<p>Like many of the other questions I answer on Earth Science SE, the answer is complicated. Based on the <a href="http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/geos.rxml" rel="nofollow noreferrer">principle of geostrophy</a>, the Coriolis force will point to the right of the wind vector in the northern hemisphere (left in the southern hemisphere).</p>
<p><strong>Can you <em>validate</em> the Coriolis force based on the wind rose?</strong> No. You would need multiple simultaneous wind observations over a wide observation space. You can see this in weather maps that show wind direction. It becomes easier to see the Coriolis effect on upper air maps, as they aren't affect by the surface. It is also easier to see further from the equator, where the Coriolis is zero. <a href="https://earth.nullschool.net/#current/wind/isobaric/500hPa/orthographic=-58.35,45.51,734" rel="nofollow noreferrer">Earth NullSchool has nifty maps that show worldwide winds</a> that can help you see the Coriolis effect (the color is the mean sea level pressure):</p>
<p><a href="https://i.sstatic.net/dOVxg.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/dOVxg.jpg" alt="enter image description here" /></a>
<a href="https://i.sstatic.net/ENs84.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ENs84.jpg" alt="enter image description here" /></a>
In your wind rose, you can approximate the Coriolis effect though, by turning it sideways.</p>
<p>The Coriolis effect can be experienced beyond the realm of meteorology (and I am not referring to the toilet-bowl fallacy). Long before Gaspard Coriolis lived, <a href="https://en.wikipedia.org/wiki/Giovanni_Battista_Riccioli#The_%22coriolis_effect%22_argument" rel="nofollow noreferrer">it was recognized that cannonballs were deflected by the Earth's rotation</a>, though it did have some inaccuracies.</p>
<p>One question I anticipate that I anticipate you might have is &quot;How far would a something (wind or a cannonball) have to travel before the Coriolis effect would have a meaningful effect?&quot; Projectiles are, in my opinion, the easiest way to show that this is a real thing. This &quot;acceleration&quot; is approximately <span class="math-container">$a_{cor}=fv$</span> where <span class="math-container">$f$</span> is the <a href="https://en.wikipedia.org/wiki/Coriolis_frequency" rel="nofollow noreferrer">latitudinally dependent Coriolis parameter</a>. and <span class="math-container">$v$</span> is the speed. <a href="https://en.wikipedia.org/wiki/Coriolis_force#Ballistic_trajectories" rel="nofollow noreferrer">You can see how this impacts ballistics through this Wikipedia entry</a>. Given various assumptions, like a constant latitude, ignoring drag, and ignoring the loss of forward momentum from the perpendicular acceleration, the distance one would be off  after firing a projectile would be: <span class="math-container">$$\Delta y = 2 \Omega \sin(\phi) (\Delta x)^2 u_0^{-1} $$</span>, where <span class="math-container">$\Omega$</span> is the  angular velocity of the earth, <span class="math-container">$\phi$</span> is the latitude, and <span class="math-container">$u$</span> is the speed, and <span class="math-container">$\Delta x$</span> is the distance to the target. So at say 45 degrees north and going 5 meters per second over a 100 meter distance, you would be off by 0.2 meters, <a href="http://academickids.com/encyclopedia/index.php/Paris_Gun" rel="nofollow noreferrer">but if you are the Paris gun with a horizontal velocity of 1009.7 m/s, trying to bombard the city 120 km away at a latitude of 49 degrees north, you'll be 1.3 km off if you don't account for the Coriolis effect</a>.</p>
<p>And because I know I will get someone who wants me to show my work for the previous equation here it is: <span class="math-container">$$\frac{d\vec{v}}{dt}=-f\hat{k}\times\vec{v}$$</span>. If we consider the initial conditions <span class="math-container">$\vec{v}(t=0)=u(t=0)\hat{i}+v(t=0)\hat{j}=u_0\hat{i}+0\hat{j}$</span>, then we have the system <span class="math-container">$$\frac{du}{dt}=fv$$</span> <span class="math-container">$$\frac{dv}{dt}=-fu$$</span>. Since I made the assumption that momentum along the trajectory is not altered by the Coriolis force, the system reduces to  <span class="math-container">$$\frac{du}{dt}=0$$</span> <span class="math-container">$$\frac{dv}{dt}=-fu$$</span>. If we think that wind is the changing coordinates (<span class="math-container">$u=\frac{dx}{dt}$</span>, <span class="math-container">$v=\frac{dy}{dt}$</span>), then the system can be represented as <span class="math-container">$$\frac{d^2x}{dt^2}=0$$</span> <span class="math-container">$$\frac{d^2y}{dt^2}=-fu=-fu_0$$</span>. Therefore <span class="math-container">$$\Delta y= fu_0t^2$$</span>. And since the projectile must travel distance <span class="math-container">$\Delta x$</span> at velocity <span class="math-container">$u_0$</span>, then <span class="math-container">$$t=\frac{\Delta x}{u_0}$$</span> which makes  <span class="math-container">$$\Delta y = 2 \Omega \sin(\phi) (\Delta x)^2 u_0^{-1} $$</span></p>

