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?
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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.
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$\begingroup$ do longitude circles become shorter as you move away from the equator or latitude circles ? Longitude circles are great circles. $\endgroup$– user1066Jan 24, 2016 at 6:20
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$\begingroup$ Yes, valid point. Fixed my wording, thank you $\endgroup$ Jan 24, 2016 at 7:22
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$\begingroup$ What are the practical consequences, then, in converting for example a CRU time series of 0.5 degrees latitude/longitude grid cells (see doi.wiley.com/10.1002/joc.3711) into a 5km grid? $\endgroup$ Sep 25, 2019 at 23:05
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$\begingroup$ 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 $\endgroup$ Sep 26, 2019 at 7:18
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$\begingroup$ 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. $\endgroup$ Sep 26, 2019 at 9:29