Transcript
Rohit Potla
SC12B044
Atmospheric Flight Mechanics AE321
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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;
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