Acceleration Modelling of EV1_FR8695

Name: Siddhesh Ozarkar
MSME Dept. Mechanical Engineering
Wayne State University, Detroit
P: 313.330.5196
E: fr8695@wayne.edu
ACCELERATION
MODELLING
OF GM EV1
9/14/2015
Acceleration Modelling from 0 to 60mph
for GM EV1
Gm Ev1 is a small electric vehicle which has played important
role in the recent developments of electric vehicles. This report
contains simple calculations to predict the acceleration of EV1
computed from the given data. Simple differential equations
solved on MATLAB.

Acceleration Modelling of GM EV1
Page 1
---Equation 2
} Acceleration governing equations
for mass m= 1560kg
---Equation 1
A C C E L E R AT I O N M O D E L L I N G F R O M 0 T O 6 0 M P H F O R G M E V 1
Part 1: Acceleration Performance Simulation of EV1 using MATLAB script.
Case 1: Mass of vehicle m =1560kg.
The acceleration of the vehicle is computed in two phases. Phase one is considered when the torque
supplied by the motor linearly increases with speed. This is true for most types of motors that are used to
propel electric drive vehicles. The torque of the motor used in ev1 increase linearly till 140 Nm till motor
reaches critical speed ωc. In the later phase this torque reduces in simple relation with velocity & motor
operates at constant power of 102kW replicating the WOT conditions.
Tmax=140Nm ωc=733 rad/s
Torque is maximum until velocity v=19.8m/s (v=71.3km/h)
As velocity reaches 19.8m/s the motor operates at constant power this is the condition of WOT test
T =
10200
37
∗ v
For initial phase of acceleration when T=Tmax
Case: Now considering mass of the vehicle with two passengers each weighing 70 kg we have the overall
mass of the vehicle as 1560kg.
Now substituting the values of constants from table 1 in the Equation 2 & considering Equation 1
Phase 1 of Acceleration: 3.117-0.000136v2=(
𝑑𝑣
𝑑𝑡
)
Phase 2 of acceleration:
62.1
𝑣
-0.046-0.000136v2

Page 2
Table 1
Parameter Description Value & units
Cd Air Drag(Ultra Low Air Drag
Co-efficient as from vehicle
data)
0.19
μrr Rolling Resistance 0.0048
m Mass of vehicle
m=1560kg
m=1326kg(-15% mass)
m=1794kg(+15% mass)
1560kg
1326kg(Case 1)
1794kg(Case 2)
G Gear Ratio 11:1 11
r Radius of Tyre 0.30m
A Frontal Area of vehicle 1.8m2
ηg Efficiency of transmission 0.95

Page 3
Description of Graph 1
At 60Mph the time required to
reach the speed is 9.1
seconds.
The values of acceleration are
respectable.
The velocity is observed to be
constant 128kmph.
MATLAB script for obtaining acceleration m=1560kg:
% Acceleration Performance simulation of GM EV1
t=linspace(0,15,151); % time from 0 to 15 seconds with 0.1 second timestep
v=zeros(1,151); % 150 readings of velocity
dT=0.1; %0.1second time step
for n=1:150
if v(n)<19.8
v(n+1)=v(n)+dT*(3.107-(0.000136*(v(n)^2)));
elseif v(n)>35.8
v(n+1)=v(n);
else
v(n+1)=v(n)+dT*((62.1/v(n))-0.046-(0.000137*(v(n)^2)));
end;
end;
v=v.*3.6;
plot(t,v);
xlabel('Time/second');
ylabel('velocity/kph');
title('Acceleration Simulation of GM EV1');
legend('Acceleration of EV1','Location','northwest');
Results for the Case: Graph 1

Page 4
for mass m= 1326kg
Part 2: Acceleration Performance Simulation of EV1 using MATLAB script with reduced
and increase overall mass of the vehicle.
Now substituting the values of constants from table 1 in the Equation 2 & considering Equation 1.
𝑑𝑣
𝑑𝑡
)
73.05
𝑣
-0.047-0.0001606v2
for n=1:150
if v(n)<19.8
v(n+1)=v(n)+dT*(3.6641-(0.0001606*(v(n)^2)));
elseif v(n)>35.8
v(n+1)=v(n);
else
v(n+1)=v(n)+dT*((73.05/v(n))-0.0470-(0.0001606*(v(n)^2)));
end;
end;
v=v.*3.6;
plot(t,v);

