Clear Filters
Clear Filters

error: index exceeds matrix dimensions

1 view (last 30 days)
Shannon Hemp
Shannon Hemp on 13 Mar 2016
Commented: Guillaume on 14 Mar 2016
I have the following code that I am trying to run but keep getting the "index exceeds matrix dimensions" error for the rho_a line. Does anyone know how to fix this error?
h1=0;r=0;go=9.81;Mi_1=298000;
for t=0:100;
Pc2_RD_1(t+1) = Pc_RD*(1+(.015*dt_1)^2);
Pe_RD_1(t+1) = Pc2_RD_1(t+1)*(1+((gamma_RD-1)/2)*Me_RD^2)^(gamma_RD/(1-gamma_RD));
Isp_RD_1(t+1) = (e_RD/Pc2_RD_1(t+1))*(Pe_RD_1(t+1)-Pa));
mprop_RD1(t+1) = At_RD*Pc2_RD_1(t+1)/c_RD;
thrust_RD1(t+1) = Isp_RD_1(t+1)*g0*mprop_RD1(t+1)/1000;
dM_RD1(t+1) = mprop_RD1(t+1)*dt_1
dU1(t+1) = Isp_RD_1(t+1)*g0*dM_RD1(t+1)/Mi_1
rho_a(t+1) = 1.2*exp(((-2.9e-5)*h1(t+1)^1.15))
D(t+1) = .5*C_D*Af*rho_a(t+1)*dU1(t+1)^2
g(t+1) = g0*(r_E/(r_E+h1))^2
Uy_dt(t+1) = dU1(t+1)*cos(theta)-dt_1(t+1)*(D(t+1)/(Mi_1-dM_RD1(t+1))+g(t+1))*cos(theta)
Ux_dt(t+1) = dU1(t+1)*sin(theta(t+1))-dt_1(t+1)*(D(t+1)/(Mi_1-dM_RD1(t+1)))*sin(theta(t+1))
U(t+1) = (Ux_dt(t+1))^2+(Uy_dt(t+1))^2
h1(t+1) = h1(t+1)+Uy_dt(t+1)*dt_1(t+1)
r(t+1) = r(t+1)+Ux_dt(t+1)*dt_1(t+1)
if Ux_dt(t+1)>0
theta(t+1) = atan(Ux_dt(t+1)/Uy_dt(t+1))
else
theta(t+1) = atan(Ux_dt(t+1)/Uy_dt(t+1))+pi()
end
dt_1(t+1)=dt_1+.2;
time(t+1)=t+.2;
end
  4 Comments
Shannon Hemp
Shannon Hemp on 13 Mar 2016
Ru = 8314; % J/(kmol*K)
g0 = 9.81; % m/s^2
C_D = 0.5;
r_E = 6378e3; % m
Pa = 1.01325e5; % Pa
h1 = 0;
U = 0;
r = 0;
theta = 0;
phi = pi()/4;
Mi_1 = 284089; % kg
Df = 4.2; % m
Af = (Df^2)*pi()/4;
Tc_RD = 3600; % K
M_RD = 23.6; % kg/kmol
gamma_RD = 1.217;
Pc_RD = 25.662e6; % Pa
De_RD = 1.43; % m
eta_RD = 1.056;
lamda_RD = 0.98;
e_RD = 36.4;
At_RD = .08824; % m^2
Me_RD = 4.1;
c_RD = (eta_RD*sqrt(gamma_RD*(Ru/M_RD)*Tc_RD))/(gamma_RD*((2/(gamma_RD+1))^((gamma_RD+1)/(2*(gamma_RD-1)))));
dt_1=0.2;
for t=0:100;
Pc2_RD_1(t+1) = Pc_RD*(1+(.015*dt_1)^2)
Pe_RD_1(t+1) = Pc2_RD_1(t+1)*(1+((gamma_RD-1)/2)*Me_RD^2)^(gamma_RD/(1-gamma_RD));
Isp_RD_1(t+1) = (lamda_RD*c_RD/g0)*(gamma_RD*sqrt((2/(gamma_RD-1))*((2/(gamma_RD+1))^((gamma_RD+1)/(gamma_RD-1)))*(1-((Pe_RD_1(t+1)/Pc2_RD_1(t+1))^((gamma_RD-1)/gamma_RD))))+(e_RD/Pc2_RD_1(t+1))*(Pe_RD_1(t+1)-Pa));
mprop_RD1(t+1) = At_RD*Pc2_RD_1(t+1)/c_RD;
thrust_RD1(t+1) = Isp_RD_1(t+1)*g0*mprop_RD1(t+1)/1000;
dM_RD1(t+1) = mprop_RD1(t+1)*dt_1
dU1(t+1) = Isp_RD_1(t+1)*g0*dM_RD1(t+1)/Mi_1
rho_a(t+1) = 1.2*exp(((-2.9e-5)*h1(t+1)^1.15))
D(t+1) = .5*C_D*Af*rho_a(t+1)*dU1(t+1)^2
g(t+1) = g0*(r_E/(r_E+h1))^2
Uy_dt(t+1) = dU1(t+1)*cos(theta)-dt_1(t+1)*(D(t+1)/(Mi_1-dM_RD1(t+1))+g(t+1))*cos(theta)
Ux_dt(t+1) = dU1(t+1)*sin(theta(t+1))-dt_1(t+1)*(D(t+1)/(Mi_1-dM_RD1(t+1)))*sin(theta(t+1))
U(t+1) = (Ux_dt(t+1))^2+(Uy_dt(t+1))^2
h1(t+1) = h1(t+1)+Uy_dt(t+1)*dt_1(t+1)
r(t+1) = r(t+1)+Ux_dt(t+1)*dt_1(t+1)
if Ux_dt(t+1)>0
theta(t+1) = atan(Ux_dt(t+1)/Uy_dt(t+1))
else
theta(t+1) = atan(Ux_dt(t+1)/Uy_dt(t+1))+pi()
end
dt_1(t+1)=dt_1+.2;
time(t+1)=t+.2;
end
Image Analyst
Image Analyst on 14 Mar 2016
This uncommented code is impenetrable. It looks like alphabet soup to me. I suggest you step through it a line at a time in the debugger to figure out what it's doing.

