Atmospheric flight mechanics

Rohit Potla

SC12B044

Atmospheric Flight Mechanics AE321

____________________________________________________________________________________

Q.1 Derive the 2-DOF(Degree of Freedom) equations of motion for an earth

entry vehicle.

Ans.

Equations of Motions expressed in an inertial coordinate system:

Forces are resolved in instantaneous coordinates (i,j):

By Equating and , we get

To get the dynamic equations of motions , we need to solve for

and

Kinematic Equations are given by:-

Q2.

Ans. Code:

clear all; clc;

%%inputs given by sir mass=3000; %% Mass of vehicle in kg fpangle(1)=-2*pi/180; %% Initial Flight Path Angle in degrees v(1)=7908; %%Initial Velocity in m/s phi(1)=0; %% Elevation angle in radians h(1)=100; %% Initial Altitude in km S=6; %% Reference Area in m^2 Re=6378; %% Mean Radius of Earth in km Cd=1.1; %% Coefficient of Drag L_D=0.2; %% Lift to drag ratio %%inputs end

Cl=Cd*L_D; %% Coefficient of lift g_i=9.81; %% Accleration due to gravity in m/s^2 at earth

surface rho_i=1.225; %% Air Density in kg/m^3 at sea level rho(1)=rho_i*exp(-g_i*h(1)*1000/(287.058*300)); g(1)=g_i*(Re/(h(1)+Re))^2; D(1)=0.5*rho(1)*v(1)*v(1)*S*Cd; L(1)=D(1)*L_D; dt=0.05;

Force(1)= ((-D(1)*sin(fpangle(1))+L(1)*cos(fpangle(1))-mass*g(1)) ^2+(-

D(1)*cos(fpangle(1)) -L(1)*sin(fpangle(1)))^2)^0.5; acc(1)=Force(1)/mass;

for i=1:60000 if(h(i)<0) break end h(i+1)= h(i) + (v(i)*sin(fpangle(i)))*dt/1000; g(i+1)=g_i*(Re/(h(i+1)+Re))^2; rho(i+1)=rho_i*exp(-g_i*(h(i+1))*1000 /(287.058*300)); phi(i+1)=phi(i)+(v(i)*cos(fpangle(i))*10^-3/(h(i)+Re))*dt; v(i+1)=v(i)+(-g(i) *sin(fpangle(i))-(D(i)/mass))*dt; L(i+1)=0.5*rho(i+1)*v(i+1)*v(i+1)*S*Cl; D(i+1)=0.5*rho(i+1)*v(i+1)*v(i+1)*S*Cd; fpangle(i+1)=fpangle(i)+(v(i)*cos(fpangle(i))*10^-

3/(h(i)+Re))*dt+(((L(i)/mass)-g(i)*cos(fpangle(i)))/v(i))*dt;

Force(i+1)= ((-D(i+1)*sin(fpangle(i+1))+L(i+1)*cos(fpangle(i+1))-

mass*g(i+1)) ^2+(-D(i+1)*cos(fpangle(i+1)) -L(i+1)*sin(fpangle(i+1)))^2)^0.5; acc(i+1)=Force(i+1)/mass; end t(1)=0; for j=1:60000 t(j+1)=t(j)+dt; j=j+1; if(j==i) break; end end

figure(1); plot(t,acc); xlabel('Time(s)'); ylabel('acce(m/sec^2)'); title('Variation of acce with Time'); hold on

figure(2); plot(t,v); xlabel('Time(s)'); ylabel('velocity(m/s)'); title('Variation of velocity with Time');

figure(3); plot(t,phi); xlabel('Time(s)'); ylabel('phi(radians)'); title('Variation of phi with Time');

figure(4); plot(t,h); xlabel('Time(s)'); ylabel('height(km)'); title('Variation of height with Time');

figure(5); plot(t,fpangle); xlabel('Time(s)'); ylabel('fpangle(radians)'); title('Variation of flight path angle with Time');

Q3. For a Ballistic Flight, prove that the maximum deceleration level remains

independent of the Ballistic Parameter.

Ans.

For a Ballistic Flight the conditions are:

From here,

Therefore, from Question 1

and by substituting the conditions of Ballistic Flight, we get

Now Substituting from above, we get

To get a maximum deceleration, differentiating the above

w.r.t. time and then

equating it to '0'.

Now, putting above two equations in

, we get

Therefore, from above result of

we get to know that maximum deceleration is

independent of Ballistic Parameter.

Q4.

Ans. I plotted the maximum value of acceleration by changing the parameter mass,

thus changing . Code:

clear all; clc;

%%inputs given by sir %mass=3000; %% Mass of vehicle in kg fpangle(1)=-2*pi/180; %% Initial Flight Path Angle in degrees v(1)=7908; %%Initial Velocity in m/s phi(1)=0; %% Elevation angle in radians h(1)=100; %% Initial Altitude in km S=6; %% Reference Area in m^2 Re=6378; %% Mean Radius of Earth in km Cd=1.1; %% Coefficient of Drag L_D=0; %% Lift to drag ratio

