Matlab assignment

2. Exercise 5: Write a generalized code to generate a line between the given two end
points using DDA line algorithm (accommodate all the four + four conditions as
discussed in the theory class).
CODE:
clc;
clear all;
X_0 = input('X_0: ');
Y_0 = input ('Y_0: ');
X_1 = input('X_1: ');
Y_1 = input('Y_1: ');
dX = abs(X_0 - X_1);
dY = abs(Y_0 - Y_1);
sx = sign(X_1-X_0);
sy = sign(Y_1-Y_0);
n = max(dY,dX);
X(1) = X_0; Y(1) = Y_0; j=1;
for i=0:1:n
if (X_1==X)&(Y_1==Y)
break
end
j=j+1;
X(j) = X(j-1)+(dX/n)*sx;
Y(j) = Y(j-1)+(dY/n)*sy;
end
plot(round(X),round(Y));
OUTPUT:
X_0: -3
Y_0: 5
X_1: 6
Y_1: -8

3. Exercise 6: Write a generalized code to generate a line between the given two end
points using Bresenhams’ algorithm.
CODE:
clc;
clear all;
enter='X1: '
x1=input(enter);
enter='Y1: '
y1=input(enter);
enter='X2: '
x2=input(enter);
enter='Y2: '
y2=input(enter);
dx=abs(x2-x1);
dy=abs(y2-y1);
p=(2*dy)-dx;
i=1;
if(x1>x2)
x=x2;
y=y2;
xEnd=x1;
temp=x2;
else
x=x1;
y=y1;
xEnd=x2;
temp=x1;
end
a=zeros(abs(xEnd-temp),1);
b=zeros(abs(xEnd-temp),1);
a(1,1)=(x);
b(1,1)=(y);
while x<xEnd
x=x+1;
i=i+1;
a(i,1)=round(x);
if p<0
p=p+2*dy;
b(i,1)=round(y);
else
if((y2-y1)/(x2-x1))>0
y=y+1;
else
y=y-1;
end
p=p+(2*(dy-dx));
b(i,1)=y;
end
plot(a,b)
end

4. OUTPUT:
X1: 2
Y1: 6
X2: 9
Y2: 11
Exercise 7: Write a generalized code to generate a circle for a user specified radius and
coordinates of center point.
CODE:
clc;
clear all;
close all;
x1=input('x_centre: ');
y1=input('y_centre: ');
R=input('Radius: ');
x=0;
y=R;
p=1-R;
m=[x,y];
while(x<y)
x=x+1
if(p<0)
p=p+2*x+1
else
y=y-1
p=p+2*(x-y)+1
end
m=[m;x y]
end
m;
x=m(:,1);
y=m(:,2);
x2=x1+x;
y2=y1+y;
x3=x1-x;
y3=y1+y;

5. x4=x1+x;
y4=y1-y;
x5=x1-x;
y5=y1-y;
x6=x1+y;
y6=y1+x;
x7=x1-y;
y7=y1+x;
x8=x1+y;
y8=y1-x;
x9=x1-y;
y9=y1-x;
plot(x2,y2)
hold on
plot(x3,y3)
hold on
plot(x4,y4)
hold on
plot(x5,y5)
hold on
plot(x6,y6)
hold on
plot(x7,y7)
hold on
plot(x8,y8)
hold on
plot(x9,y9)
OUTPUT:
x_centre: 5
y_centre: 3
Radius: 8
x = 1
p = -4
m = 0 8
1 8
x = 2
p = 1
m = 0 8
1 8

6. 2 8
x = 3
y = 7
p = -6
m = 0 8
1 8
2 8
3 7
x = 4
p = 3
m = 0 8
1 8
2 8
3 7
4 7
x = 5
y = 6
p = 2
m = 0 8
1 8
2 8
3 7
4 7
5 6
x = 6
y = 5
p = 5

7. m = 0 8
1 8
2 8
3 7
4 7
5 6
6 5
Exercise 8. Write a generalized code to perform a 2D translation on user specified
points (For line, triangle, and quadrilateral). Plot the figures before and after
transformation.
CODE:
clear all;
close all;
n= input('enter number of points of the figure= ');
tx= input('insert the value of translation in x direction= ');
ty= input('insert the value of translation in y direction= ');
for i=1:n
x(i)= input('insert initial x co-ordinate of point= ');
y(i)= input('insert initial y co-ordinate of point= ');
P(i,1)= [x(i)];
P(i,2)= [y(i)];
P(i,3)= [1];
end
P(n+1,1)=P(1,1);
P(n+1,2)=P(1,2);
P(n+1,3)= [1];
A= [1 0 0; 0 1 0;tx ty 1];
B= P*A;
B(n+1,1)=B(1,1);
B(n+1,2)=B(1,2);
B(n+1,3)= [1];
plot(P(:,1),P(:,2));
hold on;
plot(B(:,1),B(:,2));

