West coast tides

I understand more or less how tides work – the moon pulls on the water (sort of), causing a swell along a meridian (and the opposite meridian as well). As the Earth spins, this swell gathers up along a coast, causing high water. As the Earth continues to spin, this water is pulled across dry land, causing massive flooding and…

OK, maybe I don’t fully understand it. Anyway, the water abates. 12 hours later, it rises again. This 12 hour cycle is delayed a bit each time because the moon itself is in orbit. But at any rate, tides are more or less predictable, as one can guess that if there was a high tide at noon today, there will be one tomorrow at around 11:10 am.

This is all well and good for the East coast. But what about the West? I understand that since all that water has to get around two continents, things get a little weird. So much so that tide clocks are pretty useless.

So how do tides work in, say, Santa Monica? Or what about the Gulf? Or what about, say, the East coast of GB?

There are local tide tables published for each local area. The time of tides depends quite a bit on the physical layout of the coastal region.

You don’t understand how tides work. You think that they are the result of water moving around the globe. Instead, they are the result of the added attraction of the moon’s gravity at the surface of the Earth compared to the attraction of the moon’s gravity on the center of the Earth. That added attraction pulls the water “up” towards the moon as it passes; the land doesn’t do the same thing since it is rigid and connected to the rest of the Earth (well, it doesn’t do it to the same degree, a better way of putting it).

So it doesn’t matter that the water is on the West Coast of North America; it still gets pulled “up” as the moon passes, rising in comparison to the shore line. 12 hours later, the land gets pulled “down” more than the water does as the moon passes “under” the Earth (relative to the surface where you are), causing the second tide of the day.

There are actually some effects from a wave of water pulled up by tides flowing around the Atlantic. But that’s not the only thing causing the tide to come in. That wave doesn’t seem to flow east to west, either- it goes from south to north in the Atlantic.

But the main mechanism for tides is what you described, not what the OP is thinking.

The moon, and to a lesser extent the sun, pull the earth into a slightly oval shape with the oval’s long axis pointing slightly ahead of a line pointing to each body. These oval shapes stand still relative to the moon and sun and the earth rotates through them giving two high and two low tides every day. However the tide time and height at any locality depends quite heavily on the configuration of the local coastline.

There is also the phenomenon of resonance[sup]1[/sup]. The extreme tides at the Bay of Fundy result from a combination of a long tapering end to the bay plus the fact that the length of the bay and some multiple of the travel time of waves from end to end is in sync with with the period of the moon.

  1. According to a Scientific American article of 20 years or so ago.

In any one given location, the effect of having the water pulled “up” can be different; the Bay of Fundy is a classic example.

But one important thing to do when answering questions here is to recognize the basic underlying concept upon which the question is formulated. The OP appears to be wondering what effect having a massive continent stretched from almost the north pole to almost Antarctica would have on the movement of the wave of water being pulled around the globe by the tide. This is the point I was addressing. Obviously, the tide behavior in Los Angeles is different from the tide behavior in Puget Sound; that goes without saying.

(related question)

Is the moons gravitational pull on us measureable?
If you had an ultrasensative scale that measured to 100ths of a pound would your weight be different when the moon is directly overhead compared to when it is on the opposite side of the earth?

Are there measureable tides in landlocked seas ? (like the Red sea, baltic Sea, Hudson’s bay,etc.)??

The predictability of tides does vary on the East and West Coast. The east coast of the U.S. experiences Diurnal Tides where the change from High to High is every 12 hours and 40 some minutes, and the tidal pattern follows a very even sine wave. On the west coast of the U.S. we have Irregular tides where the difference between high and low changes and the tidal high and low changes, the plotted wave form is not symmetric like the diurnal.

As for Landlocked Seas, the Caspian Sea, The Great Salt Lake, or the Great Lakes may be better examples of seas that are landlocked. The Red Sea is an Inlet of the India Ocean, and the Baltic is an extenstion of the North Atlantic.

But it’s not so different that it’s impossible to compile a tide table for Santa Monica. Even a place like Southampton, which has a double high tide because of the effects of the Isle of Wight, can have tide tables computed.

