Financial analysis for vehicle program

What is needed? Units Description Source

Sales demand estimate

Number of vehicles

How many vehicles can be sold?

Sales & marketing

Sales price estimate $/unit What is the customer willing to pay?

Sales & Marketing

Investment cost estimate

$$$$ Plant costTooling costEngineering costCompany overhead

Vehicle Engineering, Finance, Manufacturing & Assembly, Suppliers

Variable cost estimate

$/unit Material costProduction cost including labor

Vehicle Engineering, Finance, Manufacturing & Assembly, Suppliers

Profit Analysis

Profit = Revenue – cost

WhereRevenue = selling price*number of vehicles sold

Cost = investment cost + variable cost* number of vehicles produced

Break even volume is the number of vehicles need to be sold so that there is no loss

Break even Volume = Investment cost/(selling price – variable cost)

Examples of Successful & Unsuccessful Programs

Entity Estimate Actual Actual Actual

Sales Volume 150,000 120,000 75,000 150, 000

Sale Price ($/unit)

22,000 22,000 22,000 18,000

Investment Cost ($)

500 M 500 M 500 M 500 M

Variable Cost ($/unit)

17,000 17,000 17,000 17,000

Total Profit (M$)

250 M 100 M - 125 M -350 M

Break even volume = 500,000,000/(22,000-17,000) = 100,000 vehicles

Your Calculations

1. Estimate selling price for your car from market survey2. Estimate the number of vehicles that can be sold3. Assume variable cost to be about X% of the selling price4. Assume investment cost to be Y RM5. Figure out break even volume and profit6. Figure out a way to distribute investment and variable cost to systems

InvestmentCost

VariableCost

Body ChassisPowertrainClimate ControlElectrical

Body ChassisPowertrainClimate ControlElectrical

Weight Analysis

Curb Weight : Weight of an assembled vehicleGross Vehicle Weight (GVW) = Curb weight + passenger & cargo weight

Corner weight = weight on each suspension

CurbWeight

Body ChassisPowertrainClimate ControlElectrical

Vehicle-fixed Coordinate System

Roll

Lateral

Vertical

Yaw

Pitch

Longitudinal

Z

Y

X

pq

r

CG

• ISO (International Standards Organization) coordinate system• Defines directions with respect to the vehicle

M g

co

s

M g sin

M a

Fxf

Fxr

DAR

hxA

B

Fzf

Fzr

M g

Rhz

hh

d h b

c

h

L L/2

PM

LA

x

Moment Equations

M = F L + M a h + M g h sin - M g c cos + L + PM + R h + R d = 0A zf x AL2 hx hz hh

F = M g cos - M a - M g sin - - - R - R cL x

hL

hL 2

LAL

PM hLhxh d

Lh

zf hz

Taking moments about point B yields

Taking moments about point A

F = M g cos + M a + M g sin - + + R + R bL x

hL

hL 2

LAL

PM hLhxh d

Lh

zr hz+ L

• Gravity• Tire normal forces (loads)• Tire shear forces (driving or braking)• Aerodynamic forces and moments• D’Alembert (acceleration) forces• Trailer hitch loads

Forces Acting on a Car, Truck or Motorcycle

Static Loads

• Sitting statically on a level surface:

fs c

c cW M g W

L L rs c

b bW M g W

L L

AB

M gc

Wf Wr L

b c

Longitudinal Dynamics

• Dynamic load transfer

• Acceleration limits

• Braking limits

• Aerodynamic forces/moments

Acceleration at Low Speed

• Acceleration on a level surface with no aerodynamic reactions

c

xfs

c

xf g

a

L

hWW

g

a

L

h

L

cWW )(

c

xrs

c

xr g

a

L

hWW

g

a

L

h

L

bWW )(

Climbing a Grade

• No aerodynamic or acceleration effects

)sincos

( L

h

L

cWW f

• For small angles: cos= 1, sin =

L

hWWW fsf

)sincos

( L

h

L

bWWr

L

hWWW rsr

• = Grade angle (in radians)

