Timo Moilanen Posted January 15, 2017 Author Share Posted January 15, 2017 So you will get the wrong answer. I expect this is the main reason your "theory" is wrong. There may be others. But as you have failed to explain anything, it is not clear. I doubt it very much. The "my theory " will survive and prove right because physics do not need human acception . I do not see me allowed to keep this kind of "finding" to myself , but your "funktion" in this is for me more and more unclear -2 Link to comment Share on other sites More sharing options...
Strange Posted January 15, 2017 Share Posted January 15, 2017 The "my theory " will survive and prove right because physics do not need human acception . Physics needs correct results. As the only thing you have calculated was wrong, and you refuse to do any further tests, it is not going to be accepted. I do not see me allowed to keep this kind of "finding" to myself , but your "funktion" in this is for me more and more unclear My function is to challenge your idea, ask you to show that it works (it doesn't), and if possible help you see where you have gone wrong. Link to comment Share on other sites More sharing options...
Timo Moilanen Posted January 15, 2017 Author Share Posted January 15, 2017 Of course he did. Don't be so stupid. https://en.wikipedia.org/wiki/Shell_theorem As swansont points out you can test your equation by choosing a pair of spherical masses with any mass, any radius and any separation. Then compare your results with Newton. If your results are different then they are wrong. The result on small (laboratory) spheres will give same F(force) results as far as accuracy goes today , but " the my theory " will give a more "stable" constant not tending to change with increased mass for example as today. On the right scale (world) they will prove very different and to that I have not enough data or computing capacity and most of all astronomers skills and accumulated experience The result on small (laboratory) spheres will give same F(force) results as far as accuracy goes today , but " the my theory " will give a more "stable" constant not tending to change with increased mass for example as today. On the right scale (world) they will prove very different and to that I have not enough data or computing capacity and most of all astronomers skills and accumulated experience http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell.html Here you can insert R=1 and r=1 in the second from bottom and third from bottom formulas and get F= GmM/(2r^2) , That is absolutely my finding for the most outher shells arithmaticly done on computer . I can only say that Gauss had something little wrong or we interprete him too widely Link to comment Share on other sites More sharing options...
Sensei Posted January 15, 2017 Share Posted January 15, 2017 (edited) http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell.html Here you can insert R=1 and r=1 in the second from bottom and third from bottom formulas and get F= GmM/(2r^2) , That is absolutely my finding for the most outher shells arithmaticly done on computer . ??? How? If You have R=1 and r=1 then [math]F=-\frac{GmM}{4Rr^2}*(2R+2R)[/math] [math]F=-\frac{GmM}{4*1*1^2}*(4*1)[/math] 4*R cancels with 4*R... [math]F=-\frac{GmM}{1^2}[/math] Edited January 15, 2017 by Sensei Link to comment Share on other sites More sharing options...
Strange Posted January 15, 2017 Share Posted January 15, 2017 (edited) The result on small (laboratory) spheres will give same F(force) results as far as accuracy goes today , but " the my theory " will give a more "stable" constant not tending to change with increased mass for example as today. Are you claiming that G increases with increased mass? Please provide some evidence to support this. I do not believe that your theory will give a more "stable" constant, or even the right constant. The only evidence you have given so far proves that your equation is wrong. On the right scale (world) they will prove very different and to that I have not enough data or computing capacity and most of all astronomers skills and accumulated experience You can invent test cases with mass of any size and separation. Compare your results with Newton (which we know is correct) and see if they are the same. They won't be, proving you are wrong. http://hyperphysics....s/sphshell.html Here you can insert R=1 and r=1 in the second from bottom and third from bottom formulas and get F= GmM/(2r^2) , That is absolutely my finding for the most outher shells arithmaticly done on computer . So you agree with Newton's shell theorem? In which case your claim that "gravity need to be calculated to the midpoint of gravity field instead of midpoint of mass" is wrong. EDIT: and as Sensei points out, your algebra is wrong. I can only say that Gauss had something little wrong Then please show, in mathematical detail, exactly where Gauss's error is. Edited January 15, 2017 by Strange Link to comment Share on other sites More sharing options...
