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kevin June 8, 2006 10:00

time step and vortex shedding frequency
hi all, i found the vortex frequency are influenced by the time step i chose, what is your oppion about the time step if when want to get right vortex shedding frequency?

thanx in advance.


ganesh June 9, 2006 01:26

Re: time step and vortex shedding frequency
Dear Kevin,

Time accuracy is an important issue in unsteady flows. It is therefore imperative that you choose the right time step for a desired accuracy. You could compare your experience with the dependency of solution on grid and attempts to achieve a grid-independent solution. Therefore, you must use a time-step that gives you the acceptable temporal accuracy. In general, a physical time step of 1e-3 to 1e-4 should be good enough. For ant delt less than this, changes in shedding frequency should be minimal.

Hope this helps



kevin June 9, 2006 03:47

Re: time step and vortex shedding frequency
thanks for your detailed reply,ganesh. i agree with you, and there is another problem, sometimes i found at T/4 and 3T/4, Cl achieve its max and min, and it is almost 0 at 0T/4 and 4T/4, but when i changed the time delt to check again, i got different results.

i'm wondering maybe the low pressure induced by the vortex account for the relationship between vortex shedding frequency and Cl(what i said above)

would u please give me some comments.

best regards.


Anton Lyaskin June 9, 2006 07:03

Re: time step and vortex shedding frequency
I guess you're modelling something like flow past the cylinder? Lift oscillations have frequency twice than vortex shedding. To get trustworthy results your time step should be at least 20 times smaller than lift oscillation period (so talking about absolute values of time step is bit meaningless for this case). And may be you'd better use some high order differencing in time (Cranck-Nikolson or Runge-Kutta), but in this case you'd probably need even smaller time step to satisfy stability criterion (CFL number)

Mani June 9, 2006 15:22

Re: time step and vortex shedding frequency
I get weary repeating myself, but I see this too often to hush up. You guys may have all the experience and expertise necessary to give good advice, but sometimes you're completely missing out on common sense. What does it mean, when I talk about a physical time step of 1e-3 to 1e-4? My physics teacher in high school would ask: What is that? Is it 1e-3 seconds, 1e-3 hours, 1e-3 potatoes, or some non-dimensionalized time?

Numbers are meaningless without unit, unless you can assume that everyone knows what you mean, and that's rarely the case. The only thing that makes sense in this context is a time step non-dimensionalized by an appropriate time scale, i.e. the (approximate) vortex shedding period. However, a non-dimensional time of 1e-3 times the vortex shedding period would mean to resolve each period by 1000 time steps. That may be necessary for an explicit time marching method but is very excessive if you're using an implicit method (which you really should).

With implicit time stepping the vortex shedding frequency can be (and has been) quite accurately obtained with 30-50 time steps per vortex shedding period on a reasonably fine mesh. For laminar 2-D vortex shedding over a cylinder (47<Reynolds number<180), a second order finite volume method will need about 300-400 grid cells around the cylinder circumference, according to a study by J. Alonso, and my own experience.

This case is quite simple because you know the result (shedding frequency, lift and drag amplitudes) from ample experimental data that's available to you. If you're not getting the right result, you may not have the necessary temporal or spatial resolution.

diaw June 9, 2006 22:38

Re: time step and vortex shedding frequency
A few physics questions:

What do the *physics* & the *form* of the N-S equations tell us about the selection of a 'suitable' time step?

Should this be a fixed, or variable time-step? Why?

Is the time-step linked to spatial coordinates? If so, why?

Do we *have to* scale everything, or could we elect to work in real physical coordinates - space & time?

What is the significance of the d()/dt term in the N-S? When does it 'become active'? Are there flow situations where it may be zero/non-zero?

Does the concept of numeric stability, numeric 'error' need to take precedence in all our computations - or, does physics dictate the end-game?

ganesh June 10, 2006 05:09

Re: time step and vortex shedding frequency
Dear Mani,

Your statements are true and I apologise for having left off the units. However, in case I was talking about a physical time step which was dimensional I would have actually put down the units, and therfore the time step had to be a non-dimensional one. The non-dimensionalisation is what I missed and this in my case is t* = t/(L/U_inf), where L and U_inf are the characteristic length adn reference velocity respectievely. With an implicit time stepping procedure(I use Crank-Nicholson), the physical time step roughly translates to 75-100 time steps per vortex shedding period.



Mani June 12, 2006 08:38

Re: time step and vortex shedding frequency
Are 75+ steps really necessary, according to your experience? For which time-stepping method are you quoting these numbers (I'm assuming implicit)?

To add one question to diaw's list: In choosing an appropriate time step, should we not account for the (smallest) time scales that our spatial resolution is able to handle? Does it make sense to let delta_t go to zero and then speak of convergence, when the grid is fixed? Will there be stability issues, when the grid is too coarse to resolve the unsteady solution at a very small time step?

