<p>Well for one, this will surely be the speed of planetary rotation and the wind speeds your atmosphere can generate in North-South-direction.<br />
In my understanding those cells form as N-S winds flow and become deflected from this direction. So the number of cells will be determined by the winds that start either at the equator or the pole and the distance that they can cover before being deflected into completely zonal direction.</p>
<p>To have a rough idea about this distance I tried to estimate this (in a very quick, dirty and most likely extremely wrong way) by assuming the simplified Navier-Stokes equation in planetary coordinates on synoptic scales:<br />
<span class="math-container">$$\partial_t \vec v = 2 \Omega \sin\Theta (\vec v \times \vec z),$$</span><br />
where <span class="math-container">$\Theta$</span> is the planetary latitude, <span class="math-container">$\Omega=\Omega(\Theta)$</span> the local planetary rotation frequency at the given latitude and <span class="math-container">$v_0$</span> the local wind speed, leading to a harmonic equation for the meridional speed<br />
<span class="math-container">$$\partial_{tt} v_{\phi} = -(2\Omega \sin\Theta)^2 v_{\phi},$$</span>
which with the usual arguments of scale-analysis gives us a timescale <span class="math-container">$\tau_{\phi}$</span> for the periodic movement involved in solutions of that equation, of<br />
<span class="math-container">$$\tau_{\phi} \sim \frac{1}{2\Omega \sin\Theta}$$</span><br />
corresponding to a distance
<span class="math-container">$$ d \sim \frac{v_0}{2\Omega \sin\Theta_0}$$</span>
covered by the vortex. Of course, I cheated here by assuming the initial speed <span class="math-container">$v_0 = \text{const}$</span> and the initial Coriolis-factor constant with <span class="math-container">$\Theta = \Theta_0$</span>.<br />
As now such a &quot;structure&quot; (solution of the above eqs.) will have a size of <span class="math-container">$2d$</span> in the meridional direction, having space of <span class="math-container">$\pi R_{\text{planet}}$</span> our number of Hadley cells per hemisphere becomes simply
<span class="math-container">$$N = \frac{\pi R_{\text{planet}} \Omega \sin\Theta_0}{v_0}$$</span><br />
Where we'd need to estimate this number starting at the poles, as the Coriolis-parameter vanishes at the equator.</p>
<p>I will later (if I don't forget it) pluck in some numbers and see how bad this estimate is, or perhaps someone can try that. Anyway, I'm open for improvements of those arguments from anyone who knows something, but a quick look into Holton, <em>An Introduction to Dynamic Meteorology</em>, unfortunately didn't reveal much.</p>
<p>But to also address your question a little bit further: Of course, we'd need to have some climatic model to be able to estimate <span class="math-container">$v_0$</span> and know the speed of planetary rotation. Independent of how bad my formula is, the initial arguments of deflection vs. 'climbing the meridian' should hold for the formation of a Hadley cell.<br />
Returning winds from higher atmospheric layers, surface friction in the atmospheric boundary layer will also play a role for an exact calculation.</p>
<hr />
<p>Addendum: <a href="http://journals.cambridge.org/action/displayAbstract?fromPage=online&amp;aid=373502&amp;fileId=S0022112075001504" rel="nofollow noreferrer">Rhines (1975)</a> has however found a relation <span class="math-container">$N \sim R_{planet} \sqrt{\frac{\Omega}{v_0}}$</span>, by considering how turbulent eddies will break up energy-limited by zonal jets. Also turbulent eddies can provide a good mechanism for justifying a locally constant <span class="math-container">$v_0$</span>, that I simply assumed in my calculation above.</p>
<p>On the other hand I just noticed a trivial error in my above calculation and cleared that. The fraction that is N was the other way round. Sorry for that.</p>
<hr />
<p>I wanted to note one other thing that seems important to me (and is perhaps also to anyone on this planet..):<br />
While this rule of thumb-calculations can work out, there is more physics that play a role in shaping the exact morphology of circulation-cells on a planet. Namely the radiative equilibrium in the atmosphere, which links the exact positions of those cells to a changing climate.</p>
<p>As a <a href="http://www.nature.com/ngeo/journal/v1/n1/full/ngeo.2007.38.html" rel="nofollow noreferrer">study</a> suggests by measuring Tropical to Polar-Jet distances and Tropopause heights, those cells can expand and reshape the environment people live in.</p>

