# Need help understanding rain rate formulas from drop sizes

I have come across different formulas by which the rain rate [mm/s] or [mm/h] can be computed from the drop size distribution. I'm somehow unable to convert the different formulas into each other. It is maybe a stupid question with some unit error, but I cannot find it.

A well-known formula seems: \begin{equation}R_1 = \frac{\pi}{6} \int_0^{\infty} v(D) D^3 N(D) dD \end{equation} This should be the instantaneous rain rate in [mm/s]?? Unfortunately most publications just quote the formula without units...

I'm supposing that $D$ is in [mm], $N(D)$ in [mm$^{-1}$ m$^{-3}$], and $v(D)$ in [m s$^{-1}$].

$v(D)$ is usually approximated as $3.778\cdot D^{0.67}$.

Another formula I have come across is \begin{equation} R_2 = 6\pi \cdot 10^{-4} \int_0^{\infty} v(D) D^3 N(D) dD, \end{equation} which is said to be [mm/h], then. Also for $D$ in [mm], $N(D)$ in [mm$^{-1}$ m$^{-3}$], and $v(D)$ in [m s$^{-1}$].

Now apparently $R_1 \cdot 0.0036 = R_2$. But why?

If $R_1$ is in [mm/s], then we should multiply by 3600 to get to [mm/h]. Can someone spot where I am wrong? Is it right that $R_1$ is in [mm/s]? (given the units of D, N(D) as above)

• The '36' in 0.0036 looks like a conversion between seconds and hours - as you say. Since 0.0036 = 3600 * 10^-6, we are missing some other unit conversion, for which we need 10^6 (or 10^-6). This could be 'km <-> mm'. Might v(D) in the second equation be given in km/h? – daniel.heydebreck Feb 1 '17 at 15:58
• v(D) is -as far as I can see from publications - always in [m/s]... at the moment I'm thinking it can have something to do with a reference area? maybe the rainrates are to be understood like "[mm/h] in one square meter " or "[mm/h] in one square centimeter"... – jenna Feb 2 '17 at 8:57

Also it doesn't really matter whether $N(D)$ has units of $mm^{-1} m^{-3}$ or $cm^{-1} m^{-3}$, since that first unit cancels out in integration.
If the units are as you say they are in the first example, then the expression should read $$R_1 = 10^{-6}\frac{\pi}{6} \int_0^\infty v(D)D^3N(D)dD$$ to get a result in $mm/s$.
Multiplying this by 3600 to get a result in $mm/hr$ yields the the second expression.