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pban92
Joined: 23 Nov 2009
Posts: 38
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 Posted: Mon Apr 04, 2011 12:26 pm Post subject: Approximating time gradient |
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Hello,
I have a two-column discrete data file with 10,000 rows. The first column is the time (t) and the second is the velocity data (v) where time increment, Delta-T = 0.0001.
I am trying to find the time gradient of the velocity (dv/dt). Can someone help me with the coding below?
Many thanks!
| Code: | OPEN (UNIT=100, FILE=data.txt, STATUS='OLD', ACTION='READ', IOSTAT=ierr)
DO i = 1, size(y)
READ (3,*,IOSTAT=ierr) xdummy, vel(i)
if (ierr /=0) exit
END DO
n=i-1
tgrad = 0.
DO i = 1,n
tgrad = tgrad+(vel(i+1)-vel(i))/(0.0001)
if (i>n) exit
ENDDO
write(*,'Time gradient is ') tgrad/n*0.0001
CLOSE(UNIT=100, FILE=data.txt, STATUS='OLD', ACTION='WRITE', IOSTAT=ierr) |
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JohnCampbell
Joined: 16 Feb 2006
Posts: 2615
Location: Sydney
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| Posted: Tue Apr 05, 2011 1:32 pm Post subject: |
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Isn't the answer to the second DO loop you have coded simply
tgrad = (vel(n+1)-Vel(1)) / 0.0001
Also, does dv/dt ( = (vel(i)-vel(i-1)) / (xdummy(i)-xdummy(i-1)) ) vary with I ?
Assuming it does, you can use a forward difference calculation, in each time step to calculate an estimate of x(i)
x(i) = x(i-1) + (v(i)+v(i-1))/2 * dt + (a(i)+a(i-1))/4 * dt**2
( x = x0 + vel * dt + acc /2 *dt**2 )
You can vary the calculation, depending on the way you estimate acceleration (tgrad(i)) varying over dt. It's an interesting change to the forward difference equations.
John |
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