8. OUTPUT:
enter number of points of the figure= 4
insert the value of translation in x direction= 5
insert the value of translation in y direction= 6
insert initial x co-ordinate of point= 2
insert initial y co-ordinate of point= 2
insert initial x co-ordinate of point= 4
insert initial y co-ordinate of point= 3
insert initial x co-ordinate of point= 8
insert initial y co-ordinate of point= 9
insert initial x co-ordinate of point= 7
insert initial y co-ordinate of point= -3
8a. Demonstrate Scaling, Reflection and Rotation about the coordinate axes.
clc;
clear all;
close all;
w= input('Select type of transformation(1=scaling,2=reflection about
xaxis,3=reflection about y-axis,4=rotation)=')
switch w
case 1

9. n= input('enter number of points of the figure= ');
ax= input('insert the value of scaling in x direction= ');
dy= input('insert the value of scaling in y direction= ');
for i=1:n
x(i)= input('insert initial x co-ordinate of point= ');
y(i)= input('insert initial y co-ordinate of point= ');
P(i,1)= [x(i)];
P(i,2)= [y(i)];
P(i,3)= [1];
end
P(n+1,1)=P(1,1);
P(n+1,2)=P(1,2);
P(n+1,3)= [1];
A= [ax 0 0; 0 dy 0;0 0 1];
B= P*A;
B(n+1,1)=B(1,1);
B(n+1,2)=B(1,2);
B(n+1,3)= [1];
plot(P(:,1),P(:,2));
hold on;
plot(B(:,1),B(:,2));
case 2
n= input('enter number of points of the figure= ');
for i=1:n
x(i)= input('insert initial x co-ordinate of point= ');
y(i)= input('insert initial y co-ordinate of point= ');
P(i,1)= [x(i)];
P(i,2)= [y(i)];
P(i,3)= [1];
end
P(n+1,1)=P(1,1);
P(n+1,2)=P(1,2);
P(n+1,3)= [1];
A= [1 0 0; 0 -1 0;0 0 1];
B= P*A;
B(n+1,1)=B(1,1);
B(n+1,2)=B(1,2);
B(n+1,3)= [1];
plot(P(:,1),P(:,2));
hold on;
plot(B(:,1),B(:,2));
case 3
n= input('enter number of points of the figure= ');
for i=1:n
x(i)= input('insert initial x co-ordinate of point= ');
y(i)= input('insert initial y co-ordinate of point= ');
P(i,1)= [x(i)];
P(i,2)= [y(i)];
P(i,3)= [1];
end
P(n+1,1)=P(1,1);
P(n+1,2)=P(1,2);

10. P(n+1,3)= [1];
A= [-1 0 0; 0 1 0;0 0 1];
B= P*A;
B(n+1,1)=B(1,1);
B(n+1,2)=B(1,2);
B(n+1,3)= [1];
plot(P(:,1),P(:,2));
hold on;
plot(B(:,1),B(:,2));
case 4
n= input('enter number of points of the figure= ');
o= input('enter the angle of rotation= ')
for i=1:n
x(i)= input('insert initial x co-ordinate of point= ');
y(i)= input('insert initial y co-ordinate of point= ');
P(i,1)= [x(i)];
P(i,2)= [y(i)];
P(i,3)= [1];
end
P(n+1,1)=P(1,1);
P(n+1,2)=P(1,2);
P(n+1,3)= [1];
A= [1 0 0; 0 1 0;tx
ty 1];
B= P*A;
B(n+1,1)=B(1,1);
B(n+1,2)=B(1,2);
B(n+1,3)= [1];
plot(P(:,1),P(:,2));
hold on;
plot(B(:,1),B(:,2));
end
OUTPUT:
Select type of
transformation(1=scaling,2=reflection about xaxis,3=reflection about
y-axis,4=rotation)=1
enter number of points of the figure= 4
insert the value of scaling in x direction= 3
insert the value of scaling in y direction= 2
insert initial x co-ordinate of point= 2
insert initial y co-ordinate of point= 2
insert initial x co-ordinate of point= -3
insert initial y co-ordinate of point= 6
insert initial x co-ordinate of point= 5
insert initial y co-ordinate of point= 6

