<p>A model is a simplified representation of a system.</p>
<p>Rather than try to model everything down to the microscopic level, or as vectors, we often aggregate phenomena across geographic areas, to simplify the computation a lot, and to lose only a little accuracy and precision.</p>
<p>So if I want to model the UK, which roughly spans 8°W - 2°E, 50°N - 60°N, that's an area of 10° x 10°. I can split that up into 400 cells, each 0.5°x0.5°. So, starting in the south-west corner, I model the area 8°W-7.5°W x 50°N-50.5°N as a single homogenous area, with no variation within it. And the same for the other 399 cells (8°W-7.5°W x 50.5°N-51°N and so on all the way through to 1.5°E-2°E x 59.5°N-60°N). That would be a 0.5° grid.</p>
<p>Note that the gridcells do not overlap at all, and they completely cover the area of interest. (in mathematical terms, they are exclusive and exhaustive - every point of interest is in exactly one gridcell).</p>
<p>That can create vast amounts of data - particularly if I'm doing this for the whole world, at second-by-second resolution for ten years, for a few hundred variables of interest. And no one will ever need that amount of precision from my results. So I aggregate up to a more coarse grid: so that I take 25 grid squares that form a larger square, and aggregate each of my statistics of interest in them. So now my area 8°W-7.5°W x 50°N-50.5°N is just one of 25 cells in a square in the coarser grid which spans 8°W-5.5°W x 50°N-52.5°N, and I report statistics for 16 cells in a 2.5° grid, rather than 400 cells in a 0.5° grid.</p>
<p>Exactly how that aggregation is done, will vary for each statistic:</p>
<ul>
<li><p>If it's a minimum overnight temperature, then I'll take the minimum across all 25 0.5° gridcells, to give the minimum overnight temperature for the 2.5° grid cell that consists of those 25.</p>
</li>
<li><p>If it's an average rainfall, I'll take an average.</p>
</li>
<li><p>If it's insolation per unit area, and I'm interested in insolation per unit land area, then I'll take a weighted average of insolation, weighting by the proportion of each 0.5° gridcell that contains land.</p>
</li>
</ul>
<p>And then I'll have modelled on a 0.5° grid, and archived on a 2.5° grid</p>
<p>Here, for illustration, is Britain in a 100km grid:</p>
<p><a href="https://i.sstatic.net/9Ujua.gif" rel="noreferrer"><img src="https://i.sstatic.net/9Ujua.gif" alt="enter image description here" /></a> <sub><a href="http://www.colorado.edu/geography/gcraft/notes/coordsys/coordsys.html" rel="noreferrer">source</a></sub></p>
<p>If I wanted to aggregate that to a 200km grid, then the four squares in the bottom-left corner, SQ, SR, SV, SW, would be combined into a single grid cell. And SS, ST, SX, SY would be combined into another grid cell. And so on.</p>

CC BY-SA 3.0
2016-01-05T14:40:19.523Z
Why would you take a weighted average for insolation, but a regular average for rainfall?  You propose an equirectangular grid, which is not equal-area, so any variable for which you calculate the mean would need to be weighed.
gerrit
CC BY-SA 3.0
2016-01-05T08:05:19.137Z
2016-01-05T08:12:16.243Z
2020-06-18T08:25:19.090Z
410 gone
<p>Specifically the term </p>

<blockquote>
  <p>the values are archived on 2.5° grid</p>
</blockquote>

<p>means that the values that are being reported are being saved and are available at that resolution for historical values. The higher resolutions (1°x1°, 1/4°x1/4°) mean exactly what they say: the values are reported on a 2-dimensional grid every degree, or quarter degree, across the area of interest.</p>

CC BY-SA 3.0
2016-01-05T04:17:36.413Z
2016-01-05T04:17:36.413Z
Ben MS
<p>To add a bit more mathematical conception to the idea of a grid: think about a 2-D grid(X,Y) as a matrix: this matrix represents a solution to the system of equations you are trying to solve at grid point x,y (individual points on X,Y respectively). </p>

<p>Why is it important to make this distinction? Because knowing the resolution, span, regularity and structure of your grid will determine how you set up your numerical process in order to get reliable numerical solutions. In particular, the boundary conditions are especially sensitive to the grid, which can be difficult to properly place depending on the type of numerical process you are using. Furthermore, in heavy numerical tasks, your grid can often determine how long it takes for your model to resolve (if it even can)!</p>

<p>So yes, a grid is how spatial data is organized and represented, but more importantly, it is fundamentally where the solutions are calculated and live. A well resolved, thought out, and executed grid is fundamental to any scientific study using numerical methods in their argument. </p>

CC BY-SA 3.0
2016-01-05T08:36:09.380Z
2016-01-05T08:36:09.380Z
Neo
<p>I have just begun reading some notes about modelling and can't seem to understand what a 1°x1° grid or a 2.5° grid or 1/4° grid is. I know it is something related to the resolution of the area over which the model is run. But when a line goes like this, 'the values are archived on 2.5° grid', what exactly should I make of it?</p>

<models><atmosphere-modelling><climate-models><numerical-modelling><ocean-models>
What exactly is a model grid?
Darkgod
# What exactly is a model grid?