CC BY-SA 3.0
2017-01-25T19:24:01.347Z
What are $\Omega$ and $v_0$ and $\Theta$ in these equations?
Kate Bertelsen
CC BY-SA 3.0
2018-03-27T16:42:28.060Z
The first two paragraphs provide a nice basic intuitive explanation. But according to that explanation the most reasonable way to go forward with the calculation would be to calculate the width of the cell (the distance the N-S wind can travel before becoming E-W). But instead you assume that there must be a number N of cells of equal size. Why is that? Why all cells should be of equal size?
Camilo Rada
CC BY-SA 3.0
2018-03-27T22:05:27.797Z
@CamiloRada: Simplicity. One can of course take into account how the coriolis factor changes with planetary latitude, but then the formula for the number of cells per hemisphere becomes an implicit equation.
AtmosphericPrisonEscape
CC BY-SA 3.0
2018-03-27T22:13:50.323Z
@AtmosphericPrisonEscape I see. But it would be nice to include a explanation/justification for the assumption that all cells would be of equal latitudinal extent.
Camilo Rada
CC BY-SA 4.0
2021-08-21T23:27:13.533Z
Does a change in average planetary temperature (i.e. global warming) affect v0?  How much is it changing on Earth?
Mike Serfas
CC BY-SA 4.0
2024-03-26T00:42:31.990Z
more recent study https://www.cps-jp.org/~mosir/pub/2016/2016-11-29/01_showman/lecture03/pub-web/20161129_showman_lec03.pdf
John
CC BY-SA 4.0
2024-03-28T05:22:40.697Z
@John: That's a lecture from 2016, not a study. It furthermore doesn't contradict the Rhines length (as it doesn't really make a statement about them). But the lecture quotes e.g. Showman & Polvani 2011, whose results are in line with Rhines' scalings.
AtmosphericPrisonEscape
CC BY-SA 4.0
2014-05-25T14:38:14.720Z
2021-08-16T21:19:20.370Z
2021-08-16T21:19:20.370Z
AtmosphericPrisonEscape
<p>We know (think?) that Earth has three Hadley cells per hemisphere, but from observing gas giants such as Jupiter, we see that they have many more cells. According to a link from a comment in <a href="https://earthscience.stackexchange.com/questions/388/why-does-the-hadley-cell-descend-at-30-degrees">this question</a>, <a href="http://saturn.jpl.nasa.gov/news/cassinifeatures/feature20070601f/" rel="noreferrer">Titan might have one cell going from north to south poles</a>. </p>

<p>What factors affect the number of cells a planet has, and how? Or to put it another way, given a hypothetical planet with atmosphere, what facts would you need to know in order to guess at how many cells it has?</p>

CC BY-SA 4.0
2021-08-10T16:15:31.340Z
I thought in the lower atmosphere of Titan, there are three cells (as opposed to Earth's six), and in Titan's upper atmosphere there is a single cell: "In the lower atmosphere there may be three, or even fewer, cells compared to Earth's six, and higher up there is a single, global, Hadley cell in which gas flows directly from one pole to the other." Or so [ESA claims](https://sci.esa.int/web/cassini-huygens/-/51176-cassini-reveals-titan-s-atmosphere-in-reverse).
Alex Nelson
CC BY-SA 3.0
2014-12-10T02:56:04.060Z
Note that the cells (from equator to pole) are the Hadley cell, Ferrel cell and Polar cell (i.e. we dont have 3 Hadley cells per hemisphere, just one).
casey
CC BY-SA 3.0
2014-12-11T16:11:57.483Z
I guess everyone's well aware of that, it's just easier to steal this concept for planetary- rather than earth-science than to name new concepts, when the physics is the same.
AtmosphericPrisonEscape
<meteorology><planetary-science><atmospheric-circulation>
What factors determine the number of Hadley cells for a planet?
congusbongus
# What factors determine the number of Hadley cells for a planet?

(Asked by: username_0 on 2014-05-23)

We know (think?) that Earth has three Hadley cells per hemisphere, but from observing gas giants such as Jupiter, we see that they have many more cells. According to a link from a comment in [this question](https://earthscience.stackexchange.com/questions/388/why-does-the-hadley-cell-descend-at-30-degrees), [Titan might have one cell going from north to south poles](http://saturn.jpl.nasa.gov/news/cassinifeatures/feature20070601f/).

What factors affect the number of cells a planet has, and how? Or to put it another way, given a hypothetical planet with atmosphere, what facts would you need to know in order to guess at how many cells it has?

**Tags:** meteorology, planetary-science, atmospheric-circulation

> **Comments:**
> **username_2**: Note that the cells (from equator to pole) are the Hadley cell, Ferrel cell and Polar cell (i.e. we dont have 3 Hadley cells per hemisphere, just one).  
> **username_3**: I guess everyone's well aware of that, it's just easier to steal this concept for planetary- rather than earth-science than to name new concepts, when the physics is the same.  
> **username_1**: I thought in the lower atmosphere of Titan, there are three cells (as opposed to Earth's six), and in Titan's upper atmosphere there is a single cell: "In the lower atmosphere there may be three, or even fewer, cells compared to Earth's six, and higher up there is a single, global, Hadley cell in which gas flows directly from one pole to the other." Or so [ESA claims](https://sci.esa.int/web/cassini-huygens/-/51176-cassini-reveals-titan-s-atmosphere-in-reverse).  

