<p>I am a graduate student and, in the research project I am a part of, I am tasked to investigate the track of volcanic <span class="math-container">$SO_2$</span> emissions. I was comparing the satellite data with the tracks calculated by the HYSPLIT model, and I noticed that there are some curvature, deflection, and &quot;curling&quot; found in satellite data but is not replicated by the HYSPLIT model. After some discussion and stumbling in the dark, our best guess was that these deflections may be due to the quasi-geostrophic factors affecting the parcel of <span class="math-container">$SO_2$</span>. As far as we know (which isn't much), we think that this means the pressure gradient, Coriolis pseudo-force, as well as the inertia of the parcel are affecting its trajectory.</p>
<p>The pressure gradient and inertia are easy enough to wrap our heads around. Synoptic weather maps contain the pressure readings. Inertia can be approximated through the horizontal motion of the parcel. However, we are not sure on how to put a value on the effect of the Coriolis pseudo-force on the <span class="math-container">$SO_2$</span> parcels.</p>
<p>One of the ways we found on quantifying the Coriolis effect is through the Rossby number. The formula we found was:</p>
<p><span class="math-container">$Ro = \dfrac{U}{L \left (2 \omega \sin{\varphi} \right )}$</span></p>
<p>where <span class="math-container">$U$</span> is the speed/velocity of the parcel, <span class="math-container">$L$</span> is the length of the phenomenon, <span class="math-container">$\omega$</span> is the angular velocity of the Earth, and <span class="math-container">$\varphi$</span> is the latitude of the phenomenon.</p>
<p>This formula seems to be enough for parcels moving in the East-West axis. We've managed to calculate the Rossby number of a parcel of air moving Westward, covering 2000 kilometers in 24 hours, at a latitude of <span class="math-container">$10^\circ N$</span>. In such case, the Rossby number seems to be <span class="math-container">$\sim 0.456$</span>.</p>
<p><span class="math-container">$
L = 2000 km = 2 \times 10^6 m \\
U = \dfrac{2 \times 10^6 m}{24 hours \times \dfrac{60 minutes}{1 hour} \times \dfrac{60 seconds}{1 minute}} = 23.148 m/s \\
\omega = 7.292 \times 10^{−5} rad/s \\
\varphi = 10^\circ N = \dfrac{\pi}{18} radian \\
\sin{\varphi} = \sin{\dfrac{\pi}{18}} = 0.174 \\
Ro = \dfrac{23.148 m/s}{(2 \times 10^6m)\left [2 (7.292 \times 10^{-5} rad/s) (0.174)\right ]} = 0.456
$</span></p>
<p>Is this calculation/process correct? From what we've read so far, <span class="math-container">$Ro &lt; 1$</span> seems to indicate that the Coriolis pseudo-force has noticeable effect on the trajectory of the parcel.</p>
<p>Additionally, how does the Rossby number change for parcels moving Northward or Southward?</p>
<p>Say, for example, a parcel of air from <span class="math-container">$2^\circ N$</span> travels to <span class="math-container">$15^\circ N$</span>, covering 1100 kilometers in 13 hours.</p>
<p><span class="math-container">$
L = 1100 km = 1.1 \times 10^6 m \\
U = \dfrac{1.1 \times 10^6 m}{13 hours \times \dfrac{60 minutes}{1 hour} \times \dfrac{60 seconds}{1 minute}} = 23.504 m/s \\
\omega = 7.292 \times 10^{−5} rad/s \\
\varphi_1 = 2^\circ N = \dfrac{\pi}{90} radian \\
\sin{\varphi_1} = \sin{\dfrac{\pi}{18}} = 0.035 \\
\varphi_2 = 15^\circ N = \dfrac{\pi}{12} radian \\
\sin{\varphi_2} = \sin{\dfrac{\pi}{18}} = 0.259 \\
Ro = \dfrac{23.504 m/s}{(1.1 \times 10^6m)\left [2 (7.292 \times 10^{-5} rad/s) (?)\right ]} = ?
$</span></p>
<p>Any help, comment, or correction is appreciated. I am very new to this so references to places or resources where I could learn more is also very much appreciated. Thank you!</p>

