1. Introduction
To simulate the acceleration of a car subject to a step input of torque and air
'drag' resistance, proportional to velocity squared, the following methodology could be used:
2. Assumptions
- the engine produces a constant torque of 170 Nm
- the calculation is done assuming the car remains in one gear
- the overall gear ratio is 6:1
- the diameter of the wheels ( = 2 x radius, r) is 600 mm
- the mass of each wheel is 10 kg
- the vehicle (less wheels) mass is 1000 kg
- the moment of inertia of each wheel is 0.4 kgm^{2}
- the vehicle frontal area, A, is 2 m^{2}
- the vehicle drag coefficient, Cd, is 0.35
3. Developing the model
From the gear ratio and the engine torque, the torque at the driving wheels
is 6 x 170 Nm = 1020 Nm.
To determine the vehicle acceleration, the equivalent inertia of the car at the wheels the expression
derived in Mechanics - Engineering Science can be used to give:
acceleration = T/(r[4(I/r^{2} + m) + M])
where: T = torque acting on the driving wheels,
m = mass of each wheel,
I = moment of inertia of each wheel
M = mass of vehicle less the wheels
However the above expression is only valid if air resistance is ignored. Some part
of the total torque will be required to overcome the air resistance in this simulation.
The vehicle drag force, Fd, is given by Ro A Cd v^{2}/2, where Ro is the air density and v is the vehicle
velocity.
The torque required to overcome this air drag resistance, Td, is given by:
Td = r Fd = r Ro A Cd v^{2}/2
The net torque available to accelerate the vehicle will be T - Td, hence the above
expression for acceleration needs to be modified:
acceleration = (T - Fd r)/(r[4(I/r^{2} + m) + M])
This can be represented in a Simulink model as shown (a conventional product can be used rather than a dot product):
The gain block is given the value 1/(r[4(I/r^{2} + m) + M])
= 1/(0.3[4(0.4/0.3^{2} + 10) + 1000])
= 0.00315
The drag varies with the velocity squared and there is a feedback loop after the
acceleration has been integrated to give the velocity. The gain value in this loop is:
Ro A Cd r/2 = 1.25 x 2 x 0.35 x 0.3/2 = 0.13125
60 mph is equal to approximately 26.4 m/s.
Looking at the results indicates that the vehicle takes about 7 seconds to
reach 60 mph, which appears reasonable, but bear in mind that the (single) gear ratio is rather high
for rapid acceleration.
A Scilab Xcos model for the above is shown below:
David Grieve. Modified: 10th October 2014, 14th March 2011, original: 2nd November 2004.