Time of day with most and least land?

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In summary, the time of day when Earth´s land area is largest, and what time of day is it smallest, is dependent on the phase of tide. The largest and smallest amount of ground exposed by tides at any phase of tide is dependent on the position of Moon.
  • #1
snorkack
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What is the time of day when Earth´s land area is largest, and what time of day is it smallest? Or more importantly, at which position of Moon is land largest and when is it smallest?

The volume of ocean is conserved by waves and tides. The water volume that rises in high tide can only come from low tide somewhere else in the ocean at the same time.
But there is no reason for ocean to conserve area! If there is a high tide on a gently shelving coast which floods broad tidal flats, there might be at the same time low tide at the opposite coast being steep cliffs and exposing no projected area. Or depending on the width and depth of the oceans, there might be high tides on both coasts at the same time while low tide happens in deep midocean and has no shores or shoals to expose.

The phase of tide is different around the oceans - while some places have high tide, others have low tide but yet others are in middle of flood and yet others in the middle of ebb. So the area of the ocean and land should be continuously changing depending on the distribution of tidal phases and ranges around the coasts and the distribution of ground at various altitudes between low and high tides.

What is the total area of Earth surface between low and high tide?
And what is the largest and smallest amount of ground exposed by tides at any phase of tide?
 
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  • #2
I'm confused.

What is the total area of Earth surface between low and high tide?
You seem to be ignoring the fact that, using your model of tidally exposed/unexposed land, that high tide happens one one side of the earth, and you forget the other side is low tide (simplified take on this). Only a small contiguous fraction of the ocean is at any full high tide at anyone time, the same idea holds for low tide.

Also, tides are cyclic neap tides -> spring tides -> neap tides. So only at the height of a spring tide can the most land area be exposed by low tide on two "sides" of the Earth (actually a band), while least land area is exposed at high tide (two bulges: sublunar and antipodal). Is this what you are driving at in your question?

Plus "time" has nothing in particular to do with the biggest spring tide - just the position of the Earth relative to sun and moon, plus the moon at apogee. A new moon at lunar apogee and Earth apogee (closest to the sun) would be the strongest possible spring tide.

Maybe what I said is not relevant to what you are asking, so please try again... and I just do not get it.

This explains things with pictures. http://en.wikipedia.org/wiki/Tide
 
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  • #3
jim mcnamara said:
You seem to be ignoring the fact that, using your model of tidally exposed/unexposed land, that high tide happens one one side of the earth, and you forget the other side is low tide

Is that true? I thought the other side was on high tide too?

300px-Tide_overview.svg.png


Anyway, are you suggesting that the tides preserve area? Because this is clearly not true. Consider an Earth with an ocean of 10 km deep everywhere with only one mountain that barely sticks out of the ocean at low tide. Then as the tides change, the entire Earth is covered with water. So in this earth, the area of land covered by water does change during the day. So I see no reason as to why the area of land does not change during the day in this earth.
 
  • #4
No, I cited the same picture. High tides are two bulges. I thought I conveyed that idea, my bad if not.

I do not know a really good answer. My attempt at an answer:

An answer is to look for an aggregation of geographic locations like Minas Bay http://www.thehighesttides.com/ and use tide charts to figure out when low and or high tides are in majority of them. Your island example is correct in an abstract sense, but tides around islands are minimal. The Bay of Fundy can, in your abstract sense, be likened to water sloshing in a bathtub.

What we are discussing is the delta in the exposed area of the littoral (intertidal zone) http://en.wikipedia.org/wiki/Intertidal_zone worldwide.

What snorkack is actually asking I think: is there a way to determine a worlwide exposed littoral maximum and minimum and how to predict the UTC time? I cannot find an authoritative answer, but this is a really interesting question. He would require the largest possible spring tide.

Anybody?

Note:
From http://en.wikipedia.org/wiki/Tide#Phase_and_amplitude
The shape of the shoreline and the ocean floor changes the way that tides propagate, so there is no simple, general rule that predicts the time of high water from the Moon's position in the sky. Coastal characteristics such as underwater bathymetry and coastline shape mean that individual location characteristics affect tide forecasting;
 
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  • #5
jim mcnamara said:
No, I cited the same picture. High tides are two bulges. I thought I conveyed that idea, my bad if not.

Ah yes, then I just misinterpreted your previous answer. I already thought I misinterpreted you, but I wanted to make sure :smile:
 
  • #6
It would be interesting (and challenging) to try to estimate this for a smaller region where all the data (maps showing mean high and low water around the coast, plus relative times of high and low water) is available - e.g. the UK.

As Jim mcnamara implied, the simple model of tides on a planet completely covered with deep water is completely irrelevant to this question. For example on the east coast of the UK, the high tide is essentially a traveling wave of water traveling south into the relatively shallow water of the North Sea, and there is 5 or 6 hours time difference between high tide at the north and south ends of the coastline, over a distance of only about 1000 miles.

