Calculations Of Hill Climbing Force
AIM:
To Analysis TATA ULTRA T.12 TRUCK by equating various calculation and observing the ratio of hill climbing power,gradeability, and to create Model the WOT (Wide Open Throttle) response of GM EV1 car.
OBJECTIVE:
- To find the ratio of hill climbing power required by fully loaded tata ultra truck to the half loaded one.
- To find how much can be the minimum length of the flyover, considering the minimum height at the summit should be 5.5 meters
- To Prepare a new drive cycle excel file having third column representing hill climbing angle corresponding to speed and time. Revise the model file “Driver_Glider.slx” such that it can calculate gradient force from drive cycle.
- Model the WOT (Wide Open Throttle) response of GM EV1 car. Compare it with the official figures of the performance of a real vehicle. Prepare a table showing variation in acceleration time with 5% change in vehicle parameters.
TATA ULTRA T.12 TRUCK:
Tata Motors Limited is India's largest automobile company and a market leader in the Country's commercial vehicles segment. As the world's fifth largest truck manufacturer, Tata Motors has crossed almost seven decades of automotive excellence in a journey marked with significant milestones in excellence and leadership. Using state-of-the-art technology in its world-class manufacturing units, the Company started exporting vehicles in 1961. Since then, Tata Motors has created a name globally, and completed 50 years of its international business in 2011.
Tata Motors was the first company to foray into the segment of indigenously developed Light Commercial Vehicles with the launch of Tata 407 and Tata 608 in 1986. Today, Tata Motors offers a wide range of light commercial vehicles (LCVs) and intermediate commercial vehicles (ICVs) widely used for inter-city and intra-city goods transport or passenger transport, with specific applications. Its range of revolutionary Light Trucks deliver superior performance, reliability and durability at low operating cost.
STUDY-1:
1. To find the ratio of hill climbing power required by fully loaded tata ultra truck to the half loaded one.
GIVEN DATA:
A data's are collected by the TATA websitehttps://light-trucks.tatamotors.com/light-trucks/tata-ultra/specifications/tata-ultra-t-12-light-trucks-specifications.aspx
FORMULA USED:
HILL CLIMBING FORCE:
Fhc = m x g x sin`theta`
where, m = Total mass
g = Gravity
POWER:
Power is a rate at which work is done, or energy is used. It is equal to the amount of work done divided by the time it takes to do the work. The unit of power is the Watt (W), which is equal to a Joule per second (J/s).
Total power = Total force x speed
SOLUTION:
Gross vehicle weight = Kerb weight + Pay load
Gross vehicle weight = 11990kg
gravity = 9.8m/s^2
Gradeability = 25.6%
Gradeability = 15 degree
`theta`= 15 degree
Force = m x g x sin `theta`
F = 11990x9.8xsin(15)
F = 11990 x 9.8 x 0.6503
F = 117502 x 0.6503
F = 76411.5506N
Total power = Total force x speed
speed = 30KMPH
By converting the unit speed = 30x5/18 = 8.33
power = 76411.5506 x 8.33
power = 636762.922 W
The half loaded truck,
Pay load = 8255
= 8255/2
= 4127.5 kg
kerb weight = 3735 kg
Gross vehicle weight = Kerb weight + Pay load
Gross vehicle weight = 3735+4127.5
= 7862.5 kg
Force = m x g x sin `theta`
F = 7862.5x9.8xsin(15)
F = 7862.5 x 9.8 x 0.6503
F = 77052.5 x 0.6503
F = 50107.241 N
Total power = Total force x speed
speed = 30KMPH
By converting the unit speed = 30x5/18 = 8.33
power = 50107.241 x 8.33
power = 417393.318 W
The ratio of hill climbing power required by fully loaded tata ultra truck to the half loaded one,
Power Ratio = Power of fully loaded truck/Power of half loaded truck
Power Ratio = 636762.922/417393.31
The ratio of hill climbing power = 1.53.
STUDY-2:
A flyover is planned to be constructed on a national highway. Keeping in mind about the gradeability for trucks, how much can be the minimum length of the flyover, considering the minimum height at the summit should be 5.5 meters? State your assumptions.
Gradeability:
A Truck gradability is its ability to climb slopes. Gradability is measured either in degrees or percentage. Gradability is dependent on engine power, drivetrain type, gear ratio, weight, weight distribution, car's centre of gravity and traction.