Maybe measurable, but very small. The Baltic and Mediterranean seas only have very small tides because their connection to the ocean is so narrow.

The moon is 1/80 as massive as the earth, and its center of mass is about 60 times farther away from us than the center of the earth. Ergo, the moon’s gravitational pull on us is (1/80) * (1/3600) = 1/288,000 of the Earth’s pull. Whether this is measurable, I do not know.

The moon’s distance from earth varies from 222,000 to 254,000 miles. The earth’s radius is only 4,000 miles. So you see that the moon’s position in its orbit has more impact on the distance between yourself and the moon than does the rotation of the earth.

When the moon is below you and at apogee (258,000 miles away), it exerts only 71% of the pull on you than it does when it’s above you at perigee (218,000 miles). But again, this is 71% of 1/288,000. We’re somewhere in the range of ten-thousandths of a pound. So don’t blame the moon if you’re gaining weight!

AND the gravitational field is lower on the far side (relative to sun or moon) than the field at earth center, thus the second tidal hump, and a 12 hour cycle instead of 24. In this case the earth as a whole is being more strongly pulled toward the moon that the water on the far side.

The gravity gradient is key to understanding tides. Even though the sun has a stronger gravitational pull on the earth, the long distance means that the gravitational gradient is not a large as that due to lunar gravity, thus the stronger effect of the moon on tides.

As I think about this some more, the rotation of the earth is more important than perigee versus apogee, because the overhead-moon pulls opposite to the earth and decreases your weight, whereas the underneath-moon pulls in the same direction and adds to your weight. The moon’s distance from earth affects the magnitude of the diurnal swing–greater at perigee, lesser at apogee.

So I estimate the diurnal swing at around one part in 144,000–in the range of a thousandth of a pound for a human being. Provided, of course, that you live in the tropics and the moon happens to be exactly overhead and underneath on a particular day.

Now if you were to stand on Mount Kilimanjaro during a total eclipse when both the moon and sun were directly overhead–mmmm, it beats dieting.

(how about a follow up question?)

How about weight difference while standing on the moon? Noticeable difference when the earth is overhead compared to when your on the opposite side of the moon?

Well, let’s see. The Earth is about 240 moon-radii away from the moon, and is still 80 times as massive . . . so as you stand on the moon, the earth’s gravity equals 80/240^2 = 1/720 of the moon’s gravity. That influence can either be added to or subtracted from the moon’s pull, resulting a weight change of about 0.28%.

Mind you, that’s off of a smaller weight to begin with, since the moon’s surface gravity is only one-sixth of the earth’s. And mind you also, you’ll have to hop a commuter rocket and fly to the other side to observe the difference, because no spot on the moon will ever rotate from directly under to directly over the earth!

Seems pretty close, actually, according to this. I understand the folly I presented in my OP when I asked about water going around two continents.

Still, while close to diurnal, they are not exact. What accounts for this, and how can it be predicted?

I believe, if you read carefully, I said that. :wink:

And you’re both wrong. Or at least misleading. The Moon’s gravity pulls on the oceans just as much as the land or Earth as a whole.

The Earth is in orbit around the Moon just as much as the Moon is in orbit around the Earth. But the Earth is not a point object but rather has a radius of 6000+ Km. So consider an object on the surface of the Earth nearest the Moon. If it were not constrained by the Earth’s gravity, it would go into a different orbit about the Moon than the Earth since it’s 6000+ Km closer to the Moon but has the same orbital velocity as the Earth. By the same token, an object on the opposite side will also want to go into a different orbit, since it also has the same orbital velocity as the Earth but is 6000+ Km further from the Moon. This is what causes tides on the opposite side of the Earth.

The difference between the ocean and land in response to tides has to do with the fact that land has structural integrity (i.e. it’s a solid) that keeps it from responding as much as the free-form ocean. Tides in land are on the order of half a meter or so, while in the ocean they are more (I forget the amount, 2 or 3 meters or something like that). Note I’m talking about the open ocean here; tides at any particular piece of shoreline will vary considerably depending on local conditions (as indicated in other posts in this thread).

If you live on the east coast of Vancouver Island you’d know that tides are the result of water moving around.