milli360 wroteThe menu on my website has always provided access to a page describing the constants and equations, and a page giving the calculated data used to plot the graphs. Direct links to the equations and quantities are now conveniently available in my reply to Roy Batty above, posted: Tue Sep 07, 2004 8:10 pm.

To make it easier for you to determine whether my calculations are right or not, here is a copy of my Tide Program used for the calculations. It is written in MS-DOS QBASIC, a very generic computer language. Be aware that finite-bit-length representation of numbers limits computer accuracy in calculating angle theta in radians, which precludes obtaining zero values for cos(theta) at 90 and 270 degrees. Very small numbers are returned instead, giving forces ranging from 1E-11 through 1E-20 with the single precision numbers used in the program. In the table of data I replaced those insignificant small numbers with zeroes.

I look forward to learning of any errors that would invalidate the model, the equations, or their implementation in the program.

milli360 also wrote:When water is pumped into a water tower, a pump provides the force to raise the water from a lower potential to a higher potential. In that last sentence the word ‘raise’ means ‘moved farther from the center of the earth.’ The water is pumped uphill, so to speak. Similarly, the Mississippi flows from a region of lower potential to a region of higher potential, pumped by the centrifugal force of the earth’s rotation. The ocean potential is higher at the equator by virtue of the centrifugal force that brings water there and holds it there. If the earth stopped spinning, the ocean would begin to run downhill toward the poles because of its higher potential at the equator, making the Mississippi flow downhill and northward. And if it would run downhill with no centrifugal force, it is centrifugal force that now makes it run uphill.

Since the earth's rotation is relatively constant, its effect on the oceans is constant, ie no tidal variation. Also, since the earth formed and cooled in its oblate spheroid shape, it makes no sense to use a spherical datum. "Sea level" wouldn't mean a whole lot if you did. While your approach in that case is valid (as long as you are careful about defining your terms), it is also needlessly complicated. Since the earth's rotation is not going to change today, there won't be any change in the flow of the Mississippi river either - which is good, since the MS river isn't tidal (except where it meets the ocean).

So the addition of a term for the earth's rotation is a useless addition to tidal predictions. If you are saying, however, that the earth's rotation somehow effects tidal variation, that's wrong: A constant force cannot cause periodic changes.

Does your model predict tidal variation in the flow of the MS river?

Also, what does your model predict about the tidally induced shape of an object that is not rotating (or is tidally locked with its companion)? What would it predict about the behavior of a group of objects traveling near an object with much larger mass (for example, Comet SL-9 and Jupiter)? How would the Earth's shape be different if there were no moon (or another planet that rotates but has no moon...)? Is the earth's oblatenes inclined? Are the tides inclined? These are only a few of many cases where your model would conflict with what is observed. Have you even tried to apply your model to other objects besides Earth?

edit: reading your website, I found a large number of errors in your understanding of how tides work (some discussed above, some not), both in how the accepted explanation works and what the implications are of what you are suggesting. Quite frankly, it seems like you were unable to get your arms around the conventional explanation and rather than put more effort into learning it, you invented something that makes sense to you. I admire your tenacity, but thats an awful waste of effort.

Not in the vernacular.

"Raise" would mean to follow the line of the plumb bob, but if you were to continue the line of the plumb bob, it might miss the center of the Earth by tens of kilometers, depending upon latitude. So, raising something is to move at an angle to the line to the center of the Earth.
Is that two different questions? If an object is tidally locked to its companion, it will be rotating, once per revolution.

That was the point that I made earlier, and I think Richard went back and fixed that up.

Well, one question with a clarification, I guess.

I'm not sure what "fixed" means - does his model make accurate predictions now? Did it before? It couldn't both before and now. Since that's his main misunderstanding of the accepted explanation as shown here: ...eliminating that misunderstanding should have (but clearly hasn't) eliminated his objection to the accepted theory.

Good. I wasn't sure if you thought those were too different scenarios, or two versions of the same scenario.
His simplified model ignores too much to make accurate predictions in either case. Plus, just computing a force will not translate directly into a displacement. Still, there are obvious errors.

still, the question remains, "what causes tides?" everybody seams happy with the gravity model but i'm not so sure.