Aerodynamic Resistance Load

Aerodynamic drag load

DA = 0.5 ρ V2 CD A

Where:

CD = Aerodynamic drag Coefficient

ρ = Air density

A = Frontal Area of the vehicle

Tire Rolling Resistance LoadRolling resistance load

Rx = Rxf + Rxr = fr Wf + fr Wr

Where:

fr = Rolling Resistance Coefficient

Wr = Rear axle load

Wf = Front axle load

fr = 0.015 or 0.01*(1+ V/160)

Where, V is vehicle speed in km/h

Powertrain Applications

• Powertrain development– Architecture evaluation (FWD, RWD, 4WD)– Acceleration (0-100 kph, passing), top speed– Tuning (engine, torque converter,

transmission matching)– Traction limits– Fuel economy

Powertrain Architecture

Front wheel drive Rear wheel drive

Four wheel drive

• Traction-limited acceleration depends on loads on the drive wheels

• I.e., x zF F

Powertrain Architecture• Components in a solid axle rear drive

Engine Dyno Performance

• Steady speed; Wide Open Throttle (WOT)

Gasoline Engine

SPEED (rpm)

SPECIFIC FUEL CONSUMPTION

POWER

TORQUE

kWhp

140

120

100

80

60

40

20

0

100

80

60

40

20

02000 4000 6000

0.32

0.30

0.28

200

180

160

0.52

0.50

0.48

0.46

SPECIFIC FUEL CONSUMPTIONkg/kW-h lb/hp-h

N-m ft-lb

TORQUE

150

130

110

PO

WE

R

SPEED (rpm)

SPECIFIC FUEL CONSUMPTION

POWER

TORQUE

kWhp

140

120

100

80

60

40

20

0

100

80

60

40

20

02000 4000 6000

0.32

0.30

0.28

200

180

160

0.52

0.50

0.48

0.46

SPECIFIC FUEL CONSUMPTIONkg/kW-h lb/hp-h

N-m ft-lb

TORQUE

150

130

110

PO

WE

R

SPEED (rpm)

hp kW

N-m

kg/kW-h lb/hp-h

ft-lb

17001300

1200

0.40

0.35

0.30

0.25

0.20

1200 1500 1800 2100

375

350

325

300

275

PO

WE

R

TORQUE

SPECIFIC FUEL CONSUMPTION

TORQUE

POWER

SPECIFIC FUEL CONSUMPTION

1000

1100

1300

1500

200

225

250

275

Diesel Engine

Basic Acceleration Model.30

.25

.20

.15

.10

.05

0 10 20 30 40 50 60Speed (mph)

a

gx

Typical Heavy Truck, 250 lb/hp

10% Passenger Car, 40 lb/hp and x x x x

c

WF M a a P F V

g

=c cx x

g gPa F

W V W

1x

c

a P

g V W

• Simple acceleration model used by highway engineers• Acceleration is:

– Proportional to power to weight ratio– Inversely proportional to speed

Example of Simple Model

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 20 40 60 80 100 120 140 160 180 200

Speed (km/h)

Acc

eler

aito

n (

g)

100% Efficient

50% Efficient

Simulated Vehicle (250 kw, 1833 kg, with all losses)

• Simple acceleration model used by highway engineers– It over-predicts performance with actual P/W ratios

– Models are calibrated with effective P/W ratio

Tractive Force Performance

• Multiple gears approximate constant engine power• Continuously variable transmission (CVT) can follow constant engine

power curve

12010080604020000

500

1000

1500

2000

2500

3000

Speed (mph)

Tra

ctiv

e F

orc

e (

lb)