CC BY-SA 4.0
2024-11-08T11:45:29.947Z
The Coriolis has plenty of affect, but I'd think any reasonable modeling would account for that fine, it's a basic term of the fundamental equations, even quite simplified ones.  First key check: is the HYSPLIT plotting projection masking the curvature (some projections minimize it I believe?)  Very few paths should move straight N/S, because the advection follows the wind, which follows isobars/isoheights aloft, which tend towards east-west (due to the Earth's temperature gradient).  Believe only the sides of large lows/highs will there be a straight N-S path.
JeopardyTempest
CC BY-SA 4.0
2024-11-08T11:57:10.177Z
Coriolis itself isn't a QG idea (Coriolis balance with pressure is the definition of geostrophic... Ro is 0 in the very simplest of models)... in fact large Coriolis means the denominator is large -> very small Ro, means less deviation from geostrophic.  Differences in Coriolis over the Earth are quasigeostrphic (in Beta plane) (and straightup ageostrophic (additional terms), but is well accounted for in weather models that I'd think HYSPLIT would be using (and if it's not using a model, it's not accounting for any changing in weather over time)
JeopardyTempest
CC BY-SA 4.0
2024-11-08T11:58:16.140Z
Whether your HYSPLIT in just based off initial conditions, or uses a weather model, there'd be curvature.  Would think small localized curling that isn't included may be a factor of wind analysis that's too coarse, or too large of a time step in the HYSPLIT trajectory calculating.  That said, you ask an interesting question (Coriolis = 0 at the equator means it definitely changes!)... just not sure if it'll be that important to your work as you've framed it!
JeopardyTempest
CC BY-SA 4.0
2024-11-08T12:38:57.980Z
It took me quite a while to fully understand your comments, but I am so glad you laid it out for me. This isn't at all sarcasm. I am a neophyte, thrust suddenly onto this field. It isn't an understatement to say that I am fumbling in the dark on this. You've just given me ideas on where I did wrong/was not so rigorous in my methodology. I think for the HYSPLIT results, as I've only started learning the model for just a little over a week now (maybe 24 hours cumulative actual experience), I could tack it to errors on my part. Weather is "time-evolving" and I think I did not account for that.
J. Velonta
CC BY-SA 4.0
2024-11-08T12:43:59.560Z
As for the curvature, these are not the actual numbers but they reflect the data. From the HYSPLIT results, volcanic SO2 from Mt. Ruang (Lat.: 2 degrees north) tended to travel in the East-West line. However, there is a significant amount which travelled around 1000km in 13 hours (around 13 degrees difference in latitude) North and the said "off shoot" curved clockwise.  We suspected this was due to the Coriolis effect, but might this also be due to pressure and temperature difference? I ask this again because you mention the temperature difference driving air trajectory.
J. Velonta
CC BY-SA 4.0
2024-11-08T12:46:59.677Z
I have to add that in the comment right above this, the "offshoot" was observed using satellite data. It wasn't present in the HYSPLIT results. I ran out of space to add this. In any case, I have a lot of things to study. If it isn't too much, may I ask any references that you can recommend for me to self-learn this?
J. Velonta
CC BY-SA 4.0
2024-11-08T14:41:54.767Z
Not seen as sarcasm at all.  Glad it helped.  Wish I knew more than cursory familiarity with HYSPLIT, but haven't really messed with it.  That said, the numerical models I'd expect HYSPLIT to work with (perhaps something from the GFS class?) would incorporate all of those effects... meteorologically nothing is "because of Coriolis" or "because of temperature", but always a varying combination of all factors, whichever ones are driving the weather in the area.  The actual movement likewise wouldn't be due to any one thing.  Now why doesn't HYSPLIT pick it up? ...
JeopardyTempest
CC BY-SA 4.0
2024-11-08T14:41:59.597Z
... I'm back to my main thoughts (timestep of HYSPLIT?, maybe model resolution).  Plus of course pollutants/molecules disperse gradually, taking spreading paths typically, so perhaps calculating more tracks, starting from the intermediate points of the initial path, and tweaking slightly, if more tracks from the initial volcano aren't possible/changing... might see more of the varying paths?  Maybe something from turbulence (which typical models probably don't have the resolution to handle??) But I'm no expert on the trajectories/HYSPLIT, just enough background in meteorology to be trouble :-p
JeopardyTempest
<geophysics><atmospheric-circulation>
How does the Rossby number of an air parcel change as it travels northwards or southwards?
J. Velonta
# How does the Rossby number of an air parcel change as it travels northwards or southwards?

