I can not understand completely why finite volume method (FVM) is more popular than finite difference method (FDM) in simulation of high-Re, especially for incompressible flows although FVM is less accurate and has a complex formulism. I think maybe, it is because FVM contains more physics than FDM. Is it right?

Hi there,

Finite difference methods are based simply on representing the derivatives of the variables using Taylor expansions. This is an approximation of the equations. In the FVM the advection of the variables (density, momenta) through the grid cells (volumes) is taken directly into account in the writting of the numerical equations. Thus FVM conserves mass and momentum better than the FD methods. When the flow is incompressible and inviscid the conservation depends only on how advection of the variables is solved (since there is no viscous dissipation and density does not change with time). So basically you are right, FVM contains more physics, or more precisely the physics it contains is to conserve advection of the variables through the cells, and advection is the dominant process in incompressible inviscid flows.

Cheers, Patrick.

Hi Patrick,

just a bit of nomenclature and clarification:

1. As you said, if the flow is invicid and incompressible, there are no diffusive fluxes NOT viscous dissipation. Viscous dissipation (mu * Phi) is the volume source to the thermal energy equation due the viscous stresses...which is also zero for the invicid equations but not part of the topic.

2. If the flow is viscous, a conservative treatment of the diffusive fluxes is also possible and is quite easily done with a second order CDS.

3. The main difference between FDM and FVM is:

FDM starts with the differential equation and makes appoximations to the derrivatives either by taylor series or passing curves through the nodes and then taking the derrivative of the profile.

FVM starts with the exact integral equation in conservation form (derivatives are left in terms of the divergence of a flux and not expanded eg. for advection flux of momentum component "i":

dif(rho*u_j*u_i)/dif(x_j) NOT rho*u_j*dif(u_i)/dif(x_j).

The FVM method then applies GAUSS DIVERGENCE Theorem to convert the volume integral of a divergenge term to a surface integral. Approximations are then made on the surface (flux terms) and volume (source and transient terms) integrations in 2 stages.

1. Some quadrature rule for discrete integration at integration points.

2. Closure of the integration point fluxes in terms of the nodal values (which are what we are solving for) for each discrete volume.

If the evaluation is such that the approximation of the fluxes (advective and diffusive) leaving a volume on one side of a surface are the same as those entering the other side......then we have a conservative scheme.

Regards........................................... ...Duane

Hi Duane,

OK let me clarify the situation:

The Dissipation Function (i.e. the source of energy due to dissipation of kinetic energy in the energy equation) itself is obatined by multiplying the viscosity force (in the velocity equation) by the momentum and then adding for each dimension (that is to say that one uses the three momentum equations to generate the source term in the energy equation). Thus, the term viscous dissipation is the general physical process by which one defines the viscous process, and when I wrote no viscous dissipation, I wrote it in that sens (since it would be absurde to have viscous diffusion of momentum - transfer of momentum due to viscosity - if there is no viscous dissipation of the kinetic energy and vice versa). The fact is that in many cases people are just neglecting the treatment of the kinetic energy. However the kinetic energy (EVEN WHEN ONE DOES NOT SOLVE FOR THE ENERGY EQUATION OR DOES NOT INCLUDE THE SOURCE TERM IN THE ENERGY EQUATION) is always dissipated. If this source of energy is not included in the energy equation, then one actually assumes that all this energy is instantly radiated away instead of being absrbed by the flow. However, the viscous dissipation of the kinetic energy always happens since this is what the velocity (momentum) equations implies directly. Thus saying that there is no viscous dissipation implies that there is no viscous force (and therefore no diffusive terms in the momentum equations). I think that the confusion comes from the fact that the SOURCE term in the energy equation is called the DISSIPATION FUNCTION. It is first a source of energy, and its actual value is set to match the loss of kinetic energy that has been dissipated due to the viscosity. I hope I made myself clear.

Thanks for giving more details on FDM and FVm since I am actually using a different method, and my knowledge there is limited.