3rd4th

2nd

Constant Engine Power

1st Gear

• Tractive force vs. speed:– Reflects engine torque

curve– Depends on gear

• Low (1st) gear– High tractive effort– Limited speed range

• Higher gears expand the speed range but reduce tractive effort

Acceleration Performance(M+Mr) ax = T Ngf ηgf/r – Rx – DA – Rhx – W sinθ

Where

M = vehicle mass = W/g

Mr = equivalent mass of rotating components

ax = longitudinal acceleration

T = Engine Toerque

Ngf = combined ratio of transmission & final drive

ηgf = combined efficiency of transmission & final drive

Rx = Rolling resistance forces

DA = Aerodynamic forces

Rhx = Hitch forces

θ = Inclination angle

Mr = [(Ie+It)Ngf2 + IdNf

2 + Iw]/r2 or Mr/M = 0.04Ngf+0.0025Ngf2

and Ie,It and Iw are engine, transmission, axle inertias

Top Speed CalculationT Ngf ηgf/r >= Rx + DA + Rhx + W sinθ

If LHS > RHS, acceleration to higher speed is possible

LHS = RHS corresponds to top speed in that gear

Powertrain System Design

Vehicle

•Engine torque/power•Transmission Gear Ratios•Final Drive Gear Ratio•Torque Converter•Tire Size•Tire Traction Limit•Axle Roll

Aerodynamic DragRolling Resistance

Climbing GradeMass, Driveline Inertias

Gear Inefficiencies

AccelerationTop Speed

Design Specifications

UncontrolledVariables

What is needed?

• Procedure for calculating top speed and time to reach 100 km/h from 0

• Procedure to calculate top speed

• spreadsheet

Torque Converter• Fluid coupling between engine

and transmission• Stator:

– Deflects return flow in direction of the impeller

– Adds to torque of impeller– Turbine torque > engine torque

• Zero output/input speed ratio is “stall”

• Turbine input to transmission is typically two times engine torque

80

60

40

20

0

100

1.0

2.0

1.5

Ou

tpu

t/In

pu

t To

rqu

e R

atio

Effi

cie

ncy

(%

)

Output/Input Speed Ratio0 0.2 0.4 0.6 0.8 1.0

0

0.5

Lockup

Differential

Differential Rules

Rules for Free Differentials

Tleft Tright Tcarrier

2 left right

2carrier

Rules for Locking Differential

Tleft Tright Tcarrier

left right carrier

Axle Shaft

Pinion Gear

Ring Gear

Carrier

Carrier Gear

Side Gear

Axle Shaft

Torques on a Chassis

Traction Limits

Solid rear axle, non-locking differential:

Fx max =

W bL

1 - hL

+ 2 rNf t

K fK

Static Load

Longitudinal Load Transfer

Lateral Load Transfer

Solid rear axle, locking differential:

Fx max =

W bL

1 - hL

Independent front drive:

Fx max =

W cL

1 + hL

Fx max =

W cL

1 + hL

rNf t

K rK

Solid front drive axle, non-locking differential:

max 21

xf

f

bWLFKh r

L N t K

max 21

xr

f

cWLFKh r

L N t K

max

1x

cWLFhL

max

1x

bWLFhL

Differential Performance

Left - Right Wheel Speed (RPM)

Le

ft -

RIg

ht T

orq

ue

(N

-m)

Free with friction

Viscous

Hydraulic

Lock

ed

Brake Systems Applications

• Proportioning evaluation– Weight variations (curb weight to GVWR)– High and low friction

• Testing for regulatory compliance (FMVSS 105, 121..)

• Stability in braking (e.g., split mu, FMVSS 135)

• Evaluating effect of partial system failures

Typical Braking System

Parking Brake

Master CylinderRear brake lines

BrakePedal

VacuumAssist

CombinationValve

Rear Brake

Front Brake

Front

brake

lines

Tire Slip

Contact Length

Tire

Vertical Load

Friction Force

Relative Slip

V

Slip (S) = V - r

V

0 20 40 60 80 1000

0.2

0.4

0.6

0.8

Wheel Slip (%)

Bra

kin

g C

oe

ffici

en

t

Hysteresis

Dry

Wet

30 mph

30 mph

s

p

Adhesion

Wheel Lockup

• Front wheel lockup will cause loss of ability to steer the vehicle

• With rear wheel lockup, any yaw disturbance will initiate rotation of the vehicle making it unstable

• Brake proportioning strategy should allow the front brakes to lock first if ABS is not provided

