How would I integrate different values into 4 tables?
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I have this code
clc
clear
close
a = 40*(pi/180);
v = 1200;
g = 32.2;
k=[0,2e-6,10e-6,20e-6];
k=0;
dt=.5;
t = 0:dt:55;
v = zeros(length(t),4);
vx = zeros(length(t),4);
vy= zeros(length(t),4);
x = zeros(length(t),4);
y = zeros(length(t),4);
v(1,1) = 1200;
vx(1,1) = v(1,1)*cos(a);
vy(1,1) = v(1,1)*sin(a);
x(1,1)=0;
y(1,1)=0;
for i=2:length (t)
x(i,1) = x(i-1,1)+vx(i-1,1)*dt-.5*k*v(i-1,1).^2*cos(a)*dt.^2;
y(i,1) = y(i-1,1)+vy(i-1,1)*dt-.5*k*v(i-1,1).^2*sin(a)*dt^2-0.5*g*dt.^2;
vx(i,1) = vx(i-1,1)-k*v(i-1,1).^2*cos(a)*dt;
vy(i,1) = vy(i-1,1)-k*v(i-1,1).^2*sin(a)*dt-g*dt;
a=atan(vy(i,1)./vx(i,1));
v(i,1)=sqrt(vx(i,1).^2+vy(i,1).^2);
end
vx(1:round(5/dt):end,:)
vy(1:round(5/dt):end,:)
x(1:round(5/dt):end,:)
y(1:round(5/dt):end,:)
That results in
ans =
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
ans =
771.35 0 0 0
610.35 0 0 0
449.35 0 0 0
288.35 0 0 0
127.35 0 0 0
-33.65 0 0 0
-194.65 0 0 0
-355.65 0 0 0
-516.65 0 0 0
-677.65 0 0 0
-838.65 0 0 0
-999.65 0 0 0
ans =
0 0 0 0
4596.27 0 0 0
9192.53 0 0 0
13788.80 0 0 0
18385.07 0 0 0
22981.33 0 0 0
27577.60 0 0 0
32173.87 0 0 0
36770.13 0 0 0
41366.40 0 0 0
45962.67 0 0 0
50558.93 0 0 0
ans =
0 0 0 0
3454.23 0 0 0
6103.45 0 0 0
7947.68 0 0 0
8986.90 0 0 0
9221.13 0 0 0
8650.35 0 0 0
7274.58 0 0 0
5093.81 0 0 0
2108.03 0 0 0
-1682.74 0 0 0
-6278.52 0 0 0
My issues is whenever I change the K value to [2e-6,10e-6, or 20e-6] it results in the first column changing to the respective k values. However, if I change the x(i,1) positions to say x(i,2) for [2e-6] (and all subsequent notations) then I get every k = 2e-6 value in the 2nd column and the first column becomes zero. Question is how do I get the answers to add on to the table made from previous calculations?
2 comentarios
BALAJI KARTHEEK
el 20 de Abr. de 2020
Just give sample image of the table (what ur expecting in the result), so it will easy to understand and modify the code written by u..
Michael Sabol
el 20 de Abr. de 2020
Respuesta aceptada
Más respuestas (1)
darova
el 20 de Abr. de 2020
0 votos
Try this solution

2 comentarios
Michael Sabol
el 20 de Abr. de 2020
darova
el 20 de Abr. de 2020
It doesn't matter. More reasonable be inside first
for j=1:length (k)
k=K1(j);
for i=2:length (t)
% code
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