Why is my graph of my falling object going random?

Question

Jelle 2013-10-16

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https://ww2.mathworks.cn/matlabcentral/answers/90392-why-is-my-graph-of-my-falling-object-going-random

评论： dpb 2013-10-16

采纳的回答： dpb

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Dear,

i made this for loop for an object going down a hill. I specified all the variables above this part:

      for t = t:t_stap:endtime
          speed = v0+a*t; %Speed at time t, m/s
          D = Cd*p*0.5*speed^2*A; %Drag force, N
          a = (Fg-D-Fsnow)/m; %Acceleration, m/s^2
          height = x0-((speed^2)/g)*log(cosh((g*t)/speed)); %Distance fallen, speed/time graph integrated over time, m
          lastspeed = speed;
          hold('all')
          plot(t,height,'r')
          plot(t,speed,'g')
          plot(t,a,'b')
          axis([0 20 0 100])
          drawnow
          if height <= 0
              break
          end
      end

Now when i run this it shows me a perfect graph of all the results, just as intended. But after about t = 3 it starts going random. All 3 graphs go up and down in random sense. It does not make any sense. It should just keep going on like normal.

What am I doing wrong/why is matlab suddenly interpreting my code wrongly? My best guess is that

speed = v0+a*t

is wrong. I know the formula does not hold, because this assummes constant acceleration, but i do not know anything better. Is the problem here?

Thanking you in advance

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Jelle 2013-10-16

Well, my graph makes sense now and it does exactly as expected, so you solved it! Thank you!

dpb 2013-10-16

Now, as "exercise for the student", write the underlying differential equations and integrate them and compare your solution... :P

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Answer 1

dpb 2013-10-16

1
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/90392-why-is-my-graph-of-my-falling-object-going-random#answer_99881

Instead of using a*t for the instantaneous speed, you need to compute the speed at the time step and increment the current speed by the differential timestep at the acceleration computed for the time step. Your solution presumes the same acceleration has held from t=0.