Not to mention the fact that at some places on Earth and at some phases of the moon, there are not even two high tides per day. There may only be one, or (for example at Southampton on the south coast of the UK) four.
 
  • #7
  • #8
micromass said:
Is that true? I thought the other side was on high tide too?

300px-Tide_overview.svg.png
That isn't true. There is no such thing as a tidal bulge. That was Newton's idea, and it is one of the few places where Newton was wrong. It's Laplace's dynamic theory of the tides that provides a much more accurate picture of how the tides work.

At any given time of day, one can always find a spot somewhere along the coast of the North Sea where it's high tide, and some other spot where it's low tide. The same applies to New Zealand, Patagonia, and Hudson Bay.
 
  • #9
D H said:
That isn't true. There is no such thing as a tidal bulge. That was Newton's idea, and it is one of the few places where Newton was wrong. It's Laplace's dynamic theory of the tides that provides a much more accurate picture of how the tides work.

At any given time of day, one can always find a spot somewhere along the coast of the North Sea where it's high tide, and some other spot where it's low tide. The same applies to New Zealand and Patagonia.

Then why do you see Newton's theory everywhere? Including wikipedia and in classrooms where children are actually taught that? This is very much a surprise for me.
 
  • #10
micromass said:
Then why do you see Newton's theory everywhere? Including wikipedia and in classrooms where children are actually taught that? This is very much a surprise for me.
Lies to children, maybe?

Newton's tidal theory does yield the correct tidal forcing functions. It does not yield the correct responses to these forcings. I've posted on this before, multiple times. Here's one:
D H said:
Both the inertial and Earth-fixed points of view yield valid explanations of the tidal forcing functions. They do not explain the tidal bulges for the simple reason that Newton's tidal bulges do not exist. They can't exist for a number reasons:
  • Newton's equilibrium theory of the tides dictate that high tide occurs when the Moon is closest to zenith / closest to nadir and that low tide occurs when the Moon is on the horizon. This happens only rarely, and it's sheer dumb luck when it does. One problem is that there are two huge north-south barriers to this supposed tidal bulge, the Americas and Africa/Eurasia. If Newton's equilibrium theory of the tides was correct, the timing and magnitudes of the tides at the Pacific and Atlantic sides of the Panama canal would be more or less the same. They're anything but. The tides on the Pacific side of the Panama canal are huge, over an order of magnitude larger than predicted by Newton's equilibrium theory of the tides. The tides on the Atlantic side are rather small, smaller even than the smallish tides predicted by the equilibrium theory. The tides on the Pacific and Atlantic sides of the canal also differ significantly in timing. The tides don't magically pick up where they would have been had Panama not existed.

    Another place of interest is the North Sea. The North Sea is rather small; per Newton's equilibrium theory of the tides, high tide should occur more or less simultaneously across all of the North Sea. That's not what happens. Instead, regardless of time of day, you can always find some point in the North Sea where it's high tide. At the exact same time, you can also find some other point in the North Sea where it's low tide. The equilibrium theory of the tides cannot come close to explaining the tides in the North Sea.

  • Suppose those continental barriers didn't exist, with the Earth covered everywhere by four kilometer deep oceans (the median depth of the oceans). You still wouldn't get Newton's tidal bulges. The problem is the Earth's rotation. The tidal bulges would be an ocean wave with a wavelength equal to half the circumference of the Earth. This wavelength is much, much greater than the four kilometer depth. The tidal bulge (if it existed) would be a shallow wave. Shallow waves travel at a speed of ##\sqrt gd##, or about 200 km/second in the case of a 4 km deep ocean. The tidal bulge on the other hand would need to travel at 465 km/second. The tidal bulge cannot exist even in a water world with oceans only 4 km deep.

  • The oceans would have to be over 20 to 30 km deep to support Newton's tidal bulge, and even with that Newton's tidal bulge still wouldn't exist except near the equator. The problem this time is the coriolis effect. The tidal bulge can't exist as a world-wide phenomenon thanks to Kelvin waves. Any bumps in the bottom of this deep, deep ocean will eventually result in amphidromic systems forming, even in this very deep ocean.

The correct theory of the ocean tides started with Laplace, about 100 years after Newton's time, and culminated in the early 20th century with the works of George Darwin and A T Doodson. This dynamic theory of the tides accounts for the issues described above. There is no tidal bulge, at least not in the ocean tides. Newton's equilibrium theory does work (to some extent) for the Earth tides.
 
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  • #11
D H said:
That isn't true. There is no such thing as a tidal bulge. That was Newton's idea, and it is one of the few places where Newton was wrong.

I wonder if the experimental data Newton quoted in "Principia" was selectively chosen to support his theory, or whether being honest with the limited amount of data he actually new. An uncritical reading certainly gives you the idea that he had explained "everything".

Of course the seafarers of that time would have known a lot of accurate empirical data about tides (especially around an area like the North Sea), but scientists didn't have much access to that level of information.
 
