Using a non-linear value for pause

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Abdallah 2014-4-8

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回答： Abdallah 2014-4-8

I am trying to set a certain object to rotate, but the rotation is meant to be decaying with time. So I assume I will need an exponential function, that I can use in the pause command to do this. However I seem to still get linear rotation, even though my pause function is set to be exponential. Here is my code:

 %%%%%I just posted the related part of the code
 for trib=1:1:length(ang);             %%%ang was used before
    for ang=-0.1366*pi:.02:1.1366*pi;
         xx=log(4-ang/4).*-10.96;
          %%%%figure setting
          %%%%plotting command
          pause(xx);
    end
end

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Answer 1

Walter Roberson 2014-4-8

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Your x values are all negative. ang would have to be at least 3.819718633 * pi in order for (4-ang/4) to be as low as 1 and thus generate a negative log() to multiply by the negative 10.96 in order to generate a positive pause time.

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Abdallah 2014-4-8

在 MATLAB Online 中打开

I noticed i wrote it in the wrong way actually. It is supposed to be:

xx=log((4-ang)/4).*-10.96

but that still doesn't give me an exponential pause. I am trying to do a rotation similar to a step response, where time changes rapidly at first and gets slower later. That doesn't seem to work here for some reason.

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Answer 2

Joseph Cheng 2014-4-8

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Try exaggerating the the non-linearity and see if you see a bigger difference.

you can see how things are performing if you insert some debugging code

for ang=-0.1366*pi:.02:1.1366*pi;
     tic,
     xx=log((4-ang)/4).*-10.96;
     pause(xx)
     disp([toc xx]);
end

the displayed toc should be close to your xx value.

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Joseph Cheng 2014-4-8

wouldn't it also be more practical to add a term to the angular rotation equation to vary the angular rotation over time rather than keeping the angular rotation constant but stretching time?

Abdallah 2014-4-8

I have the first order equeation of 10.96dv/dt+v=4 .. assuming v is ang (the angle)

4 is the steady-state angle in radians

I am trying to get an equation for time.

the ODE can be converted into this form: v=4-4*exp(-t/10.96)

then I did a rearranging and I got t in this from t=ln((4-V)/4).*-10.96

in other words t=ln((4-ang)/4).*-10.96

but since it is inside in a for loop I imagine the ang should change therefore t is changing as well. Is there another way I can manage to get time in a more tidy and probably accurate way?

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Answer 3

Abdallah 2014-4-8

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Ok now I am very certain of my mathematical model, but I am not sure how to deal with the loops.

My equation is in the form of:

 for k=1:1:length(ang); %%ang previously defined as -0.1*pi:0.0174:1.103*pi;
 for ang=-0.1*pi:0.0174:1.103*pi;
 ggg=log((3.33-ang))/3.759);
 t=ggg/-0.09;
 figure(1); clf reset 
 plot whatever
 pause(t);
 end 
 end

My new equation for t has no chance of giving a negative value because I made sure angles are within boundary. However this time no rotation at all is happening. Can I please get help on this one.