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 Commenti
BALAJI KARTHEEK
il 20 Apr 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
il 20 Apr 2020
Risposta accettata
Più risposte (1)
darova
il 20 Apr 2020
0 voti
Try this solution

2 Commenti
Michael Sabol
il 20 Apr 2020
darova
il 20 Apr 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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