Timo Moilanen Posted January 15, 2017 Author Share Posted January 15, 2017 Perhaps you need to explain why you think these measurements do not exist. Then we can explain why you are wrong. There are for me useful "measurements to set const. Ti . Cavendish experiments with spheric masses .but I need details of the masses radiuses and r the distance . That might not seem very important to labs. who do experiments , otherwice than publicing them is like giving away their quite expensive job for free . G = Ti* f(R1,R2, r) as I see it Link to comment Share on other sites More sharing options...
Strange Posted January 15, 2017 Share Posted January 15, 2017 There are for me useful "measurements to set const. Ti . Cavendish experiments with spheric masses .but I need details of the masses radiuses and r the distance . Have you even looked for this data? Presumably not. See posts 26 and 31 to get you started. http://rsta.royalsocietypublishing.org/content/372/2026/20140022 https://scholar.google.co.uk/scholar?q=measuring+the+gravitational+constant If you are too lazy to find support for your "theory" then no one is going to take you seriously. Link to comment Share on other sites More sharing options...
swansont Posted January 15, 2017 Share Posted January 15, 2017 I do not have data on thousands of satellites and the mass of earth and other planets and sun I predict are not right You miss the point. You can generate any Newtonian orbit you wish, since we know the equations work with those values. Link to comment Share on other sites More sharing options...
Timo Moilanen Posted January 16, 2017 Author Share Posted January 16, 2017 There is more astronomical data out there than masses. Do you disagree with orbital periods, too? How about distances? Just masses and constant G much wrong ??? How? If You have R=1 and r=1 then [math]F=-\frac{GmM}{4Rr^2}*(2R+2R)[/math] [math]F=-\frac{GmM}{4*1*1^2}*(4*1)[/math] 4*R cancels with 4*R... [math]F=-\frac{GmM}{1^2}[/math] I should have said third and forth formula from end Link to comment Share on other sites More sharing options...
Sensei Posted January 16, 2017 Share Posted January 16, 2017 I should have said third and forth formula from end That does not matter. They are the same equation, just before transformation.. Link to comment Share on other sites More sharing options...
Timo Moilanen Posted January 16, 2017 Author Share Posted January 16, 2017 Have you even looked for this data? Presumably not. See posts 26 and 31 to get you started. http://rsta.royalsocietypublishing.org/content/372/2026/20140022 https://scholar.google.co.uk/scholar?q=measuring+the+gravitational+constant If you are too lazy to find support for your "theory" then no one is going to take you seriously. After sniffing the air here I'm happy to see that my theory holds all arguments so far it's only wrong -3 Link to comment Share on other sites More sharing options...
Klaynos Posted January 16, 2017 Share Posted January 16, 2017 After sniffing the air here I'm happy to see that my theory holds all arguments so far it's only wrong You're one prediction was wrong. Therefore the idea is wrong. I'm struggling with your post here because you seem to have ignored everything everyone has said to you. Do you not understand or are you not willing to understand? Link to comment Share on other sites More sharing options...
Timo Moilanen Posted January 16, 2017 Author Share Posted January 16, 2017 After sniffing the air here I'm happy to see that my theory holds all arguments so far it's only wrong G measurements with newer methods and old theory will end up in G being 8,05*10^-11 Nm2/kg2 , when gap between masses goes to near 0 and R2 to small(free atoms ) .Cylinders give smaller ca. (8,05-6,67)/3+ 6,67 = 7,1 *10^-11 -2 Link to comment Share on other sites More sharing options...
swansont Posted January 16, 2017 Share Posted January 16, 2017 Just masses and constant G much wrong Then you have no excuse for not working this out. Link to comment Share on other sites More sharing options...
