0
$\begingroup$

Given 6-hour rainfal data with units in meters(m), how can one convert the 6-hourly rainfall depth in meters(m) to 1-hourly rainfall intensity or rate in kg/m^2/s?

Below is a sample data value to convert.

rainfall depth (m)= 0.0014061

This is what I have tried but not sure it is correct:

To convert the 6h value to the hourly rate, I applied a multiplicative factor:

factor=0.046296296 kg/m2/s

based on the reasoning that the equivalent to:

m (in 6 hours) to mm/s = 1000/21600 (i.e. 0.0462963)

How do I use the multiplicative factor (0.046296296 kg/m2/s) to convert 6-hour rainfall depth to hourly?

Maybe I am thinking too much here.

Any thoughts?

Thank you.

$\endgroup$
4
  • 1
    $\begingroup$ i do not understand your question.rainfall is measured in mm per hour this is the amount of water needed to cover the ground one millimeter deep. $\endgroup$ Jul 8, 2020 at 17:02
  • $\begingroup$ @trondhansen in this case rainfall is issued at 6-hour resolution and in meters(m). How can I convert the rainfall depth to hourly rate in kg/m2/s? $\endgroup$
    – code123
    Jul 8, 2020 at 17:59
  • 1
    $\begingroup$ fao.org/3/r4082e/r4082e05.htm $\endgroup$ Jul 8, 2020 at 21:27
  • 2
    $\begingroup$ As others hint at, it's worth noting that doing this calculation is at best like using the average speed from a roadtrip to estimate speed at any point (your calc is weak but potentially slightly representative for estimating longterm stratiform precipitation)... and quite possibly more like using a 6 hour average speed for Usain Bolt to estimate his sprint speed (i.e. situations dealing with stronger convective precipitation). The whole idea with precision (or sci not) is you don't try to get things from data of greater precision than you have. Really only a 6 hr rate makes useful sense. $\endgroup$ Jul 12, 2020 at 23:42

1 Answer 1

4
$\begingroup$

Your 6-hour rainfall data represent the depth of water fallen in 6 hours. In order to have the average rainfall intensity in m/s corresponding to that interval of time, you have to divide the rainfall depth for the number of seconds in the time interval or for mm/s you multiply for 1000/21600. If you want the intensity in m/h you divide for 6 hours.

You have only the cumulative rainfall in 6 hours so you can estimate the intensity dividing equally the volume of rainfall over the 6 hours. The information remains the same. The real intensity was not constant over the 6 hours, obviously. You cannot, therefore, obtain the real 1-hour rainfall depth for each hours from 6-hours rainfall depth but only an average value that assumes it constant in each hour. If you have the measured 1-hour rainfall depth data in meter and you calculate the intensity in m/s from that data the results will be different.

I hope to have correctly understood your doubts and that this made things clearer.

$\endgroup$
2
  • $\begingroup$ For mm/s do you imply that I should multiply the 6hour rainfall with 1000/21600? Once I get this value I will simply duplicate it at hourly timesteps? $\endgroup$
    – code123
    Jul 9, 2020 at 19:24
  • $\begingroup$ @code123 why do you use meters and kilos per second of rainfall,you are making this a lot more complicated than it needs to be.if it falls 60mm of rain in 6hrs=10mm on average each hour. $\endgroup$ Jul 10, 2020 at 4:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.