It's possible to predict tide at a given location and at a given time reasonably well (e.g. predicting low tide). And there are tools available that make it possible to do this calculation for any location on earth at any time, such as OTPS.

Given that, it's possible calculate the annual maximum tide by calculating (sub)hourly data, and then taking the maximum. However, this is a pretty expensive way to do things - you're calculating 8.7k heights just to calculate a single maximum, and the calculations themselves are not particularly cheap. This quickly becomes a problem if you want to calculate a lot of locations.

Annual maximum tide is not the same year to year, and from memory, the tides have a long-term cyclic nature of around 18 years (that is the lowest common multiple of the periods of all of the cyclic components).

Is there a way of either analytically calculate the maximum tide height on a given day, or for a given year; or to calculate/estimate the time of that maximum (from which it would be possible to calculate the height)?

I'm only interested in the long-term predictable component of tides - astronomical tides (and perhaps other things, like local orographic effects?). Not interested in storm surge or wave effects, or other things that are hard to predict in the long term.

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    $\begingroup$ You are looking for an easy solution to the local maxima of $$\sum_i a_i \sin(\theta_i + \omega_i t)$$ While there are many ways to find such maxima that vary in computational complexity and computational cost, there is no nice and easy way to find such maxima. $\endgroup$ – David Hammen Jul 28 at 10:28
  • $\begingroup$ @DavidHammen: While the other answers are interesting and potentially useful, I guess your comment is the actually correct answer to the question. Post it as an answer, and I'll accept it. $\endgroup$ – naught101 Jul 29 at 23:54
  • $\begingroup$ Yes, that's the very basis, for realistic values for a given time and place a lot of correction is needed to find and refine the constituting elements. It'll be a brave deed :-) $\endgroup$ – user20217 Jul 30 at 10:17

You are looking for an easy solution to the local maxima of $$\sum_i a_i \sin(\theta_i+\omega_i t)$$ This means solving for solutions to $$\sum_i a_i \omega_i \cos(\theta_i+\omega_i t) = 0$$ This is a transcendental equation with multiple terms with different frequencies. While there are many ways to find such maxima that vary in computational complexity and computational cost, there typically is no nice and easy way to find such maxima.

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  • $\begingroup$ When you talk about "ways to find [local] maxima", are you talking about numerical solutions like Newton's method etc? $\endgroup$ – naught101 Jul 30 at 2:30

Tidal tables are regularly calculated by several oceanographic institutes worldwide and handed out e.g. for navigation. Influencing factors (superficially spoken) are positions of celestial bodies like Sun, Moon, Venus, Mars, Jupiter and others, modelling of the propagation of the tidal bulges over the ocean basins, and topography and coastal features to calculate individual heights and, equally important, strength and timing of currents. These calculations are done for certain points of interest, like ports or coastal features or areas that could pose a risk to navigation. It is calculated and published for a year in advance.

Modern electronic equipment displays values directly, else one would have to do it old style and search in the tables for adjacent places and a time interval and do a little interpolation manually.

The tables are usually to the minute, to the tenth of a knot in current, and a decimeter in height, but of course small scale local features and meteorology and weather can have huge influence and significally alter the precomputed values. But weather prediciton has its limits which are days.

Examples for tide tables:



You can check tidal figures for individual times and places yourself with the help of a nautical chart viewer (e.g. navionics.com chartviewer, the red diamonds with a T inside), or by buying the tables if one needs a copy, e.g. the ATT. I fear they cost a little something.

All that said, there is the concept of the "Highest Astronomical Tide" HAT, which is part of the calculations. Simply spoken the highest possible tide if all the celestial bodies pull in the same direction. But meteorological influences (e.g. a storm) cannot be calculated beyond a few days.

So, generally, this can not be done with exactness because of unforeseeable meteorological influences. If we ignore these (not advisable for nautical purposes) and for average meteorological conditions, the highest (or lowest) tide for a time and place can be taken from these tables.

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  • $\begingroup$ Good answer. I should have specified in the question that I was only interested in astronomical tides/the long-term predictable component of tides. $\endgroup$ – naught101 Jul 27 at 14:43
  • $\begingroup$ Ok, then i wouldn't bother doing the calculations myself and use such a product as the tide tables, eg the ATT if you're not afarid about the weather. Those guys know what they're doing and have refined the methods over decades. Some countries (Germany for example) even have stopped doing tidal predictions themselves and use publications of others. Btw. topography is the reason why a tide can rise 2m on an Atlantic island and 12m in the Channel. And that are no extermes. Just sayin' ... $\endgroup$ – user20217 Jul 27 at 15:34
  • $\begingroup$ Tide tables don't exist everywhere that I want them so i guess I will just have to calculate them myself. Yes, I'm aware of bathymetric effects (was including that in "long-term predictable components). The OTPS/TXPO9 solution does include at least some of these effects. $\endgroup$ – naught101 Jul 28 at 0:13

It depends on what you are trying to achieve.

If you need exact figures for the maximum tidal height each day/year at a given location, there are no simple analytical solutions due to the complex nature of tides.

If you are interested in ballpark figures (which are good enough for most recreational, but also some engineering applications), the maximum annual tidal height can be approximated by tidal datums, either the Mean High Water Springs (MHWS) or Mean Higher High Water (MHHW) depending on the tidal regime. If you would want to be extremely conservative, you could use Highest Astronomical Tide (HAT). Also taking the mean of MHWS/MHHW and HAT could be an option. This approximation would be suitable for 'regular' coastal and offshore locations. Complex bathymetry (e.g. funnel-shaped estuary) is trickier.

This could potentially help you save some of the computational effort, as you would need to calculate the tidal datums once per location (rather than once per year per location).

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  • $\begingroup$ +1 Oh, yeah, I didn't think of simply using MHWS for the location :-) $\endgroup$ – user20217 Jul 27 at 14:10
  • $\begingroup$ Clarified the question: only interested in astronomical tides (local bathymetric effects would be nice, but not essential). I'm interested in making a dataset that contains distribution parameters for annual maximum high tide for many locations around the world. So I only need that one number per year. HAT is not useful, because it only occurs every few of decades. I guess I could maybe infer the distribution from MHWS/MHHW and HAT though? But then, I'd also need to know those values, and I'm not sure if they're calculable analytically either. $\endgroup$ – naught101 Jul 27 at 14:52
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    $\begingroup$ You can calculate tidal constituents/harmonics from water level time series. Check (t_tide). Also the software you linked in your initial post (OTPS) seems to provide tidal constituents from gridded model output. To calculate MHWS etc, see formulas on page 6 of this R package documentation (sorry for the weird reference). This is a brute force method though :D. $\endgroup$ – Ingvar Lukas Jul 27 at 18:16
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    $\begingroup$ Yeah, OK. But even then, I'd still have to figure out how to infer a maximum annual tide height distribution from those, and that seems pretty non-obvious. I guess I'd have to do it empirically/numerically anyway. Seems like my best bet is really going to be to DIY.. $\endgroup$ – naught101 Jul 28 at 0:50

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