|
"Thom Stewart" wrote in message ... Donal, I hope you're satisfied! For a damned 1/4 of a point!? All this differential and centrifugally, how does this Knowledge help a sailor to ride the tides? That was the original question. Remember? IIRC, I gave him my honest opinion. I hope you're satisfied (g) I'll admit that I'm pleased. Now, I wonder if Scot did any Tide Riding while he has been on Vacation Cruise? It will be interesting to find out how his trip went. Does anybody know when we can expect to hear from him? Regards Donal -- |
No, you never said it wasn't needed, but you did minimize its importance, claiming
that it can't be used alone to predict the tides. You said: "Even worse, the site then goes on to use the _differential_ expression to calculate the ratio of forces between the moon and sun!" You even produced a bogus formula to support your claim that centrifugal force varies across the Earth. This would predict a tidal force 65 times stronger than differential gravity. If that isn't minimalizing differential gravity, I don't know what is. You demonstrated a further lack of understanding with: If the moon stopped its rotation around the earth and the earth and the moon was "falling" toward each other, there would still be two bulges. "YES but how big would they be (Hint: Smaller than the tides?)" Actually, the tides would be the same. One fundamental difference that we have is that you insist that taking centrifugal force into account is the *only* way to look at the problem. As I've said a number of times, centrifugal force is a "fictional force" that is only needed if you wish to work in the accelerating Earth-centric reference frame. In fact, it is required that there must be an alternate approach that does not use "fictional" forces. It is quite easy to explain the tides without Centrifugal force - I mentioned the approach in one of my first posts. There is a gravitational force pulling the Earth and Moon together. This creates an acceleration such that the Earth is in free fall, and no net force is felt at the Earth's center. However, there is a larger force on the Moon side, and a smaller force on the far side. Since the average force has been accounted for it must be subtracted from these two forces, and the smaller force ends up being the same magnitude as the larger, but in the opposite direction. Hence, two tidal bulges, both caused solely by gravity. Another fundamental difference we have is that I agree with the traditional value for the tidal force. Ignoring minor effects, the result predicted by Differential Gravity (whether or not you use centrifugal force as part of the explanation) that is about 2 feet for both the near and far side bulges. (The Sun's contribution is about half of the Moon's.) The land masses and shallow water tends to "pile up" the water to create tides that are somewhat higher. You, however, have claimed that the variation of the centrifugal force from the near side to the far creates a force that dominates the tides. Your formula predicts a force 65 times greater than the differential formula, which would seem to create tides 100 feet or more. Your explanation that land masses dampen these tides runs counter to common experience. And, your formula predicts the Sun's contribution is only 1% of the Moon's, which is clearly not the case. As to your claim that the rotation around the barycenter "powers" the tides, well, that's not true. In fact, even if the Earth and Moon were not rotating around the barycenter, the tides would be the same, at least, until the Earth and Moon collided. Frankly, it clear that you still do not accept the fact that CF is constant, exactly equals the average gravitational force, and thus has no interesting contribution to the tides. "Nav" wrote in message ... Now why try to distort the truth Jeff? I never ever said differential gravity was not needed. I always said that it's the difference between gravity and centrifugal forces. You do understand the connotations of the DIFFERENCE between forces don't you? It does not mean that either component is zero and actually implies that both are important. Shesh! Still it's nice to see that you now agree that centrifugal forces should not be ignored (as they are in the gravity only model). As I've said so many times, the key to understanding is that the system rotates about the barycenter and it is not just a gravity field problem. The rotation actually provides the energy needed to power the daily tides -think about it OK? Cheers Jeff Morris wrote: "Nav" wrote in message ... Jeff Morris wrote: "Nav" wrote in message ... F= mr omega^2. The distance from the barycenter to all points on earth is NOT the same. As the site expalins in the next paragraph, only the center of the Earth rotates around the barycenter. Other points rotate around neighboring points. Anyway, the site clearly shows in Fig. 2 that it is the _DIFFERENCE_ between gravity and centrifugal force that makes the tides, not gravity alone. We never disagreed on this point. HOLY BACKPEDAL!!!!!! I'm