CC BY-SA 4.0
2021-06-17T20:06:53.797Z
2021-06-17T20:06:53.797Z
BarocliniCplusplus
<p>Yes, your figure is correct: in the mid-latitudes we have cyclones and anticyclones. They can't exist on equator because the Coriolis force there is too small to keep them stable. But the cyclones or anticyclones are present all around your country and their effects cancel each other. So, you can't determine the presence of Coriolis force on such micro scale. Otherwise, the effect of spining anticyclones and cyclones isn't specific to your country, but is (as previously stated) present over whole mid-latitudes. Also, the Coriolis effect isn't acting on microswirls because the Coriolis parameter is so small (due to too slow Earth's rotation).</p>

CC BY-SA 4.0
2021-05-17T15:48:28.450Z
2021-06-17T07:27:41.033Z
2021-06-17T07:27:41.033Z
User123
<p>I once learned in high school that the area where I used to live experiences the Coriolis effects of the air movement above the city. I never paid attention to this before, but I recently heard about it again, and I have no idea if it is true or not. There is absolutely no way I can validate that theory by using the wider definition of the Coriolis effect, but I do have access to a wind rose which looks like in the graph below:</p>
<p><a href="https://i.sstatic.net/a1am6.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/a1am6.png" alt="enter image description here" /></a></p>
<p>Is there any way to understand if there are any Coriolis-like winds by using this graph or a similar one? I found a nice sketch of this effect online, but I fail to understand how to correlate these two.</p>
<p><a href="https://i.sstatic.net/jitkN.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/jitkN.png" alt="http://www.pas.rochester.edu/~blackman/ast104/coriolis.html" /></a></p>
<p><strong>Is there anyone who could help me to understand if one can get this effect from a wind-rose?</strong></p>

CC BY-SA 4.0
2021-05-17T15:28:17.547Z
Where did you get that wind rose? I found this (https://world-weather.info/archive/romania/medias/) wind rose and there is much more westerly winds.
User123
CC BY-SA 4.0
2021-05-17T15:30:51.383Z
I'm not sure if this has any relevance to this specific question, but you can get rose wind-diagrams from calitateaer.ro. My question was more general
CuriousPaul
CC BY-SA 4.0
2021-05-17T15:35:22.220Z
Ok, but the data should have more westerly winds, I believe.
User123
CC BY-SA 4.0
2021-05-17T15:38:14.370Z
You would expect that at a greater scale due to the earth's rotation, right? But I'm wondering if there are local swirling effects caused by a micro-climate that has a Coriolis-like effect. Somehow I would have expected to see a more uniform distribution among all the coordinates, considering that it is like a swirling effect, but is it? I'm afraid that I don't understand it enough and I draw the wrong conclusions orrrrrrrrrr the Coriolis-effect theory on that micro-climate was a lie (or a guess)
CuriousPaul
<meteorology><geography><wind><coriolis>
Is it possible to detect Coriolis effect using a Wind Rose?
CuriousPaul
# Is it possible to detect Coriolis effect using a Wind Rose?

(Asked by: username_0 on 2021-05-17)

I once learned in high school that the area where I used to live experiences the Coriolis effects of the air movement above the city. I never paid attention to this before, but I recently heard about it again, and I have no idea if it is true or not. There is absolutely no way I can validate that theory by using the wider definition of the Coriolis effect, but I do have access to a wind rose which looks like in the graph below:

[![enter image description here](https://i.sstatic.net/a1am6.png)](https://i.sstatic.net/a1am6.png)

Is there any way to understand if there are any Coriolis-like winds by using this graph or a similar one? I found a nice sketch of this effect online, but I fail to understand how to correlate these two.

[![http://www.pas.rochester.edu/~blackman/ast104/coriolis.html](https://i.sstatic.net/jitkN.png)](https://i.sstatic.net/jitkN.png)

**Is there anyone who could help me to understand if one can get this effect from a wind-rose?**

**Tags:** meteorology, geography, wind, coriolis

> **Comments:**
> **username_1**: Where did you get that wind rose? I found this (https://world-weather.info/archive/romania/medias/) wind rose and there is much more westerly winds.  
> **username_0**: I'm not sure if this has any relevance to this specific question, but you can get rose wind-diagrams from calitateaer.ro. My question was more general  
> **username_1**: Ok, but the data should have more westerly winds, I believe.  
> **username_0**: You would expect that at a greater scale due to the earth's rotation, right? But I'm wondering if there are local swirling effects caused by a micro-climate that has a Coriolis-like effect. Somehow I would have expected to see a more uniform distribution among all the coordinates, considering that it is like a swirling effect, but is it? I'm afraid that I don't understand it enough and I draw the wrong conclusions orrrrrrrrrr the Coriolis-effect theory on that micro-climate was a lie (or a guess)  

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## 2 answers

### Answer by username_2 on 2021-06-17. Score: 2 | Accepted answer

Like many of the other questions I answer on Earth Science SE, the answer is complicated. Based on the [principle of geostrophy](http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/geos.rxml), the Coriolis force will point to the right of the wind vector in the northern hemisphere (left in the southern hemisphere).