Anti-lock BrakesW

he

el S

pe

ed

1 2 3 4 5 6 7Time (sec)

0

Vehicle Speed

LRRR

LFRF

12

3

0 20 40 60 80 1000

0.2

0.4

0.6

0.8

Wheel Slip (%)

Bra

kin

g C

oe

ffic

ien

t

1

2

3Cycling

Ap

plic

atio

n

FMVSS Regulatory Requirements1. A fully loaded passenger car with new brakes will stop from

speeds 30/60 mph in distance with average deceleration of 17/18 ft/s^2

2. A fully loaded passenger car with burnished brakes will stop from speeds 30/60/80 mph in distance with average deceleration of 17/19/18 ft/s^2

3. A lightly loaded passenger car with burnished brakes will stop from speeds 60 mph in distance with average deceleration of 20 ft/s^2

4. A fully and lightly loaded passenger car with brake failure will stop from speeds 60 mph in distance with average deceleration of 8.5 ft/s^2

Brake ProportioningMaximum brake force an axle can carry without locking

μp(Wfs + Fxr*h/L)

Front Axle Fxmf = -------------------------

1 – μp*h/L

μp(Wrs - Fxf*h/L)

Rear Axle Fxmr = -------------------------

1 + μp*h/L

Where Fxf and Fxr are front and rear brake forces

Wfs and Wrs are front and rear static weights

μp is the peak brake coefficient

h is the c.g. height L is the wheelbase

Brake Proportioning

Brake Force Fx = Tb/r = G Pa/r

Where

Fx is front or rear brake force (N)

Tb is front or rear brake torque (Nm)

r is the tire rolling radius (m)G is front or rear brake gain (N.m/MPa)

Pa is brake application pressure

What is needed

• Explanation on how to draw braking limits on the chart

• How to draw FMVSS requirement

• How to draw applied brake force diagram

• Brake pressure/brake torque relation

• Brake proportion strategy graph

Performance Triangles

1

p fs

p

W

hL

1

p rs

p

W

hL

Fro

nt

Bra

ke F

orc

e

Rear Brake Force

1

p

p

hLSlopehL

1

p

p

hLSlopehL

ProportioningRange

Idea

l Pro

po

rtio

nin

g

Front Lockup Boundary

Rear L

ocku

p B

ou

nd

ary

2000

1000

1500

500

00 500 1000 1500 2000

Brake Proportioning1st Effectiveness

2nd Effectiveness

3rd Effectiveness

500 1000 1500 2000

Rear Brake Force (lb)

500

1000

1500

2000F

ron

t Bra

ke F

orc

e (

lb)

= 0.3, lightly loaded

= 0.3, GVWR

Proportioning Line

Braking Efficiency

Eb = Dx/μp

Where

Eb is the braking efficiency

Dx is the actual deceleration

μp is the braking coefficient

Braking Efficiency Calculation1. Assume front and rear brake proportioning strategy such as

Pf = Pa and Pr = 0.8 Pa

2. Calculate front and rear axle brake forces

Fxf = 2Gf*Pf/r and Fxr = 2Gr*Pr/r

3. Calculate deceleration Dx

Dx = (Fxf+Fxr)/W

4. Calculate front and rear axle loads

5. Wf = Wfs + (h/L)(W/g)Dx

Wr = Wrs – (h/L)(W/g)Dx

5. Calculate braking coefficients μf and μr

μf = Fxf/Wf and μr = Fxr/Wr

6. Calculate braking efficiency Eb

Eb = Dx/ (higher of μf or μr)

7. Increase Pa till desired level of Dx is reached

Brake System Design

Vehicle

•Brake Pressure•Brake Torque Gains•Brake Proportioning•Tire Size•Tire Friction Limit

Aerodynamic DragRolling Resistance

Mass, C.G., wheelbase

DecelerationEfficiencyLocking Strategy

Design Specifications

UncontrolledVariables

Energy/Power Absorption

Energy and power absorbed by the brake system during braking

E = MV2/2

P = MV2/(2ts)

Where

M is the mass of the vehicleV is the initial speed

ts is the time to stop