not sure you really want to go back over this thread - your record is rather shaky. Mine, however, has been quite consistent. Remember, I started by posting sites with differing approaches to show that this problem can be looked at in different ways. I then made my first comment about Centrifugal force with: "Remember that Centrifugal Force may be a handy explanation, but it is a "fictional force" that only appears real to an observer in an accelerating frame of reference. Therefore, whenever it is used to explain something, there must be another explanation that works in a non-accelerating frame." but then you started claiming that differential gravity wasn't needed, I responded with: "Before I thought you were just arguing philosophically how much we should credit centrifugal force, but now it appears you haven't really looked at the math at all. The reason why "differential gravity" is invoked is because it represents the differing pull of the Moon on differing parts of the Earth. Although this force is all obviously towards the Moon, when you subtract off the centrifugal force this is what is left. It is this differing pull that causes the two tides." a few posts later: "Given that, your argument falls apart. The centrifugal force is exactly the same on all points of the Earth, and (not by coincidence) is exactly opposite the net gravitational force. What is left over is the differential gravity." The bottom line here is that the tides are properly described by the differential gravity equation. Centrifugal force can be used to explain how an outward force can be generated, but it is not needed, and it does not yield the equation that describes the tides. Frankly, your the one who started this by claiming that the traditional explanation of tides is fundamentally flawed, and that the differential force normally cited is not what causes the tides. You really haven't produced any coherent evidence to support this claim. |
"Jeff Morris" wrote in message ... "Donal" wrote in message ... "Martin Baxter" wrote in message Donal wrote: So why does the moon seem to have a greater impact on the tides? Well duh! Remember F=G*(m'*m")/(d^2), Emmm... Huh? What the hell does that mean in English? Did you not take physics in school? Sadly, I didn't! With hindsight, I suspect that I had a poor teacher who managed to make the subject appear much duller than it really is. My high school physics teacher was possibly the worst teacher I ever had - a true nut case who shouldn't have been left alone with children. Fortunately I found much better teachers in college. I've still got a suspicion that if we expand your equation, we will find that the sun has a greater gravitational influence on the earth than the moon does. Yes, its does. The direct gravitational pull of the Sun is enormous, much larger than the Moon's. However, the tides are caused by the difference in pull between the near side and the far side. Sorry! This doesn't make any sense at all. How does the water on the far side(of the earth) know that there is a different pull on the other side? Even if you were correct, then there would be a high tide facing the moon, a low tide at right angles to the moon, and a much lower *high* tide opposite the moon. The reality is that the HW opposite the moon is only fractionally smaller. Centrigugal force explains why there is a high tide on the oppisite side of the Earth from the moon - if you consider that the two bodies are rotating around a common centre. Since the Moon is a lot closer, that difference is more significant. If you remember any calculus, you'd know that differentiating a 1/r^2 function yeilds a 1/r^3 function. The inverse cube of the Sun's distance becomes a tiny number compared to the Moon's. The net result is that the Moon's effect on the tides is 2.2 stronger than the Sun's. Tsk...tsk. The moon only has a stronger effect on tides because the Earth and Moon are an orbiting pair. Where's Gilligan when you need him? Regards Donal -- |
"Donal" wrote in message ... "Jeff Morris" wrote in message I've still got a suspicion that if we expand your equation, we will find that the sun has a greater gravitational influence on the earth than the moon does. Yes, its does. The direct gravitational pull of the Sun is enormous, much larger than the Moon's. However, the tides are caused by the difference in pull between the near side and the far side. Sorry! This doesn't make any sense at all. How does the water on the far side(of the earth) know that there is a different pull on the other side? It doesn't "know" anything. Because the Earth and Moon are an "orbiting pair," as you say, they are falling towards each other. Because the gravitational field varies, the near side falls faster than the middle; and the far side falls slower. Hence, they bulge out from the middle. That's actually all that is needed to explain the tides; its so simple a lot of people have trouble getting it. Even if you were correct, then there would be a high tide facing the moon, a low tide at right