(Asked by: username_0 on 2016-01-05)

I have just begun reading some notes about modelling and can't seem to understand what a 1°x1° grid or a 2.5° grid or 1/4° grid is. I know it is something related to the resolution of the area over which the model is run. But when a line goes like this, 'the values are archived on 2.5° grid', what exactly should I make of it?

**Tags:** models, atmosphere-modelling, climate-models, numerical-modelling, ocean-models


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## 3 answers

### Answer by username_1 on 2016-01-05. Score: 11

A model is a simplified representation of a system.

Rather than try to model everything down to the microscopic level, or as vectors, we often aggregate phenomena across geographic areas, to simplify the computation a lot, and to lose only a little accuracy and precision.

So if I want to model the UK, which roughly spans 8°W - 2°E, 50°N - 60°N, that's an area of 10° x 10°. I can split that up into 400 cells, each 0.5°x0.5°. So, starting in the south-west corner, I model the area 8°W-7.5°W x 50°N-50.5°N as a single homogenous area, with no variation within it. And the same for the other 399 cells (8°W-7.5°W x 50.5°N-51°N and so on all the way through to 1.5°E-2°E x 59.5°N-60°N). That would be a 0.5° grid.

Note that the gridcells do not overlap at all, and they completely cover the area of interest. (in mathematical terms, they are exclusive and exhaustive - every point of interest is in exactly one gridcell).

That can create vast amounts of data - particularly if I'm doing this for the whole world, at second-by-second resolution for ten years, for a few hundred variables of interest. And no one will ever need that amount of precision from my results. So I aggregate up to a more coarse grid: so that I take 25 grid squares that form a larger square, and aggregate each of my statistics of interest in them. So now my area 8°W-7.5°W x 50°N-50.5°N is just one of 25 cells in a square in the coarser grid which spans 8°W-5.5°W x 50°N-52.5°N, and I report statistics for 16 cells in a 2.5° grid, rather than 400 cells in a 0.5° grid.

Exactly how that aggregation is done, will vary for each statistic:

* If it's a minimum overnight temperature, then I'll take the minimum across all 25 0.5° gridcells, to give the minimum overnight temperature for the 2.5° grid cell that consists of those 25.
* If it's an average rainfall, I'll take an average.
* If it's insolation per unit area, and I'm interested in insolation per unit land area, then I'll take a weighted average of insolation, weighting by the proportion of each 0.5° gridcell that contains land.

And then I'll have modelled on a 0.5° grid, and archived on a 2.5° grid

Here, for illustration, is Britain in a 100km grid:

[![enter image description here](https://i.sstatic.net/9Ujua.gif)](https://i.sstatic.net/9Ujua.gif) [source](http://www.colorado.edu/geography/gcraft/notes/coordsys/coordsys.html)

If I wanted to aggregate that to a 200km grid, then the four squares in the bottom-left corner, SQ, SR, SV, SW, would be combined into a single grid cell. And SS, ST, SX, SY would be combined into another grid cell. And so on.

> **Comments:**
> **username_2**: Why would you take a weighted average for insolation, but a regular average for rainfall?  You propose an equirectangular grid, which is not equal-area, so any variable for which you calculate the mean would need to be weighed.  

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### Answer by username_3 on 2016-01-05. Score: 5

Specifically the term

> the values are archived on 2.5° grid

means that the values that are being reported are being saved and are available at that resolution for historical values. The higher resolutions (1°x1°, 1/4°x1/4°) mean exactly what they say: the values are reported on a 2-dimensional grid every degree, or quarter degree, across the area of interest.


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### Answer by username_4 on 2016-01-05. Score: 5

To add a bit more mathematical conception to the idea of a grid: think about a 2-D grid(X,Y) as a matrix: this matrix represents a solution to the system of equations you are trying to solve at grid point x,y (individual points on X,Y respectively).

Why is it important to make this distinction? Because knowing the resolution, span, regularity and structure of your grid will determine how you set up your numerical process in order to get reliable numerical solutions. In particular, the boundary conditions are especially sensitive to the grid, which can be difficult to properly place depending on the type of numerical process you are using. Furthermore, in heavy numerical tasks, your grid can often determine how long it takes for your model to resolve (if it even can)!

So yes, a grid is how spatial data is organized and represented, but more importantly, it is fundamentally where the solutions are calculated and live. A well resolved, thought out, and executed grid is fundamental to any scientific study using numerical methods in their argument.


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