%%inputs end w=1; for mass=3000:200:4000 Cl=Cd*L_D; %% Coefficient of lift g_i=9.81; %% Accleration due to gravity in m/s^2 at earth

surface rho_i=1.225; %% Air Density in kg/m^3 at sea level rho(1)=rho_i*exp(-g_i*h(1)*1000/(287.058*300)); g(1)=g_i*(Re/(h(1)+Re))^2; D(1)=0.5*rho(1)*v(1)*v(1)*S*Cd; L(1)=D(1)*L_D; dt=0.05;

Force(1)= ((-D(1)*sin(fpangle(1))+L(1)*cos(fpangle(1))-mass*g(1)) ^2+(-

D(1)*cos(fpangle(1)) -L(1)*sin(fpangle(1)))^2)^0.5; acc(1)=Force(1)/mass;

for i=1:60000 if(h(i)<0) break end h(i+1)= h(i) + (v(i)*sin(fpangle(i)))*dt/1000; g(i+1)=g_i*(Re/(h(i+1)+Re))^2; rho(i+1)=rho_i*exp(-g_i*(h(i+1))*1000 /(287.058*300)); phi(i+1)=phi(i)+(v(i)*cos(fpangle(i))*10^-3/(h(i)+Re))*dt; v(i+1)=v(i)+(-g(i) *sin(fpangle(i))-(D(i)/mass))*dt; L(i+1)=0.5*rho(i+1)*v(i+1)*v(i+1)*S*Cl; D(i+1)=0.5*rho(i+1)*v(i+1)*v(i+1)*S*Cd; fpangle(i+1)=fpangle(i)+(v(i)*cos(fpangle(i))*10^-

3/(h(i)+Re))*dt+(((L(i)/mass)-g(i)*cos(fpangle(i)))/v(i))*dt; Force(i+1)= ((-D(i+1)*sin(fpangle(i+1))+L(i+1)*cos(fpangle(i+1))-

mass*g(i+1)) ^2+(-D(i+1)*cos(fpangle(i+1)) -L(i+1)*sin(fpangle(i+1)))^2)^0.5; acc(i+1)=Force(i+1)/mass; end maximum(w)=max(acc); beta(w)=mass/(Cd*S); w=w+1; end plot(beta,maximum); xlabel('beta(kg/m^2)'); ylabel('max acce(m/sec^2)');

Q5.

Ans. The code used was same as code used in Q2, only the value of L/D was set to

0. The blue curve is for L/D=0.2 and green is for L/D=0;

Q6.

Ans. For getting variation with beta, the parameter Cd and Cl were changed. Code:

clc; clear all

%%inputs given by sir color={'r','b','y','g','k'}; mass=3000; %% Mass of vehicle in kg fpangle1(1)=-2*pi/180; %% Initial Flight Path Angle in degrees v1(1)=7908; %%Initial Velocity in m/s phi1(1)=0; %% Elevation angle in radians h1(1)=100; %% Initial Altitude in km S=6; %% Reference Area in m^2 Re=6378; %% Mean Radius of Earth in km % Cd=1.1; %% Coefficient of Drag L_D=0.2; %% Lift to drag ratio %%inputs end g_i=9.81; %% Accleration due to gravity in m/s^2 at earth

surface rho_i=1.225; %% Air Density in kg/m^3 at sea level rho(1)=rho_i*exp(-g_i*h1(1)*1000/(287.058*300)); g(1)=g_i*(Re/(h1(1)+Re))^2; dt=0.05; w=1; for Cd=0.5:0.5:2.5 %%changing beta by changing Cd Cl=Cd*L_D; %% Coefficient of lift D(1)=0.5*rho(1)*v1(1)*v1(1)*S*Cd; L(1)=D(1)*L_D;

Force(1)= ((-D(1)*sin(fpangle1(1))+L(1)*cos(fpangle1(1))-mass*g(1)) ^2+(-

D(1)*cos(fpangle1(1)) -L(1)*sin(fpangle1(1)))^2)^0.5; acc1(1)=Force(1)/mass; for i=1:60000 if(h1(i)<0) break end h1(i+1)= h1(i) + (v1(i)*sin(fpangle1(i)))*dt/1000; g(i+1)=g_i*(Re/(h1(i+1)+Re))^2; rho(i+1)=rho_i*exp(-g_i*(h1(i+1))*1000 /(287.058*300)); phi1(i+1)=phi1(i)+(v1(i)*cos(fpangle1(i))*10^-3/(h1(i)+Re))*dt; v1(i+1)=v1(i)+(-g(i) *sin(fpangle1(i))-(D(i)/mass))*dt; L(i+1)=0.5*rho(i+1)*v1(i+1)*v1(i+1)*S*Cl; D(i+1)=0.5*rho(i+1)*v1(i+1)*v1(i+1)*S*Cd; fpangle1(i+1)=fpangle1(i)+(v1(i)*cos(fpangle1(i))*10^-

3/(h1(i)+Re))*dt+(((L(i)/mass)-g(i)*cos(fpangle1(i)))/v1(i))*dt; Force(i+1)= ((-D(i+1)*sin(fpangle1(i+1))+L(i+1)*cos(fpangle1(i+1))-

mass*g(i+1)) ^2+(-D(i+1)*cos(fpangle1(i+1)) -

L(i+1)*sin(fpangle1(i+1)))^2)^0.5; acc1(i+1)=Force(i+1)/mass; end t1(1)=0; for j=1:i t1(j)=dt*j;

end plot (t1,h1,color{w}); hold on w=w+1; end