Patrick

(1). The finite volume method was based on the control volume approach.( you have to know the control volume method first though) (2). So, in principle, you should be able to use only one cell to compute the flow field, because it conserves mass, momentum and energy over that one cell. In this case, the inlet velocity, pressure, temperature etc will be uniform so are the exit variables. ( it could be tough to apply the real wall boundary conditions). (3).In old days, people were limited by the computer memory available ( 512k core memory for main frame computers, and 64k for early PC . Radio-shack model-III has only 48k user's memory). So, you are talking about a 2-D problem of 41x41 size for main frame and 21x21 size for PC. Those mesh sizes were commomly used in research and publications. You could push the mesh size to the limit but at same time it took longer to run. The speed of computer was another limitation in those days. (4). It became easier to use finite volume method to satisfy the global mass conservation. Unless you use the stream function-vorticity approach, there is no guarantee that the mass will be conserved between the exit and the inlet. (5). In those days, the mesh independent solution was not possible. (6). In 90's, when commercial CFD software became widely available ( relatively speaking), it was necessary to focus on the general purpose CFD codes in order to survive in every application fields. It is easier to use finite volume approach to handle complex geometry without coordinate transformation factors, it is also easier to put these cells in the unstructured form. These cells are totally independent of one another before they are assembled together. (similar to Lego toy) It takes a long time just to solve a set of 3-D elliptic equations in order to get a smooth 3-D mesh. (7). So, the finite-volume method development comes from the real world limitation on computer and the survival of the commercial CFD software. (8) No one was really interested in THE SOLUTION OF NAVIER-STOKES EQUATIONS. ( using commercial CFD software is normally one way to cut the CFD development budget. To cut the computing time of a CFD job is another way to save money and to save time to get the product to market. As a result, engineers can only solve a problem with small mesh size. ) (9). This problem has been recognized by ASME and AIAA journals, and for technical publication, the paper must demonstrate the mesh independent properties of the CFD solutions. ( the main concern in this case is not the profit made) (10). So, there are conflicting goals in the real world of CFD. (11). CFD does not cut costs, but it will reveal more detail problems which, when solved, can be very beneficial to the quality of the products.

I feel further clarification of the clarification may be in order because some of the statements seem a bit wonky. I would guess this has arisen by not distinguishing between mechanical and thermal energy but I may be wrong.

You can have viscous forces without dissipation if there is no velocity in the direction of the viscous force. At the wall in a boundary layer there is viscous shear but no viscous dissipation because the fluid is stationary.

An equation for the transport of mechanical energy ( 0.5 rho v^2 ) can be derived by taking the scalar product of velocity and the momentum equation. The term involving the viscous stress, -V.del(T), (the sum of the viscous forces * the (rate of) distance moved in the direction of the viscous forces) represents the transfer of mechanical energy to/from the mean flow due to the action of viscosity. It has two components; a reversible component, -del.[T.V], and an irreversible component, T:del(V). The latter is often termed the dissipation function and represents the transfer of energy to heat. As stated by Duane, it is the transport equation for the thermal component of energy which has the dissipation function as the only source term involving viscosity.

One also cannot state that kinetic energy is always locally reduced (dissipated) by the action of viscosity. Although rare, there are regions of laminar flows where viscosity locally increases the mechanical energy. Perhaps the most common is when a boundary layer encounters an adverse pressure gradient. The peak shear stress, normally located on the wall, moves slightly inboard and the mean flow can gain energy in the region between the wall and the peak shear stress. The effect is more pronounced in turbulent flows and can even energise the flow sufficiently for a noticable bump in the mean flow to develop next to the wall for strongly curved turbulent flows. The fact that viscosity does not always dissipate kinetic energy tends to upset people but is nonetheless true.

To state that a finite volume method contains more physics than a finite difference method is not defensible. For example, consider deriving a discrete representation of the viscous term in the mechanical equation above. Although I have used symbolic notation to split the term into two (a "flux" term and a "source" term to use a very loose terminology) it is the casting of the terms into two that has revealed the physics. If a bit of common sense is used in the differencing to maintain strict numerical conservation for the "flux" term it does not matter whether a finite volume or finite difference approach is used.

To state that a straightforward application of the FV method numerically conserves quantities that are contained within del() is fine (but are these always the quantities you most want to conserve?) To add that this is often obtained at the expense of formal accuracy (in the TSE sense) would be balanced. To say that accuracy in the TSE sense is only relevant for smoothly varying fields and that high Reynolds numbers flows are not noted for their smoothness would also be fair.