Sign in to comment.

Answers (2)

Image Analyst
Image Analyst on 13 Mar 2016
You set h1 to be zero, a scalar. So why do you think there should be a (t+1)'th element as if it were an array? It's not an array.
  1 Comment
Shannon Hemp
Shannon Hemp on 14 Mar 2016
Ok I see what you're saying. What my problem is is I initially want to solve the rho at h1=0 and have that go through the iteration. Then at the end I want to find a new h1 and go through the iteration again. For some reason I can't figure out how to do this.

Sign in to comment.


Guillaume
Guillaume on 13 Mar 2016
What's puzzling is why you can't see the problem.
rho(t+1) depends on h1(t+1), which does not exists yet, hence why you get the error. h1(t+1) is calculated later but since it depends on rho(t+1), you've got a circular reference. Assuming the circular reference is a mistake, only you can tell which of the equations is wrong.
  2 Comments
Shannon Hemp
Shannon Hemp on 14 Mar 2016
Ok I see what you're saying. What my problem is is I initially want to solve the rho at h1=0 and have that go through the iteration. Then at the end I want to find a new h1 and go through the iteration again. For some reason I can't figure out how to do this.
Guillaume
Guillaume on 14 Mar 2016
The problem is that you're using t both as an index and as your time. Ideally, the two should be decoupled, and you'd have an index going from 1 to numel(t) with t a vector of times (e.g 0:100 for example).
t = 0:100;
h1 = [h0, zeros(size(t))]; %h1(1) is initial condition
PC2_RD = [PC_RD, zeros(size(t))]; %PC_RD is initial condition
%... etc. for all other arrays. They all should have numel(t)+1 elements
for iter = 1:numel(t)
PC2_RD(iter+1) = PC2_RD(iter) * (1+(.015*dt_1)^2); %is dt_1 supposed to change during the iteration?
%...
rho_a(iter+1) = 1.2*exp(((-2.9e-5)*h1(iter)^1.15))
%...
end

Sign in to comment.

Categories

Find more on MATLAB in Help Center and File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!