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Jim
1582 Posts |
 Posted - 03 Jan 2003 : 11:51:47
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I want to get away from all the distractions at the paradox forum and focus on the issue of gravity effects as TVF, Makis and I were chatting about. TVF says the current model correctly applies math to the acceleration of both the sun and Earth. And Makis and I have a different opinion here mainly because acceleration is not focce and it not equal for both bodies. To be cont,
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Jim
1582 Posts |
Posted - 03 Jan 2003 : 12:27:33
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Picking up on the remarks by TVF on the dynamics of the Earth/sun orbit-I see several perplexing issues. The acceleration of the Earth by the sun is not equal to the acceleration of the sun by the Earth so the fact that two bodies fall as the same rate seems to be applied in different ways for both bodies. The force of gravity is greater on the sun than on Earth so this is also applied different for both bodies. There are different ways the math is applied to the two bodies. I think you(TVF) say the sun and Earth orbit a barycenter and also say this is not real. This is still another way math is applied in different ways. It is my opinion that the mass of the sun is not being pushed around a barycenter and the model requires this action. Another example of misuse of math to suit the end result.
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tvanflandern
USA
2793 Posts |
Posted - 03 Jan 2003 : 13:11:59
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The acceleration of the Earth by the sun is not equal to the acceleration of the sun by the Earth so the fact that two bodies fall as the same rate seems to be applied in different ways for both bodies.
No, it is applied in identical ways to the two bodies. Only the amounts differ, just as you say.
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The force of gravity is greater on the sun than on Earth so this is also applied different for both bodies.
Is that a typo? In any case, in gravitation, forces are not observable, and are therefore a theoretical distraction. Only accelerations matter for determining orbits. And obviously, the Sun's acceleration due to Earth is much smaller than Earth's acceleration due to the Sun.
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There are different ways the math is applied to the two bodies.
No, the math is applied identically. Only the amounts differ.
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I think you (TVF) say the sun and Earth orbit a barycenter and also say this is not real.
A barycenter is also a mathematical abstraction, and is not observable. It is useful for some purposes, but is just a distraction in this discussion. Forget barycenters here.
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This is still another way math is applied in different ways.
I have no idea what the basis for this claim might be, or even what it means.
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It is my opinion that the mass of the sun is not being pushed around a barycenter and the model requires this action. Another example of misuse of math to suit the end result.
I would say "another example of your not understanding dynamics", because you continue to state that dynamics makes absurd claims. Claiming something absurd, then knocking the studding out of it, is called a "straw man argument".
Barycenters have no forces or accelerations, and nothing gets "pushed around a barycenter". Eliminate barycenters from your thinking. You have to understand the basics first. Those basics consist of Newton's acceleration law: acc. = GM/r^2, applied to both bodies in the system. (M = Sun's mass to get Earth's acceleration, or Earth's mass to get Sun's acceleration.)
What problems do you have with this basic picture? -|Tom|-
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Jim
1582 Posts |
Posted - 03 Jan 2003 : 20:00:13
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There is data at JPL Horizons that clearly shows the barycenter model in operation. It is the Earth/Moon orbit and the Earth is nearer the sun at Full Moon than at New Moon. This can be drawn on paper also. This is the one of the problems and I think the generator of the data- that is the barycenter model needs to push or how else do the results occur? I'm quite willing to forget this model for this forum, however. So onward-The force is as you say F=ma. Doing the math here I get a greater force exerted on the Earth from the sun than the Earth exerts on the sun. The force can be calculated and I suspect observed if you look in the right place. F=ma for the Earth is very small compared to the force exerted by the sun. Is this OK so far? If so-I think the sun is not moved by the Earth. The force is too small and tidal effects absorb the force. I know you tell me the motion is measured so I am wrong here. Thats fine if the data exists that indicates the motion then I'll rethink my idea. I'm very open to anything that makes sense and rebel against unproven theory. I also know Newton designed most of this stuff and I'm aware of LaPlace, Einstein and others.