Mani June 12, 2006 09:04

Re: time step and vortex shedding frequency
fixed or variable time step:

I suppose this topic is equivalent to the question: fixed or adaptive grid. yes, you could go through the trouble to adapt your spatial and temporal resolution to the 'current' requirements throughout your computation. this may give you a performance benefit, as long as your adaptation is both efficient and effective.

linkage between spatial and temporal resolution:

under certain circumstances you have to take account of that. I posed a similar question in response to ganesh. I recall discussing a transonic flutter case, where a shock wave oscillates on an airfoil. if you keep decreasing the time step you will eventually get to a point where the solution from one time level to the next does not change, because the shock displacement is within the cell size. it makes no sense to further reduce the time step.

scaling or no scaling:

if you're smart you'll scale for two reasons: a) non-dimensional computation and data interpretation make your life easy for well known reasons, b) multiple scales lead to stiff systems of equations, and scaling may help to reduce that stiffness.

in response to the more fundamental questions on numerics versus physics:

you'll have to live with the physics you select to model. the challenge is to resolve the physical phenomena you are interested in, without exceeding the necessary resolution (time and cost). hence, your choice of equations, solution method, spatial and temporal resolution, etc., all depend on what you want to get out of it. as an example: if you want to test your unsteady method by evaluating the vortex shedding frequency for a cylinder, your time step should be chosen to resolve that frequency. will that give you all the physics, i.e. resolve all the vortices, get the correct wake dissipation rate, give you an accurate flow field at all points in space-time? not necessarily, but that's not what you were looking for.

ganesh June 12, 2006 09:38

Re: time step and vortex shedding frequency
Dear Mani,

I make use of a Dual time stepping method, which solves the pseudo-steady equations in pseudo-time, which makes things more easier (I can make use of convergence accelaeration devices here, icluding implicit stepping, residual smoothing etc.. , although I do not make use of all available ones). The physical time step which is important, is generally held constant. This time step is the one that decides the temporal accuracy. When I say 75+ steps, I mean 75 outer iterations or in other words 75 physical time steps (The DTS performs what I call "inner iterations" where the implicit time stepping can be employed). This choice of the physical time step has a bearing on both the temporal accuracy and stability and must therefore be chosen with care. This choice depends on the problem and the mesh resolution available. What possibly makes things easy for the standard problems like vortex shedding past cylinder is the knowledge if the shedding frequency, this should not be expected for all real-life problems.

As far as the spatial and temporal resolution are related, my thoughts are as follows. If we consider the very common linear convective equation (which is an unsteady case), and use a simple explicit scheme, we end up with the CFL criterion for stability. With the convective velocity and grid size known, the delt can be obtained. Any delt smaller than this quantity guarantees stability. So, there is nothing wrong when I speak of "temporal convergence" (the choice of time step that guarantees me a given accuracy at the least computational cost). On a given grid, for the transonic flutter problem, decreasing the time step does not any way guarantee "spatial accuracy"(the shock will be well within the grid size if the spactial resolution is good enough or you make use of an adaptive strategy in time), as you seem to point out, all that happens is that after sufficient number of iterations, the solution achieves a certain periodicity. The periodicity is what gets affected if there is a loss in temporal resolution. A grid too coarse to see no features is not useful, and a grid coarse enough to detect some significant activity with a small time step should not pose any stability problem, to my knowledge. In fact, on coarser grids, you can have a larger physical time step compared to the finer grid, much as you would expect for the linear convective problem, delt ~ delx. The time step for the inner iterations for the DTS is same as that while handling any steady-state problem.



diaw June 12, 2006 10:13

Re: time step and vortex shedding frequency
Thanks Mani for the excellent & very complete reply.


diaw June 12, 2006 10:22

Re: time step and vortex shedding frequency
Thanks Mani - excellent thought.

Mani wrote:

To add one question to diaw's list: In choosing an appropriate time step, should we not account for the (smallest) time scales that our spatial resolution is able to handle? Does it make sense to let delta_t go to zero and then speak of convergence, when the grid is fixed? Will there be stability issues, when the grid is too coarse to resolve the unsteady solution at a very small time step?

diaw adds:

If I could add a few pre-questions to Mani's question.

Is there an appropriate time-step for a certain mesh/grid configuration? What happens if our selected time step is too large, or too small for this mesh?

Then afterwards: Is the grid spatial mesh the limiting factor in the simulation, or time-step, or both acting simultaneously?

If the mesh is refined during the simulation, does the time step have to be altered?


I'm going to add in a small twist - "It depends"...

ganesh June 12, 2006 12:38

Re: time step and vortex shedding frequency
Dear Diaw,

The "appropriate" time step for a given grid and proble, should be the one that is stable, yet gives results to desired accuracy. The choice, in general could therefore be a difficult one. A too large time step would result in a lower temporal accuracy, reflected in the results or worse could cause stability issues. A very small time step is undesirable, purely because you spend more time than required in actually getting the solution. The choice for the "optimum" or "appropriate" time step is a tricky issue, though in some cases available information( such as frequency of vortex shedding etc..) could help us make a good decision.

