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In a Chinese book, I read about the relationship between the numerical model spatial resolution and the computation cost.

It states: when the spatial resolution of model is finer, the result will be preciser. But to achieve a fine grid resolution, the computation cost will increase. For example, when the grid resolution in a 3-D model framework halves (e.g. from 10 km to 5 km), the computation cost will increase by a factor 24.

Why is this? How can I determine the increase of computation costs when the spatial grid resolution changes?

Update

The model here is a 3-d numerical model for weather forcasting like WRF, MM5, etc.

Update 2

Here is an schematic representation of the impact of model resolution on the calculation of chemical oxidants enter image description here

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  • $\begingroup$ Can you give us specific details about the model being discussed? $\endgroup$ – Isopycnal Oscillation Apr 29 '16 at 18:30
  • $\begingroup$ Are you talking about WRF? If so, then asking the question in forum.wrfforum.com might be easier. $\endgroup$ – arkaia Apr 30 '16 at 2:10
  • $\begingroup$ Thanks for your reply. I think the body of mathematic of general numerical atmospheric model seems to be similar. The spatial resolution halve twice, the computation should consider the variation of 4 dimension(horizontal grid X Y, vertical grid Z and the time period t). $\endgroup$ – Han Zhengzu Apr 30 '16 at 2:52
  • $\begingroup$ For the same simulation area, the computation space keep the same. But the computation frequency would be changed. I think the key factor lieing in the primitive equations. But I can't find any clue to think about yet. $\endgroup$ – Han Zhengzu Apr 30 '16 at 2:58
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I can't answer why there's a 24x increase, according to the textbook - but it may well be the case for a specific model. In general,

If you double the resolution of a 3D model in each dimension, you are multiplying the number of cells (or elements) in the model by $2^3 = 8$.

Typically the maximum length of the time step that a model may be run at and remain stable is proportional to $\frac{\Delta x}{u}$, where $\Delta x$ is the size of a grid cell and $u$ is the speed of the flow in that cell (see info on the CFL condition) (for simplicity I'm assuming that the flow is parallel to one grid axis, which makes this one-dimensional; but the same result applies if not). This means that if you double the spatial resolution, you will probably also have to halve the timestep to maintain numerical stability. This gives an extra factor of two on the computation time, taking us to $2^4 = 16$ times.

This will vary in its details from model to model, and it is entirely feasible that additional non-linear processes in a given model push that to 24 times.

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  • $\begingroup$ Further to Simon's answer, also bear in mind that it is not just the plan grid size that influences the computational time, it is also the number of interactive atmospheric layers in the model. Some of the more sophisticated global climatic models have about 30 vertical layers, adding massively to the number crunching time. $\endgroup$ – Gordon Stanger May 2 '16 at 12:01
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Further to Simon's answer, also bear in mind that it is not just the plan grid size that influences the computational time, it is also the number of interactive atmospheric layers in the model. Some of the more sophisticated global climatic models have about 30 vertical layers, adding massively to the number crunching time.

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  • $\begingroup$ This isn't really an answer. It would be better as a comment on Simon's answer, or perhaps even an edit to it. If you want to keep it, please at least link to Simon's answer, to preserve the thread. $\endgroup$ – kwinkunks May 1 '16 at 23:27
  • $\begingroup$ Also, this is already included in Simon's answer, which discusses doubling the resolution in all three physical dimensions. $\endgroup$ – naught101 May 2 '16 at 5:56

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