Does one express pressure gradients in "FVM" form? How does one treat discrete volumes in a cell centred scheme when evaluating diffusive fluxes? How does one evaluate face normals? Why use QUICK for FD schemes when one can get higher order accuracy with the same stencil? A careful consideration of the numerical properties of the individual terms can only lead to the conclusion that it does not really matter how you derive the discrete form of the equations. It is the properties of the discrete form that matters.

I seem to have ended up in rant mode - apologies.

Only, if the irreversible term you mentioned is zero, in some particular configuartion in an entire flow, then you can say that viscosity does not dissipate kinetice energy. However, if this term is not zero (and most of the time it isn't), then you DO dissipate kinetic energy (even you if don't solve any energy equation and your flow is e.g. polytropic, then it just means that the dissipated kinetid energy is radiated away instantly: it will not affect the pressure but it does affect the velocities). So if you don't want to accept the term viscous dissipation, fine with me.

IN your example with the flow on a wall, I am not sure I follow you. The direction of the viscous force is parallel to the wall ('the wall is trying to slow down the flow') and so is the velocity of the flow. Work is done by the wall by trying to slow down the flow in the boundary layer. Also kinetic energy is introduced at the boundaries of the domain where the imposed boundary conditions imply that one pushes the flow (inflow) on one side and pulls it (outflow) at the other end.

I am not sure I fully grasp all your points or where the residual confusion still lies.

Dissipating kinetic energy does not usually involve any noticable process of radiation except, possibly, at high temperatures and then only indirectly. It is simply a local transfer of energy between components.

The work done by the mean flow against the stresses can be determined from the stress and velocity fields but relatively few numerical schemes are written to numerically conserve it.

You cannot, in general, have a process which affects the velocity field and does not affect the pressure field. Consider the pressure in an incompressible flow which is fully determined (to within a constant) by the velocity field. Anything that changes the velocity field must change the pressure field.

One would normal refer to viscous dissipation of something and, if careful, specify whether that meant locally or globally. In addition, if very careful, it would be made clear whether dissipation was being used to describe the strict reduction of the quantity or a term which was predominantly a sink. So when you state "viscous dissipation" do you mean viscous dissipation of thermal energy or viscous dissipation of mechanical energy (or some other energy component or even a different quantity all together)? In a global or local sense? Pure sink or mainly a sink? Used in isolation I would guess it referred to the local sink of thermal energy. In context I might make some readjustments if the use was consistent.

If something (i.e. wall or fluid) does not move in the direction of an applied force it is not working against (or with) that force. No work is done by a stationary wall to "slow down" a boundary layer. Work is done by the flow against the stress in the boundary layer.

Hi, I feel that slowly we are getting convergence, maybe after another few iterations we'll finally understand each other.

1) When one solves only the velocity equations and the conservation of mass (without an energy equation) it is very usefull to compute the total energy of the system, (for example the global kinetic - integral of rho*v**2 over the entire domain and the same for the thermal energy) to check how good the scheme is to conserve energy, especially for time dependent problems. (It is wise to stop the simulations if you loose more than a few percent). It is right that few schemes are conserving energy, but then one has to be careful about the interpretation of his results if he loses most of its energy because of the scheme. Spectral methods for example are very good at conserving energy (e.g. Cho and Polvani, 1996, Physics of Fluids, vol.8, n.6, pp. 1531-1552, "The emergence of jets and vortices in freely evolving, shallow-water turbulence on a sphere").

2) In the same case as (1) you can also compute the irreversible term (from the 'missing' energy equation). This term will let you know if you dissipate the kinetic energy by viscous processes. Though you are not solving the energy equation, you will find out that you are still dissipating the kinetic energy of the system due to the viscosity - mu.

3) When you are dissipating the kinetic energy of the system, this energy is suppose to go to 'heat' (a source term in the 'missing' energy equation) back the system. If you do not solve for the energy equation (or if you solve for the energy equation but do not include the source term due to the dissipation of the kinetic energy) then you ACTUALLY ASSUME THAT ALL THIS ENERGY IS LOST. This is why I wrote 'radiated away' because this is also a good way to explain it in many physical systems. It is usually called an ADIABATIC process, i.e. it is a quasistatic change of the gas during which not heat is added to the system (dQ=0, in Thermodynamics).

4) The work done by the flow against the stress should be computed, since it a part of total energy that one usually wants to compute to see how good his scheme performs, and how accurate are the results.

