Cumulus Support

It doesn't matter to Cumulus what you have the console set to, the console always supplies the data in metric and Cumulus converts as required.

Cool beans! Thanks for the answers.

mcrossley wrote:Still think mround() will do it...

Code: Select all

mm	inches	mm	round(mm)	mround(mm,0.3)
0.0	0.00	0.0000 	0.0	0.0
0.3	0.01	7.6200 	0.3	0.3
0.6	0.02	15.2400	0.5	0.6
0.9	0.04	22.8600	1.0	0.9
1.2	0.05	30.4800	1.3	1.2
1.5	0.06	38.1000	1.5	1.5
1.8	0.07	45.7200	1.8	1.8
2.1	0.08	53.3400	2.0	2.1
2.4	0.09	60.9600	2.3	2.4
2.7	0.11	68.5800	2.8	2.7
3.0	0.12	76.2000	3.0	3.0
3.3	0.13	83.8200	3.3	3.3

I am not sure how to use this properly in a spreadsheet. I figured I could copy/paste a column from a log file into a new spreadsheet in Column A then do MROUND in Column B.

B1
=MROUND(A1;0.3)

B2
=MROUND(A2;0.3)

B3
=MROUND(A3;0.3)

And so forth. I am trying this with the rainfall rates and it doesn't like 0.05, 0.09, 0.12, or similar. Maybe I need to convert Column B to bucket tips to determine mm by taking A * 100 then * 0.3.

B1
=((A1*100)*0.3)

B2
=((A2*100)*0.3)

B3
=((A3*100)*0.3)

These seem to work for converting in to bucket tips to mm.

I am not sure yet how to convert in/hr to mm/hr.

EDIT:

No I cannot do it that way as the in values are calculated so perhaps 3.54" is not really 354 bucket tips.

Yes, you need to convert from inches to mm first, then mround.

So if you had inches in column A, in column B you would have (in Excel)

mcrossley wrote:Yes, you need to convert from inches to mm first, then mround.

So if you had inches in column A, in column B you would have (in Excel)

Code: Select all

    A       B
1   inch    mm
2   0.09    =mround(A2*25.4, 0.3)
3   0.12    =mround(A3*25.4, 0.3)

Would this also work correctly for converting in/hr to mm/hr?