Financial analysis for vehicle program
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Transcript of Financial analysis for vehicle program
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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
$/unit Material costProduction cost including labor
Vehicle Engineering, Finance, Manufacturing & Assembly, Suppliers
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Profit Analysis
Profit = Revenue – cost
WhereRevenue = selling price*number of vehicles sold
Cost = investment cost + variable cost* number of vehicles produced
Break even volume is the number of vehicles need to be sold so that there is no loss
Break even Volume = Investment cost/(selling price – variable cost)
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Examples of Successful & Unsuccessful Programs
Entity Estimate Actual Actual Actual
Sales Volume 150,000 120,000 75,000 150, 000
Sale Price ($/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
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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
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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
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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
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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
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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
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Longitudinal Dynamics
• Dynamic load transfer
• Acceleration limits
• Braking limits
• Aerodynamic forces/moments
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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 )(
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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)
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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
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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
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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
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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
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Powertrain Architecture• Components in a solid axle rear drive
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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
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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
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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
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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
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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
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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
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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
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What is needed?
• Procedure for calculating top speed and time to reach 100 km/h from 0
• Procedure to calculate top speed
• spreadsheet
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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
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Differential
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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
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Torques on a Chassis
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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
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Differential Performance
Left - Right Wheel Speed (RPM)
Le
ft -
RIg
ht T
orq
ue
(N
-m)
Free with friction
Viscous
Hydraulic
Lock
ed
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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
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Typical Braking System
Parking Brake
Master CylinderRear brake lines
BrakePedal
VacuumAssist
CombinationValve
Rear Brake
Front Brake
Front
brake
lines
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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
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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
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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
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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
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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
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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
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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
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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
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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
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Braking Efficiency
Eb = Dx/μp
Where
Eb is the braking efficiency
Dx is the actual deceleration
μp is the braking coefficient
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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
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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
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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