11. insert initial x co-ordinate of point= 9
insert initial y co-ordinate of point= 11
Exercise 9: Write a generalized code to perform a 2D Rotation about an user specified
point on user specified entities (For line, triangle, and quadrilateral). Plot the figures
before and after transformation.
CODE:
clc;
clear all;
close all;
n=input('enter no of points on the figure ');
a=input('enter x coordinate o point about which entity is to be
rotated');
b=input('enter y coordinate o point about which entity is to be
rotated');
p=zeros(n,3);
for i=1:n
x(i)=input('enter x co-ordinate of point ');
y(i)=input('enter y co-ordinate of point ');
p(i,1)=[x(i)];
p(i,2)=[y(i)];
p(i,3)=[1];
end
p(n+1,1)=p(1,1);
p(n+1,2)=p(1,2);
d=input('angle to be rotated ');
rd=[cosd(d) sind(d) 0;-sind(d) cosd(d) 0;0 0 1];
t=[1 0 0;0 1 0;-a -b 1];
u=[1 0 0;0 1 0;a b 1];
q=p*t*rd*u
for j=1:n
r(j,1)=q(j,1);
r(j,2)=q(j,2);
end
r(n+1,1)=r(1,1);
r(n+1,2)=r(1,2);
plot(p(:,1),p(:,2))
hold on
plot(r(:,1),r(:,2))
OUTPUT:
enter no of points on the figure 3
enter x coordinate o point about which entity is to be rotated5
enter y coordinate o point about which entity is to be rotated5
enter x co-ordinate of point 1

12. enter y co-ordinate of point 2
enter x co-ordinate of point 3
enter y co-ordinate of point 3
enter x co-ordinate of point 9
enter y co-ordinate of point 11
angle to be rotated 270
Exercise 10. Write a generalized code to demonstrate that the 3D Rotation is not
commutative. Use a simple rectangular parallelepiped to prove the same by plotting the
results.
function ret = rotate3(data,theta,axis)
%ROTATE3 Rotate points[data] in 3D about X, Y orZ axis by 'theta'
radians in CCW dir.
% Input: set of points, theta[angle of rotation] and axis
abbr.['x','y' or 'z'] about
% which the pionts are to be rotated
if nargin~=3
error('Enter set of points, angle of roation, and the axis to
ratate about');
end

13. %creating matrix to work on
matrix = [data ones(size(data,1),1)];
%for easy use
ct = cos(theta);
st = sin(theta);
%deciding the matrix to use according to given parameter of axis
switch axis
case {'x','X'}
m_trans = [1 0 0 0; 0 ct -st 0; 0 st ct 0; 0 0 0 1];
case {'y', 'Y'}
m_trans = [ct 0 st 0; 0 1 0 0; -st 0 ct 0; 0 0 0 1];
case {'z', 'Z'}
m_trans = [ct -st 0 0; st ct 0 0; 0 0 1 0; 0 0 0 1];
otherwise
error('Choose axis from X Y or Z only!!')
end
%Calculating the multiplication and returning the data
ret = matrix*m_trans;
ret = ret(:,[1:3]);
end
Exercise 11: Design Problems
1. Develop a Matlab program with following details:
Design problem: Shaft
Input parameters: Power (KW), rpm of shaft, Allowable shear stress, factor of safety,
length of shaft
Output: diameter of shaft, weight of shaft.
CODE:
function FinalDimensions =
designShaft(power,rev_speed,tau,dia_ratio,length,rho)
%Calculating the torque first
power=power*1000;%kW to W
t = (60*power)/(2*pi*rev_speed);
t=t*1000;% Nm to Nmm
%From Strength criterion
FinalDimensions.OD = ((16*t)/(tau*pi*(1-dia_ratio^4)))^(1/3);%in mm
FinalDimensions.OD = ceil(FinalDimensions.OD); %rounding off
FinalDimensions.ID = floor(dia_ratio*FinalDimensions.OD);
FinalDimensions.wt = rho*pi*FinalDimensions.OD*FinalDimensions.OD*(1-
dia_ratio^2)*length;
FinalDimensions.wt = FinalDimensions.wt/10^9;%normalising to kg due to
OD taken in mm instead of m