Strange Posted January 16, 2017 Share Posted January 16, 2017 After sniffing the air here I'm happy to see that my theory holds all arguments so far it's only wrong It is wrong in all respects. So I can't see why you are happy. Unless you mean you are happy to abandon the idea as it has been proved wrong. G measurements with newer methods and old theory will end up in G being 8,05*10^-11 Nm2/kg2 , when gap between masses goes to near 0 and R2 to small(free atoms ) .Cylinders give smaller ca. (8,05-6,67)/3+ 6,67 = 7,1 *10^-11 Please provide the evidence to support this claim. Please give references to the papers to show that using cylindrical masses gives a different value for G than using spherical masses. Or admit that you are just making stuff up, now. Link to comment Share on other sites More sharing options...
Timo Moilanen Posted January 16, 2017 Author Share Posted January 16, 2017 That does not matter. They are the same equation, just before transformation.. And yet give different answer ?. One mainpoint is that now M*G = Ti *M -1 Link to comment Share on other sites More sharing options...
Strange Posted January 16, 2017 Share Posted January 16, 2017 And yet give different answer ?. Of course not. How could they? Why do you think that? One mainpoint is that now M*G = Ti *M Therefore Ti = G. Link to comment Share on other sites More sharing options...
Timo Moilanen Posted January 17, 2017 Author Share Posted January 17, 2017 After sniffing the air here I'm happy to see that my theory holds all arguments so far it's only wrong And who says the theory found no support ? It is anyhow without (magic ) and chaos, simply classical mechanics and some vectors and a reasonable simple logic . However the gravity particle look like , their qualities need to be summed up to form a gravity of "macro" physical size Link to comment Share on other sites More sharing options...
Strange Posted January 17, 2017 Share Posted January 17, 2017 And who says the theory found no support ? You: the only prediction you have made was wrong. So: 1 wrong prediction; zero correct predictions = no support. Link to comment Share on other sites More sharing options...
Timo Moilanen Posted January 17, 2017 Author Share Posted January 17, 2017 Of course not. How could they? Why do you think that? Therefore Ti = G. No because on shorter distances g = Ti M/(p-R^2/p)^2 must be right and that do not fit with G . So the only solution is that the "known" mass is much bigger . And G is supposed to be proven by the masses and the masses is calculated from G , how conwinient but in fact the same must also apply to Ti because we still have [kg] . The distance unit thou is[ R ] and distance therefor n*R =p Link to comment Share on other sites More sharing options...
Strange Posted January 17, 2017 Share Posted January 17, 2017 No because on shorter distances g = Ti M/(p-R^2/p)^2 must be right and that do not fit with G . So the only solution is that the "known" mass is much bigger . You need to provide some evidence of this. Link to comment Share on other sites More sharing options...
Timo Moilanen Posted January 17, 2017 Author Share Posted January 17, 2017 Of course not. How could they? Why do you think that? Therefore Ti = G. You need to insert r=1 before reducing r away obviously F=-GmM/(4Rr^2)*{ 2R-(r^2-R^2)*[a.s.o.]} => F=-GmM/4R*{2R-0*[ ] } => F=-GmM /2 -2 Link to comment Share on other sites More sharing options...
Strange Posted January 17, 2017 Share Posted January 17, 2017 You need to insert r=1 before reducing r away obviously F=-GmM/(4Rr^2)*{ 2R-(r^2-R^2)*[a.s.o.]} => F=-GmM/4R*{2R-0*[ ] } => F=-GmM /2 That doesn't make much sense. Please show your working in detail. Link to comment Share on other sites More sharing options...
Timo Moilanen Posted January 17, 2017 Author Share Posted January 17, 2017 You: the only prediction you have made was wrong. So: 1 wrong prediction; zero correct predictions = no support. So you mean that this single site is where support /no support and thereby right/wrong is decided (by Strange) Link to comment Share on other sites More sharing options...
Strange Posted January 17, 2017 Share Posted January 17, 2017 (edited) So you mean that this single site is where support /no support and thereby right/wrong is decided (by Strange) If your model produces only incorrect results, then the model is wrong. That is nothing to do with this website or me. Why would anyone use an equation that gives the wrong results in 100% of cases? Edited January 17, 2017 by Strange Link to comment Share on other sites More sharing options...
Recommended Posts