Cumulus Support

Does anyone know what the exact bucket tip amount is? It cannot be both.

If it is 0.01" per bucket tip, then that is 0.254 mm. Wouldn't that mean each bucket tip is short 0.046 mm for those measuring in millimeters?

If it is 0.3 mm per bucket tip, then that is 1/64 of an inch or 0.01181102362206". Wouldn't that mean each bucket tip is over 0.00181102362206" for those measuring in inches?

I went to Fine Offset's webpage and using this common model, it says:

1) "Rainfall data (inches or millimeters): 1-hour, 24-hour, one week, one month and total since last reset."

2) "Rain volume display: 0 - 9999mm (show OFL if outside range)

Resolution : 0.3mm (if rain volume < 1000mm)

1mm (if rain volume > 1000mm)"

So wouldn't that mean each bucket tip is 0.3 mm? If so, then those reporting in inches need to do an offset of 0.00181102362206" right?

Like John Harrison, I am obsessed with accuracy.

A bucket tip is (nominally) 0.3mm. For people using inches, Cumulus converts the total for any given period by multiplying by 0.0393700787, which is effectively 0.01181102361 in per tip, and then rounding to 2 dp. No calibration setting is required.

I have been slowly converting my logs over to metric values on a separate install of Cumulus. I fire up the other install so it gets imperial units 2-3 times a day. Anyway, I have been taking inches * 25.4 to get mm. Will my rain totals originally in inches be accurate rounded to the tenth?

If I understand what you're doing correctly, you can't get back exactly to mm to 1 dp from inches to 2 dp. For example, 0.01 in is 0.254 mm, which rounds to 0.3 mm, but 0.02 in is 0.508 mm, which rounds to 0.5 mm rather than 0.6.

Maybe this would help - easy to convert to an Excel function.
http://support.microsoft.com/kb/155696

Actually Excel already has a function to do this MROUND()

You need the analysis pack enabled to use this function.

Mark, the issue isn't rounding - it's un-rounding a previously rounded number and then multiplying to amplify the error even further.
I think you need the original tip count. You can calculate that as an intermediate step and use Excel to round to an integer, but it will be +/- 1 tip, which unfortunately doubles the original error tolerance.

Still think mround() will do it...

Let me see if I have this right. Each bucket tip is 0.3 mm and minutely bigger than 0.1" right?

I see MROUND is available in OpenOffice.org Calc so I might try this again with what I have done so far which is just the dayfile.txt and this month's log files.

Are wind run miles and mph values also calculated in a certain way from km? Fahrenheit and Celsius?

1 mile = 1.609344 kilometers
ºC = (ºF - 32) / 1.8

From:

1 kilometer = 0.621371192 miles
ºF = ºC * 1.8 + 32

If it is going to be a lot of work, maybe instead I will just start fresh in metric.

The station supplies wind speed in metres/second and temperature in degrees C. Cumulus converts as required.

steve wrote:The station supplies wind speed in metres/second and temperature in degrees C. Cumulus converts as required.

Using the formulas in my last post?

For temperature, it multiplies by 1.8 and adds 32. For wind speed, it converts from m/s to mph by multiplying by 2.23693629. Wind run is then calculated from the mph figure.

steve wrote:For temperature, it multiplies by 1.8 and adds 32. For wind speed, it converts from m/s to mph by multiplying by 2.23693629. Wind run is then calculated from the mph figure.

So to get the wind right, I would need to divide mph by 2.23693629 to get back to m/s then convert to km/s then multiply by 60 to get the correct km/h reading? There is no time conversion from m/s to mph then? Just the multiplication of the m/s reading to go directly to mph?

To convert from mph to kph directly, just multiply by 1.60934.

A few more questions which relate to the original subject and what I added on to it:

Does it matter what I have the Fine Offset display displaying? If the display is displaying °F, mph, and inches, does this affect Cumulus if it is set to °C, km/h, and mm or vice versa? Does the Fine Offset display actually use metric values, but simply converts them internally to imperial values if selected?