14. struct2table(FinalDimensions);
end
2. Develop a Matlab program by assuming same data as in problem 1 to find the
material saving if hollow shaft is used instead of solid shaft
CODE:
function [ output_args ] = Excercise11Question2( input_args )
%UNTITLED9 Summary of this function goes here
% Detailed explanation goes here
clc;
clear all;
P = input('Power (kW): ');
N = input('Speed (rpm): ');
Smax = input('Allowable Shear Stress (MPa): ');
FOS = input('Factor of safety: ');
L = input('Length of shaft (m):');
D = input('Density of the shaft material (kg/m^3): ');
k = input('Ratio of outer to inner diameter: ');
T = 60000*P/(2*pi*N);
d = ((16*T*FOS/(pi*Smax*1000000))^(1/3))*1000
d2 = ((16*k*T*FOS/(pi*Smax*(k^4-1)*1000000))^(1/3))*1000
d1 = k*d2
Weight_hollow = pi*((d1/1000)^2 - (d2/1000)^2)*L*D/4
Weight_Solid = pi*(d/1000)^2*L*D/4
Percentage_Material_Saving = (Weight_Solid-
Weight_hollow)*100/Weight_Solid
display '%';
end
3. Develop a Matlab program to design a cotter joint with following details:
Input: Material properties, load applied on cotter joint (tension and compression), factor
of safety for different parts
Output: All dimensions of cotter joint
CODE:
function FinalDimensions = designCotter(P)
%P is in kN
clc;
load matlab.mat
fprintf('nChoose a Material')
ff=MaterialProperties1(:,1);
%Make a selectable list assigning the values of Syt
Syt = 400; %N/mm^2
fosR = 6; %for spigot, socket and Rod
fosC = 4; %for Cotter
%permissible stresses for Rod
RsigmaT = Syt/fosR;
RsigmaC = 2*Syt/fosR;

15. Rtau = Syt*0.5/fosR;
%permissible stresses for Cotter
CsigmaT = Syt/fosC;
CsigmaC = 2*Syt/fosC;
Ctau = Syt*0.5/fosC;
CsigmaB = CsigmaT;
%Calculation of Dimensions
d = ceil(sqrt(4*P*1000/(pi*RsigmaT)))+1; %Dia of rods
t = ceil(0.31*d); %thk. of cotter
% P = [pi/4 d2^2 - d2*t]sigmaT
d2 = ceil(max(roots([pi/4,-t,-P*1000/RsigmaT])))+1; %Dia of Spigot
d1 = ceil(max(roots([pi/4,-t,(-P*1000/RsigmaT)+(-
pi*0.25*d2^2)+(t*d2)])))+3;%Dia of Socket outside
d3 = ceil(1.5*d); d4 = ceil(2.4*d)+3;%Spigot Collar d3 and Socket
Collar d4
a = ceil(.75*d); c = a;
b = ceil(max((P*1000/(2*Ctau*t)),sqrt((((d4-
d2)/6)+(d2/4))*3*P*1000/t/CsigmaB)));%Width of cotter (Shear vs
Bending)
%Cotter Length ??!!
l= 2*d4;
%Verification for crushing and shearing in spigot
flag=1;
if RsigmaC <= (P*1000/t/d2)
fprintf('nSpigot Failing under CRUSHING!')
flag = 0;
end
if Rtau <= (P*1000/2/a/d2)
fprintf('nSpigot Failing under SHEARING!')
flag = 0;
end
%Verification for crushing and shearing in socket
if RsigmaC <= (P*1000/t/(d4-d2))
fprintf('nSocket Failing under CRUSHING!')
flag = 0;
end
if Rtau <= (P*1000/2/c/(d4-d2))
fprintf('nSocket Failing under SHEARING!')
flag = 0;
end
%Spigot collar thk.
t1 = ceil(.45*d);

