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wrote...
Posts: 825
4 weeks ago
Referring to following articles, for surfing, I would like to know on whether formula is available or not to calculate the tides' conditions on any given date.

https://surfsimply.com/science-skepticism/calculating-the-tides-the-rule-of-twelfths/

Does anyone have any suggestions?
Thanks in advance for any suggestions
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wrote...
Educator
4 weeks ago
Are we discussing tidal force?

Tidal forces are based on the gravitational attractive force. With regard to tidal forces on the Earth, the distance between two objects usually is more critical than their masses. Tidal generating forces vary inversely as the cube of the distance from the tide generating object. Gravitational attractive forces only vary inversely to the square of the distance between the objects. The effect of distance on tidal forces is seen in the relationship between the sun, the moon, and the Earth’s waters.

Our sun is 27 million times larger than our moon. Based on its mass, the sun's gravitational attraction to the Earth is more than 177 times greater than that of the moon to the Earth. If tidal forces were based solely on comparative masses, the sun should have a tide-generating force that is 27 million times greater than that of the moon. However, the sun is 390 times further from the Earth than is the moon. Thus, its tide-generating force is reduced by 3903, or about 59 million times less than the moon. Because of these conditions, the sun’s tide-generating force is about half that of the moon.

Here's a better article pertaining to the "rule of twelfth" -- https://en.wikipedia.org/wiki/Rule_of_twelfths. Mind you, it's my first time reading about this, so I'll need to research it further to understand it better.
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wrote...
4 weeks ago
by using the rule of twelfths the tide is calculated like that: take the difference in height between the high and low tide on that day, and divide that by 12 equal chunks. For example, if high tide was 8ft above chart datum, and low is 2ft above chart datum – there is a 6ft difference, divided that into 12 even chunks; equaling 0.5ft per chunk. Simple.
What we then do is assume that there is a 6 hour gap between high and low tide; although not completely accurate, it is close enough for what we are attempting here.
For each hour of those 6, the tide will move at a fixed rate.
So starting at low tide:
The first hour, 1 twelfth of the tide would rise; so one chunk would come in, 0.5 ft.
The second hour, 2 twelfths of the total will come in, therefore 1 ft.
The third hour, the tide is starting to move quickly, and so 3 twelfths will come on, therefore 1.5ft.
At this point we are now passing mid tide.
The fourth hour again is 3 twelfths, so another 1.5ft.
The fifth hour the tide slows down again, 2 twelfths, therefore 1ft.
The sixth and final hour, 1 twelfth would come in, and therefore 0.5ft.
The tide would then drop following the same pattern. This is a way of putting a numerical value on the interim period between high and low tide, where the tide is running particularly fast.
wrote...
4 weeks ago
Thanks, to everyone very much for suggestions (^v^)
wrote...
Staff Member
3 weeks ago
If a tide takes on cyclical pattern, it can probably be modelled using a trigonometric function, like sine or cosine.
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wrote...
3 weeks ago
Thanks, to everyone very much for suggestions (^v^)
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