<p>They are missing $g$ in equations 4,5 and 6. When they go from the hydrostatic equation to equation 4, they forget to include gravity in their equations.
All the dimensions except from $g$ and $f$ compensate each other and then $g/f$ produces the right velocity dimensions $m/s$.</p>

CC BY-SA 3.0
2018-01-23T12:23:54.627Z
2018-01-23T12:23:54.627Z
arkaia
<p>I am puzzled by the dimensions of the formulas given <a href="http://www.rsmas.miami.edu/users/lbeal/MPO603/Lecture%2012.xhtml" rel="nofollow noreferrer">here</a> for the baroclinic component of ocean currents (first terms on the right-hand side of formulas 5 and 6).</p>

<p>If I am not wrong:</p>

<p>[f]=1/s</p>

<p>[$\rho$]=kg/m3</p>

<p>[$\frac{\partial}{\partial y}$]=1/m</p>

<p>[dz]=m</p>

<p>therefore:
$$[\frac{1}{f\rho}\frac{\partial}{\partial y}\int_{-z}^0\rho(z)dz]=s\frac{m^3}{kg}\frac{1}{m}\frac{kg}{m^3}m = s$$</p>

<p>What am I missing?</p>

<oceanography>
baroclinic current - dimensional analysis
shamalaia
# baroclinic current - dimensional analysis

(Asked by: username_0 on 2018-01-23)

I am puzzled by the dimensions of the formulas given [here](http://www.rsmas.miami.edu/users/lbeal/MPO603/Lecture%2012.xhtml) for the baroclinic component of ocean currents (first terms on the right-hand side of formulas 5 and 6).

If I am not wrong:

[f]=1/s

[$\rho$]=kg/m3

[$\frac{\partial}{\partial y}$]=1/m

[dz]=m

therefore:
$$[\frac{1}{f\rho}\frac{\partial}{\partial y}\int\_{-z}^0\rho(z)dz]=s\frac{m^3}{kg}\frac{1}{m}\frac{kg}{m^3}m = s$$

What am I missing?

**Tags:** oceanography


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## 1 answers

### Answer by username_1 on 2018-01-23. Score: 2 | Accepted answer

They are missing $g$ in equations 4,5 and 6. When they go from the hydrostatic equation to equation 4, they forget to include gravity in their equations.
All the dimensions except from $g$ and $f$ compensate each other and then $g/f$ produces the right velocity dimensions $m/s$.


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