16. if flag == 1
FinalDimensions.Parameter = {'Force Acting'; 'Diameter of Each
Rod'; 'Outside Diameter of Socket'; 'Diameter of Spigot or inside
diameter of Socket'; 'Diameter of Spigot-collar'; 'Diameter of Socket-
collar'; 'Distance from end of slot to the end of Spigot on Rod-B';
'Mean width of Cotter'; 'Axial distance from slot to end of Socket-
collar'; 'Thickness of Cotter'; 'Thickness of Spigot-collar'; 'Length
of Cotter'};
FinalDimensions.Value = [P; d; d1; d2; d3; d4; a; b; c; t; t1;
l];
FinalDimensions.Unit = {'(kN)'; '(mm)'; '(mm)'; '(mm)';
'(mm)'; '(mm)'; '(mm)'; '(mm)'; '(mm)'; '(mm)'; '(mm)'; '(mm)'};
FinalDimensions = struct2table(FinalDimensions);
end
fprintf('n')
4. Develop a Matlab code for the following data:
Objective: Selection of single row deep groove ball bearing
Input data: Radial load, axial load, expected life in hours, diameter of shaft
Output: Bearing designation
CODE:
clc;
clear all;
d=input('Enter the inner diameter of the shaft:');
Fr=input('Enter the radial load on bearing (kN):');
Fa=input('Enter the axial load on bearing (kN):');
Lh=input('Enter the expected life (hours):');
n=input('Enter the RPM:');
%the load factors assumed to be 1 each for both Fr and Fa
k=3;
P=Fr+Fa;
L=60*n*Lh;
co=(P*(L/(10^6))^(1/k));
switch d
case 25
if (co<4.36)
disp('The Bearing code is: 61805')
elseif (co>=4.36)&&(co<7.02)
disp('The Bearing code is: 61905')
elseif (co>=7.02)&&(co<8.06)
disp('The Bearing code is: 16005')
elseif (co>=8.06)&&(co<10.6)
disp('The Bearing code is: 68205')
elseif (co>=10.6)&&(co<11.9)
disp('The Bearing code is: 6005')
elseif (co>=11.9)&&(co<14.8)
disp('The Bearing code is: 6205')
elseif (co>=14.8)&&(co<17.8)
disp('The Bearing code is: 6205 ETN9')

17. elseif (co>=17.8)&&(co<23.4)
disp('The Bearing code is: 6305')
elseif (co>=23.4)&&(co<26)
disp('The Bearing code is: 6305 ETN9')
elseif (co>=26)&&(co<35.8)
disp('The Bearing code is: 6405')
else
disp('No bearings available for the given load and diameter')
end
case 28
if (co<16.8)
disp('The Bearing code is: 62/28')
elseif (co>=16.8)&&(co<25.1)
disp('The Bearing code is: 63/28')
else
disp('No Bearings available for the given load and
diameter.')
end
case 30
if (co<4.49)
disp('The Bearing code is: 61806')
elseif (co>=4.49)&&(co<7.28)
disp('The Bearing code is: 61906')
elseif (co>=7.28)&&(co<11.9)
disp('The Bearing code is: 16006')
elseif (co>=11.9)&&(co<13.8)
disp('The Bearing code is: 6006')
elseif (co>=13.8)&&(co<15.9)
disp('The Bearing code is: 98206')
elseif (co>=15.9)&&(co<20.3)
disp('The Bearing code is: 6206')
elseif (co>=20.3)&&(co<23.4)
disp('The Bearing code is: 6206 ETN9')
elseif (co>=23.4)&&(co<29.6)
disp('The Bearing code is: 6306 ')
elseif (co>=29.6)&&(co<32.5)
disp('The Bearing code is: 6306 ETN9')
elseif (co>=32.5)&&(co<43.6)
disp('The Bearing code is: 6406')
else
disp('There are no bearings available for the given load
carrying capacity and diameter.')
end
case 35
if (co<4.75)
disp('The Bearing code is: 61807')
elseif (co>=4.75)&&(co<9.56)
disp('The Bearing code is: 61907')
elseif (co>=9.56)&&(co<13)
disp('The Bearing code is: 16007')
elseif (co>=13)&&(co<16.8)
disp('The Bearing code is: 6007')

18. elseif (co>=16.8)&&(co<27)
disp('The Bearing code is: 6207')
elseif (co>=27)&&(co<31.2)
disp('The Bearing code is: 6207 ETN9')
elseif (co>=31.2)&&(co<35.1)
disp('The Bearing code is: 6307')
elseif (co>=35.1)&&(co<55.3)
disp('The Bearing code is: 6407')
else
disp('There are no bearings available for the given load
carrying capacity and diameter.')
end
otherwise
disp('Please enter a diameter from the above given options.')
end