While reading on Wikipedia, I read the following

The Froude number is defined as:

$$\mathrm{Fr} = \frac{v}{c}$$

where $v$ is a characteristic velocity, and $c$ is a characteristic water wave propagation >velocity. The Froude number is thus analogous to the Mach number. The greater the Froude >number, the greater the resistance.

While reading the following paper on shallow water waves, I could not understand how/why the Froude number was given to be

$$\mathrm{F} = \frac{gt_0^2}{L}$$

Where $t_0$ is the time scale, and $L$ is the length scale along the X and Z axes. Note that the scale for the amplitude of the surface above the mean depth is some other $A$ and not $L$.

Please explain how the author has defined the Froude number. Is it the same as the one given on Wikipedia? If so, then please provide a step-by-step reasoning to show that they are indeed the same. If not, then please explain what the Froude number mentioned in the paper represents.

EDIT: I'd like to add another minor question as it is related to the abovementioned paper. It mentions that the surface tension is accounted for but I do not understand how that is possible given that there is no term containing the surface tension $\gamma$ explicitly.


2 Answers 2


To answer the question in the title: What does the Froude number represent?

The Froude ($Fr$) number is a non-dimensional value that (typically) is used to quantify the degree of linearity/nonlinearity of a gravity wave through the ratio of a characteristic fluid velocity ($v$) and the gravity wave speed ($c$). As shown below, the measure is analogous to wave steepness, $s = ak$, where $a$ is the wave amplitude and $k$ is the wavenumber. Consider a fluid particle embedded in the wave, its orbital (characteristic) velocity, $v$, is given by $\omega a$, where $\omega$ is the wave frequency. This makes sense because as the wave advances fluid particles (at first order) move around in circles with a spatial scale given by the wave amplitude $a$ and a temporal scale given by $\omega^{-1}$. Moreover, we know that the wave speed is given by, $c = \omega/k$, so:

$Fr = (v)/(c) = (\omega a)/(\omega/k) = ak = s$. Hence, the Froude number is analogous to wave steepness.

The importance of this result is that, when you nondimensionalize the Navier-Stokes equations, wave steepness (or $Fr$) ends up multiplying the convective acceleration term $\mathbf{u} \cdot \nabla \mathbf{u}$. As you know, this term is responsible for much of the beautiful fluid dynamics arising from nonlinear wave breaking such as turbulence. You can read a bit more in this other post.

So in essence, the $Fr$ number tells you if a wave is big or not. In theory, if it is much less than 1 then the wave is linear (small), if it is larger than 1 then the wave will break because the fluid speed $v>c$ and the orbital fluid particle velocity exceeds the wave speed. That is, the wave topples over and breaks.

The Froude number is also used in other contexts but this is strictly for surface gravity waves.

Also, what the author uses in the text is fine, as long as you are consistent with the scaling throughout your analysis you can use any sensible nondimensional number. It may not the most conventional way of representing it but it is not wrong.


Since $L$ is the characteristic length of the fluid and $t_0$ the characteristic time, then the characteristic speed is $c=L/t_0$. Since this term is in the denominator, then $$ \frac{1}{c} = \frac{t_0}{L} $$

The paper is considering gravity waves, so I would imagine that the characteristic velocity is then $v=gt_0$ (which appears to be consistent with the Newtonian relation $v_f=v_i+gt$). Thus, $$ F=\frac{v}{c} = gt_0\cdot\frac{t_0}{L} = \frac{gt_0^2}{L} $$

So yes, they are do appear to be the same. However, I usually see the Froude number defined as $F=v/\sqrt{gH}$ where $H$ is the height above the ocean floor, so I do agree that this form of it is strange.

  • $\begingroup$ But hasn't the velocity been scaled using $A/t_0$ in the paper? Are we justified in using $gt_0$? And shouldn't $c$ be scaled using $\sqrt{(p/\rho)}=\sqrt{(AL/t_0^2)}$. Could you also answer the second part of the question, i.e., what part of the equations accounts for the surface tension? $\endgroup$ Aug 29, 2013 at 14:09
  • $\begingroup$ Yes, velocity was scaled using $A/t_0$, but that is not necessarily the characteristic speed, is it? Using $v=gt_0$ helps eliminate dimensions from the problem. I agree, I do not know why $c$ is defined as such, but most dimensionless numbers used are specific to a problem. $\endgroup$
    – Kyle Kanos
    Aug 29, 2013 at 16:01
  • $\begingroup$ I presume that the surface tension is probably accounted for in a boundary condition, but I am not 100% sure. $\endgroup$
    – Kyle Kanos
    Aug 29, 2013 at 16:07

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