To study the effect of center of gravity of vehicle on traction force induced in front and rear wheels and based on that give suggestion for a type of drive for passenger car, commercial vehicles/trucks, and racing car.

Problem statement: To study the effect of center of gravity of vehicle on traction force induced in front and rear wheels and based on that give suggestion for a type of drive for passenger car, commercial vehicles/trucks, and racing car.

Theory:

whenever any vehicle is moving on the road, that time it is being opposed by aerodynamic drag, rolling resistance, and gradient of the road which is known as 'tractive resistance'. as well as vehicle, also demands additional force to achieve require a degree of acceleration.

As vehicle generates rotational torque in an engine and it is being amplified by transmission and differential gearbox while reducing speed and it is given to the wheel which then applied at the tire and road interface as traction force.

This Traction force has to overcome tractive resistance to accelerate the vehicle, hence we can write

Traction force=F_traction=F_aerodynamic_drag+F_rolling_resistance+F_road_gradient+F_inertia_due_to_acceleration

F_traction=Tractive resistance+F_inertia_due to acceleration

Here traction force applied at tire-road interface varies proportional to load on the tire and that is equivalent to normal reaction experienced by the wheel.

here if we get shear stress which is created between road and tire due to traction force as

Shear stress=Traction force/Area of contact between road and tire........(1)

whereas pressure force is created in the same area due to weight or normal reaction and it ensures proper grip of the tire over the road surface.

pressure= Normal reaction or weight shared by that wheel/Area of contact between road and tire........(2)

So we can say that from equation 1 and 2 as

Traction force=(shear stress/Pressure)*(Normal reaction or weight of vehicle shared by that wheel)

As shear stress and pressure have the same units then we can say that it is a dimensionless quantity and can be considered as a constant of proportionality though it changes with the material of road and tire and road condition.

So we can say that

Traction force=K*normal reaction on the wheel

F_traction=K*Wf or K*Wr

Wf, Wr are normal reactions on front and rear wheels respectively.

Hereby considering quasi_static condition for analysis, we can assume the following data

m=2000;% kg mass of the vehicle
g=9.81;% acceleration due to gravity
W=m*g;
L=4; % m, length of the vehicle from front end to the rear end
l=3;% m wheelbase
a=0.25;%m distance between front end to the front axle
d=0.75;% m distance between rear axle to the rear end
d1=L-d;
h=0.25;% m height of C.G.along Z-axis
theta=(10*180)/pi; %gradient of road as 10 degree

b= distance of C.G. from front axle
c= distance of C.G.from rear axle
Normal reaction at front wheels

W_total=Wf+Wr

for x(i)>>a && x(i)<<d1

which acts at C.G.of vehicle.

Wf(i)=(W*c*cos(theta)+W*h*sin(theta))/(l).......................(3)

if we use front-wheel drive then traction force would be

F_tractionf=K*Wf

The normal reaction at rear wheels

Wr(i)=-(W*h*sin(theta)-W*b*cos(theta))/(l)........................(4)

if we use rear-wheel drive then traction force would be

F_tractionr=K*Wr

Hence we can say that if we calculate the value of normal reaction for various positions of C.G. along the length of the vehicle then we can find an optimum position of C.G. to achieve

1) Proper balancing of a vehicle to avoid rolling, pitching and yawing while steering

2) Proper require acceleration to meet the performance requirement

3) space availability

4) Powertrain sizing and space for powertrain

Let's do simulation using Matlab program to find optimum value of C.G. to meet the performance requirement

Matlab code:

clc;
clear all;
close all;
m=1000;% kg mass of vehicle
g=9.81;% acceleration due to gravity
W=m*g;
L=4; % m,length of vehicle from front end to rear end
l=3;% m wheel base 
a=0.25;%m distance between front end to front axle
d=0.75;% m distance between rear axle to rear end
d1=L-d;
h=0.25;% m height of C.G.along Z axis
theta=(10*pi)/180; %gradient of road as 10 degree
nx=100;
x=linspace(0,L,nx);
z=W*cos(theta);
for i=1:length(x)
if x(i)=a
b=x(i)-a;% distance of C.G. from front axle
c=l-b;% distance of C.G.from rear axle
Wf(i)=(W*c*cos(theta)+W*h*sin(theta))/(l);% Normal reaction at front wheels
Wr(i)=-(W*h*sin(theta)-W*b*cos(theta))/(l);% Normal reaction at rear wheels
W_total(i)=Wf(i)+Wr(i);
else

