MATLAB: ODE45 Code Application

functionode45plot

M = 100/6*pi*0.04^3; g=9.8; A=0.25*pi*0.04^2;
Re = @(U) 1.25*0.04*U/0.000015;
CD = @(Re) 24./Re+(2.6*Re/5)./(1+(Re/5).^1.52)+(0.411*(Re/263000).^(-7.94))./(1+(Re/263000).^(-8))+(0.25*Re/10^6)./(1+Re/10^6);
function Udot = Velocity(t,U)
    Udot = g -(0.5*1.25*U^2*A*CD)/M;
    [t,U]=ode45 (@Velocity, [0, 20], 0);
    plot(t,U)
end
I'm in agony with this coding for 5hours….. I can solve easy function but it is very complicate function for me. I'm confused how to make up 'Re' 'CD' coding clearly.
Thanks for reading.. how could I get right plotting?

Best Answer

Hi James,
[t,U] = ode45(@Velocity, [0, 20], .001);
plot(t,U)
function Udot = Velocity(t,U)
M = 100/6*pi*0.04^3;
g=9.8;
A=0.25*pi*0.04^2;
Re = 1.25*0.04*U/0.000015;
CD  = 24./Re+(2.6*Re/5)./(1+(Re/5).^1.52)+(0.411*(Re/263000).^(-7.94))./(1+(Re/263000).^(-8))+(0.25*Re/10^6)./(1+Re/10^6);
Udot = g -(0.5*1.25*U^2*A*CD)/M;
end
You need to get ode45 outside of the functon definition. And you can calculate Re and CD without defining functions for them. If you do define Re and CD as functions, then calculating CD would require using Re(U) everwhere instead of just Re, and similarly CD would have to supplied with an argument. Easier here to not do that.
I changed the initial value of U to .001 since the 24/Re term blows up if you use zero. Changing the initial value does not seem to make much difference as long as it's small.
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