GIVEN DATA,
Minimum height = 5.5m
`theta`= 15 degree
tan`theta`x 100 = gradeability
Percentage of gradeability = Rise/Runx100

TO FIND: The minimum length of the flyover
SOLUTION
sin`theta`= opp/hyp
sin(`theta`) = opp/hyp
sin(`theta`) = 5.5/hyp
0.258 = 5.5/hyp
hyp = 5.5/0.258
hyp = 21.32
hyp = 21.3
Minimum length of flyover = 21.3m
By pythagoras theorem,

Run =?
a^2+b^2 = c^2
(5.5)^2+b^2 = (21.3)^2
b^2 = (21.3)^2-(5.5)^2
b = 15.8m
Minimum Run of flyover = 15.8m
Total Run of flyover = 31.6m
STUDY-3:
To Prepare a new drive cycle excel file having third column representing hill climbing angle corresponding to speed and time. Revise the model file “Driver_Glider.slx” such that it can calculate gradient force from drive cycle.
EXCEL SHEET FOR NEW DRIVE CYCLE:
Here the input values are assigned and entered in excel sheet. Time and velocity are assumed that data are inserted in driver glider.
https://drive.google.com/file/d/1a4JlZb5qL8uTj3jnFr1Lz9s6xIHme3TP/view?usp=sharing
EXPLANATION OF THIS STUDY:
- In driver glider the velocity are assigned in excel sheet to find an acceleration of vehicle. The inputs values are plotted by graph as time and velocity.
- The unit conversion for veloicity is initiate in the new table and that value is taken as input of velocity.
- A speed of vehicle is varies with acceleration changes that will be calculated by time.
- Data's are inserted in MATLAB and that values are intiate in the velocity block to plot the acceleration of vehicle.
VELOCITY GRAPH:
Thus the acceleration of vehicle are deterimned in the graph as the values of given excel sheet.
Gradient Force From Drive Cycle:
To find the gradient force, an inclination slope angle are assigned in the excel sheet for run time 50s. The degree are converted as radian in excel to insert in grade force block.
Then the value of radian in entered in the incilantion angle that are converted from degree. Maximum slope angle is taken as the inclination angle of truck in slope.
Display block is used to obatin the output value of gradient force N that value are given below.
Gradient Force From Drive Cycle = 7737N.
STUDY-4:
Model the WOT (Wide Open Throttle) response of GM EV1 car. Compare it with the official figures of the performance of a real vehicle. Prepare a table showing variation in acceleration time with 5% change in vehicle parameters.
WIDE OPEN THROTTLE:
Wide open throttle (WOT) refers to an internal combustion engine's maximum intake of air and fuel that occurs when the throttle plates inside the carburetor or throttle body are "wide open", providing the least resistance to the incoming air. In the case of an automobile, WOT is when the accelerator is depressed fully.
In the case of a diesel engine, which does not have a throttle valve, WOT is the point at which the maximum amount of fuel is being injected relative to the amount of air pumped by the engine, generally in order to bring the fuel-air mixture up to the stoichiometeric point.
Given data of a real vehicle,
m = 1540 kg
T max = 140 N m
T = 102000/37 × v = 2756/v
ωc = 733 rad s−1
gear ratio is 11:1, thus G = 11.
