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From: Edoardo on 3 Mar 2010 06:48
Hi, I'm new to this forum and I hope to find some answers to my problems.
First of all I'm a begginer in MatLab/Simulink and all I've learned has been learned by myself, no teacher, so probably I lack in some easy fundamentals.

By now I'm developing a very simple model, based on first principles to study the dynamics of an helicopter.
The final step should be calculate the trim values of my model.
But still I can't understand how does this function work.
First of all I'm using the standard 6DoF aerospace blockset as the unique "dynamic" block in my model, all the other blocks do not have any dynamic components (integrators,derivative etc etc) , only simple calculation.

I'm quite aware of the grammar of the trim function... but still I found very strange how the results are shown.
To test the work of the trim function I decided to test it only using the 6DoF block, with no step between, so the "In" blocks are directly attached to a Mux block and then to the 6DoF, and the exits are directly attache to the "Out" block.

When I ask for the trim I usually use this formulas :
[x,u,y]=trim('name of my sys',x0,[],y0) where the x0 are the values of the state state in the initial condition and y0 the requested one...
The fundamental test is to provide x0 with the same values of y0... so the force to have constant speed are equal to 0.
The results are correct from the aspect of the u (inputs...equal to 0), but in the x and y I got two vector of values very close to 0... in theory if a put a an initial velocity different from 0 and I ask for that velocity as output the final state vector x and the requested output y should be equal to the requested values y0... (I usually put also and additional constraint over y0 using iy).

Why the trim gives my as output only values close to zero and not the one I ask for?
Is due to the use of the 6DoF block?

To give and example :
[x,u,y]=trim('name of my sys',x0,[],y0)
x0=[10;0;0;0;0;0;0;0;0;0;0;0]
y0=[10;0;0;0;0;0] output of Vbody and P,Q,R rotational speed.
and I got
x=1.0e-021*[0.7;0;0;0;0;0;0;0;0;0;0;0]
y=1.0e-021*[0.7;0;0;0;0;0]

A sincere thenk to anyone who will help me.
Edoardo

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