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)