not able to plot

close all
clear all
clc
d = 0.1;                            % Diameter of the rocket (m)
A = pi*(d/2)^2;                     % Area of the rocket (m^2)
Cd = 0.7;                           % Drag coefficient
Rho = 1;                            % Air density (kg/m^3)
g = 9.81;                           % Acceleration due to gravity (m/s^2)
Isp = 250;                          % Specific impulse of motor and propellant (s)
Ve = Isp*g;                         % Exhaust velocity (m/s)
m0 = 100;                           % Initial mass (kg)
me = 25;                            % Empty mass (kg)
mp = me;                            % Propellant mass (kg)
mf = m0-me;                         % Final mass (kg)
E = mp/(mp+m0);                     % Propellant mass fraction
MR = 1-E                            % Mass ratio

MR = 0.8000

l = log(mf/me)

l = 1.0986

C = Isp*g;                          % Effective exhaust velocity
u = Isp*g*l                         % Vehicle velocity

u = 2.6943e+03

D = (Rho^2)*(u^2)*A*Cd/2            % Drag force

D = 1.9956e+04

t = 0                               % t = t0 = 0

t = 0

%
tp = 9;                             % Burn time
for i = 1:141
    tp = tp + 1;
    F = Isp*m0*g/tp;
    a = (C*E/tp)/(1-E*(t/tp))-g-(Cd*Rho*(u^2)*A/m0)/(1-E*(t/tp));
    hbo = g*Isp*tp*(mf*log(mf/m0)/(m0-mf)+1)-g*(tp^2)/2;
end
plot (hbo,tp)

plot (a,tp)

Answer 1

[Walter Roberson]

d = 0.1;                            % Diameter of the rocket (m)
A = pi*(d/2)^2;                     % Area of the rocket (m^2)
Cd = 0.7;                           % Drag coefficient
Rho = 1;                            % Air density (kg/m^3)
g = 9.81;                           % Acceleration due to gravity (m/s^2)
Isp = 250;                          % Specific impulse of motor and propellant (s)
Ve = Isp*g;                         % Exhaust velocity (m/s)
m0 = 100;                           % Initial mass (kg)
me = 25;                            % Empty mass (kg)
mp = me;                            % Propellant mass (kg)
mf = m0-me;                         % Final mass (kg)
E = mp/(mp+m0);                     % Propellant mass fraction
MR = 1-E                            % Mass ratio

MR = 0.8000

l = log(mf/me)

l = 1.0986

C = Isp*g;                          % Effective exhaust velocity
u = Isp*g*l                         % Vehicle velocity

u = 2.6943e+03

D = (Rho^2)*(u^2)*A*Cd/2            % Drag force

D = 1.9956e+04

t = 0                               % t = t0 = 0

t = 0

%
tp = 9;                             % Burn time
for i = 1:141
    tp = tp + 1;
    tps(i) = tp;
    F = Isp*m0*g/tp;
    a(i) = (C*E/tp)/(1-E*(t/tp))-g-(Cd*Rho*(u^2)*A/m0)/(1-E*(t/tp));
    hbo(i) = g*Isp*tp*(mf*log(mf/m0)/(m0-mf)+1)-g*(tp^2)/2;
end
plot (hbo,tps)

plot (a,tps)

Question: why are you using the dependent variable as the x axes, and the independent variable as the y axes?

[Suchita Gupta]

hi, thankyou for helping. because im working in upward direction and wanted to keep the confusion away

[Walter Roberson]

You might want to use

set(gca, 'YDir', 'reverse')