How can I account for the area of latitude longitude grid cells in a matrix for precipitation data?

Hi everyone!

I am trying to look at precipitation over South America and need to account for the size of the latitude longitude grid box size. My dataset is 0.25 degree by 0.25 degree resolution, but different areas of would mean that 1 degree covers a larger area. Does anyone know what I can do to account for this? From what I can remember I may need to multiple my data by cos...but I cannot remember exactly what I need to do.

If you have any ideas or need more information from me please let me know!

Thank you!

[Mathieu NOE]

Sorry , I am not sure I get it , is there a missing piece in your second sentence ?

but different areas of what ? would mean that 1 degree covers a larger area.

does your question relates to normalize precipitation qty per unit surface ?

[James Karden]

I'm sorry I wasn't clearer in my first message! But yes I am trying to normalize precipitation qty per unit surface. Because each 0.25 by 0.25 degree box would mean different surface area (i.e. in sq km) I am trying to find the specific area of each lat x lon grid cell and multiply it to my precipitation data to make sure each area is counted as the proportional size...if that makes sense!

[Mathieu NOE]

sure

if your data are sampled by long and lat angles (in deg) , and you would like to compute the area to normalize the precipitation data,

area dS = product of dx by dy

knowing the radius R of earth , if your deg resolution is 0.25 in both directions (must be first converted in radians) :

dx = 0.25*pi/180 * R

dy = 0.25*pi/180 * R

dS = dx * dy;

[James Karden]

How could I do this for a lat x lon matrix that is 240 x 203?

[Mathieu NOE]

what is the deg resolution in each direction ?

[James Karden]

0.25 by 0.25 degrees

[Mathieu NOE]

the size of your matrix has no importance , as you normalize the data by dS which is scalar

example below (dummy data) :

rain_qty = 100*rand(240,203);   % rain qty (units TBD) ; lat x lon matrix
R = 6371; % earth radius (km)

% knowing the radius R of earth ,
% if your deg resolution is 0.25 in both directions (must be first converted in radians) : 
dx = 0.25*pi/180 * R;
dy = 0.25*pi/180 * R;
dS = dx * dy;   % unit surface in km²

% rain qty per unit surface 
rain_qty_per_unit_surface = rain_qty/dS

[James Karden]

Great, thank you very much!

[Mathieu NOE]

you're welcome