0

I am familiar with non-dimensionalizing a differential equation; however I am not familiar with a method to non-dimensionalize a non-differential equation. I just need to arrive to a non-dimensionalized formula. I have attempted it as shown here.

The equation i am trying to non-dimensionalize is as follows.

$$U_{orb} = \frac{\pi H_{rms}}{T_{p}sinh(2kh))}$$

Thank you for the response.

What I am looking for is U_{orb} to be completely dimensionless. so initially what I did to non dimensionalize it was the following:

$$U_{orb} \times\dfrac{T_p}{H_{mo}}$$

the unit of Uorb is $$\dfrac{m}{s }$$

the unit of T_p is $$s$$

the unit of H_rms is $$m$$

the unit of h is $$m$$

the unit of L is $$m$$

$$ U_{orb} = \frac{\pi H_{rms}}{T_{p}sinh(2(\frac{2\pi}{L}))h))}\\$$

$$ H_{rms} = m \;\;[meters]\\$$ $$ T_{p} = s \;\; [seconds]\\$$ $$ h = m \;\;[meters]\\$$ $$ L = m \;\;[meters]\\$$

$$ U_{orb} = \frac{\pi [m]}{[s]sinh(2(\frac{2\pi}{\cancel{[m]}}))\cancel{[m]}))}\\$$ $$ U_{orb} = \frac{\pi [m]}{[s]sinh(4\pi)}\\$$

$$ U_{orb} \times \frac{T_{p}}{H}= \frac{\pi H_{rms}}{T_{p}sinh(4\pi)}\times \frac{T_{p}}{H}\\$$

$$ U_{orb} \times \frac{T_{p}}{H}= \frac{\pi \cancel{[m]}}{\cancel{[s]}sinh(4\pi)}\times \frac{\cancel{[s]}}{\cancel{[m]}}\\$$

$$ U_{orb} \times \frac{T_{p}}{H}= \frac{\pi}{sinh(4\pi)}\\ $$

This is how I was thinking of doing it.

Happy Dov
  • 1
  • FYI - If you place your Latex in between $U_{orb} = \frac{\pi H_{rms}}{T_{p}sinh(2kh))} $, it will be formatted in-line. Double the $ to get display formatting. $$U_{orb} = \frac{\pi H_{rms}}{T_{p}sinh(2kh))}$$ yields $$U_{orb} = \frac{\pi H_{rms}}{T_{p}sinh(2kh))} $$ – Paul Sinclair Mar 15 '20 at 00:40
  • And while we are really happy that you are familiar with non-dimensionalizing a differential equation, would you mind sharing what that is with the rest of us? Why does everything in sight turn into $[L]$, except for the one $[T]$? what do these mean? – Paul Sinclair Mar 15 '20 at 00:45
  • I want a unitless number. U_orb is the orbital velocity of a propagating wave. which is a function of the wave height, wave period and wave length. While T_p and Hrms are the wave height and period of said wave. what i am trying to achieve is a dimensionless orbital velocity. – Happy Dov Mar 15 '20 at 07:34
  • I used [L] and [T] to represent length and time – Happy Dov Mar 15 '20 at 07:37
  • So that is what you meant after all. "Dimension" has multiple meanings, and quite frankly, there is some funny business going on in your calculations that made me doubt this interpretation. Apparently $k=\frac {2\pi}L$, but is $h = L$? (Not just the dimensions, but the actual values.) What is the difference between $H_{rms}$ and $H_{mo}$, and just plain $H$? – Paul Sinclair Mar 15 '20 at 15:03
  • $H_{rms}$ is the root-mean square wave height, $H_{mo}$ is the wave height in first moment of energy (where waves break), and H is measured wave height. $k=2π/L$ where L is wave length. it was my bad to show h as L – Happy Dov Mar 15 '20 at 15:38
  • What is also "your bad" is that you treated $H_{rms} = H_{mo} = H$, which is not true. And possibly $T = T_p$, but maybe there you just dropped the subscript. Just like $h$ and $L$, just because they have the same dimensions does not mean they have the same value, and you can cancel them out. Your result should be $$U_{orb} \times \frac{T_p}{H_{mo}} = \left(\frac {H_{rms}}{H_{mo}}\right) \dfrac{\pi}{\sinh\left(\frac {4\pi h}L\right)}$$ If you happen to know what $ \left(\frac {H_{rms}}{H_{mo}}\right)$ is equal to, then you can replace that with the numeric value, but otherwise, this is it. – Paul Sinclair Mar 15 '20 at 17:03

0 Answers0