5) If you do a work against the wall, because of the law of action and reaction, the wall does the same work against the flow. Since the wall is attached to the earth, the wall-earth system will almost not move. However, if you take the ocean and the friction of the ocean on the surface of the earth (this simple picture is more complicated because of the tidal force of the moon and sun, but let us not go too far), the earth is slowed down by a signficant fraction and the days are actually shorter (significant on the time scale of the life of the earth of course). All this to tell you that the BOUNDARIES CAN BE A SOURCE OR A SINK OF ENERGY for the flow, and that when you calculate the total energy of the system (to check the numerical scheme) you must take these sources/sinks into account.

6) and finally you are right that the change of the velocity will affect the pressure. So if you dissipate kinetic energy, you first affect the velocities, which then affect the pressure (the pressure is not affected directly). This is just a matter of definition, since in the (simulated) flow the change of P happens one time sterp after the change of v. The change due to the source term, however, happens at the same step for P and v. So it is just a matter of how you look at things. I think, from the point of view of Physics, it gives more insight if you look at first the change in v and then in P.

I hope we are getting close to convergence, do you? Cheers, Patrick.

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Hi, I feel that slowly we are getting convergence, maybe after another few iterations we'll finally understand each other.

>>>>>>>>>>>>

We can but hope.

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1) When one solves only the velocity equations and the conservation of mass (without an energy equation) it is very usefull to compute the total energy of the system, (for example the global kinetic - integral of rho*v**2 over the entire domain and the same for the thermal energy) to check how good the scheme is to conserve energy, especially for time dependent problems. (It is wise to stop the simulations if you loose more than a few percent). It is right that few schemes are conserving energy, but then one has to be careful about the interpretation of his results if he loses most of its energy because of the scheme. Spectral methods for example are very good at conserving energy (e.g. Cho and Polvani, 1996, Physics of Fluids, vol.8, n.6, pp. 1531-1552, "The emergence of jets and vortices in freely evolving, shallow-water turbulence on a sphere").

>>>>>>>>>>>>

It can be an excellent check for numerical error if you evaluate the total loss of work in two different ways and compare them. Curiously, it is not a check that commercial packages seem to provide. It may be wrong to read too much into that although I suspect most CFD users performing 3D predictions feel more comfortable without an estimate of the numerical error in their solution.

Not sure about the few percent. Energy is supposed to be dissipated. If you are careful an estimated error around 10% (normalised by inflow dynamic pressure) is pretty good for a complex turbulent flow field. Although it obviously depends the amount of true dissipation.

It is quite straightforward for time varying flows to numerically conserve mass, momentum, vorticity and energy if you restrict yourself to uniform Cartesian grids. There is no need to use spectral methods. These numerical properties are considered in depth by the authors of LES and DNS codes.

>>>>>>>>>>>>

2) In the same case as (1) you can also compute the irreversible term (from the 'missing' energy equation). This term will let you know if you dissipate the kinetic energy by viscous processes. Though you are not solving the energy equation, you will find out that you are still dissipating the kinetic energy of the system due to the viscosity - mu.

>>>>>>>>>>>>

This is one component. Work is also usually done against numerical stresses, turbulent stresses and, depending on the problem, other stresses. Kinetic energy can also be exchanged for pressure rise and it can be transported.

>>>>>>>>>>>>

3) When you are dissipating the kinetic energy of the system, this energy is suppose to go to 'heat' (a source term in the 'missing' energy equation) back the system. If you do not solve for the energy equation (or if you solve for the energy equation but do not include the source term due to the dissipation of the kinetic energy) then you ACTUALLY ASSUME THAT ALL THIS ENERGY IS LOST. This is why I wrote 'radiated away' because this is also a good way to explain it in many physical systems. It is usually called an ADIABATIC process, i.e. it is a quasistatic change of the gas during which not heat is added to the system (dQ=0, in Thermodynamics).

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Within a solution domain, "lost" kinetic energy does not only go to heat (see above) it can be transfered to other components of energy. One does not have to make any assumptions concerning energy loss. The transfer of energy can be determined from the velocity and stress field.

>>>>>>>>>>>>

4) The work done by the flow against the stress should be computed, since it a part of total energy that one usually wants to compute to see how good his scheme performs, and how accurate are the results.

