

Wlil hl~ Front shy 21()) Ib Rear - 1850 Ib Total shy 3950 Ib

G h i middothl 210 in Wheelbase - 108 in

(I ffic i nl or liiction 062 Tread shy 590 in

1I llal drive ratio 290 Tire size - 130 in

Roll sliffnesscs Fronl- 1150 ft-Ibdeg Rear shy 280 ft-Ibldeg

SlIllIUon

Ill equation for the maximum tractive force of a solid axle rear-drive ( 1 willI t non locking differential was given in Eg (2-23)

(2-23)

III this lquation W bL is just the rear axle weight which is known othl dor W do not have to find the value for the parameter b Likewise all

l il t tlth or t rfllS arc known and can be substituted into the equation to obtain

(062) 1850 IbI ~ x ilia x =

1 _ 21 0 62 2 (062) 13 in 1150 108 + 29 59 in 1430

1147lb 1147 - 1 - 0121 + 00758 =0 9548 = 1201 Ib

1201 Iha x I o ()4 1 g s - 97() flIt) (1111

s (c

Wilh a lOCking Jifferential the thinj term in the dlnominator disappears CEq (2 24raquo so that we obta in

F _ (062) 1850 Ib 11471b = 1147 = 1305 lb xmax - 21 1-0121 0879

1 - 108 062

a -~ xmax = 1305 lb =0 330 g s = 10 64 ~ x- Mg 39501b sec 2

Notes

a) For both cases the numerator term is the weight on the drive axle times the coefficient of friction which is equivalent to 1147 Ib of tractive force

b) Similarly the dynamic load transfer onto the rear (drive) axle from acceleration is accounted for by the second term in the denominator which diminishes the magnitude of the denominator by 121 thereby increasing the tractive force by an equivalent percentage

c) The lateral load transfer effect appears in the third term of the denominator increasing its value by approximately 76 which has the effect of decreasing the tractive force by about the same percentage Comparing the two answers the loss from lateral load transfer on the drive axle with a nonshylocking differential is 104 lb On higher friction surfaces a higher loss would be seen

2) Find the traction-limited performance of a front-wheel-drive vehicle under the same road conditions as the problem above The essential data are

Weights Front - 1950 Rear - 1150 Total- 3100

CG Height 190 in Wheelbase - 105 in

Coefficient of friction 062 Tread - 60 inches

Final drive ratio 370 Tire size - 1259 inches

Roll sliffnesses Front - 950 ft-Ibdeg Rear - 620 ft-lbdeg

llJNJ)AMENTALS OF VEHICLE DYNAMICS

Solution

Most front-wheel-drive vehicles have an independent front suspension Ihll s til equation for maximum tractive effort is given by Eq (2-26) and we 111 i Illat all the data required to calculate lateral load transfer on the axle are ( Ill middotJeu The maximum tractive force is calculated by substituting in the j1llalillll as follows

~ We I - L (2-26)xmax shy

l+hJlL

I _ (062) 1950 lb = 1209lb xmax - 19 1 + 01122 = 1087 lb

1 + 105 062

F xmax 1087 Ib ft a x = Mg = 3100Ib = 03506 g s= 1129shy

sec 2

Nj11

I) e ven though the front-wheel-drive vehicle has a much higher percentshyIe of its weight on the drive axle its performance is not proportionately better lb r ~ason is the loss of load on the front (dri ve) axle due to longitudinal weight LI ullsf J during acceleration

I~EFEI~ENCES

I Gillespie TD MethodsofPredicting Truck Speed Loss on Grades Th ~ University of Michigan Transportation Research Institute Reshyport No UM-85-39 November 1986 J69 p

SL John AD and Kobett DR Grade Effects on Traffic Flow Staoility alld aracity Interim Report National Cooperative lIighway RlS anh Pro ram Project 3- 19 December 1972 173 p

I Mnlslla ll 111 M llli llllllll 11111 PrlIhlhl 111 I Jo onolll) of JlItomoshyhi k s S A I ~ 111 WI Nil HOII I I I)HO H p

CHAPTER 2 - ACCELERATION PERFORMANCE

4 Cole D Elementary Vehicle Dynamics course notes in Mechanishycal Engineering The University ofMichigan Ann Arbor Michigan

1972

5 Smith GL Commercial Vehicle Performance and Fuel Economy SAE Paper SP-355 1970 23 p

6 Buck RE A Computer Program (HEVSIM) for Heavy Duty Vehicle Fuel Economy and Performance Simulation US Departshyment of Transportation Research and Special Projects Administrashytion Transportation Systems Center Report No DOT -HS-805-912

September 1981 26 p

Zub RW A Computer Program (VEHSIM) for Vehicle Fuel 7 Economy and Performance Simulation (Automobiles and Light Trucks) US Department ofTransportation Research and Special Projects Administration Transportation Systems Center Report No DOT-HS-806-040 October 198150 p

8 Phillips AW Assanis DN and Badgley P Development and Use of a Vehicle Powertrain Simulation for Fuel Economy and Performance Studies SAE Paper No 900619 1990 14 p

llJNJ)AMENTALS OF VEHICLE DYNAMICS

Solution

Most front-wheel-drive vehicles have an independent front suspension Ihll s til equation for maximum tractive effort is given by Eq (2-26) and we 111 i Illat all the data required to calculate lateral load transfer on the axle are ( Ill middotJeu The maximum tractive force is calculated by substituting in the j1llalillll as follows

~ We I - L (2-26)xmax shy

l+hJlL

I _ (062) 1950 lb = 1209lb xmax - 19 1 + 01122 = 1087 lb

1 + 105 062

F xmax 1087 Ib ft a x = Mg = 3100Ib = 03506 g s= 1129shy

sec 2

Nj11

I) e ven though the front-wheel-drive vehicle has a much higher percentshyIe of its weight on the drive axle its performance is not proportionately better lb r ~ason is the loss of load on the front (dri ve) axle due to longitudinal weight LI ullsf J during acceleration

I~EFEI~ENCES

I Gillespie TD MethodsofPredicting Truck Speed Loss on Grades Th ~ University of Michigan Transportation Research Institute Reshyport No UM-85-39 November 1986 J69 p

SL John AD and Kobett DR Grade Effects on Traffic Flow Staoility alld aracity Interim Report National Cooperative lIighway RlS anh Pro ram Project 3- 19 December 1972 173 p

I Mnlslla ll 111 M llli llllllll 11111 PrlIhlhl 111 I Jo onolll) of JlItomoshyhi k s S A I ~ 111 WI Nil HOII I I I)HO H p

CHAPTER 2 - ACCELERATION PERFORMANCE

4 Cole D Elementary Vehicle Dynamics course notes in Mechanishycal Engineering The University ofMichigan Ann Arbor Michigan

1972

5 Smith GL Commercial Vehicle Performance and Fuel Economy SAE Paper SP-355 1970 23 p

6 Buck RE A Computer Program (HEVSIM) for Heavy Duty Vehicle Fuel Economy and Performance Simulation US Departshyment of Transportation Research and Special Projects Administrashytion Transportation Systems Center Report No DOT -HS-805-912

September 1981 26 p

Zub RW A Computer Program (VEHSIM) for Vehicle Fuel 7 Economy and Performance Simulation (Automobiles and Light Trucks) US Department ofTransportation Research and Special Projects Administration Transportation Systems Center Report No DOT-HS-806-040 October 198150 p

8 Phillips AW Assanis DN and Badgley P Development and Use of a Vehicle Powertrain Simulation for Fuel Economy and Performance Studies SAE Paper No 900619 1990 14 p