---

## 1 answers

### Answer by username_3 on 2014-05-25. Score: 21 | Accepted answer

Well for one, this will surely be the speed of planetary rotation and the wind speeds your atmosphere can generate in North-South-direction.  
In my understanding those cells form as N-S winds flow and become deflected from this direction. So the number of cells will be determined by the winds that start either at the equator or the pole and the distance that they can cover before being deflected into completely zonal direction.

To have a rough idea about this distance I tried to estimate this (in a very quick, dirty and most likely extremely wrong way) by assuming the simplified Navier-Stokes equation in planetary coordinates on synoptic scales:  
$$\partial\_t \vec v = 2 \Omega \sin\Theta (\vec v \times \vec z),$$  
where $\Theta$ is the planetary latitude, $\Omega=\Omega(\Theta)$ the local planetary rotation frequency at the given latitude and $v\_0$ the local wind speed, leading to a harmonic equation for the meridional speed  
$$\partial\_{tt} v\_{\phi} = -(2\Omega \sin\Theta)^2 v\_{\phi},$$
which with the usual arguments of scale-analysis gives us a timescale $\tau\_{\phi}$ for the periodic movement involved in solutions of that equation, of  
$$\tau\_{\phi} \sim \frac{1}{2\Omega \sin\Theta}$$  
corresponding to a distance
$$ d \sim \frac{v\_0}{2\Omega \sin\Theta\_0}$$
covered by the vortex. Of course, I cheated here by assuming the initial speed $v\_0 = \text{const}$ and the initial Coriolis-factor constant with $\Theta = \Theta\_0$.  
As now such a "structure" (solution of the above eqs.) will have a size of $2d$ in the meridional direction, having space of $\pi R\_{\text{planet}}$ our number of Hadley cells per hemisphere becomes simply
$$N = \frac{\pi R\_{\text{planet}} \Omega \sin\Theta\_0}{v\_0}$$  
Where we'd need to estimate this number starting at the poles, as the Coriolis-parameter vanishes at the equator.

I will later (if I don't forget it) pluck in some numbers and see how bad this estimate is, or perhaps someone can try that. Anyway, I'm open for improvements of those arguments from anyone who knows something, but a quick look into Holton, *An Introduction to Dynamic Meteorology*, unfortunately didn't reveal much.

But to also address your question a little bit further: Of course, we'd need to have some climatic model to be able to estimate $v\_0$ and know the speed of planetary rotation. Independent of how bad my formula is, the initial arguments of deflection vs. 'climbing the meridian' should hold for the formation of a Hadley cell.  
Returning winds from higher atmospheric layers, surface friction in the atmospheric boundary layer will also play a role for an exact calculation.

---

Addendum: [Rhines (1975)](http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=373502&fileId=S0022112075001504) has however found a relation $N \sim R\_{planet} \sqrt{\frac{\Omega}{v\_0}}$, by considering how turbulent eddies will break up energy-limited by zonal jets. Also turbulent eddies can provide a good mechanism for justifying a locally constant $v\_0$, that I simply assumed in my calculation above.

On the other hand I just noticed a trivial error in my above calculation and cleared that. The fraction that is N was the other way round. Sorry for that.

---

I wanted to note one other thing that seems important to me (and is perhaps also to anyone on this planet..):  
While this rule of thumb-calculations can work out, there is more physics that play a role in shaping the exact morphology of circulation-cells on a planet. Namely the radiative equilibrium in the atmosphere, which links the exact positions of those cells to a changing climate.

As a [study](http://www.nature.com/ngeo/journal/v1/n1/full/ngeo.2007.38.html) suggests by measuring Tropical to Polar-Jet distances and Tropopause heights, those cells can expand and reshape the environment people live in.

> **Comments:**
> **username_4**: What are $\Omega$ and $v_0$ and $\Theta$ in these equations?  
> **username_5**: The first two paragraphs provide a nice basic intuitive explanation. But according to that explanation the most reasonable way to go forward with the calculation would be to calculate the width of the cell (the distance the N-S wind can travel before becoming E-W). But instead you assume that there must be a number N of cells of equal size. Why is that? Why all cells should be of equal size?  
> **username_3**: @username_5: Simplicity. One can of course take into account how the coriolis factor changes with planetary latitude, but then the formula for the number of cells per hemisphere becomes an implicit equation.  
> **username_5**: @username_3 I see. But it would be nice to include a explanation/justification for the assumption that all cells would be of equal latitudinal extent.  
> **username_6**: Does a change in average planetary temperature (i.e. global warming) affect v0?  How much is it changing on Earth?  
> **username_7**: more recent study https://www.cps-jp.org/~mosir/pub/2016/2016-11-29/01_showman/lecture03/pub-web/20161129_showman_lec03.pdf  
> **username_3**: @username_7: That's a lecture from 2016, not a study. It furthermore doesn't contradict the Rhines length (as it doesn't really make a statement about them). But the lecture quotes e.g. Showman & Polvani 2011, whose results are in line with Rhines' scalings.  

---

