<p>Omega, ω is closely related to w in meteorology.
It can be moved towards w using the chain (<a href="http://www.sjsu.edu/faculty/watkins/omega.htm" rel="noreferrer">this reminder</a> from Watkins at SJSU helped)...</p>

<pre><code>ω = dp/dt = (dp/dz)(dz/dt) = (dp/dz)w 
</code></pre>

<p><sup><em>[where p is pressure, t is time, z is height, and w is vertical motion in height coordinates]</em></sup></p>

<p>Often in general meteorology you can estimate dp/dz using the <a href="https://www.e-education.psu.edu/meteo300/node/7" rel="noreferrer">hydrostatic approximation</a>, which assumes gravity and buoyancy approximately balance.  This allows the equations:</p>

<pre><code>dp/dz = −ρg  
</code></pre>

<p><sup><em>[where ρ is air density and g is gravitational acceleration]</em></sup></p>

<p>And so would mean you can continue</p>

<pre><code>ω = (dp/dz)w = −ρgw
</code></pre>

<p>which results in <strong>w = −ω/ρg</strong>.</p>

<p><strong>Note that this is not always valid.</strong>  As <a href="https://www.google.com/url?sa=t&amp;rct=j&amp;q=&amp;esrc=s&amp;source=web&amp;cd=1&amp;cad=rja&amp;uact=8&amp;ved=0ahUKEwjJmqzPudDRAhUJLSYKHcarChQQFggdMAA&amp;url=https%3A%2F%2Fwww.amazon.com%2FIntroduction-Dynamic-Meteorology-International-Geophysics%2Fdp%2F0123848660&amp;usg=AFQjCNGdSbK_Pfo2P8HbJIyT5NckdVTUpg&amp;sig2=7CCPxD8f00nU6vubKHTuJw" rel="noreferrer">Holton's An Introduction to Dynamic Meteorology, Volume 1</a> suggests on <a href="https://books.google.com/books?id=-ePQ6x6VbjgC&amp;pg=PA202&amp;lpg=PA202&amp;dq=estimating%20w%20from%20omega%20qg&amp;source=bl&amp;ots=l3lT-YIsiq&amp;sig=aOb6ZXDWv7f9Vp_IGUgxtqYDgWM&amp;hl=en&amp;sa=X&amp;ved=0ahUKEwjk3rnptdDRAhXLRiYKHTzRA1Q4ChDoAQgoMAI#v=onepage&amp;q=estimating%20w%20from%20omega%20qg&amp;f=false" rel="noreferrer">page 20</a>:</p>

<blockquote>
  <p>This condition of hydrostatic balance provides an excellent
  approximation for the vertical dependence of the pressure field in the
  real atmosphere.  Only for intense small-scale systems such as squall
  lines and tornadoes is it necessary to consider departures from
  hydrostatic balance.</p>
</blockquote>

<p>That's a loose idea.  It's been a while, but basically as I remember it, hydrostatic is strongly valid when vertical acceleration (and thus vertical motion when looking near the surface) is much smaller in scale compared to the horizontal motions/accelerations.  So indeed, this is not true in tornadoes and in convection.  So you get very correct values for strong thunderstorms.  But it will help you diagnose regions of rising motion.</p>

<p>You can look a bit more towards breaking it down into more advanced detail and the mathematics of hydrostatic and how it fails in Doswell and Markowski's <a href="http://www.cimms.ou.edu/~doswell/relbuoy/relbuoy.html" rel="noreferrer">paper</a> on buoyancy.</p>

<p>But, with such limited model output, not sure that you're looking to get into all of that detail, as I don't believe you'll find any extra help going those avenues with the variables you have.</p>

<p>With just slightly more information, most importantly temperature, you could use the <a href="https://en.wikipedia.org/wiki/Ideal_gas_law" rel="noreferrer">ideal gas law</a>:</p>

<pre><code> p = ρRT
</code></pre>

<p><sup><em>[where R is the ideal gas constant of the gas under consideration and T is temperature in Kelvin]</em></sup></p>

<p>This would allow you replace the unknown density, and you'd then end up with:</p>

<p><strong>w = −ωRT/pg</strong> </p>

<p>Which you could then either further approximate R to be that for dry air = Rd = 287 J/Kkg, or even work up the adjustment your moisture content variable causes (though don't think doing so would offer much useful improvement as you've already got greater errors due to the failings of the hydrostatic approximation).</p>

<p>Since you don't have temperature, it seems your only remaining option would be to plug in the density approximation for the level of interest from the approximated "Standard Atmosphere" <a href="http://www.engineeringtoolbox.com/standard-atmosphere-d_604.html" rel="noreferrer">here</a> (use the table where ρ is in units of kg/m^3).</p>