Is that because you think the moon is always over the low tide?

russ_watters wrote:I agree. My model does not contain of a term for the earth's rotation. It does contain terms for the gravitational effects of the moon and sun on the earth and terms for the centrifugal forces caused by the earth’s orbiting of both the sun and the earth-moon barycenter.

russ_watters also wrote:My extremely simplified model deals only with the relationships of gravitational and centrifugal forces to ocean tides. As I wrote in the article
My model does predict that on the plane of the earth-moon orbit (equator in the model) there should be a high tide on the side of the earth facing the moon and another high tide on the opposite side of the earth. It also predicts that the moon contributes about 60 percent of the tidal forces and the sun the remaining 40 percent.


milli360 wrote:As I wrote above Especially obvious ones.

First two things are your insistence that the "usual" explanation must be in error ("The old idea that the second high tide is caused by gravitational attraction of the moon just doesn't hold water." from the Ocean Tidal Model (fixed) page), and your use of kilograms as a unit of force (Constants and Equations page). I know you've mentioned these before, and tried to justify them, but I don't think that there is any justification for them, really.

When I look at your Constants and Equations page, it still has references to equations and formulas that aren't presented ("with the distances from sun and moon appropriately changed to reflect distances to that particular location") and so cannot really be evaluated.

And it appears that you are using mass and distance parameters from some source, but changing them to fit your simplified equation? Is that true? The chart at the bottom of the Constants and Equations page and the following discussion seems to say that you used a modified earth/moon distance. Of course, the earth to moon distance is not constant anyway--and the orbital distance, even assuming that the orbits are circular, is to the barycenter, whereas the distance for the calculation of the gravitational force is from center of earth to center of moon.

I also note that your value for the earth/moon distance is off by a factor of ten, although you seem to have used the correct value in your calculations.

Right, so using only the gravitational component (which should be twice as simple), what shape would it predict?

You imply in your discussion of the existing theory that the existing theory should predict only one high tide (even though everyone else says the existing theory predicts two). Did you do the math on that? Doing the math one component at a time (which, it appears you tried to do anyway) should (according to your theory) show that the existing interpretation is wrong while showing the shape your theory should predict for non-rotating (tidally locked or free-fall) systems.

Ie, your calculations must show the relationship between the forces on a 1kg object on each side and at the center of the earth as caused by the moon. Then you must correctly interpret what these numbers are telling you. What you should find, if you are correct, is that the net force on an object on the far side of the earth is still directed toward the moon, meaning the lunar gravitational force on a 1kg object on the far side of the earth must be greater then the force on a 1kg object at the center of the earth. Is it?

I suspect you have already calculated the answer and simply don't understand it. A thought experiment: you have 3 lead balls of 1kg each, held stationary, at even intervals apart, at some distance above the earth. If you drop all 3 at the same time, how will they move relative to each other and why?

milli360 wrote:Prior to my statement that the second high tide is not caused by gravitational attraction I gave the reasons for that claim. Are any of those reasons in error? In any event, even if I am not justified to make that claim, that is not an error in my model.

The only reference to kilograms in my page on Constants and Equations is in the first sentence of the second paragraph as “w is the weight of an earthbound object (say 1 kg, of ocean water)”. In the immediately preceding equation weight w is divided by g (the acceleration of gravity) to convert it to mass. I do not use kilograms as a unit of force on that page.

milli360 also wrote:The equations and formulas for the calculations were indeed presented in my Tide Program page in my post of Thu Sep 09, 2004 12:06 pm.

milli360 wrote:That is not true. Parameters were not changed to fit an equation. In The revised model I wrote “Therefore, to make meaningful tidal force calculations the constants in the equations used must be accurate to a sufficient number of significant places. They must also assure a sufficient approximation of equality between a body's centrifugal force and the gravitational force acting on it at its intersection with the line of the orbit.”

If the constants were consistent the gravitational attractive force between the earth and sun would exactly equal the centrifugal force of the earth in its solar orbit. The literature values I used did not give sufficient approximation of that equality, hence the search for which one of four constants to modify in regard to solar interactions and which one of five constants to modify in regard to lunar interactions.