In a unsteady flow simulation, both the spatial and temporal accuracies are important, spatial acuracy would resolve the flow features better and since the problem is time--dependent, the temporal accuracy would have its effect on the simulation.

For a dynamic adaptation, the spatial resolution gets changed during adaptation, I believe that the temporal resolution should also be altered accordingly, though I have not seen this actually being done. This is because the chosen physial time step for the initial mesh could prove unstable for the refined mesh after a few levels of refinement and the decision of optimally altring the time step I believe could be important, which takes us back to square one, the choice of "appropriate time step". For a general problem, the initial choice possibly comes out of prior experience and a general idea of the problem in hand.



Praveen. C June 12, 2006 23:16

Re: time step and vortex shedding frequency
  1. Convergence results are of the form:

    <blockquote> ... the scheme converges to the exact solution as dt->0, dx->0 while the CFL number is held fixed ... </blockquote>

    For an unsteady problem, the error depends both on dt and dx, and it does not make sense to talk of "spatial convergence" and "temporal convergence" separately.
  2. You can talk of temporal resolution alone provided you have taken a sufficiently fine spatial grid. The CFL condition puts an upper bound on the allowable time-step "t<sub>s</sub>. Any time-step smaller than this is fine but as Ganesh wrote, taking a very small time-step is wasteful since then the spatial errors will dominate and you do not gain anything. The problem also has a physical time-scale "t<sub>p</sub>. Your time-step must be smaller than both

    "t &le; min("t<sub>s</sub>, "t<sub>p</sub>)
  3. Of course, "t<sub>p</sub> will vary from problem to problem. Just as we do grid convergence studies, we should also do "t convergence studies. What is an acceptable time step will depend on what quantities you are interested in (integral quantities like lift/drag/heat transfer or detailed spatio-temporal dynamics and structures).
  4. It is better to use a higher order time-integration scheme than to reduce the time-step to a very small value.

diaw June 13, 2006 06:42

Re: time step and vortex shedding frequency
Excellent overviews, Mani, Ganesh & Praveen... good points.

I've tended to work in a sequence as follows:

1. Determine reasonable spatial scale; 2. Select Courant no. of choice; 3. Determine dt upper bound based on courant, dx,dy,dz;

If the dx etc change during run, then determine new appropriate dt.


I agree entirely that all dx, dy, dz, dt must travel to zero - in the limit. I am convinced that along that path, they should also do so in *direct proportion* to each other, if true dynamic relationships are to be retained. If not, then term biase comes into play & we no longer solve the original pde, but a weighted version. In other words, a global scaling rule.


Praveen, you refer to a dt,p. How would you pre-determine this for the problem at hand?


With all the above in mind, & with all dx,dy,dz,dt moving in lockstep... *How* do we determine an appropriate dx, dy, dz for the model at hand?

I'd value debate around this last point particularly, as it is currently something I have been grappling with.


Mani June 13, 2006 12:29

Re: time step and vortex shedding frequency
Good discussion.

Just one more comment:

Praveen (and everyone else, including me) said something like:

> What is an acceptable time step will depend on what quantities you are interested in (integral quantities like lift/drag/heat transfer or detailed spatio-temporal dynamics and structures)

... and that could be a textbook comment. However, to be fair, it's not as easy as we make it seem at first sight. It's hard to tell how much detail you need to predict in a flow in order to even get the integral quantities correctly. In the cylinder example, the drag is linked to the rate at which energy is pumped into the flow and then dissipated. How much detail (in both space and time) is required to get the vortex street right, and predict the right drag? In practice, you will find that not just the grid locally around the cylinder, but also the resolution in the whole wake region, has an impact on the integral quantities obtained on the cylinder surface.

Likewise, the drag will depend on the time step that you choose to resolve bigger or smaller vortex interaction events and dissipation at considerable distance downstream. In the turbulent case, all this becomes more complicated, as you need to model the turbulent dissipation rate accurately (not just in the boundary layers, but in the wake). In the extreme case, i.e. if you were to solve the steady-state equations for this problem, you would end up with a completely wrong result for drag (steady-state is not equal to time-average!), because you neglect all unsteady transport of energy through vortex shedding.

That's why, when I say 30-50 time steps per period are enough, it's not meant as universally applicable rule of thumb for unsteady problems, just specific experience on the laminar circular cylinder with a second-order implicit method.

alessios123 November 19, 2009 07:00

Hi, i would like that some holy man could solve this simple problem: i' m simulating the motion of a fluid post a cilinder of 13 mm diameter. my problem is that i can't create vortex shedding under reynold number of 60000, but we know the the vortex shedding phenomenon begins and it shows better at 80 untill 160 Re...

I put also an initial disturb like a initial velocity in direction perpendicular to the fluid motion, but nothing i can't create the vortex shedding..

i thank everybody who help me..

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