How to add known value of position to an ode45 acceleration problem

Hi there,

Once again i am struggling with MATLABS ODE functions. I am using the differential solvers to find the accelerations of a mass spring damper problem, which is a simplification of a motorcycle front end. I have written some code which solves from the parameters i have given as a test, and it solves it with no errors and gives the expected response shape. However, in this problem i know the change in position of the input, aka the road, which is what i really want to solve for. However, because i have had to convert to a first order system, the equations are in the form;

y(1) = xs - xu
y(2) = xs' - xu'
y(3) = xu - Z
y(4) = xu' - Z'

    dydt = [y(2); -(Ks/ms)*y(1) - (Cs/ms)*y(2); y(4); +(Ks/ms)*y(1) + (Cs/ms)*y(2) - (Ku/mu)*y(3) - (Cu/mu) *y(4);]

My problem is that i know the value of Z, but can't find a way to input it, as that then causes issues with undefined xu and xs. My only thought is that i can put it in the initial values, but then i would still have the problem with undefined xs ad xu.

as an aside, i was told to place the load() function outside of the function, but then i get the error

Attempt to execute SCRIPT Quarter_Car_5 as a function:
C:\Users\user\Dropbox\Ross Hanna Project\Matlab\Diff Solving\Quarter_Car_5.m

Error in odearguments (line 90)
f0 = feval(ode,t0,y0,args{:});   % ODE15I sets args{1} to yp0.

Error in ode15s (line 150)
    odearguments(FcnHandlesUsed, solver_name, ode, tspan, y0, options, varargin);

Error in ODE_Solver_Quarter_Car (line 8)
[t,y]=ode15s(@Quarter_Car_5, tspan, y0)

Could someone show me the correct way to achieve this, or let me know if i need to start a new thread.

My code is below, any help would be greatly appreciated.

function dydt = mss(t,y)

load('Standard_European_Bump_Data.mat','length','height')
height1 = height/1000
length1 = length/1000

%h_vs_l = plot(length1, height1)

ms = 100
mu = 18
Ks = 20000
Ku = 167000
Cs = 1200
Cu = 50

Z = height1

% y(1) = (xs - xu)
% y(3) = (xu - Z)

% y1' = y(2)              % = (xs' - xu')
% y2' = (Ks/ms)*y(1) + (Cs/ms)*y(2)
% y3' = y(4)              % = (xu - Z)
% y4' = -(Ks/ms)*y(1) - (Cs/ms) * y(2) + (Ku/mu)*y(3) + (Cu/mu) * y(4)

dydt = [y(2); -(Ks/ms)*y(1) - (Cs/ms)*y(2); y(4); +(Ks/ms)*y(1) + (Cs/ms)*y(2) - (Ku/mu)*y(3) - (Cu/mu) *y(4);]

end

y0 = [0.7 0.1 0.5 0.3]
tspan = [0 10]

[t,y]=ode15s(@Quarter_Car_5, tspan, y0)

where everything following y0 is in a separate file.

Thanks

[Walter Roberson]

Do not use load in the ode routine. See https://www.mathworks.com/help/matlab/math/parameterizing-functions.html

Also your ode45 call should be to mss not Quarter_car_5

[Ross Hanna]

I removed load for the moment while i get used to parametizing functions. The data is just a 1-(cos(x))^2 function anyway. However, i am still unsure how to use this function in the ODE? Can you see anything simple i am missing. Been trying to research it, but i must not be looking in the right place. Thanks

Answer 1

[Walter Roberson]

file_struct = load('Standard_European_Bump_Data.mat','length','height');
bump_length = file_struct.length;  %let us avoid overwriting the length() function
bump_height = file_struct.height;

y0 = [0.7 0.1 0.5 0.3]
tspan = [0 10]

[t, y] = ode15s(@(t,y) mss(t, y, bump_length, bump_height), tspan, y0)

function dydt = mss(t, y, bump_length, bump_height)

height1 = bump_height/1000;
length1 = bump_length/1000;

ms = 100;
mu = 18;
Ks = 20000;
Ku = 167000;
Cs = 1200;
Cu = 50;

%Z = height1;

% y(1) = (xs - xu)
% y(3) = (xu - Z)

% y1' = y(2)              % = (xs' - xu')
% y2' = (Ks/ms)*y(1) + (Cs/ms)*y(2)
% y3' = y(4)              % = (xu - Z)
% y4' = -(Ks/ms)*y(1) - (Cs/ms) * y(2) + (Ku/mu)*y(3) + (Cu/mu) * y(4)

dydt = [y(2);
        -(Ks/ms)*y(1) - (Cs/ms)*y(2);
        y(4);
        +(Ks/ms)*y(1) + (Cs/ms)*y(2) - (Ku/mu)*y(3) - (Cu/mu) *y(4);];

end