**Can you *validate* the Coriolis force based on the wind rose?** No. You would need multiple simultaneous wind observations over a wide observation space. You can see this in weather maps that show wind direction. It becomes easier to see the Coriolis effect on upper air maps, as they aren't affect by the surface. It is also easier to see further from the equator, where the Coriolis is zero. [Earth NullSchool has nifty maps that show worldwide winds](https://earth.nullschool.net/#current/wind/isobaric/500hPa/orthographic=-58.35,45.51,734) that can help you see the Coriolis effect (the color is the mean sea level pressure):

[![enter image description here](https://i.sstatic.net/dOVxg.jpg)](https://i.sstatic.net/dOVxg.jpg)
[![enter image description here](https://i.sstatic.net/ENs84.jpg)](https://i.sstatic.net/ENs84.jpg)
In your wind rose, you can approximate the Coriolis effect though, by turning it sideways.

The Coriolis effect can be experienced beyond the realm of meteorology (and I am not referring to the toilet-bowl fallacy). Long before Gaspard Coriolis lived, [it was recognized that cannonballs were deflected by the Earth's rotation](https://en.wikipedia.org/wiki/Giovanni_Battista_Riccioli#The_%22coriolis_effect%22_argument), though it did have some inaccuracies.

One question I anticipate that I anticipate you might have is "How far would a something (wind or a cannonball) have to travel before the Coriolis effect would have a meaningful effect?" Projectiles are, in my opinion, the easiest way to show that this is a real thing. This "acceleration" is approximately $a\_{cor}=fv$ where $f$ is the [latitudinally dependent Coriolis parameter](https://en.wikipedia.org/wiki/Coriolis_frequency). and $v$ is the speed. [You can see how this impacts ballistics through this Wikipedia entry](https://en.wikipedia.org/wiki/Coriolis_force#Ballistic_trajectories). Given various assumptions, like a constant latitude, ignoring drag, and ignoring the loss of forward momentum from the perpendicular acceleration, the distance one would be off after firing a projectile would be: $$\Delta y = 2 \Omega \sin(\phi) (\Delta x)^2 u\_0^{-1} $$, where $\Omega$ is the angular velocity of the earth, $\phi$ is the latitude, and $u$ is the speed, and $\Delta x$ is the distance to the target. So at say 45 degrees north and going 5 meters per second over a 100 meter distance, you would be off by 0.2 meters, [but if you are the Paris gun with a horizontal velocity of 1009.7 m/s, trying to bombard the city 120 km away at a latitude of 49 degrees north, you'll be 1.3 km off if you don't account for the Coriolis effect](http://academickids.com/encyclopedia/index.php/Paris_Gun).

And because I know I will get someone who wants me to show my work for the previous equation here it is: $$\frac{d\vec{v}}{dt}=-f\hat{k}\times\vec{v}$$. If we consider the initial conditions $\vec{v}(t=0)=u(t=0)\hat{i}+v(t=0)\hat{j}=u\_0\hat{i}+0\hat{j}$, then we have the system $$\frac{du}{dt}=fv$$ $$\frac{dv}{dt}=-fu$$. Since I made the assumption that momentum along the trajectory is not altered by the Coriolis force, the system reduces to $$\frac{du}{dt}=0$$ $$\frac{dv}{dt}=-fu$$. If we think that wind is the changing coordinates ($u=\frac{dx}{dt}$, $v=\frac{dy}{dt}$), then the system can be represented as $$\frac{d^2x}{dt^2}=0$$ $$\frac{d^2y}{dt^2}=-fu=-fu\_0$$. Therefore $$\Delta y= fu\_0t^2$$. And since the projectile must travel distance $\Delta x$ at velocity $u\_0$, then $$t=\frac{\Delta x}{u\_0}$$ which makes $$\Delta y = 2 \Omega \sin(\phi) (\Delta x)^2 u\_0^{-1} $$


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### Answer by username_1 on 2021-05-17. Score: 0

Yes, your figure is correct: in the mid-latitudes we have cyclones and anticyclones. They can't exist on equator because the Coriolis force there is too small to keep them stable. But the cyclones or anticyclones are present all around your country and their effects cancel each other. So, you can't determine the presence of Coriolis force on such micro scale. Otherwise, the effect of spining anticyclones and cyclones isn't specific to your country, but is (as previously stated) present over whole mid-latitudes. Also, the Coriolis effect isn't acting on microswirls because the Coriolis parameter is so small (due to too slow Earth's rotation).


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