How to save "y" and "dy" in the same vector A each time the loop WHILE is incremented?
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function cam()
clc
clear all
close all
N_i = input('N de intervalos:');
k=1;
while(k<=N_i)
k
beta_i(k) = input('Beta do intervalo:');
L(k) = input('Lift do intervalo:');
mov(k) = input('Movimento do intervalo:');
k=k+1;
end
R_0 = input('Raio base:');
teta=0;
passo=30;
n=1;
T(n)=teta;
j=1;
while(j<=N_i)
y=0;
dy=0;
teta=0;
T=0;
j
while(teta<=beta_i(j))
switch mov(j)
case 1
y=0.5*L(j)*(1-cos(pi*teta/beta_i(j)));
dy=((pi*L(j))/(2*beta_i(j)))*sin(pi*teta/beta_i(j));
u=(R_0+y)*sin(teta)+dy*cos(teta);
v=(R_0+y)*cos(teta)-dy*sin(teta);
T(n)=teta;
case 2
y=L(j)*((teta/beta_i(j))-0.5*(pi^-1)*sin(2*pi*(teta/beta_i(j))));
dy=(L(j)/beta_i(j))*(1-cos(2*pi*(teta/beta_i(j))));
u=(R_0+y)*sin(teta)+dy*cos(teta);
v=(R_0+y)*cos(teta)-dy*sin(teta);
T(n)=teta;
case 3
y=0.5*L(j)*((1-cos(pi*teta/beta_i(j)))-0.25*(1-cos(2*pi*teta/beta_i(j))));
dy=0.5*pi*(L(j)/beta_i(j))*(sin(pi*teta/beta_i(j))-0.5*sin(2*pi*teta/beta_i(j)));
u=(R_0+y)*sin(teta)+dy*cos(teta);
v=(R_0+y)*cos(teta)-dy*sin(teta);
T(n)=teta;
case 4
y=L(j)*(1-cos(0.5*pi*teta/beta_i(j)));
dy=((pi*L(j))/(2*beta_i(j)))*sin(0.5*pi*teta/beta_i(j));
u=(R_0+y)*sin(teta)+dy*cos(teta);
v=(R_0+y)*cos(teta)-dy*sin(teta);
T(n)=teta;
case 5
y=L(j)*sin(pi*0.5*teta/beta_i(j));
dy=0.5*pi*(L(j)/beta_i(j))*cos(0.5*pi*teta/beta_i(j));
u=(R_0+y)*sin(teta)+dy*cos(teta);
v=(R_0+y)*cos(teta)-dy*sin(teta);
T(n)=teta;
case 6
y=L(j)*((teta/beta_i(j))-(pi^-1)*sin(pi*(teta/beta_i(j))));
dy=(L(j)/beta_i(j))*(1-cos(pi*(teta/beta_i(j))));
u=(R_0+y)*sin(teta)+dy*cos(teta);
v=(R_0+y)*cos(teta)-dy*sin(teta);
T(n)=teta;
case 7
y=L(j)*((teta/beta_i(j))+(pi^-1)*sin(pi*(teta/beta_i(j))));
dy=(L(j)/beta_i(j))*(1+cos(pi*(teta/beta_i(j))));
u=(R_0+y)*sin(teta)+dy*cos(teta);
v=(R_0+y)*cos(teta)-dy*sin(teta);
T(n)=teta;
case 8
y=0.5*L(j)*(1+cos(pi*teta/beta_i(j)));
dy=-((pi*L(j))/(2*beta_i(j)))*sin(pi*teta/beta_i(j));
u=(R_0+y)*sin(teta)+dy*cos(teta);
v=(R_0+y)*cos(teta)-dy*sin(teta);
T(n)=teta;
case 9
y=L(j)*(1-(teta/beta_i(j))-0.5*(pi^-1)*sin(2*pi*(teta/beta_i(j))));
dy=-(L(j)/beta_i(j))*(1-cos(2*pi*(teta/beta_i(j))));
u=(R_0+y)*sin(teta)+dy*cos(teta);
v=(R_0+y)*cos(teta)-dy*sin(teta);
T(n)=teta;
case 10
y=0.5*L(j)*((1+cos(pi*teta/beta_i(j)))-0.25*(1-cos(2*pi*teta/beta_i(j))));
dy=-0.5*pi*(L(j)/beta_i(j))*(sin(pi*teta/beta_i(j))+0.5*sin(2*pi*teta/beta_i(j)));
u=(R_0+y)*sin(teta)+dy*cos(teta);
v=(R_0+y)*cos(teta)-dy*sin(teta);
T(n)=teta;
case 11
y=L(j)*(cos(0.5*pi*teta/beta_i(j)));
dy=-((pi*L(j))/(2*beta_i(j)))*sin(0.5*pi*teta/beta_i(j));
u=(R_0+y)*sin(teta)+dy*cos(teta);
v=(R_0+y)*cos(teta)-dy*sin(teta);
T(n)=teta;
case 12
y=L(j)*(1-sin(pi*0.5*teta/beta_i(j)));
dy=-0.5*pi*(L(j)/beta_i(j))*cos(0.5*pi*teta/beta_i(j));
u=(R_0+y)*sin(teta)+dy*cos(teta);