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tvanflandern
USA
2793 Posts |
Posted - 03 Jan 2003 : 21:09:33
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There is data at JPL Horizons that clearly shows the barycenter model in operation. It is the Earth/Moon orbit and the Earth is nearer the sun at Full Moon than at New Moon. This can be drawn on paper also. This is the one of the problems and I think the generator of the data- that is the barycenter model needs to push or how else do the results occur?
Forget the barycenter and just look at the accelerations actually occuring. During two weeks of each month, the Moon pulls Earth in the same direction as the Sun. That eventually results in the Earth getting closer to the Sun than it would if only the Sun were pulling it. Right? Two forces are stronger than one, right?
Of course, for the other two weeks, the Moon pulls opposite to the Sun, which eventually results in the Earth getting farther away from the Sun than it would be if there were no Moon. Right?
So without any barycenter or any pushes or centrifugal forces or anything at all expect simple Newtonian gravity, we get the observed behavior shown in the models.
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I'm quite willing to forget this model for this forum, however. So onward-The force is as you say F=ma. Doing the math here I get a greater force exerted on the Earth from the sun than the Earth exerts on the sun. The force can be calculated and I suspect observed if you look in the right place. F=ma for the Earth is very small compared to the force exerted by the sun. Is this OK so far?
Solar system gravitational forces are not observable and cannot be observed, even in principle. So let's stick to observable accelerations or we will not be able to explain the basics. Forces and barycenters are unnecessary, advanced concepts introduced for special purposes where they are useful. They cannot help us understand the fundamentals.
The acceleration law is not F = ma. (Forget you ever learned that. It has no value for gravitation.) The acceleration law for any target body in a gravitational field is acc. = GM/r^2, where M is the mass of a source of gravity and r is the distance of a target body from the source mass. Because every mass in a system can be a source mass, every body in the system must take its turn as a target body, and that acceleration formula must be applied to it. So everything moves, no matter how big or small it is.
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If so-I think the sun is not moved by the Earth. The force is too small and tidal effects absorb the force. I know you tell me the motion is measured so I am wrong here. Thats fine if the data exists that indicates the motion then I'll rethink my idea. I'm very open to anything that makes sense and rebel against unproven theory.
We all rebel against unproven theory. But most of us accept theory that has proved itself.
** 300 years ago, the acceleration law was new, and it was amazing to see that the planets and their moons all obeyed it so well.
** 200 years ago, comets and asteroids could be added to the list of successful predictions, even though observations were an order of magnitude better.
** By 100 years ago, the acceleration law was known to apply to many double stars and to certain laboratory experiments; and observations had become another two orders of magnitude more accurate. The acceleration law had been used to predict Neptune, leading to its discovery in 1845. Remaining anomalies indicated that at least one additional planet remained to be found.
** By 50 years ago, the acceleration law was verified to eight decimal places of accuracy. But is was found inaccurate (finally) in going from the eigth to the ninth decimal. This led to Einstein's correction to explain (among other things) why Mercury's elliptical orbit had a slight precession that the acceleration law did not predict. The acceleration law was extended to stellar clusters and even galaxy dynamics.
** In the last 50 years, we have added another four orders of magnitude to the accuracy of the observations. GPS, VLBI, laser ranging, and spacecraft data are accurate to meters, or even centimeters in some cases, even over distances of billions of meters. Yet the acceleration law (except for Einstein's small corrections) has still lived up to its name: the universal law of gravitation.
Instead of just thinking about these matters, why not get a few books and read up on the state of the field today? That is really the only way to get up to speed. If you want to make a contribution, you will first have to see where everyone else is and how they got there. Then you can lead the pack in some new direction. But you cannot just go off in a new direction and expect anyone in the pack to follow. The route to where the pack is now is straightforward, easy to follow, and known to be safe because it was forged by geniuses. To the pack, those who go off in other directions before they get to the safe place where the pack is appear "mad" because all other directions have been explored many times before, and no one who went that way has every returned.
So either enjoy your adventure -- alone -- or learn the basics so you can communicate your original ideas to others. -|Tom|-
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Jim
1582 Posts |
Posted - 04 Jan 2003 : 13:22:34
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The issue is still the same and here at least there a a tiny window that is open to the issue. Lets not bring the baggage from the other forum here- that is why I left the paradox forum and started this one. I want to know dynamics-not dogma-not how egnorant "they" are. Do that at the other forum. So, TVF-you say F=ma is not an issue in orbital dynamics? I ask you,"is that a typo"?
