Larger moon = Larger Tides = No oceanic ships?

It would not discourage oceanic shipping, but it would make the times it can be done more selective.

If you have a port that is 20m deep and full at high tide, with our Moon you should still have at least 10m left at low tide (depending on the tides in that area). That's plenty to operate big ships in.

If your Moon is closer and your tides bigger, your 20m harbor might be empty at low tide. This means ship captains need to plan more accurately: they need to be in and out of a harbor this shallow within a couple of hours. Delays in voyages would lead to big delays at the destination waiting for the next tide.

The alternative is that harbors are built bigger, which costs a lot more in both money and time.

posted about 9 years ago

ArtOfCode‭ staff

101 reputation 4 65 10 3

If you want a detailed analysis, simply solve Laplace's tidal equations. These are not easy to do. If you're curious, I can get you part of the way to what might be an answer. Also, I'm curious as to what the math will turn up.

Achille Hui has a very helpful spoiler (regular ones don't work for $\LaTeX$, and take up too much space), which I've used to cover up the math. Click on it to show some of my work, or skip it to read more.

$$\require{action} \toggle{ \begin{array}{cl} & \bbox[2pt,color:black;border-radius:3px;box-shadow:4px 4px 8px black]{ \verb/Click to show math/} \end{array} }{ \begin{array}{cl} A=\frac{1}{a\cos\varphi}\frac{\partial}{\partial\lambda}(g \zeta+U) \\ B=\frac{1}{a}\frac{\partial}{\partial\varphi}(g \zeta+U) \\ C=2\Omega\sin\varphi \\ \\ \text{We now have} \\ \\ \frac{du}{dt}-vC+A=0 \\ \frac{dv}{dt}+uC+B=0 \\ \\ \text{This leads to} \\ \\ \text{I have no idea how to solve this. Wolfram Alpha}^1\text{ could help, but I don't} \\ \text{have a subscription. Something out there will help you.} \\ \\ \text{When you have that, substitute in for $A$, $B$ and $C$ and plug it all in to } \\ \\ \frac{\partial\zeta}{\partial t}+\frac{1}{a\cos\varphi}\left[\frac{\partial}{\partial\lambda}(uD)+\frac{\partial}{\partial\varphi(vD\cos\varphi)} \right]=0 \\ \\ \text{Solve for } \zeta \text{ and enjoy.} \\ \end{array} } \endtoggle$$

^{1 You can find a potential starting point here.}

Note:

This is just to let you know what the math is. There are other answers that cover everything else, so I've decided to leave this as is, because I couldn't add anything else.

posted about 9 years ago

HDE 226868‭

586 reputation 61 463 77 49