<p>Assuming spherical shape, Earth's circumference is about $40,000\mathrm{km}$. These are split into $360^\circ$, so one degree is a bit more than $100\mathrm{km}$. This is true for latitudes. For longitudes, the circles of constant latitude are shorter as you move away from the equator. Therefore you need to multiply that number by $\cos(latitude)$. To summarize, if $R_e$ is Earth's radius, we got</p>

<p>$$\Delta y=2\pi R_e\frac{\Delta lat}{360^\circ}$$</p>

<p>and</p>

<p>$$\Delta x=2\pi R_e\cos(lat)\frac{\Delta lon}{360^\circ}.$$
Note that here I am using $\Delta x$ and $\Delta y$ for the resolution in a cartesian frame with $y$ pointing south-north and $x$ pointing west-east. For your numbers, think about it as around $50\mathrm{km}$ resolution.</p>

CC BY-SA 3.0
2016-01-24T07:22:04.610Z
Yes, valid point. Fixed my wording, thank you
Christoph
CC BY-SA 4.0
2019-09-25T23:05:02.427Z
What are the practical consequences, then, in converting for example a CRU time series of 0.5 degrees latitude/longitude grid cells (see http://doi.wiley.com/10.1002/joc.3711) into a 5km grid?
Nikos Alexandris
CC BY-SA 4.0
2019-09-26T07:18:01.903Z
That depends on what you want to do with the result. Only 1 in about 100 grid cells will contain new information. So keep that in mind when analysing the data
Christoph
CC BY-SA 4.0
2019-09-26T09:29:22.927Z
In fact, I want to remain in the latitude/longitude world. The task is to go from a 360 x 720 to a 2987 x 7200 (rows, columns) grid. Maybe I am going off-topic and should investigate elsewhere, since this _is_ an interpolation question. But I'd still would like to better grasp the "1 in about 100 grid cells will contain new information" statement.
Nikos Alexandris
CC BY-SA 4.0
2019-09-26T09:42:10.350Z
Well, simple interpolation does not give you more information. It just makes things look smoother. So going from 50x50km to 5x5km leaves you with that statement. There are some intelligent approaches to the interpolation problem (like taking terrain into account) that can improve your situation there. But that is a whole different story
Christoph
CC BY-SA 4.0
2019-09-26T10:23:47.943Z
Thank you @Christoph. Initially I was curious about potential other implications (than interpolation issues), which I named "practical consequences". I was then, however, curious (especially) about the "new information" part. Going from ~50km to ~5km, no matter the approach, will always come with a loss/cost/alteration of the original data. Maybe you meant to write "noise" or "artefacts", or instead write "loss" (or "bias" or similar) of information"?
Nikos Alexandris
CC BY-SA 3.0
2016-01-23T17:19:40.883Z
2016-01-24T07:21:29.463Z
2016-01-24T07:21:29.463Z
Christoph
<p>I am curious how can one have an idea of grid resolution. How much is the distance resolution in km of a grid with resolution of 0.5 deg x 0.666 deg?</p>

<models><grid-spacing>
Converting grid resolution from degrees to kilometers
Jahangir
# Converting grid resolution from degrees to kilometers

(Asked by: username_0 on 2016-01-23)

I am curious how can one have an idea of grid resolution. How much is the distance resolution in km of a grid with resolution of 0.5 deg x 0.666 deg?

**Tags:** models, grid-spacing


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## 1 answers

### Answer by username_1 on 2016-01-23. Score: 11 | Accepted answer

Assuming spherical shape, Earth's circumference is about $40,000\mathrm{km}$. These are split into $360^\circ$, so one degree is a bit more than $100\mathrm{km}$. This is true for latitudes. For longitudes, the circles of constant latitude are shorter as you move away from the equator. Therefore you need to multiply that number by $\cos(latitude)$. To summarize, if $R\_e$ is Earth's radius, we got

$$\Delta y=2\pi R\_e\frac{\Delta lat}{360^\circ}$$

and

$$\Delta x=2\pi R\_e\cos(lat)\frac{\Delta lon}{360^\circ}.$$
Note that here I am using $\Delta x$ and $\Delta y$ for the resolution in a cartesian frame with $y$ pointing south-north and $x$ pointing west-east. For your numbers, think about it as around $50\mathrm{km}$ resolution.

> **Comments:**
> **username_1**: Yes, valid point. Fixed my wording, thank you  
> **username_2**: What are the practical consequences, then, in converting for example a CRU time series of 0.5 degrees latitude/longitude grid cells (see http://doi.wiley.com/10.1002/joc.3711) into a 5km grid?  
> **username_1**: That depends on what you want to do with the result. Only 1 in about 100 grid cells will contain new information. So keep that in mind when analysing the data  
> **username_2**: In fact, I want to remain in the latitude/longitude world. The task is to go from a 360 x 720 to a 2987 x 7200 (rows, columns) grid. Maybe I am going off-topic and should investigate elsewhere, since this _is_ an interpolation question. But I'd still would like to better grasp the "1 in about 100 grid cells will contain new information" statement.  
> **username_1**: Well, simple interpolation does not give you more information. It just makes things look smoother. So going from 50x50km to 5x5km leaves you with that statement. There are some intelligent approaches to the interpolation problem (like taking terrain into account) that can improve your situation there. But that is a whole different story  
> **username_2**: Thank you @username_1. Initially I was curious about potential other implications (than interpolation issues), which I named "practical consequences". I was then, however, curious (especially) about the "new information" part. Going from ~50km to ~5km, no matter the approach, will always come with a loss/cost/alteration of the original data. Maybe you meant to write "noise" or "artefacts", or instead write "loss" (or "bias" or similar) of information"?  

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