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tvanflandern
USA
2793 Posts |
Posted - 04 Jan 2003 : 13:38:26
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[Jim]: you say F=ma is not an issue in orbital dynamics? I ask you, "is that a typo"?
No, it is not a typo, but I'm glad you asked. In astronomy, all of dynamics can be done with the acceleration law I quoted. Forces are never needed and cannot be observed, so it can be quite confusing to invoke them. In the case in point, your understanding of concentional dynamics would be greatly enhanced if you eliminated all thought of forces and Newton's 2nd and 3rd laws of motion from your thinking. That throws out the equivalence principle too, in case that matters to you.
What is left is the pure motions of bodies -- their accelerations -- and the single, simple rule that governs them. Once you understand what these motions are, and how the rule governing them has been verified (that list of observation techniques I gave in my last message), you will understand the basics of conventional dynamics. And once you understand conventional dynamics, you will be in a position to criticize it if there is something not right. -|Tom|-
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Jim
1582 Posts |
Posted - 04 Jan 2003 : 15:24:42
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Then Gm=rv2 must be the right road. Is this true? And if so how is this rule equally applied to large and small bodies when the moon is only moving the Earth according to its mass ratio of 81:1? The equivalance of m and v2 is evident since the other factors are exactly the same quantities.
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tvanflandern
USA
2793 Posts |
Posted - 04 Jan 2003 : 15:42:36
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[Jim]: Then Gm=rv2 must be the right road. Is this true?
No, that is another red herring. It is true under special circumstances only, such as circular orbits in the 2-body problem. But that is an "advanced topic", and should be forgotten for now until we get the basics right.
The only law that matters for 99% of all gravitational phenomena is acceleration = GM/r^2, where M is the mass of the source of gravity and r is the distance from source mass to target body.
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And if so how is this rule equally applied to large and small bodies when the moon is only moving the Earth according to its mass ratio of 81:1?
When the acceleration law is applied to source mass Earth and target body Moon, we get the Moon's normal orbit around Earth. When it is applied to source mass Moon and target body Earth, we get the Earth's much smaller (by a factor of 81) orbital motion around the Moon. Earth keeps trying to fall toward the Moon, just as the Moon tries to fall toward the Earth. The difference is that the Earth doesn't get very far before the Moon moves on and starts pulling the other way.
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The equivalance of m and v2 is evident since the other factors are exactly the same quantities.
GM/r = v^2 only for circular orbits, and only when there are just two bodies involved. Forget these complicated advanced matters until we get the basic motions straight. Nothing in this so-called "virial theorem" will help you understand the basics. -|Tom|-
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Jim
1582 Posts |
Posted - 04 Jan 2003 : 17:08:10
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OK I'll do this with a=Gm/r2. a for the moon is 1/81 if Earth a=1. The r2 is the same for both bodies. Gm=ar2. a=v2/r for circular orbits if that is ok with you? so v2/r=Gm/r2 to v2=Gm/r to Gm=rv2. It looks like the same idea to me. (I hope my math is right)
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tvanflandern
USA
2793 Posts |
Posted - 04 Jan 2003 : 19:05:21
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OK I'll do this with a=Gm/r2. a for the moon is 1/81 if Earth a=1. The r2 is the same for both bodies. Gm=ar2.
Okay so far.
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a=v2/r for circular orbits if that is ok with you?
The relation is correct for 2-body circular orbits. But I advise against going there. Stick with formulas that tell you something about the motion. It is too easy to get into paradoxes when you switch from motion equations to energy equations, which is where you appear to be headed.
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so v2/r=Gm/r2 to v2=Gm/r to Gm=rv2. It looks like the same idea to me. (I hope my math is right)
Yes, the math is right, and so is the relationship, with the previous caveats still in effect. Your second-last form is an energy equation, and no longer contains information about how any single body moves. Instead, it connects the kinetic and potential energies for one body in a circular orbit. The last formula tells you that the product rv^2 is constant for all the planets. But it would be a different constant for the Moon and Earth satellites because they are orbiting a different source mass.
