Plot velocity vector vs time

Hi there,

I have the following task and I don't know how to implement it:

So I have the general solution for the vector

v(t)=(u(t); v(t); w(t))

which is

v=-inv(phiRV)*phiRR*dr0

where phiRV, phiRR are time-dependent 3x3-matrices and dr0 is another 3x1-vector.

Now I have to calculate v for each t=1:1200:1200.

As you can guess, so far I haven't succeeded. One problem is that I don't really have an idea what the plot should look like, I have never done something like this before. And then there's the implementation code-wise, of course.

I hope you can help me with that, thank you! :)

Answer 1

[Image Analyst]

You don't do this:

   t=1:1200:1200;

you do this:

   t=1:1200;

Then to plot v vs. t, do this:

plot(t, v, 'b-', 'LineWidth', 2);
grid on;
title('v vs. t', 'FontSize', 20);
xlabel('t', 'FontSize', 20);
ylabel('v', 'FontSize', 20);

Assuming your equation for v is really delta v. Otherwise we need to know what the delta is from, like v(0) or is it the prior v, like you'd get from diff(v).

[Backtobasics]

Thank you for your help! What I named v here is delta v, yes. When I try your code, it leads to "Data must be numeric, datetime, duration or an array convertible to double."

Here's my code in full detail (n is a constant determined before):

dr0=dr;
phiRV=[1/n*sin(n*t) 2/n*(1-cos(n*t)) 0;
    2/n*(cos(n*t)-1) 1/n*(4*sin(n*t)-3*n*t) 0;
    0 0 1/n*sin(n*t)];
phiRR=[4-3*cos(n*t) 0 0;
    6*(sin(n*t)-n*t) 1 0;
    0 0 cos(n*t)];
dv0=-inv(phiRV)*phiRR*dr0
t=1:1200;
plot(t, dv0, 'b-', 'LineWidth', 2);

[Image Analyst]

Please give the full code so I can run it, including where you define dr and n. Also, you'll need to define t before you use it in line 2.

Here's some attempt with made up values:

t=1:1200;
dr = [1;2;3];
n = 2
dr0=dr;
dv0 = zeros(length(t), 3);
for k = 1 : length(t)
  this_t = t(k);
  phiRV=[1 /n *sin(n*this_t), 2 / n * (1-cos(n*this_t)), 0; ...
    2 / n * (cos(n*this_t)-1), 1 / n * (4*sin(n*this_t)-3*n*this_t), 0; ...
    0 0 1 / n * sin(n*this_t)];
  phiRR = [4 - 3*cos(n*this_t), 0, 0; ...
    6*(sin(n*this_t) - n*this_t), 1, 0; ...
    0, 0, cos(n*this_t)];
  dv0(k, :) = -inv(phiRV) * phiRR * dr0;
end
% Plot second column
plot(t, dv0(:, 2), 'b-', 'LineWidth', 2);
grid on;
title('delta v vs. t', 'FontSize', 20);
xlabel('t', 'FontSize', 20);
ylabel('Delta v', 'FontSize', 20);

[Backtobasics]

Probably you need the full task as well:

clear all; close all; clc;
syms t

 % given parameters
  mu=398600;
  r=6378+590;
  n=sqrt(mu/(r^3));
  dr0=[0; 0; 0];
  dv0=[0.02; -0.04; -0.1];
  phiVR=[3*n*sin(n*t) 0 0;
      6*n*(cos(n*t)-1) 0 0;
      0 0 -n*sin(n*t)];
  phiVV=[cos(n*t) 2*sin(n*t) 0
      -2*sin(n*t) 4*cos(n*t)-3 0
      0 0 cos(n*t)];
  phiRR=[4-3*cos(n*t) 0 0;
      6*(sin(n*t)-n*t) 1 0;
      0 0 cos(n*t)];
  phiRV=[1/n*sin(n*t) 2/n*(1-cos(n*t)) 0;
      2/n*(cos(n*t)-1) 1/n*(4*sin(n*t)-3*n*t) 0;
      0 0 1/n*sin(n*t)];

 % distance to HST at t=600s
  dr=simplify(subs(phiRR*dr0+phiRV*dv0, t, 600));

 % deltaV necessary to reach HST after 10min:
  dr0=dr; %distance to HST at new time t=0, start of maneuver
  dv0=-inv(phiRV)*phiRR*dr0 %dv0 is equal to delta v from task

% plot deltaV:
t=1:1200;
plot(t, dv, 'b-', 'LineWidth', 2);

[Backtobasics]

I finally made it now! The main problem was that I always thought it should be some kind of 3D plot etc. because its a 3D vector. And that I said "syms t" in the beginning so here's my little workaround:

k=linspace(1, 1200, 1200);
semilogx(k, subs(dv0, t, k))

Thank you so much for your support!