v=(R_0+y)*cos(teta)-dy*sin(teta);
T(n)=teta;
case 13
y=L(j)*(1-(teta/beta_i(j))+(pi^-1)*sin(pi*(teta/beta_i(j))));
dy=-(L(j)/beta_i(j))*(1-cos(pi*(teta/beta_i(j))));
u=(R_0+y)*sin(teta)+dy*cos(teta);
v=(R_0+y)*cos(teta)-dy*sin(teta);
T(n)=teta;
case 14
y=L(j)*(1-(teta/beta_i(j))-(pi^-1)*sin(pi*(teta/beta_i(j))));
dy=-(L(j)/beta_i(j))*(1+cos(pi*(teta/beta_i(j))));
u=(R_0+y)*sin(teta)+dy*cos(teta);
v=(R_0+y)*cos(teta)-dy*sin(teta);
T(n)=teta;
case 15
y=L(j);
dy=0;
u=(R_0+y)*sin(teta)+dy*cos(teta);
v=(R_0+y)*cos(teta)-dy*sin(teta);
T(n)=teta;
end
A(n,:,j)=[y,dy,j]
P(n,:,j)=[u,v,j]
n=n+1;
teta=teta+passo;
end
figure
plot(T,A(:,1,j))
figure
plot(P(:,1,j),P(:,2,j))
j=j+1;
end
end
Given the following input:
N de intervalos:2
k =
1
Beta do intervalo:180
Lift do intervalo:2
Movimento do intervalo:1
k =
2
Beta do intervalo:180
Lift do intervalo:2
Movimento do intervalo:8
Raio base:0.025
I got as result for vector A:
A(:,:,1) =
0 0 1.0000
0.1340 0.0087 1.0000
0.5000 0.0151 1.0000
1.0000 0.0175 1.0000
1.5000 0.0151 1.0000
1.8660 0.0087 1.0000
2.0000 0.0000 1.0000
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
A(:,:,2) =
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
2.0000 0 2.0000
1.8660 -0.0087 2.0000
1.5000 -0.0151 2.0000
1.0000 -0.0175 2.0000
0.5000 -0.0151 2.0000
0.1340 -0.0087 2.0000
0 -0.0000 2.0000
The problem is when j=2 (the second time the loop runs) the vector A seems to clear the values computed when j=1 (the first time the loop runs). How can I solve this problem?
I am just crashing into a wall over and over again.
Accepted Answer
Star Strider
on 4 Nov 2016
It actually is not clearing them. Think of ‘A’ as a deck of cards. Here, ‘A(:,:,1)’ is the first card, ‘A(:,:,2)’ is the second card, and so for the rest. All the values are still there.
Experiment with preallocating a new matrix ‘Ae’ and then concatenating to it in each loop to see if that works better for you:
Ae = [];
:
:
Ae = [Ae; y,dy,j];
5 Comments
Miguel Viegas Leal
on 4 Nov 2016
Star Strider
on 4 Nov 2016
Unfortunately, that tells me nothing useful.
What did it do?
What about it didn’t work?
Also, to clarify, this assignment goes before your second while loop:
Ae = [];
and this goes next to your ‘A’ assignment:
Ae = [Ae; y,dy,j];
Miguel Viegas Leal
on 5 Nov 2016
Star Strider
on 5 Nov 2016
I have no idea what you’re doing. I don’t understand your indexing.
In your plot call, the index will exceed matrix dimensions because ‘Ae’ is (Nx3) — a 2-dimensional matrix — not a 3-dimensional matrix. You need to re-write your plot call.
Apparently the preallocation assignment:
Ae = [];
is being reset. Put it before the first while loop and see it that works.
Walter Roberson
on 5 Nov 2016
Why are you using a while instead of a for ?
More Answers (0)
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