VEHICLE DYNAMICS: 3 ASSIGNMENT SERIES

ASSIGNMENT 1:

Refer to below link for vehicle modeling using Simulink:

https://in.mathworks.com/matlabcentral/fileexchange/63823-matlab-and-simulink-racing-lounge-vehicle-modeling

Prepare a new drive cycle excel file having a third column representing a hill-climbing angle corresponding to speed and time. Revise the model file “Driver_Glider.slx” such that it can calculate gradient force from the drive cycle.

SOLUTION:

ORIGINAL MODEL-

The block named 'DrCycles.US06' is a From workspace block which uses 'Drcycles' data saves in model workspace. This block inputs the velocity corresponding to the defined drive cycle.

THE GRAPH OF THE DRIVE CYCLE:

Model workspace defined in the original model

The 'Glider' block consists of following subblocks

Now defining a new excel sheet that contains 3 columns to replace the 'DrCycles.US06' From workspace block

https://docs.google.com/spreadsheets/d/1EOhDIzxh2-d7SWXmbCrGVuiFR6AgwZuH9P4BSN2cHto/edit#gid=0

a glimpse of the above data which has been generated for 100 seconds.

This file is saved in the current folder in .xlsx form

This file is imported into the Driver_Glider.slx using From Spreadsheet Block to replace the 'DrCycles.US06' From workspace block, so that our input block contains both velocity and readability input.

The specialty of the FROM SPREADSHEET BLOCK, is it automatically scans the first row of the excel file for the names of the titles, also scans the 1st column for time values against which velocity and gradient are taken.

The revised model:

We note that our from spreadsheet block has two signal outputs which correspond to velocity and gradability as it has automatically fixed time signal as a reference input. Of these, the velocity signal is sent to the driver block and the gradability is sent to the Glider block. There has been no change undergone inside the driver block. But we have altered the glider block contents so that our model takes into consideration the gradability values.

MODIFIED GLIDER BLOCK:

We have replaced a constant block containing variables defined in the model workspace by a series of mathematical operators to incorporate the values of gradability.

INITIALLY HILL CLIMBING FORCE CALCULATION DONE BY A CONSTANT BLOCK.

The value of this block was always zero initially as we initialized the inclination angle as 0 in model workspace.

This entire setup has been replaced by a new set of arithmetic operations.

Here we have used the 'inport' block to taken in values of gradability inputted to the glider block.

A D2R block has been used here as the sin block calculates values for radian inputs.

And the resulting value has been multiplied to massVeh and gravity values both already defined in the model workspace.

The resulting value is the HILL CLIMBING FORCE for different values of gradability defined in the excel sheet.

HENCE WE HAVE GENERATED OUR OWN EXCEL WHICH HAS BEEN USED AS INPUT TO THE CALCULATION OF THE HILL CLIMBING FORCE.

THE RESULTS ARE:

THE PLOT OF USER DEFINED GRADABILITY WITH RESPECT TO TIME IN DEGREES:

Plot of user defined velocity with respect to time in (m/s):

Plot of HILL CLIMBING FORCE calculated by the model

Hence we have incorporated the climbing force calculation into the model by feeding in a new set of gradability values generated using excel sheet, that has replaced the inclination angle term which always took the value 0 in our original model.

ASSIGNMENT 2(a):

What should be the maximum speed of the motor used in an electric scooter capable to run at 90 kmph, if the fixed gear ratio is 7 and tire size is 90/100 R10 53J?

Solution:

TIRE CODE:

Therefore 90/100 R10 53J means

Section width:90 mm

Aspect Ratio:100% `implies` Height of the tire sidewall:90mm

Rim Diameter:10 inches `implies` 254mm

FORMULA:

`omega=G .(v/r)` ..................................................1

v=linear velocity of the vehicle or the tire=90kmph=25m/s(given)

G=gear ratio=7(given)

r=radius of the tire

From the tire specifications, we can calculate the actual radius of the tire

`r=((rim diameter)+(2*height of sidewall))/2`

From the calculations we get

r=217 mm=0.217m

substituting all the values in equation 1.

