# How can I convert a CERA-20C total precipitation dataset of "monthly means of daily means' (edmo) to mm/month?

I downloaded an edmo dataset of 1 ensemble of the cera-20c model. I want to convert these data, which are in [m], to mm/month. Further specifications: 0.125/0.125 grid, -29/30/-30/31 area. Example of the data (see picture) of total precipitation January 1980:

My method: Taking the average of these numbers multiplying by 31 (days in January) and 1000 (from m to mm) gives me 299 mm for January. But this does not seem right. It seems overestimated. I hope you can help me with the right answer. I read something about multiplying by the scale factor and adding the add_offset. But the scale factor is so small (10E-7) that all numbers approximately similar.

Thanks a lot in advance for any ideas. Louwrens Timmer

• Retrieving the necessary values should be straightforward, but a few additional details are needed to answer this. Presumably these are the values for some subset of the lat/lon grid? Also, you mention the scale factor, but what is the add_offset? This is a pretty common method of storing meteorological data, but is there any other documentation listed describing the scaling or data structure? – dplmmr Jan 14 at 1:21
• May be helpful to both future searchers and to anyone trying to help with the question if you provide a direct link to the dataset you're using to aid with clarity? :-) – JeopardyTempest Jan 14 at 3:31
• A more or less similar question was asked earlier: earthscience.stackexchange.com/questions/7055/… – L. Timmer Jan 14 at 8:37
• @JeopardyTempest ecmwf.int/en/forecasts/datasets/archive-datasets/… this is the link to the documentation page of the CERA-20C. – L. Timmer Jan 14 at 9:35
• @dplmmr The dataset is downloaded as a netcdf file. But I extracted the data in an excel file format to see how the data is structured and to make the first calculations. Further describing documentation: dimensions: longitude = 9; latitude = 9; time = UNLIMITED variables: double tp(time=12, latitude=9, longitude=9); :scale_factor = 2.0893531189079316E-7; // double :add_offset = 0.006947661364936322; // double :_FillValue = -32767S; // short :missing_value = -32767S; // short :units = "m"; :long_name = "Total precipitation"; – L. Timmer Jan 14 at 11:40