<p>It has been a long while since I've done this stuff, but I don't see any better options.  Unless somehow there's a way to work from the geopotential, maybe apply something like mass continuity in pressure coordinates, and get to something more complex to return something more correct for w.  But if there is, it outpaces me!</p>

<p><strong>So, long story short, it appears that your best hope is w = −ωRT/pg, but it won't be 100% reliable, particularly in strong convection.</strong></p>

CC BY-SA 3.0
2017-01-21T00:49:06.830Z
Been a while, like I noted... are you suggesting that the model would need a balance to have calculated omega?  He said he's got omega.  If you are saying the model is using that, I could see that being true given the limited output... but I'm thinking you can get omega from a full primitive equations model in pressure coordinates just fine without any balance assumptions?
JeopardyTempest
CC BY-SA 3.0
2017-01-20T11:07:59.483Z
2017-01-21T00:52:39.723Z
2017-01-21T00:52:39.723Z
JeopardyTempest
<p>My model (AGCM) does not compute w-wind (the vertical component of the wind velocity). However, I do have the following available variables from AGCM output:</p>
<pre><code>u-wind, v-wind, geopotential, omega and some other variables (e.g.,
temperature, specific humidity) at 23 pressure levels.
</code></pre>
<p>where u-wind and v-wind are the horizontal components of the wind velocity, <a href="https://en.wikipedia.org/wiki/Geopotential" rel="nofollow noreferrer">geopotential</a> is the potential of Earth's gravity, and omega is the solution to the <a href="https://en.wikipedia.org/wiki/Omega_equation" rel="nofollow noreferrer">omega equation</a>.</p>
<p>I would like to compute w-wind by using u-wind, v-wind, geopotential, omega. As suggested by <em>JeopardyTempest</em> in the answer below, I can compute w-wind using the equation <strong>w = −ω/ρg</strong>  . How reliable it would be if I don't use <strong>w = −ωRT/pg</strong>?</p>

CC BY-SA 3.0
2017-01-20T07:26:21.743Z
Please add more context such as describing the model that you are using and what physical process you are attempting to model.
Isopycnal Oscillation
CC BY-SA 3.0
2017-01-23T23:59:58.510Z
I'm not sure you'll find sharper answers (but you will), but like I suggested, regionally, −ωRT/pg should be very solid, it just will have large local areas where it overlooks the greater velocities in stronger convective updrafts.
JeopardyTempest
CC BY-SA 3.0
2017-01-24T00:02:59.170Z
In terms of using the standard atmosphere ρ instead of T from the model... since it's in Kelvin, it's not going to be a drastic impact.  I'd think T is unlikely to vary by more than about 30 or 40 C.  So that's maybe 20% of the standard atmosphere.  And that'd be in very rare circumstances.  You're probably talking a typical 5-10% error maybe more regularly?  Hope that helps make sense of it, I'm certainly no modeler, I know there must be people who can give better answers
JeopardyTempest
<climate><models><wind><climate-models>
How to calculate w-wind?
Kay
# How to calculate w-wind?

(Asked by: username_0 on 2017-01-20)

My model (AGCM) does not compute w-wind (the vertical component of the wind velocity). However, I do have the following available variables from AGCM output:

```
u-wind, v-wind, geopotential, omega and some other variables (e.g.,
temperature, specific humidity) at 23 pressure levels.

```

where u-wind and v-wind are the horizontal components of the wind velocity, [geopotential](https://en.wikipedia.org/wiki/Geopotential) is the potential of Earth's gravity, and omega is the solution to the [omega equation](https://en.wikipedia.org/wiki/Omega_equation).

I would like to compute w-wind by using u-wind, v-wind, geopotential, omega. As suggested by *username_2* in the answer below, I can compute w-wind using the equation **w = −ω/ρg** . How reliable it would be if I don't use **w = −ωRT/pg**?