Perhaps by ‘simplified equation’ you meant Equation 1 or Equation 4 in my Constants and Equations page. Those equations merely express the relationships that must exist between the constants used in the formulas for gravitational and centrifugal forces if those forces are to be equal at the center of the subject bodies. I believe the presentation of the derivation of those equations is straightforward.

milli360 wrote:In The Revised Model I stated that “In calculating the forces, the distances between the centers of the sun and moon from the center of the earth are taken as their average distances.” As I said, the model is a snapshot in time, not a dynamic model. Bear in mind that the only purpose of my model was to determine if it would explain two simultaneous high tides.

b]milli360[/b] wroteAt last, another thing we do agree on. That is exactly how it is given regarding earth-moon interactions in my Constants and Equations page.

Thanks for alerting me to a typo in the exponent for the earth-moon distance on the Constants and Equations page. It has been corrected.

russ_watters wrote:I assume that by ‘far side of the earth’ you mean the side of the earth that is one earth radius farther from the moon than is the earth center. According to Newton’s law of gravitation, the gravitational attractive force between two bodies varies as the inverse square of the distance between them. That means the greater the distance the smaller the force and the smaller the distance the greater the force. The side of the earth facing away from the moon is one earth radius more distant from the moon than the center of the earth is. At the side of the earth facing away from the moon, then, the lunar gravitational force must be less than at the center of the earth. Your statement that “...meaning the lunar gravitational force on a 1kg object on the far side of the earth must be greater then the force on a 1kg object at the center of the earth” contradicts Newton’s law of gravitation.

Now let’s consider your claim “that the net force on an object on the far side of the earth is still directed toward the moon.” A net force is the sum of all force vectors acting at a point. The lunar gravitational and centrifugal forces are equal and oppositely directed for a 1kg object at the center of the earth; the net force there is zero. The same is true for the solar gravitational and centrifugal forces acting on any object at the center of the earth.

At any point on earth there is a centrifugal force having the same magnitude and same direction as that arising from the earth center revolving around the earth-moon barycenter. Only for points on earth at the same distance from the moon as the center of the earth does that centrifugal force equal the gravitational attractive force of the moon. On the far side of the earth the gravitational attractive force is weaker than at the center of the earth in accordance with Newton’s law.

On the far side while the gravitational attractive force is directed toward the moon and toward the earth’s center, the centrifugal force is directed away from the moon and away from the earth’s center. On that far side, because gravitational force is weaker than at the center of the earth and because their directions are opposite, centrifugal force dominates. The sum of the forces leaves a small net force in the direction of the centrifugal force, i.e. away from the center of the earth and away from the moon. The net force on an object on the far side of the earth is not directed toward the moon as you have claimed.

No, but it is an obvious error.


I thought by that you were using w=1 kg. I assumed that you were still using kilograms as force, as you said in your "fish scale mentality" comment.
Exactly. The literature values are probably accurate, but they did not "work" in your equations, so you modified them. Rather than being a test of how accurate your approach was, you assumed that the values needed to be changed.
They are in error, as I pointed out. Equation 4 assumes that the orbits are circular with a radius d--which is the distance from the moon center to the earth center--but the moon actually orbits the barycenter.
I was trying to point out that it is not how it is given.
You're welcome.

I realized that I skipped over this question, and didn't answer it.

From your page, the reasons given include the following three paragraphs:
The center of the earth does move closer to the moon than it would had gravity not been acting. The ocean does follow the ocean bed, but not in complete lockstep.
It is not the additional gravity of the moon on the ocean that causes the higher tide on the opposite side, it is the even higher gravity on the earth. The relative sizes of the forces don't matter so much as the potential, as may be obvious from my answer to the next paragraph.
Consider a section of an ideal river--constant width, constant depth, frictionless in a constant gravity field. The surface and bottom pressures will be the same, as will the forces, but it will still flow, because the regions downstream are at a lower potential.

Hey, I second whoever way up above pointed out that our host's book covers it pretty well. It does. Buy the guy's book. It isn't expensive and it is chock full of interesting stuff. And when it is on your shelf, you can refer to it whenever one of these discussions pops up. Or when an ignoramus friend comes up with a new theory.

That's exactly my point: your understanding of tides is based on that contradiction of Newton's law. Correction: that's your claim. You're the one who said there would be no high tide on the far side.