So where do you plan to go with this formula? -|Tom|-
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Jim
1582 Posts |
Posted - 04 Jan 2003 : 19:50:02
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I want to resolve the energy effect of mass in orbital motion it that makes sense. You lost me on the last post when you divided the math relationships into two types-Energy and Motion. I have never run into that before and will need to ponder it some. I guess tidal effects are one or the other and energy is needed to perform either type. Back to the topic-I just did a bit of math with acceleration and mass since the mass of the sun moon and Earth are known and the acceleration is proportional to mass. Now, I know you are not going to agree but I get a very large difference in the force of these three bodies have on each other. And I am now more sure the moon does not move Earth and Earth does not move the sun. Also tides are for sure a gravity effect that is currently modeled wrongly. I say this because you asked where I want to get to with this link not to belabor my point. I'm looking for truth and find some once in a while.
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tvanflandern
USA
2793 Posts |
Posted - 04 Jan 2003 : 21:28:18
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I want to resolve the energy effect of mass in orbital motion if that makes sense.
The sentence is not self-explanatory.
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You lost me on the last post when you divided the math relationships into two types-Energy and Motion. I have never run into that before and will need to ponder it some. I guess tidal effects are one or the other and energy is needed to perform either type.
Tidal effects are a type of "non-gravitational force", even though gravity is involved. But we can learn little or nothing about the nature of gravity from tides because they disrupt normal orbital motion.
And I don't know what your interest in energy is. It is an effect of gravity, not a cause.
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Back to the topic-I just did a bit of math with acceleration and mass since the mass of the sun moon and Earth are known and the acceleration is proportional to mass. Now, I know you are not going to agree but I get a very large difference in the force of these three bodies have on each other.
We agreed to leave forces out of the discussion because they are mathematical fictions. Stick with accelerations -- that is what drives orbital motion. There certainly is a large difference in the accelerations produced by Sun, Earth, and Moon on one another. Why wouldn't I agree? Only the accelerations of Earth and Moon produced by the Sun are nearly equal. All other accelerations are very different. The acceleration formula is easy to apply and easily gives those results.
Examples:
If acceleration of Sun caused by Earth is 1,
then acceleration of Earth caused by Moon is 1900,
and acceleration of Moon caused by Earth is 150,000,
and acceleration of Earth or Moon caused by Sun is 330,000.
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And I am now more sure the moon does not move Earth and Earth does not move the sun.
Regarding logical consistency, you just jumped off a cliff.  If each body applies an acceleration to each other body, that means all the bodies move by definition of the word "acceleration". How did our discussion derail so badly, out of the blue?
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Also tides are for sure a gravity effect that is currently modeled wrongly.
From your descriptions, I have to doubt that you know what a "tidal effect" really is. But let's get gravity straight first, and leave more complicated matters for chapter two. -|Tom|-
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Jim
1582 Posts |
Posted - 05 Jan 2003 : 16:29:27
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There are no cliffs here-you asked me where I want this to go and I told you. My views and yours are quite different and it is not a problem for me. The model as you have so far described is good as are the several other models I'm aware of. This one assumes acceleration is a primary factor I think you say and that is ok by me. I look forward to the day when force, energy and the tidal effect are factored in. It is good to learn that tides are a gravity effect and that must mean energy is part of the overall modeling. So once I have the other part understood I'll be ready for that.
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Jim
1582 Posts |
Posted - 05 Jan 2003 : 16:38:30
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A correction- tides are a non-gravitional effect that force orbital problems to occur.
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tvanflandern
USA
2793 Posts |
Posted - 05 Jan 2003 : 17:21:14
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I look forward to the day when force, energy and the tidal effect are factored in.
Force is not going to appear in gravitation for macroscopic bodies. Force occurs only at a sub-sub-quantum level for individual gravitons operating on individual matter ingredients. See the book Pushing Gravity for a complete model of how gravity works. Macroscopic bodies simply accelerate. "Forces" for macroscopic bodies would imply that inertia existed for these, which it does not. In gravitation, all bodies, large and small, accelerate at the same rate in a given gravitational field. No bodies resist acceleration, so there is no inertia. {An article in the current MRB explains this.]
