Talk:Rossby parameter

Equation Change
Per the following discussion posted at WP:RD/S (May 24, 2007):

More atmospheric physics for you! Our article on Rossby waves says that the wave speed is given by


 * $$ c = u - \frac{\beta}{k^2+l^2}$$

where c is the wave speed, u is the mean westerly flow, $$\beta$$ is the Rossby parameter, and k and l are the longitudinal and latitudinal wavenumbers. The Rossby parameter is given as


 * $$\beta = \frac{2\omega cos\phi}{a}$$

where $$\phi$$ is the latitude, $$\omega$$ is the angular speed of the Earth's rotation, and a is the mean radius of the Earth.

Given the above, I cannot see how c is anything other than tiny. Yet it is not.

For example, if I put in u = 3, k = 5, l = 3 and calculate $$\omega$$ for a latitude of 60 degrees, I get an answer of order $$10^{-13}$$, which is far too small.

I am surely misunderstanding something. Can someone check through this and see where I am going wrong? Many thanks, →Ollie (talk • contribs) 01:22, 24 May 2007 (UTC)


 * I agree with your assessment. Something must be awry!  My understanding of the units in the problem suggest that $$\beta$$ must be in units of velocity (meters/sec).  However, because it is defined as $$\omega / a = sec^{-1} / meters$$, this does not work out.  So, perhaps it ought to be defined as:
 * $$\beta = 2a\omega cos\phi$$
 * This would make beta quite a bit larger (~1012x, which solves our issue quite nicely). I will try to find an alternative source for this equation which may verify my belief.  Nimur 15:08, 24 May 2007 (UTC)


 * This lecture note seems to corroborate my belief that earth-radius should be multiplied, not divided. I will edit the articles in question.  Nimur 15:16, 24 May 2007 (UTC)

End of copied discussion

Please discuss any issues below this line. Nimur 15:21, 24 May 2007 (UTC)

Yes, the radius a should be in the denominator, as traditionally defined. You can verify here:

http://amsglossary.allenpress.com/glossary/search?id=rossby-parameter1

Or any GFD texts would tell you the same thing.

Pkamostai (talk) 13:23, 20 December 2007 (UTC)

Equation
Is the current expression of the Rossby parameter really correct? The above equation
 * $$ c = u - \frac{\beta}{k^2+l^2}$$

suggests $$\beta$$ has a dimension $$[m^{-1}s^{-1}]$$, since dimension of $$k$$ is $$[m^{-1}]$$ and $$c$$'s is $$[ms^{-1}]$$. So the previous equation
 * $$\beta = \frac{2\omega cos\phi}{a}$$

seems correct to me.

Am I missing something? 210.174.33.192 19:12, 13 November 2007 (UTC)

Yes, the radius a should be in the denominator, as traditionally defined. You can verify here: http://amsglossary.allenpress.com/glossary/search?id=rossby-parameter1 Or any GFD texts would tell you the same thing. —Preceding unsigned comment added by Pkamostai (talk • contribs) 13:21, 20 December 2007 (UTC)


 * This was the source of confusion some time ago (see the above discussion). I will defer to your more recent edits since I'm not an expert in the field... Nimur (talk) 06:36, 22 December 2007 (UTC)

No φ in the equation?
I'm missing something too. The text says φ represents latitude, but there is not φ in the formula. Or am I just going blind? // Jontew (talk) 13:29, 26 November 2010 (UTC)

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