angles to the moon, and a much lower *high* tide opposite the moon. The reality is that the HW opposite the moon is only fractionally smaller. Well, you're right that there are low tides at right angles, but the way the math works out the far side high tides are virtually the same. The magnitude of the differing pull 4000 miles closer to the Moon is about the same as 4000 miles further out. (Though I'm curious now just how much they do differ from each other ...) Centrigugal force explains why there is a high tide on the oppisite side of the Earth from the moon - if you consider that the two bodies are rotating around a common centre. OK. Centrifugal force is the explanation for children. Its kind of like the Tooth Fairy of physics. The problem is that while CF can explain the "outward force" needed for the far bulge, its still the differing gravitational force that defines the size and shape of the tidal force. And CF isn't needed at all if you can accept the "free fall) argument. Since the Moon is a lot closer, that difference is more significant. If you remember any calculus, you'd know that differentiating a 1/r^2 function yeilds a 1/r^3 function. The inverse cube of the Sun's distance becomes a tiny number compared to the Moon's. The net result is that the Moon's effect on the tides is 2.2 stronger than the Sun's. Tsk...tsk. The moon only has a stronger effect on tides because the Earth and Moon are an orbiting pair. The Earth and Sun are also an orbiting pair. There is no qualitative difference, only quantitative. Where's Gilligan when you need him? Gilly is an educated man. I'm sure he agrees with me. |
|
|
"Jeff Morris" wrote in message ... "Donal" wrote in message ... "Jeff Morris" wrote in message I've still got a suspicion that if we expand your equation, we will find that the sun has a greater gravitational influence on the earth than the moon does. Yes, its does. The direct gravitational pull of the Sun is enormous, much larger than the Moon's. However, the tides are caused by the difference in pull between the near side and the far side. Sorry! This doesn't make any sense at all. How does the water on the far side(of the earth) know that there is a different pull on the other side? It doesn't "know" anything. Because the Earth and Moon are an "orbiting pair," as you say, they are falling towards each other. Because the gravitational field varies, the near side falls faster than the middle; and the far side falls slower. Very neat! However, your view seems to be a little bit simplistic. Why should a solid fall more slowly than a fluid in a gravitational field? If your theory was correct, then there wouldn't be any tide at all. You seem to be ignoring momentum. Regards Donal -- |
Jeff,
Remember that I first posted that very same sentiment, and even provided a graphic. I still believe that to be true, but have modified my internal model, giving allowance for the centrifugal force. I'm not a physicist, but the way I'm seeing it, there is a middle ground in this discussion. I'm curious to know if you're discounting centrifugal force as a contributor to the far bulge. Scout "Jeff Morris" wrote in message ... "Donal" wrote in message ... "Jeff Morris" wrote in message I've still got a suspicion that if we expand your equation, we will find that the sun has a greater gravitational influence on the earth than the moon does. Yes, its does. The direct gravitational pull of the Sun is enormous, much larger than the Moon's. However, the tides are caused by the difference in pull between the near side and the far side. Sorry! This doesn't make any sense at all. How does the water on the far side(of the earth) know that there is a different pull on the other side? It doesn't "know" anything. Because the Earth and Moon are an "orbiting pair," as you say, they are falling towards each other. Because the gravitational field varies, the near side falls faster than the middle; and the far side falls slower. Hence, they bulge out from the middle. That's actually all that is needed to explain the tides; its so simple a lot of people have trouble getting it. Even if you were correct, then there would be a high tide facing the moon, a low tide at right angles to the moon, and a much lower *high* tide opposite the moon. The reality is that the HW opposite the moon is only fractionally smaller. Well, you're right that there are low tides at right angles, but the way the math works out the far side high tides are virtually the same. The magnitude of the differing pull 4000 miles closer to the Moon is about the same as 4000 miles further out. (Though I'm curious now just how much they do differ from each other ...) Centrigugal force explains why there is a high tide on the oppisite side of the Earth from the moon - if you consider that the two bodies are rotating around a common centre. OK. Centrifugal force is the explanation for children. Its kind of like the Tooth Fairy of physics. The problem is that while CF can explain the "outward force" needed for the far bulge, its still