>>>>>>>>>>>>

Agreed. Although, evaluating it by two alternatives methods is even better since it gives a measure of numerical error.

>>>>>>>>>>>>

5) If you do a work against the wall, because of the law of action and reaction, the wall does the same work against the flow. Since the wall is attached to the earth, the wall-earth system will almost not move. However, if you take the ocean and the friction of the ocean on the surface of the earth (this simple picture is more complicated because of the tidal force of the moon and sun, but let us not go too far), the earth is slowed down by a signficant fraction and the days are actually shorter (significant on the time scale of the life of the earth of course). All this to tell you that the BOUNDARIES CAN BE A SOURCE OR A SINK OF ENERGY for the flow, and that when you calculate the total energy of the system (to check the numerical scheme) you must take these sources/sinks into account.

>>>>>>>>>>>>

Ummm. Not really. There appears to be a confusion between force and work.

If you push against a wall it will exert a force back. However, if it does not move then no work has been exchanged no matter how hard you sweat and how much of your energy you use up (dissipated as heat no doubt). Unlike you, the wall is still in the same state and with the same ability to perform work as at the start. No exchange has occured. If you wish to transfer some of your energy to the wall you have got to get it moving.

Work is the product of force and the distance moved in the direction of the force (if the wall moves sideways you are not doing any work on it either). In fluids one tends to consider "rate of this" and "rate of that" so, normally, one considers rate of work (power) as the product of force times rate of distance moved (velocity) in the direction of the force.

You are correct that boundaries are a possible source of work. For example, the belt in a driven cavity exerts work on the fluid. One also needs to account for the difference between the inflow and outflow in an energy balance.

>>>>>>>>>>>>

6) and finally you are right that the change of the velocity will affect the pressure. So if you dissipate kinetic energy, you first affect the velocities, which then affect the pressure (the pressure is not affected directly). This is just a matter of definition, since in the (simulated) flow the change of P happens one time sterp after the change of v. The change due to the source term, however, happens at the same step for P and v. So it is just a matter of how you look at things. I think, from the point of view of Physics, it gives more insight if you look at first the change in v and then in P.

>>>>>>>>>>>>

Ummm.

For incompressible flow, the pressure field at every instant is fully determinable from the velocity field at that instant (within a constant). Any formulation which lags the evaluation of pressure is simply wrong for a time accurate scheme (and a similar view is likely to be faulty as well). For compressible flow things are a bit different if the velocities are comparable to the speed of pressure wave propagation.

A more fruitful line of thought may follow from considering what happens to the applied force and energy components when the fluid moves in the direction of the applied force.

>>>>>>>>>>>>

I hope we are getting close to convergence, do you? Cheers, Patrick.

>>>>>>>>>>>>

I think we are in full agreement.

Sorry, in (3)the flow is truely adiabatic when the dissipated kinetic energy is kept in the flow (only then there is no exchange of energy with the outside, dQ=0). Here it is the opposite, dQ= - the energy dissipated by the viscosity.

Also, in (5) what I really meant was not action reaction but rather conservation of the center of mass of the system flow-container. So if the flow moves in one direction, the container must move in the other, and therefore work is done in this direction.

For the rest I agree in general.

The main point was: there is viscous dissipation of kinetic energy even if you don't solve for the energy equation. Do you agree with that?

Yes. (so long as "dissipation" does not have its usual meaning as a pure sink because the action of viscosity can locally increase the kinetic energy component). With a great effort of will, I shall refrain from commenting on the first two paragraphs.

Unlike Andy, I am unable to let this one pass!

Conservation of WHAT???? The center of mass of the system and container!!!

Can you please draw a free-body diagram for us and apply a conservation law or something that is basic physics to explain this (first year physics)!

It is either explained by the conservation laws in simple terms (if the problem is analyzed properly) or not! There is never a need to invent some new term or law of conservation of quadrumilquellenbits.....leave that for the late night vacuum cleaner and oil addative commercials!

