Acceleration to displacement data based on accelerometer

Hi,

Can anyone help to guide me on how to find the displacement value based on the acceleration result from accelerometer ? I already try to use several method that have been found but it seems not provide the good result and not understand much the given method. The attach excel file is acceleration output value from a wheel that have rotation of 400 rpm and my previous code that I used.

Thanks in advance !

num=csvread("Book5.csv");                                                                            
Fn = Fs/2;                                                     
Tr = linspace(num(1,1), num(1,end), size(num,1));             
Dr = resample(num(:,2), Tr);                               
Dr_mc  = Dr - mean(Dr,1);                                       
FDr_mc = fft(Dr_mc, [], 1);                                     
Fv = linspace(0, 1, fix(size(FDr_mc,1)/2)+1)*Fn;                
Iv = 1:numel(Fv);                                              
figure
plot(Fv, abs(FDr_mc(Iv,:))*2)
grid
xlabel('Frequency (Hz)')
ylabel('Amplitude')

Answer 1

[Chunru]

num=readtable("https://www.mathworks.com/matlabcentral/answers/uploaded_files/1352749/Book5.csv");  
head(num);

     Var1      Var2       Var3       Var4  
    ______    _______    _______    _______

         0    -1.7658    -0.2943    -0.5886
    0.0238    -1.8639    -0.2943    -0.4905
    0.0476          0    -0.1962    -0.4905
    0.0714    -1.2753    -0.1962    -0.4905
    0.0952    -1.3734    -0.1962    -0.4905
     0.119    -1.2753    -0.1962    -0.4905
    0.1429    -1.3734    -0.1962    -0.4905
    0.1667    -1.3734    -0.1962    -0.4905

% initialize the velocity and position
v0 = [0 0 0]; 
s0 = [0 0 0];  % 3d
vt = v0 + cumtrapz(num{:, 1}, num{:, 2:end});

st = s0 + cumtrapz(num{:, 1}, vt);
whos

  Name          Size            Bytes  Class     Attributes

  cmdout        1x33               66  char                
  num         612x4             21215  table               
  s0            1x3                24  double              
  st          612x3             14688  double              
  v0            1x3                24  double              
  vt          612x3             14688  double              

plot3(st(:,1), st(:,2), st(:,3), 'r.')
box on; grid on;
xlabel("x"); ylabel("y"); zlabel("z")

figure
plot(num{:, 1}, num{:, 2:end})

figure
plot(num{:, 1}, st); 

figure;
plot(num{:, 1}, vt); 

mean(num{:, 2:end}, 'omitnan')

ans = 1×3

-1.2512 -0.2125 -0.5284

[MARLIANA HAFIZA]

Thank you for your help. But, I quite confuse regarding the displacement value. That means based on the output given, the displacement keeps decreasing downwards even the acceleration result got upside down waveform ?

[Chunru]

See the update above. The general trend is determined by the mean value of the acceleration data, which are all negative as shown above. You can see the velocity is negative though fluctuating (due to the acceleration up and down)