CFD Online Logo CFD Online URL
www.cfd-online.com
[Sponsors]
Home > Forums > Main CFD Forum

richardson extrapolation

Register Blogs Members List Search Today's Posts Mark Forums Read

Reply
 
LinkBack Thread Tools Display Modes
Old   March 11, 2013, 16:31
Default
  #21
Senior Member
 
flotus1's Avatar
 
Alex
Join Date: Jun 2012
Location: Germany
Posts: 1,157
Rep Power: 19
flotus1 will become famous soon enoughflotus1 will become famous soon enough
Either you hava a look at Ferzigers book (not freely available online as far as I know) or you can derive the formula yourself.

Use the Ansatz F(h_i) = F(h=0) + a*(h_i)^p

(where h_i is the grid spacing, f(h_i) is the value obtained on this grid, F(h=0) is the solution without discretization error, a is a constant and p is the error order.

With solutions on three different grids with the same refinement ratio r (r= h_2/h_1 = h_3/h_2) you get 3 equations.

(I) F(h_1) = F(h=0) + a*(h_1)^p
(II) F(h_2) = F(h=0) + a*(h_2)^p
(III) F(h_3) = F(h=0) + a*(h_3)^p

(I')=(I)-(II) F(h_1)-F(h_2)= a*[(h_1)^p-(h_2)^p]
(II')=(I)-(III) F(h_1)-F(h_3)= a*[(h_1)^p-(h_3)^p]

(I'')=(I')/(II') [F(h_1)-F(h_2)]/[F(h_1)-F(h_3)]=[(h_1)^p-(h_2)^p]/[(h_1)^p-(h_3)^p]

now using the definition of r, the right hand side becomes (1-r^p)/(1-r^2p)=1/(1+r^p)

after some more algebra we arrive at
p=log[F(h_1)-F(h_3)]/[F(h_1)-F(h_2)-1]/log(r)
flotus1 is offline   Reply With Quote

Old   March 12, 2013, 11:30
Default
  #22
New Member
 
Albert Puig
Join Date: Jul 2012
Posts: 24
Rep Power: 5
Puigar is on a distinguished road
Hi folks again

first of all, thanks for the responses! ther were really useful to get a plain understanding of what is the richardson extrapolation and how to apply it to my problem.

I have another doubt. If I only have the values of f(h20),f(30) and f(40) because the coarsened grid (f10) does not work, can I manage to get the order of error "p"?

for example. in some case I can run all teh simulations with all of the grid, including the coarsened. But there are some conditions that when I apply them to the model, they make it more unstable, to the point that the coarsened grid (f10) falls off.

for example
case 1
all of them work: I get p and F for richardson extrapolated value
x y
h -> F_h
h/2-> F_h/2
h/4-> F_h/4

case 2
Can I use the value "p" for richardson extrapolated value of CASE 1 if the trends of the two results of F and G are more or less the same? how can I assess the similarity of the trends of F and G to validate the value of the order of the error "p"?
x y
(h doesn't work)
h/2-> G_h/2
h/4-> G_h/4

thank you again for paying attention! It has been a great favour
Puigar is offline   Reply With Quote

Old   March 12, 2013, 11:49
Default
  #23
Senior Member
 
flotus1's Avatar
 
Alex
Join Date: Jun 2012
Location: Germany
Posts: 1,157
Rep Power: 19
flotus1 will become famous soon enoughflotus1 will become famous soon enough
With solutions on only two different grids, there is no way in estimating the error order.
Even worse, if the convergence is not monotonic and you just guess a value for p, you end up with a completely wrong extrapolated value.

So 3 solutions (with three different element sizes) are always necessary.

If you did not refine with the same ratio, you can still try to solve [F(h_1)-F(h_2)]/[F(h_1)-F(h_3)]=[(h_1)^p-(h_2)^p]/[(h_1)^p-(h_3)^p]
some way.

The easier way is to keep the same ratio in both refinements.
Take the coarsest mesh on which you get a reasonable result as a starting point.
Now the finer mesh doesnt have to be h/2 if the solution at h/4 becomes too expensive.
A refinement factor of e.g. 1.5 is still enough to do a reasonable extrapolation in most cases.
flotus1 is offline   Reply With Quote

Old   March 12, 2013, 12:07
Default
  #24
New Member
 
Albert Puig
Join Date: Jul 2012
Posts: 24
Rep Power: 5
Puigar is on a distinguished road
Quote:
Originally Posted by flotus1 View Post
With solutions on only two different grids, there is no way in estimating the error order.
Even worse, if the convergence is not monotonic and you just guess a value for p, you end up with a completely wrong extrapolated value.