I hope not to offend but meerly sharpen the level of analysis here!

regards......................................Duane

The whole discussion started when I mentioned that viscous dissipation occurs in the flow even if the energy equation is not solved. Somehow, the discussion turned into (in part) example of a 'stationary flow' (of Andy) next to a wall with a boundary layer, where I understood that the claim of Andy was that in this case there is no loss of energy. However, the flow is doing work against the wall since the viscous forces is in the direction of the velocity of the flow (parallel to the flow). The fact is that the wall, if it was free to move and of mass comparable to the mass of the fluid, would move too. If you stand on a large plate with wheels and start to walk in one direction, the plate will roll in the other direction (this can be seen as the conservation of the velocity of the center of mass of the whole system; if initially zero, this has to remain so even after you start walking on the plate, therefore the plate has to move backwards). The same is true for the flow and the wall. Now because of the friction between the wall and the fluid some energy is dissipated and in order to keep the flow moving work is being done constantly. THe main point is that the wall is doing a work to actually slow down the flow (through viscous forces), because if you let a fluid flow on a surface it eventually stops and stagnates. The fact that the wall is usually not moving is because it is attached to the ground, and so the motion of the whole things (wall, ground and earth) becomes infinitesimal and people think then that no work is done, which is wrong. Since the flow is 'stationary' and the wall does not seem to move, I initially was refering rather to action and reaction, which is inexact.

Any more comment there? and please stay calm....

(1). I did not try to read the conversation at all. So, I probably have missed many interesting points. (2). Anyway, I know you are talking about viscous flows, and at the same time, you don't want to solve the energy equation. This could be a problem though. (3). In the momentum equations, there is a viscous term in the equation, and it is associated with the viscosity. As long as the viscosity is non-zero, you have viscous flow. (4). The same thing is true for the energy equation ( the conservation of energy ). There is also a heat conduction term in the energy equation, which is associated with the conductivity. If the conductivity is non-zero, you will have heat transfer. Heat transfer inside the fluid flow and heat transfer through the wall. (5). The effect due to viscosity will result in loss of the total pressure and an increase in entropy. The effect due to condictivity will result in energy transfer in the fluid and through the wall. The energy can be transfered into the fluid through the wall or transfered out of the fluid from the wall. ( not mentioning the inlet and exit boundaries.) (6). As you know that the viscous effect will result in the temperature increase.( you can verify this by rubbing two hands against each other quickly and put your hands on your face to feel it.) So, you can not rule out the energy equation, I mean, you have to say something about it before you can just throw it away. (7). You really need to address the situation when "even if the energy equation is not solved". I mean, is the conservation of energy still satisfied? (that is the job of energy equation.)

Thanks for taking part in the discussion. This discussion is not about a practical problem that we are trying to solve, but only the fact that even when one does not solve the energy equation, there is viscous dissipation implied by the velocity (-momentum) equations.

Andy made no such claim.

Hi, Andy,

as you see probably never converged. What is the description of the flow you mentioned (copied here below). Is that not a flow parallel to a wall with a boundary layer ? What is a stationary flow that you mentioned ?
>
>You can have viscous forces without dissipation if there is >no velocity in the direction of the viscous force. At the >wall in a boundary layer there is viscous shear but no >viscous dissipation because the fluid is stationary.
>

(1). I reviewed your last Fridays message. I think you are right. (2). Maybe, it is less confusing to say that " when there is a viscous diffusion term in the momentum equation, there is always a corresponding viscous dissipation term in the energy equation. Since the magnitude of the viscous dissipation term is relatively small except for the high speed flows, it is normally not included in the calculation of energy equation. But this does not affect the fact that the viscous dissipation (term) is always there.( the physics, but not the mathematics.)" (3). The viscous effect comes from the rotational part of the velocity field ( the vorticity). In the region when the velocity field is uniform, or the vorticity is zero, the viscous effect is zero, so is the viscous dissipation. (4). I have just learned one effective way to start a conversation, that is " trying to use the fuzzy word like implied." (5). I see nothing wrong from your point of view. You may continue your conversation.

(1). Following Clinton's approach, it looks like that the definition of the "viscous dissipation" is not universal. The " implied logic" could be used to explain this difference in opinion. (2). I strongly suggest that a universal definition should be used if "convergence" is the ultimate goal. (3). Well, in the CFD_War Principle, this could be used as "Fuzzy Logic". Black and White is not always "Black and White", I mean, in a war situation. Outside the domain of mathematics, anything can happen. That's the beauty of the forum, what you see is not always what you get. Dont't pay attention to this part, I am just talking to myself. Trying to erase what I have just read.