(Asked by: username_0 on 2024-11-08)

I am a graduate student and, in the research project I am a part of, I am tasked to investigate the track of volcanic $SO\_2$ emissions. I was comparing the satellite data with the tracks calculated by the HYSPLIT model, and I noticed that there are some curvature, deflection, and "curling" found in satellite data but is not replicated by the HYSPLIT model. After some discussion and stumbling in the dark, our best guess was that these deflections may be due to the quasi-geostrophic factors affecting the parcel of $SO\_2$. As far as we know (which isn't much), we think that this means the pressure gradient, Coriolis pseudo-force, as well as the inertia of the parcel are affecting its trajectory.

The pressure gradient and inertia are easy enough to wrap our heads around. Synoptic weather maps contain the pressure readings. Inertia can be approximated through the horizontal motion of the parcel. However, we are not sure on how to put a value on the effect of the Coriolis pseudo-force on the $SO\_2$ parcels.

One of the ways we found on quantifying the Coriolis effect is through the Rossby number. The formula we found was:

$Ro = \dfrac{U}{L \left (2 \omega \sin{\varphi} \right )}$

where $U$ is the speed/velocity of the parcel, $L$ is the length of the phenomenon, $\omega$ is the angular velocity of the Earth, and $\varphi$ is the latitude of the phenomenon.

This formula seems to be enough for parcels moving in the East-West axis. We've managed to calculate the Rossby number of a parcel of air moving Westward, covering 2000 kilometers in 24 hours, at a latitude of $10^\circ N$. In such case, the Rossby number seems to be $\sim 0.456$.

$
L = 2000 km = 2 \times 10^6 m \\
U = \dfrac{2 \times 10^6 m}{24 hours \times \dfrac{60 minutes}{1 hour} \times \dfrac{60 seconds}{1 minute}} = 23.148 m/s \\
\omega = 7.292 \times 10^{−5} rad/s \\
\varphi = 10^\circ N = \dfrac{\pi}{18} radian \\
\sin{\varphi} = \sin{\dfrac{\pi}{18}} = 0.174 \\
Ro = \dfrac{23.148 m/s}{(2 \times 10^6m)\left [2 (7.292 \times 10^{-5} rad/s) (0.174)\right ]} = 0.456
$

Is this calculation/process correct? From what we've read so far, $Ro < 1$ seems to indicate that the Coriolis pseudo-force has noticeable effect on the trajectory of the parcel.

Additionally, how does the Rossby number change for parcels moving Northward or Southward?

Say, for example, a parcel of air from $2^\circ N$ travels to $15^\circ N$, covering 1100 kilometers in 13 hours.

$
L = 1100 km = 1.1 \times 10^6 m \\
U = \dfrac{1.1 \times 10^6 m}{13 hours \times \dfrac{60 minutes}{1 hour} \times \dfrac{60 seconds}{1 minute}} = 23.504 m/s \\
\omega = 7.292 \times 10^{−5} rad/s \\
\varphi\_1 = 2^\circ N = \dfrac{\pi}{90} radian \\
\sin{\varphi\_1} = \sin{\dfrac{\pi}{18}} = 0.035 \\
\varphi\_2 = 15^\circ N = \dfrac{\pi}{12} radian \\
\sin{\varphi\_2} = \sin{\dfrac{\pi}{18}} = 0.259 \\
Ro = \dfrac{23.504 m/s}{(1.1 \times 10^6m)\left [2 (7.292 \times 10^{-5} rad/s) (?)\right ]} = ?
$

Any help, comment, or correction is appreciated. I am very new to this so references to places or resources where I could learn more is also very much appreciated. Thank you!