So 3 solutions (with three different element sizes) are always necessary.

If you did not refine with the same ratio, you can still try to solve [F(h_1)-F(h_2)]/[F(h_1)-F(h_3)]=[(h_1)^p-(h_2)^p]/[(h_1)^p-(h_3)^p]
some way.

The easier way is to keep the same ratio in both refinements.
Take the coarsest mesh on which you get a reasonable result as a starting point.
Now the finer mesh doesnt have to be h/2 if the solution at h/4 becomes too expensive.
A refinement factor of e.g. 1.5 is still enough to do a reasonable extrapolation in most cases.
so I can get a p value from taking for example h/2 h/3 and h/4? even if they are not doubled? Just I have to apply the ansatz formula again with these refinements to get "p"?
Puigar is offline   Reply With Quote

Old   March 12, 2013, 12:32
Default
  #25
Senior Member
 
flotus1's Avatar
 
Alex
Join Date: Jun 2012
Location: Germany
Posts: 1,157
Rep Power: 19
flotus1 will become famous soon enoughflotus1 will become famous soon enough
The thing is that you connot solve for p explicitly if the refinement ratios are not identical.
So h/2, h/3 and h/4 is not a good choice.
And yes, the meshes dont need to be refined with a factor of 2 (or 0.5 in the definition I used above)
Any factor far enough from 1 will do.
flotus1 is offline   Reply With Quote

Old   March 13, 2013, 10:06
Default
  #26
New Member
 
Albert Puig
Join Date: Jul 2012
Posts: 24
Rep Power: 5
Puigar is on a distinguished road
Quote:
Originally Posted by flotus1 View Post
The thing is that you connot solve for p explicitly if the refinement ratios are not identical.
So h/2, h/3 and h/4 is not a good choice.
And yes, the meshes dont need to be refined with a factor of 2 (or 0.5 in the definition I used above)
Any factor far enough from 1 will do.
So then? If it is not explicitly done, how can I solve it?

I did not get how was this step done when getting the three equation solving of "p"
Quote:
(1-r^p)/(1-r^2p)=1/(1+r^p)
Puigar is offline   Reply With Quote

Old   March 13, 2013, 10:53
Default
  #27
Senior Member
 
flotus1's Avatar
 
Alex
Join Date: Jun 2012
Location: Germany
Posts: 1,157
Rep Power: 19
flotus1 will become famous soon enoughflotus1 will become famous soon enough
1-r^{2p} = 1^{2}-(r^{p})^{2} = (1-r^{p})(1+r^{p})
flotus1 is offline   Reply With Quote

Old   March 13, 2013, 11:44
Default
  #28
New Member
 
Albert Puig
Join Date: Jul 2012
Posts: 24
Rep Power: 5
Puigar is on a distinguished road
Quote:
Originally Posted by flotus1 View Post
1-r^{2p} = 1^{2}-(r^{p})^{2} = (1-r^{p})(1+r^{p})
Sorry for being sooo lame..
I just don't get how you are introducing r here.
Puigar is offline   Reply With Quote

Old   March 13, 2013, 13:18
Default
  #29
Senior Member
 
flotus1's Avatar
 
Alex
Join Date: Jun 2012
Location: Germany
Posts: 1,157
Rep Power: 19
flotus1 will become famous soon enoughflotus1 will become famous soon enough
r is the ratio of the element sizes and was introduced a few posts ago.

r= \frac{h_2}{h_1}=\frac{h_3}{h_2}
flotus1 is offline   Reply With Quote

Reply

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are On
Pingbacks are On
Refbacks are On


Similar Threads
Thread Thread Starter Forum Replies Last Post
Richardson extrapolation pc Main CFD Forum 11 March 8, 2013 07:50
Discretization error with Richardson extrapolation. cfdstudent004 CFX 1 May 29, 2012 19:04
Richardson Extrapolation to check time and space convergence Hooman Main CFD Forum 0 December 29, 2010 09:19
Richardson extrapolation cynthia FLUENT 4 June 20, 2008 08:58
One-sided extrapolation coolblizz Main CFD Forum 2 February 8, 2006 21:24


All times are GMT -4. The time now is 11:12.