We get `omega`=806.4 radians/seconds converting this to rotations per minute we get

`omega`=7700.5 rpm

Since in equation 1 we used maximum velocity this `omega` is the maximum speed of the motor.

ASSIGNMENT 2(b):

Recalculate the same value if the tire code is 90/90 R18 51P?

Solution:

Section width:90 mm

Aspect Ratio:90% `implies` Height of the tire sidewall:81mm

Rim Diameter:18 inches `implies` 457.2mm

FORMULA:

`omega=G .(v/r)` ..................................................1

v=linear velocity of the vehicle or the tire=25m/s(given)

G=gear ratio=7(given)

r=radius of the tire

From the tire specifications, we can calculate the actual radius of the tire

`r=((rim diameter)+(2*height of sidewall))/2`

From the calculations we get

r=309.6 mm=0.3096m

substituting all the values in equation 1.

We get `omega`= 565.24 radians/seconds converting this to rotations per minute we get

`omega`=5397.6 rpm

ASSIGNMENT 2(c):

Prepare a simple excel calculator to identify vehicle propulsion power based on given inputs and outputs.

Inputs: Kerb Weight (kg), Payload (kg), Coefficient of rolling resistance, Air density (kg/m3), Width (m), Height (m), Drag coefficient, Speed

Output: Total power in kW

What are the limitations of this calculation?

SOLUTION:

ROLLING RESISTANCE :

where

`mu_(rr)`= Coefficient of rolling resistance

m= Total mass of the vehicle

g=9.81 `m/s^2`

AIR DRAG :

`rho=`Air density

A= Cross-sectional area=width*height

`C_d`= Coefficient of Drag

v= linear velocity of the car

TOTAL POWER= TOTAL FORCE*LINEAR VELOCITY OF THE VEHICLE.

We note that

• as we increase the mass of the vehicle the power required to propel the vehicle.
• as we increase the cross-sectional area the power required increases.

The main limitation in the above calculation of power required is :

• we have assumed a constant value for `mu_(rr)` while it varies with speed of the vehicle.
• we have assumed a rough value of `C_d` assuming a common geometry for the vehicle. While the actual value has to be calculated by performing a full-fledged simulation on the exact 3-D model of the vehicle.
• We have ignored the hill-climbing force, while there is practically no surface without a gradient
• We have assumed the vehicle to move in a constant velocity, while there is no practical drive cycle without acceleration and deceleration.

ASSIGNMENT 2(d):

Assuming that tire pressure reduction by 15% results in an increase of rolling resistance effect by two times, how much will be the difference in total traction power keeping all other parameters the same?

SOLUTION:

When the tire pressure reduces the surface of contact increases which in turn increases the rolling friction coefficient as determined experimentally.

We note that the rolling resistance force has doubled in both cases. Therefore the total force increases and hence the power required also is more than what was required.

ASSIGNMENT 3:

1. For a defined driving cycle, calculate the energy required for braking.

Solution:

Here we have chosen NEDC drive cycle whose characteristics looks like this

The above has been generated using below-built Simulink model

Note the velocity profile alone has been written into a separate file that is processed upon to calculate the energy spent to decelerate the vehicle whose mass is taken to be 1000Kgs using the following MATLAB code.

clear all
close all
clc
load('untitled.mat')
mass=1000;
for i=2:2361
    
    if(velocity(2,(i+1))<velocity(2,(i))) && (velocity(2,(i-1))<=velocity(2,(i)))
        x(i)=velocity(2,(i));
   elseif (velocity(2,i+1)>=velocity(2,i)) && (velocity(2,i-1)>velocity(2,i))
        y(i)=velocity(2,(i));
    end
end
e(1)=0;
for i=2:2341
    if (y(i)==0) && (i<=2307)
       e=(0.5*mass*(x(i)^2)+e);
    else
        e=e-(0.5*mass*(y(i)^2));
    end
end

The workspace for the above code

Therefore the energy required for braking or deceleration is 2454.3 KJ for the NEDC DRIVE CYCLE.