**Tags:** geophysics, atmospheric-circulation

> **Comments:**
> **username_1**: The Coriolis has plenty of affect, but I'd think any reasonable modeling would account for that fine, it's a basic term of the fundamental equations, even quite simplified ones.  First key check: is the HYSPLIT plotting projection masking the curvature (some projections minimize it I believe?)  Very few paths should move straight N/S, because the advection follows the wind, which follows isobars/isoheights aloft, which tend towards east-west (due to the Earth's temperature gradient).  Believe only the sides of large lows/highs will there be a straight N-S path.  
> **username_1**: Coriolis itself isn't a QG idea (Coriolis balance with pressure is the definition of geostrophic... Ro is 0 in the very simplest of models)... in fact large Coriolis means the denominator is large -> very small Ro, means less deviation from geostrophic.  Differences in Coriolis over the Earth are quasigeostrphic (in Beta plane) (and straightup ageostrophic (additional terms), but is well accounted for in weather models that I'd think HYSPLIT would be using (and if it's not using a model, it's not accounting for any changing in weather over time)  
> **username_1**: Whether your HYSPLIT in just based off initial conditions, or uses a weather model, there'd be curvature.  Would think small localized curling that isn't included may be a factor of wind analysis that's too coarse, or too large of a time step in the HYSPLIT trajectory calculating.  That said, you ask an interesting question (Coriolis = 0 at the equator means it definitely changes!)... just not sure if it'll be that important to your work as you've framed it!  
> **username_0**: It took me quite a while to fully understand your comments, but I am so glad you laid it out for me. This isn't at all sarcasm. I am a neophyte, thrust suddenly onto this field. It isn't an understatement to say that I am fumbling in the dark on this. You've just given me ideas on where I did wrong/was not so rigorous in my methodology. I think for the HYSPLIT results, as I've only started learning the model for just a little over a week now (maybe 24 hours cumulative actual experience), I could tack it to errors on my part. Weather is "time-evolving" and I think I did not account for that.  
> **username_0**: As for the curvature, these are not the actual numbers but they reflect the data. From the HYSPLIT results, volcanic SO2 from Mt. Ruang (Lat.: 2 degrees north) tended to travel in the East-West line. However, there is a significant amount which travelled around 1000km in 13 hours (around 13 degrees difference in latitude) North and the said "off shoot" curved clockwise.  We suspected this was due to the Coriolis effect, but might this also be due to pressure and temperature difference? I ask this again because you mention the temperature difference driving air trajectory.  
> **username_0**: I have to add that in the comment right above this, the "offshoot" was observed using satellite data. It wasn't present in the HYSPLIT results. I ran out of space to add this. In any case, I have a lot of things to study. If it isn't too much, may I ask any references that you can recommend for me to self-learn this?  
> **username_1**: Not seen as sarcasm at all.  Glad it helped.  Wish I knew more than cursory familiarity with HYSPLIT, but haven't really messed with it.  That said, the numerical models I'd expect HYSPLIT to work with (perhaps something from the GFS class?) would incorporate all of those effects... meteorologically nothing is "because of Coriolis" or "because of temperature", but always a varying combination of all factors, whichever ones are driving the weather in the area.  The actual movement likewise wouldn't be due to any one thing.  Now why doesn't HYSPLIT pick it up? ...  
> **username_1**: ... I'm back to my main thoughts (timestep of HYSPLIT?, maybe model resolution).  Plus of course pollutants/molecules disperse gradually, taking spreading paths typically, so perhaps calculating more tracks, starting from the intermediate points of the initial path, and tweaking slightly, if more tracks from the initial volcano aren't possible/changing... might see more of the varying paths?  Maybe something from turbulence (which typical models probably don't have the resolution to handle??) But I'm no expert on the trajectories/HYSPLIT, just enough background in meteorology to be trouble :-p  

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## 0 answers