Page 5
At 60Mph the time required to reach the speed is 7.8 seconds.
The values of acceleration are respectable.
The velocity is observed to be constant 128kmph.
It can be observed that the acceleration of the vehicle is quicker and the the vehicle
approaches the 60mph mark nearly 2 seconds quicker if mass is reduced by 15%.
This is simply governed by Newton’s Law
F=ma
M=mass & a=acceleration
Results for Case 1: Graph 2

Page 6
for mass m= 1794kg
Now substituting the values of constants from table 1 in the Equation 2 & considering Equation 1.
𝑑𝑣
𝑑𝑡
)
53.998
𝑣
-0.047-0.000118v2
for n=1:150
if v(n)<19.8
v(n+1)=v(n)+dT*(2.696-(0.00011872*(v(n)^2)));
elseif v(n)>35.8
v(n+1)=v(n);
else
v(n+1)=v(n)+dT*((53.998/v(n))-0.04708-(0.0001187*(v(n)^2)));
end;
end;
v=v.*3.6;
plot(t,v);

Page 7
At 60Mph the time required to reach the speed is 10.5 seconds.
The values of acceleration are respectable.
The velocity is observed to be constant 128kmph.
It can be observed that the acceleration of the vehicle is slower and the
vehicle approaches the 60mph mark nearly 1.5 seconds later if mass is
increased by 15%.
This is simply governed by Newton’s Law
F=ma
M=mass & a=acceleration
Results for Case 1: Graph 3

Page 8
Discussion & Comparison:
Mass of Vehicle (kg) Time required to reach the 60mph mark
(Seconds)
1560 9.1
1326 (-15%) 7.8
1794 (+15%) 10.5
It can be observed from the graphs obtained the effect of mass on the acceleration of the vehicle.
As mass increases the acceleration decreases and vice versa.
Part 3: Mathematical solution to find out velocity solution using Calculus.
**For this part of the report I have considered the governing acceleration for phase 1
when vehicle weighs 1560kg.
The reduced differential equation for the acceleration is as follows
3.117-0.000136v2=(
𝑑𝑣
𝑑𝑡
) where, v= velocity in m/s
 This is a First Order Non-linear Ordinary Differential Equation.
 The eqution can be reduced in an alternative form as
V’(t)=-0.00013(v(t)-154.6)(v(t)+154.6)
 Initial conditions as v (t)=0 when t=0 seconds
 On solving we get the solution to differential equation as
V (t) =154.67*(
𝑒0.0402∗𝑡−2.71824021.44𝐶
2.71824021.44𝑐+𝑒0.0402𝑡 ) where C = constant of integration.

Page 9
The verification from
MATLAB shows that velocity
at 7 seconds is 77.57km/h.
The difference in MATLAB
and hand calculation using
calculus might be due to the
number of decimals that
MATLAB uses during
computation.
at 9.1 seconds is 96.17
km/h.
Verification of Solution:
1. Velocity when time t=7 seconds
Using calculus and the solution for the differential equation we get
 v(7) = 21.61m/s = 77.79 km/h
2. Velocity when time t = 9.1 seconds
 V(9.1) = 27.9m/s = 97.2 km/h

Page 10
at 5 seconds is 55.79 km/h.
3. Velocity when time t = 5 seconds
 v(5)= 15.49 m/s = 55.77 km/h
Similar hand calculations can be performed for the phase two of the acceleration and
the obtained first order nonlinear differential equation can be solved accordingly
Conclusion:
 The curve obtained in graph 1 closely resembles the curve obtained from the manufactures data.
 The acceleration of the vehicle reduces as the mass increases as expected.
 The hand calculations done closely match the calculations done by MATLAB.
 There is a slight difference in the values of velocity obtained from hand calculations and MATLAB this
might be due to the way matlab handles the decimal numbers.
References:
J.Y. Wong “Theory of ground Vehicles 3rd edition”
Larminie & Lowry, “Electric Vehicle Technology Explained, 2nd ed.” 2012

Acceleration Modelling of EV1_FR8695

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