b=x(i)-a;% distance of C.G. from front axle
c=b-l;% distance of C.G.from rear axle
Wf(i)=(-W*c*cos(theta)+W*h*sin(theta))/(l);% Normal reaction at front wheels
Wr(i)=-(W*h*sin(theta)-W*b*cos(theta))/(l);% Normal reaction at rear wheels
W_total(i)=Wf(i)+Wr(i);
end
end
end
figure(1)
plot(x,Wf,'r')
hold on
plot(x,Wr,'g')
hold on
plot(x,W_total,'b')
xlabel('position of C.G. along longitudinal axis')
ylabel('normal reaction or proportinal traction force on front and rear wheels')
legend('Front wheel normal reaction or traction force','rear wheel normal reaction or traction force','Total reaction or weight of vehicle')
grid on
b=2.4-a;% distance of C.G. from front axle
c=l-b;% distance of C.G.from rear axle
Wff=(W*c*cos(theta)+W*h*sin(theta))/(l);% Normal reaction at front wheels
Wrr=-(W*h*sin(theta)-W*b*cos(theta))/(l);% Normal reaction at rear wheels
W_totalfr=Wff+Wrr;

Here as we can see these axles, one on the left which indicates the front axle at x=0.25m and one on left indicate the rear axle at x=3.25 m from left.

conclusions:

A) analyzing front-wheel drive:

a) analyzing red line or front-wheel normal reaction

1) As I said earlier front-wheel reaction force is proportional to front-wheel traction force, then from the graph we can say that if C.G. is above front end/bumper then maximum traction force will be produced at the front wheel, whereas for other positions value of front-wheel traction force goes on decreasing until we move to rear end.

2) If we place C.G. above front axle then the front wheel reaction is high enough to accelerate and will not cause much trouble while steering but still to counter this force we are using power steering assisted by amplification by either drag link mechanism or hydraulic actuators.

e.g.

passenger car

b) Analyzing green line or rear wheel normal reaction

here rear-wheel normal reaction is much less than front-wheel reaction and it has a negative value when C.G. is at the Front end and that is a condition for rollover or to rolling due to lateral forces on the vehicle as well to oversteering due to skidding of the rear wheel.

Conclusion: here we can conclude for front axle drive that, it is optimal to place C.G. close to the front axle on the slight rear side to it(Rear wheel reaction force=0, around x=0.8) to get higher traction force and thus acceleration as well steering will be more easily without any skidding of rear wheels but I know space constraint plays a major role along with passenger weight and thus it gets shift slight in a rear direction up to driver position but still it gives the best performance for acceleration, balancing and steering.

e.g. passenger cars, this is the reason we are using front axle drive for passenger cars as their C.G. lies close to the front axle and it gives better gas mileage, acceleration, meet space constraint, and balanced suspension without skidding of rear wheels while steering.

B) analyzing Rear-wheel drive:

a) analyzing red line or front-wheel normal reaction

1) As we can see from the graph, we can say that if C.G. is above rear end then the reaction force at the front wheels is minimum and it is negative and that is a concern as it will create steering much difficult as wheels will lose contact with the ground and thus it will cause understeer.

b) Analyzing green line or rear wheel normal reaction

As I said earlier rear-wheel traction force is proportional to rear-wheel normal reaction and from the graph we can say when C.G. was above rear end, then rear-wheel traction force is maximum, and higher traction force will give better grip and fast acceleration.

Conclusion: for rear axle drive we can conclude that even maximum traction force is possible at rear end position but it does not satisfy the condition for steering and thus for rear-wheel-drive, it should be placed at less than x=3.8 to achieve better performance while steering and accelerating.

e.g.1)commercial vehicles: This is the reason we are using it in commercial vehicles as their C.G. located in a forward direction close to the rear axle after loading and it will be around x=2.5 and thus gives the best acceleration, steering and balancing to avoid pitching of vehicle.

2)Race cars: we are using the rear-wheel-drive for racing cars as we need huge acceleration for those cars and larger traction force to win over competitors. as well here we have space constraint in front area as we need to achieve aerodynamic shape to reduce drag force on a racing car. Hence to achieve aerofoil shape in front to reduce drag and simultaneously to achieve larger traction force needed for acceleration we generally place the engine in the rear with rear-wheel drive for the racing car along with a wide tire to optimize its performance.