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gravity effects
- tvanflandern
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21 years 11 months ago #4428
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>the standard model now is no better or worse to me than the flat world model of a 1,000 yaers ago.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
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.
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>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.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
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|-
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.
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>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.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
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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21 years 11 months ago #4752
by Jim
Replied by Jim on topic Reply from
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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21 years 11 months ago #4055
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>The data is available but is not posted in a manner that is understandable if it is posted at all.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Once you have the appropriate learning background, you will find that the data archives and instructions for using them at the NSSDC meet all your needs.
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>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.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
No "force" as such and no energy input are required to maintain a dynamical system. Only the acceleration law is required. I gave you a reference to a book that explains in understandable detail exactly how the model works, even at a microscopic level.
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>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.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
There is no force required by the model, so your condition is never met. You can think of it this way. A gravitational field makes every atom that comes into it fall. It doesn't matter whether there is one atom of a gazillion atoms. The gravitational field makes each of them fall (accelerate) in the same way. That is why it doesn't matter how big the target body is, and why a force calculation will always be misleading. -|Tom|-
Once you have the appropriate learning background, you will find that the data archives and instructions for using them at the NSSDC meet all your needs.
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>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.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
No "force" as such and no energy input are required to maintain a dynamical system. Only the acceleration law is required. I gave you a reference to a book that explains in understandable detail exactly how the model works, even at a microscopic level.
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>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.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
There is no force required by the model, so your condition is never met. You can think of it this way. A gravitational field makes every atom that comes into it fall. It doesn't matter whether there is one atom of a gazillion atoms. The gravitational field makes each of them fall (accelerate) in the same way. That is why it doesn't matter how big the target body is, and why a force calculation will always be misleading. -|Tom|-
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21 years 11 months ago #4269
by Jim
Replied by Jim on topic Reply from
I will try to access the data at NSSCD as you suggest. My skill is limited and I may fail at this getting bogged down by the difficult work required to open files and such. But I'll see.
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21 years 11 months ago #4271
by Jim
Replied by Jim on topic Reply from
No "force" as such and no energy input are required to maintain a dynamical system. Only the acceleration law is required. I gave you a reference to a book that explains in understandable detail exactly how the model works, even at a microscopic level.
Do you suggest by this statement that the model is in essence a perpetual motion model? Or are these items to be factored in at a later time? I read the book a while ago and saw nothing on the matter of how the energetics of a gravity is maintained. It may be a problem on my part since my mind tends to drift off somewhat reading details.
Do you suggest by this statement that the model is in essence a perpetual motion model? Or are these items to be factored in at a later time? I read the book a while ago and saw nothing on the matter of how the energetics of a gravity is maintained. It may be a problem on my part since my mind tends to drift off somewhat reading details.
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21 years 11 months ago #4754
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>I read the book a while ago and saw nothing on the matter of how the energetics of a gravity is maintained. It may be a problem on my part since my mind tends to drift off somewhat reading details.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
?? The whole <i>Pushing Gravity</i> book is about how the energetics of gravity are maintained. The universe is filled with gravitons traveling at ultra-high speeds. These go around pushing on everything from all directions. The apple falls from the tree because more gravitons hit it from above than from below because the Earth blocks some of the gravitons from getting to the apple from below.
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>Do you suggest ... that the model is in essence a perpetual motion model?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
It is as "perpetial" as the stars and the light that travels between them. In addition, my chapter in the book mentions how gravity could be used to build a real perpetual motion machine. (It's not an original idea -- just a long-known, natural consequence of pushing gravity.) -|Tom|-
?? The whole <i>Pushing Gravity</i> book is about how the energetics of gravity are maintained. The universe is filled with gravitons traveling at ultra-high speeds. These go around pushing on everything from all directions. The apple falls from the tree because more gravitons hit it from above than from below because the Earth blocks some of the gravitons from getting to the apple from below.
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>Do you suggest ... that the model is in essence a perpetual motion model?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
It is as "perpetial" as the stars and the light that travels between them. In addition, my chapter in the book mentions how gravity could be used to build a real perpetual motion machine. (It's not an original idea -- just a long-known, natural consequence of pushing gravity.) -|Tom|-
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