Energy is an effect of gravity, but not a cause. A body falling in a gravitational field is changing potential energy into kinetic energy; and vice versa for bodies rising in the field.
Tidal effects are non-gravitational. They come in longitudinal, radial, and latitudinal flavors. These operate differently for solid, liquid, and gaseous bodies. In all cases, friction must be operating for tides to occur. There is a nice discussion of theory and applications in my book, Dark Matter, Missing Planets and New Comets. But since these effects have so little to do with gravity, why do you care? About the only local application you are likely to run into is the tidal evolution of the Moon's orbit. But that is only a few centimeters each year -- significant only over millions or billions of years.
I stress again that these are advanced concepts for someone still struggling to understand Newtonian gravitation. You need a firm foundation before attempting to build on it. -|Tom|-
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Jim
1582 Posts |
Posted - 05 Jan 2003 : 18:48:10
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So, the model requires no energy? And nothing resists acceleration? These are the two things that make the model work? I'm struggling to understand a model-you have no lock on understanding real events.
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tvanflandern
USA
2793 Posts |
Posted - 05 Jan 2003 : 19:01:44
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So, the model requires no energy? And nothing resists acceleration?
For gravitation, both of those statements are true at a macroscopic level. All you need is the acceleration formula. All of orbital dynamics flow from that one formula until you get to relativity effects, which are very, very small.
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These are the two things that make the model work?
Only the acceleration formula "makes the model work". The absence of energy in the formula and the absence of inertia are aids to understanding, not to operation. People who introduce those concepts into gravitation are often led into paradoxes. I'm trying to simplify things so you can see the essence of the standard model (not "my model").
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I'm struggling to understand a model-you have no lock on understanding real events.
I never claimed to have any "lock". But any teacher expects his/her students to understand the existing standard model or models before critiquing them and branching out. Is that reasonable enough for you? -|Tom|-
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Jim
1582 Posts |
Posted - 05 Jan 2003 : 19:16:41
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This is very reasonable as long as we are chatting about MODELS-the standard model now is no better or worse to me than the flat world model of a 1,000 yaers ago. You have posted a few items that are new to me and I want to have a good understanding these details. Even so, I would rather have data that has been measured since as I said about data posted by NASA is generated from models and not at all measured-even so, it is very good data I'm not questioning that point.
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tvanflandern
USA
2793 Posts |
Posted - 05 Jan 2003 : 19:31:46
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the standard model now is no better or worse to me than the flat world model of a 1,000 yaers ago.
Then you still have a long way to go. Remember that list of all the standard model's successes over the past three centuries? No other model in existence has been successful for so long to so many digits of precision in making predictions. You wouldn't be able to know about coming eclipses, stars disappearing at the Moon's limb, planetary conjunctions, Jupiter's moons disappearing into its shadow, transits of Mercury and Venus across the Sun, meteor storms, and many other phenomena that the standard model has predicted in advance and observers have confirmed.
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I would rather have data that has been measured since as I said about data posted by NASA is generated from models and not at all measured-even so, it is very good data I'm not questioning that point.
But you are using the words with incorrect meanings. "Data" refers to measurements in observations or experiments. The predictions of models are called "calculations". All the data at the National Space Sciences Data Center is real data, no models involved. Even in cases where data is "smoothed" using some widely agreed algorithm, the raw, "unsmoothed" data remains available. -|Tom|-
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Jim
1582 Posts |
Posted - 06 Jan 2003 : 11:39:51
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The data is available but is not posted in a manner that is understandable if it is posted at all. I know the model is very good at all the things you say. The issue I have is not about its value in these areas and I am not critical of any model. The point is perhaps better explained by using the ratios you have provided and plugging that into a formula that will clearly reveal force or energy requirements needed to maintain a system. But, you will not allow this to be done with your model. I have no such reluctance and in doing the math I find two facets of the force issue. One is the force available the other is the force required. And when the required force is greater than the available force the model you favor will not give the true result.
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gravity effects