m + I = 1638
frontal area A = 1.8 m2
tyre radius is 0.30 m
ηg = 0.95
v = 19.8 m s−1
vmax = 35.8 m s−1
Drag coefficient Cd = 0.19
μrr = 0.0048
Gear ratio = 11
FORMULA USED:
V = rx`omega`/G
G/rT = μrrmg + 0.625ACd ν^2 +(m + IxG^2/ηgr^2)dν/dt
EQUATION DERIVED FROM FORMULA,
dν/dt = 3.11 − 0.000137v^2
dν/dt = 62.1/v−0.046 − 0.000137v^2
vn+1 = vn + δt+(3.11 − 0.000137v^2n)
vn+1 = vn + δt+(62.1/v−0.046 − 0.000137v^2n)
Official figures of the performance of a real vehicle
clear all
close all
clc
t = linspace(0,20,201);
vel=0;
dt = 0.1;
for n = 1:200
if vel(n)<19.8
vel(n+1) = vel(n)+dt*(3.11-(0.000137*(vel(n)^2)));
elseif vel(n)>=35.8
vel(n+1) = vel(n);
else
vel(n+1) = vel(n)+dt*((62.1/vel(n))-0.046-(0.000137*(vel(n)^2)));
end
end
% velocity's unit changed to kmph
vel = vel.*3.6;
plot(t,vel)
hold on
%axis([0 30 0 50]);
xlabel('Time/seconds');
ylabel('velocity/kmph');
variation in acceleration time with 5% change in vehicle parameters:
DATA:2
Vehicle mass = 1617 kg
`mu`rr = 0.0048
frontal area A = 1.8 m2
Drag coefficient Cd = 0.19
Tire radius = 0.315
T max = 140 N m
T = 2938.7/v
Gear ratio = 11
`omega` = 733 rad s−1
ηg = 0.95
EQUATION DERIVED FROM FORMULA,
dν/dt = 2.83 − 0.000130v^2
dν/dt = 59.51/v−0.00474 − 0.000130v^2
vn+1 = vn + δt+(2.83 − 0.000130v^2n)
vn+1 = vn + δt+(59.51/vn−0.00474 − 0.000130v^2n)
INPUT PROGRAM:
clear all
close all
clc
t = linspace(0,20,201);
vel=0;
dt = 0.1;
for n = 1:200
if vel(n)<19.8
vel(n+1) = vel(n)+dt*(3.11-(0.000137*(vel(n)^2)));
elseif vel(n)>=35.8
vel(n+1) = vel(n);
else
vel(n+1) = vel(n)+dt*((62.1/vel(n))-0.046-(0.000137*(vel(n)^2)));
end
end
% velocity's unit changed to kmph
vel = vel.*3.6;
plot(t,vel)
hold on
%axis([0 30 0 50]);
xlabel('Time/seconds');
ylabel('velocity/kmph');
for n = 1:200
if vel(n)<20.99
vel(n+1) = vel(n)+dt*(2.83-(0.000130*(vel(n)^2)));
elseif vel(n)>=35.8
vel(n+1) = vel(n);
else
vel(n+1) = vel(n)+dt*((59.51/vel(n))-0.00474-(0.000130*(vel(n)^2)));
end
end
vel = vel.*3.6;
plot(t,vel)
hold on
%axis([0 30 0 50]);
xlabel('Time/seconds');
ylabel('velocity/kmph');
title('Full power(WOT) acceleration of gm electric car');
EXPLANATION OF PROGRAM:
DATA-1
- In this program we want to compare the official figures of the performance of a real vehicle with variation in acceleration time with 5% change in vehicle parameters respect with velocity and time
- To Model the WOT (Wide Open Throttle) response of GM EV1 car. Compare it.
- Data's are obtained from the textbook of real vehicle and the two equations obtained by the formula.
- That equations refers as the critical speed and maximum speed of 'if' ,'else'conditions in MATLAB program.
- To create a program with velocity and time and plot the obtained results on that program it shows the velocity of the vehicle.
- dt value are given by considering the split point of time.
- velocity are converted by multiplying the value of 3.6 in program
DATA-2
- Same process are followed in the data 2 but the changes are made in vehicle parameters as 5%.
- Increment of 5% is acted in the program to compare the actual velocity and time with variation in acceleration time.
- The mass,Tire radius,Frontal area are changed by sum of 5% and equated in the formula.
- Two equations are obtained in that formula and that conditions are used for 'if' 'else' statement in critical speed and maximum speed.
- An critical speed are derived from the formula V = rx`omega`/G and values are initiate in conditions.
- Plot the graph with velocity and time.
- The two velocity's are obtained by the resultant program and that should be varied in time to reached the maximum speed.
- Thus variation in acceleration time with 5% change in vehicle parameters are intiate in graph.
RESULT (GRAPH):
This graphs shows that the vehicle parameters are changed the velocity to attain the top speed will also changed due to time because the mass of vehicle will increase the speed velocity time will also increase.
In this comparison the vehicle parameters are increased by 5% as that time taken for top speed is little varied from actual vehicle time taken.
CONCLUSION:
- The ratio of hill climbing power = 1.53
- Minimum length of flyover = 15.8m
- Gradient Force From Drive Cycle = 7737N.
- variation in acceleration time with 5% change in vehicle parameters are shown below,
Therefore all the study's are analysis and derived by various calculations.
- Manohar Milkurinice bro keep it up,good explanation about given challenge