Minimum Value dependent on another Minimum Value.

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Sam Potter
Sam Potter 2020-4-1
评论： darova 2020-4-1
h(1)=150000; %initial height
a(1)=(40*10^7)/(6371+h(1))^2; %initial acceleration dependant on height
dt=5; %time step
t(1)=0; %initial time
v(1)=a(1)*t(1); %velocity
g(1)=((40*10^7)/(6371+h(1))^2); %Downward force h>100000, upward force = 0 above 100000
As= 5; %Area of spaceship
Ap = 150; %area of parachute
m=850; %Mass
c=0.7;
p(1)=-(100/71)+1.4; % Initial Air Density (Air density occurs at h=100000, from then p=(h/71)+1.4)
Fd(1)=0.5*p(1)*c*As*v^2; %Downward force h<=100000
i=1; %loop counter
while h(end)>=0
t(i+1)=t(i)+dt;
h(i+1)=h(i)-(v(i)*dt); % Find the height of previous time increment
g(i+1)=(40*10^7)/((6371+h(i+1))^2);
if h(i+1)>100000
a(i+1)= g(i+1) %Acceleration=Gravity-(Fd/m)
v(i+1)=v(i)+(a(i+1)*dt);
elseif h(i+1)>=3000
p(i+1)=-((h(i+1)/1000)/71)+1.4;
Fd(i+1)=0.5*c*(p(i+1))*As*(v(i))^2;
a(i+1)=g(i+1)-(Fd(i+1)/m); %Acceleration=Gravity-(Fd/m)
v(i+1)=v(i)+(a(i+1)*dt);
else
dt=0.1;
p(i+1)=-((h(i+1)/1000)/71)+1.4;
Fd(i+1)=0.5*c*(p(i+1))*Ap*(v(i))^2; %Air resistanace open parachute
a(i+1)=g(i+1)-(Fd(i+1)/m); %Acceleration=Gravity-(Fd/m)
v(i+1)=v(i)+(a(i+1)*dt);
end
i=i+1;
end
My code is a kinematics loop. What I need to do is find the minimum height the parachute can be opened to achieve the minimum inpact velocity. The parachute is opened normally at h=3000 when the Fd equation changes from having As to Ap. The impact velocity is the value of v when h=0. I would show what I have done already but I dont know where to start. Thanks in advance for any help.
  1 个评论
darova
darova 2020-4-1
Can you integrate in reverse direction? From earth to atmosphere?
Start from from earth with acceleration (works in opposite direction)

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