**Tags:** climate, models, wind, climate-models

> **Comments:**
> **username_1**: Please add more context such as describing the model that you are using and what physical process you are attempting to model.  
> **username_2**: I'm not sure you'll find sharper answers (but you will), but like I suggested, regionally, −ωRT/pg should be very solid, it just will have large local areas where it overlooks the greater velocities in stronger convective updrafts.  
> **username_2**: In terms of using the standard atmosphere ρ instead of T from the model... since it's in Kelvin, it's not going to be a drastic impact.  I'd think T is unlikely to vary by more than about 30 or 40 C.  So that's maybe 20% of the standard atmosphere.  And that'd be in very rare circumstances.  You're probably talking a typical 5-10% error maybe more regularly?  Hope that helps make sense of it, I'm certainly no modeler, I know there must be people who can give better answers  

---

## 1 answers

### Answer by username_2 on 2017-01-20. Score: 10 | Accepted answer

Omega, ω is closely related to w in meteorology.
It can be moved towards w using the chain ([this reminder](http://www.sjsu.edu/faculty/watkins/omega.htm) from Watkins at SJSU helped)...

```
ω = dp/dt = (dp/dz)(dz/dt) = (dp/dz)w 

```

*[where p is pressure, t is time, z is height, and w is vertical motion in height coordinates]*

Often in general meteorology you can estimate dp/dz using the [hydrostatic approximation](https://www.e-education.psu.edu/meteo300/node/7), which assumes gravity and buoyancy approximately balance. This allows the equations:

```
dp/dz = −ρg  

```

*[where ρ is air density and g is gravitational acceleration]*

And so would mean you can continue

```
ω = (dp/dz)w = −ρgw

```

which results in **w = −ω/ρg**.

**Note that this is not always valid.** As [Holton's An Introduction to Dynamic Meteorology, Volume 1](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0ahUKEwjJmqzPudDRAhUJLSYKHcarChQQFggdMAA&url=https%3A%2F%2Fwww.amazon.com%2FIntroduction-Dynamic-Meteorology-International-Geophysics%2Fdp%2F0123848660&usg=AFQjCNGdSbK_Pfo2P8HbJIyT5NckdVTUpg&sig2=7CCPxD8f00nU6vubKHTuJw) suggests on [page 20](https://books.google.com/books?id=-ePQ6x6VbjgC&pg=PA202&lpg=PA202&dq=estimating%20w%20from%20omega%20qg&source=bl&ots=l3lT-YIsiq&sig=aOb6ZXDWv7f9Vp_IGUgxtqYDgWM&hl=en&sa=X&ved=0ahUKEwjk3rnptdDRAhXLRiYKHTzRA1Q4ChDoAQgoMAI#v=onepage&q=estimating%20w%20from%20omega%20qg&f=false):

> This condition of hydrostatic balance provides an excellent
> approximation for the vertical dependence of the pressure field in the
> real atmosphere. Only for intense small-scale systems such as squall
> lines and tornadoes is it necessary to consider departures from
> hydrostatic balance.

That's a loose idea. It's been a while, but basically as I remember it, hydrostatic is strongly valid when vertical acceleration (and thus vertical motion when looking near the surface) is much smaller in scale compared to the horizontal motions/accelerations. So indeed, this is not true in tornadoes and in convection. So you get very correct values for strong thunderstorms. But it will help you diagnose regions of rising motion.

You can look a bit more towards breaking it down into more advanced detail and the mathematics of hydrostatic and how it fails in Doswell and Markowski's [paper](http://www.cimms.ou.edu/~doswell/relbuoy/relbuoy.html) on buoyancy.

But, with such limited model output, not sure that you're looking to get into all of that detail, as I don't believe you'll find any extra help going those avenues with the variables you have.

With just slightly more information, most importantly temperature, you could use the [ideal gas law](https://en.wikipedia.org/wiki/Ideal_gas_law):

```
 p = ρRT

```

*[where R is the ideal gas constant of the gas under consideration and T is temperature in Kelvin]*

This would allow you replace the unknown density, and you'd then end up with:

**w = −ωRT/pg**

Which you could then either further approximate R to be that for dry air = Rd = 287 J/Kkg, or even work up the adjustment your moisture content variable causes (though don't think doing so would offer much useful improvement as you've already got greater errors due to the failings of the hydrostatic approximation).

Since you don't have temperature, it seems your only remaining option would be to plug in the density approximation for the level of interest from the approximated "Standard Atmosphere" [here](http://www.engineeringtoolbox.com/standard-atmosphere-d_604.html) (use the table where ρ is in units of kg/m^3).

It has been a long while since I've done this stuff, but I don't see any better options. Unless somehow there's a way to work from the geopotential, maybe apply something like mass continuity in pressure coordinates, and get to something more complex to return something more correct for w. But if there is, it outpaces me!

**So, long story short, it appears that your best hope is w = −ωRT/pg, but it won't be 100% reliable, particularly in strong convection.**

> **Comments:**
> **username_2**: Been a while, like I noted... are you suggesting that the model would need a balance to have calculated omega?  He said he's got omega.  If you are saying the model is using that, I could see that being true given the limited output... but I'm thinking you can get omega from a full primitive equations model in pressure coordinates just fine without any balance assumptions?  

---