  • #12
DH. thanks for tutorial! I didn't realize the extent of the mismatch between tidal forcing and real world results.
How significant are the contributions of SST, wind and currents to the resulting water level? Surely, ice-pack complicates this; are seasonal influences significant? In my ignorance, I don't even know how wave action on the shoreline is incorporated into determination of the tide "line".
I got my info from Wikipedia, it is too bad that that presentation isn't as clear about how poorly the Newton mean depth model fits.
-=-=-=-
Would the most correct answer to snorkack's question be that the total area exposed at a given instant is most likely chaotic and unpredictable? Would it be better to consider the (19 year) mean lower low and higher high?
 
  • #13
DH
Also a thankyou from me
I had no idea there was a better explanation

cheers
Dave
 
  • #14
The tidal forcing functions that Newton derived are correct. It's the overly simplistic two bulge equilibrium response to those forcings that is incorrect.

With regard to understanding the tides, here are a couple of online resources. There are plenty more.
http://www.oberlin.edu/faculty/swojtal/SFWpage/161Stuff/161Lect17/sld001.htm
http://faculty.washington.edu/luanne/pages/ocean420/notes/tidedynamics.pdf

With regard to how tides are measured, see http://tidesandcurrents.noaa.gov/levelhow.html.Physicists and engineers study the response of a damped oscillator to oscillating inputs. Did you ever make waves in the bath tub when you were a kid? (I got in a good deal of trouble for soaking the floor.) The oceans are big huge bathtubs with complicated responses to driving inputs, and the driving inputs have a number of different characteristic frequencies. The response to anyone of these input frequencies will be an amphidromic system. For example, here's the response to the M2 (principal lunar semi-diurnal) component of the tidal forcing function:

attachment.php?attachmentid=68711&stc=1&d=1397657207.gif


There are curved white lines emanating from the the centers of the dark blue areas in this image. Those centers are "amphidromic points", places where the response to the roughly twice daily component of the tidal forcing function due to the Moon is zero. Those curved white lines are cotidal lines, places where the phase delay to the M2 tidal component are equal. If the M2 tidal component was the only thing that caused the tides, high and low tide would occur at the same time along points on one of those cotidal lines. The response to other driving frequencies (there are a number of them) will be rather different, but it's the twice daily response to the Moon, that governs the tides at most places on the Earth covered by water.

Here's the M2 tidal response in the North Sea:

attachment.php?attachmentid=68702&d=1397636275.gif


The red points in this image are the amphidromic points. There are two in the North Sea and (at least in this image) one just on-shore in Norway (others have that point just off-shore). The cotidal lines are in red, spaced by about one hour. The blue curves represent tidal responses. From this image one can see that its always high tide somewhere in the North Sea.Explaining the tides becomes rather ad hoc at the local level. If you look for a tidal station (e.g., Key West, Florida), you'll inevitably see harmonic constituents to a number of different aspects of the overall tidal forcing function. For example, at Key West, Florida, the lunar semi-diurnal (M2) tidal coefficient is the dominant cause of the tides, but there are several other frequencies that come into play. Understanding this as a physicist is easy: Just think of it in terms of Fourier analysis.With regard to your original question, snorkack, I suspect it was based on the idea that the equilibrium tidal bulge is what governs the tides.
 

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  • #15
D H said:
With regard to your original question, snorkack, I suspect it was based on the idea that the equilibrium tidal bulge is what governs the tides.

No. See the quote from the first post:
Or depending on the width and depth of the oceans, there might be high tides on both coasts at the same time while low tide happens in deep midocean and has no shores or shoals to expose.

If there were an equilibrium tidal bulge, the phase of tides across the ocean or in the middle would depend only on the width of the ocean and be independent of the depth of the ocean, as I mentioned from the start.
 

FAQ: Time of day with most and least land?

What is the time of day with the most land?

The time of day with the most land is typically around midday, between 12pm and 2pm. This is because the sun is at its highest point in the sky, providing the most direct sunlight and warmth to the landmasses.

What is the time of day with the least land?

The time of day with the least land is typically during the early morning hours, between 3am and 5am. This is because the sun is at its lowest point in the sky, causing less direct sunlight and cooler temperatures for the landmasses.

Does the time of day with the most and least land vary by location?

Yes, the time of day with the most and least land can vary by location. This is due to differences in time zones and the tilt of the Earth's axis, which affects the amount of sunlight and warmth received by different regions at different times.

How does daylight saving time affect the time of day with the most and least land?

Daylight saving time can affect the time of day with the most and least land, as it shifts the time by one hour. This can cause the peak and lowest land times to occur an hour earlier or later, depending on the location and time of year.

Is there a specific time of year when the time of day with the most and least land changes?

Yes, the time of year when the time of day with the most and least land changes is during the solstices in June and December. During the summer solstice, the time of day with the most land shifts to the evening, while during the winter solstice, it shifts to the morning.

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