the differing gravitational force that defines the size and shape of the tidal force. And CF isn't needed at all if you can accept the "free fall) argument. Since the Moon is a lot closer, that difference is more significant. If you remember any calculus, you'd know that differentiating a 1/r^2 function yeilds a 1/r^3 function. The inverse cube of the Sun's distance becomes a tiny number compared to the Moon's. The net result is that the Moon's effect on the tides is 2.2 stronger than the Sun's. Tsk...tsk. The moon only has a stronger effect on tides because the Earth and Moon are an orbiting pair. The Earth and Sun are also an orbiting pair. There is no qualitative difference, only quantitative. Where's Gilligan when you need him? Gilly is an educated man. I'm sure he agrees with me. |
"Scout" wrote in message
... Jeff, Remember that I first posted that very same sentiment, and even provided a graphic. I still believe that to be true, but have modified my internal model, giving allowance for the centrifugal force. I'm not a physicist, but the way I'm seeing it, there is a middle ground in this discussion. I'm curious to know if you're discounting centrifugal force as a contributor to the far bulge. Scout I've always said that Centrifugal Force can be used as part of the explanation, as long as you end up with the same answer. There are several different ways of looking at this, all valid. (I hope I can get through this without mangling the terms too badly ...) The problem with Centrifugal Force is that it is a "fictional force." It is only needed if you work in a non-inertial, or accelerating reference frame. If you are in a car going around a curve, your reference frame is accelerating towards the center of the curve, and thus you feel a Centrifugal Force in the opposite direction. To an outside observer, the CF doesn't exist, the only force is the car pulling the passenger around the turn. The outside observer can analyze the situation completely without invoking CF. (The passenger feels CF push him outward, the observer sees the car pull the passenger inward.) In the Earth-Moon system there is gravity pulling both the Earth and Moon around curves. Because the gravity acts on all objects, we don't notice ourselves being pulled around. The magnitude of the Centrifugal force is to small to notice, but in that reference frame it exists. To the outside observer, we're just in freefall, being pulled inward by gravity. The problem with CF arises when you look carefully at the math. One pitfall Nav fell into was trying to calculate CF as a function that varies with the distance to the barycenter. However, all points on the Earth do not rotate around the barycenter, only the center does. Other points describe the same circle around nearby points, so that all points on Earth feel the same Centrifugal Force. (This is a tough concept to explain in words; its easier to do it graphically. Consider a plate wobbling around a point but with no rotation - each point on the plate describes the same circle.) BTW, Nav provided two commonly used formulas, one for gravity and the other for CF. Although they look quite different, you should appreciate that they are the same, since the angular velocity is determined by the gravitational force. The CF will be the same (with the opposite sign) as the gravitational pull at the Earth's center. Since the CF is a constant force, it can't describe the two bulges in opposite directions. It is gravity itself that varies with distance. The differential force can be derived either by subtracting the average gravitational force which causes the freefall at the center of the Earth, or it can be derived by adding the centrifugal force. Since the two are the same, except for the sign, the math is identical. So take your pick, either explanation works, and I'm sure there are others. However, I hope you can appreciate that explanations like "gravity creates the inner bulge, centrifugal force creates the outer bulge" makes physicists wince! |
"Jeff Morris" wrote
[snip] So take your pick, either explanation works, and I'm sure there are others. However, I hope you can appreciate that explanations like "gravity creates the inner bulge, centrifugal force creates the outer bulge" makes physicists wince! Yes, I can, as I've watched my physics teaching partner wince quite a bit this week as we discussed this thread. He was quick to cover our whiteboard with formulas and drawings. It's an interesting thread though, and notwithstanding my oversimplified analogies, I've learned a lot from it. By the way, I saw that same wince from a black history professor when I suggested that the Civil War was fought to free the slaves, and then again when I suggested to an ancient lit professor that The Odyssey has all the earmarks of an Arnold Schwarzenegger movie. Probably explains why I like a good fart joke. Scout |
| All times are GMT +1. The time now is 03:42 PM. |
Powered by vBulletin® Copyright ©2000 - 2025, Jelsoft Enterprises Ltd.
Copyright ©2004 - 2014 BoatBanter.com