<p>The first equation does not match up with the units. But first things first... What you're calculating when you calculate rain rate is an accumulation rate, which is volume flux through a plane (mass flux divided by density) divided by the area of that plane. For this reason, it doesn't matter whether it is computed over a square meter or a square kilometer -- since volume flux is proportional to the area, it cancels out</p>

<p>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. </p>

<p>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 $mm/s$.</p>

<p>Multiplying this by 3600 to get a result in $mm/hr$ yields the the second expression. </p>

CC BY-SA 3.0
2017-02-02T16:53:40.063Z
2017-02-02T16:53:40.063Z
ssorg
<p>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.</p>

<p>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...</p>

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

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

<p>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$^{-1}$].</p>

<p>Now apparently $R_1 \cdot 0.0036 = R_2$. But why?</p>

<p>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)</p>

CC BY-SA 3.0
2017-02-01T15:58:14.723Z
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
CC BY-SA 3.0
2017-02-02T08:57:51.447Z
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
<meteorology><precipitation><rainfall><radar>
Need help understanding rain rate formulas from drop sizes
jenna
# Need help understanding rain rate formulas from drop sizes

(Asked by: username_0 on 2017-02-01)

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)

**Tags:** meteorology, precipitation, rainfall, radar

> **Comments:**
> **username_1**: 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?  
> **username_0**: 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"...  

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## 1 answers

### Answer by username_2 on 2017-02-02. Score: 3 | Accepted answer

The first equation does not match up with the units. But first things first... What you're calculating when you calculate rain rate is an accumulation rate, which is volume flux through a plane (mass flux divided by density) divided by the area of that plane. For this reason, it doesn't matter whether it is computed over a square meter or a square kilometer -- since volume flux is proportional to the area, it cancels out

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.


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