Professional Strawman Posted March 8, 2015 Share Posted March 8, 2015 Hi all. I am looking for a graphing software that can plot parallelogram law summations the way the diagram in the link below shows. As you can see the orbits traced with this technique begins to precess on its own, that is, without the need for an additional central force. I want to obtain the rate of precession per revolution based on the distance [inverse square]. Does anyone know how to obtain this information? http://3.bp.blogspot.com/-tFDkItTrHWY/Ug-bh0x_NRI/AAAAAAAAE6U/PKx81419uBA/s1600/Kepler's+Second+Law_4.bmpPS. If someone wants to know a little background behind this question please see here: url removed Link to comment Share on other sites More sharing options...
swansont Posted March 8, 2015 Share Posted March 8, 2015 ! Moderator Note Just like that site you linked to (and which has been deleted), we have a policy that if you want to discuss something, you need to discuss it here. Link to comment Share on other sites More sharing options...
studiot Posted March 8, 2015 Share Posted March 8, 2015 There was a program called Design View, written for Windows 3.1 that could do what you ask. Mathsoft had a somewhat less capable program, linked to MathCAD, called Imagination Engineer. There are AutoCAD plug-ins that allow AutoCAD to display moving vector resultants, but I can't remember the name. You should reasearch "Linking spreadsheets to drafting programs" Link to comment Share on other sites More sharing options...
Professional Strawman Posted March 8, 2015 Author Share Posted March 8, 2015 Studiot, I will, thanks. If you find something that is easier to use and is accurate then please let me know. I am trying to relate this data to explain Mercury's anomalous precession without using an additional central force that Newton and Einstein do in their work. If the software is not accurate then it wouldn't serve my purpose. Link to comment Share on other sites More sharing options...
studiot Posted March 8, 2015 Share Posted March 8, 2015 CAD software is as accurate as your computer can be. Link to comment Share on other sites More sharing options...
Professional Strawman Posted March 8, 2015 Author Share Posted March 8, 2015 I am not familiar with CAD. I meant that the software should allow for very small step sizes when it computes the summations. Is it realistically possible to ascertain the data I am looking for via CAD? Hi all. Also is it possible to obtain the precession angle per revolution without using a simulator? Are there mathematical techniques that can be used to ascertain the data I am looking for? Link to comment Share on other sites More sharing options...
swansont Posted March 8, 2015 Share Posted March 8, 2015 If I am understanding your drawing correctly, it also introduces errors by summing the motion rather than integrating it. You have to be sure any "precession" is not an artifact of these errors. Link to comment Share on other sites More sharing options...
Professional Strawman Posted March 9, 2015 Author Share Posted March 9, 2015 (edited) A user on a different forum wrote a program to compute orbits using my method and the orbits were not closed loops. So the method is correct and it does introduce a precession on its own. The user on the forum used very small step sizes in his program and he said it took his computer hours to make a single orbit. Unfortunately that user quit after a few days and took out the program and the graphs he had uploaded. I am not sure I understand what errors you're referring to. The summation method, uses the initial velocity v, and the vector g, at position 1 [shown in the drawing] which is obtained by Newton's force law. Then it's just a matter of computing the resultant and using it again at the next parallelogram, shown at 2 and so on. When I use smaller steps, I essentially get the same orbit, only it looks smoother. Edited March 9, 2015 by Goeton Link to comment Share on other sites More sharing options...
swansont Posted March 9, 2015 Share Posted March 9, 2015 A user on a different forum wrote a program to compute orbits using my method and the orbits were not closed loops. So the method is correct and it does introduce a precession on its own. The user on the forum used very small step sizes in his program and he said it took his computer hours to make a single orbit. Unfortunately that user quit after a few days and took out the program and the graphs he had uploaded. I am not sure I understand what errors you're referring to. The summation method, uses the initial velocity v, and the vector g, at position 1 [shown in the drawing] which is obtained by Newton's force law. Then it's just a matter of computing the resultant and using it again at the next parallelogram, shown at 2 and so on. When I use smaller steps, I essentially get the same orbit, only it looks smoother. Does it end up at the exact same point, regardless of step size? Does it work with a circular orbit? Link to comment Share on other sites More sharing options...
Professional Strawman Posted March 10, 2015 Author Share Posted March 10, 2015 (edited) Here's how my work traces the path of a projectile. http://4.bp.blogspot.com/-yln6gBrG-K0/UhCPyG7wqDI/AAAAAAAAE6s/ZjuryiEf8DU/s1600/Projectile_1.bmp This should work for all kinds of orbits. It only depends on the initial velocity, v and the gravity vector, g. But you're right if the step size is big, it may not end up at the exact point. But this is not what I am referring to. In the drawing there is a clear precession effect, like the one shown in the links below. http://www.rkm.com.au/ANIMATIONS/animation-graphics/Black-Hole-Rosette-Orbit-label.png http://www.relativity.li/en/epstein2/read/i0_en/i1_en/ I was wondering if there is an algebraic method that is equivalent to my summation method that can predict the rate of precession without relying on a simulator. Are you aware of such an algebraic technique? Edited March 10, 2015 by Goeton Link to comment Share on other sites More sharing options...
swansont Posted March 10, 2015 Share Posted March 10, 2015 Here's how my work traces the path of a projectile. http://4.bp.blogspot.com/-yln6gBrG-K0/UhCPyG7wqDI/AAAAAAAAE6s/ZjuryiEf8DU/s1600/Projectile_1.bmp This should work for all kinds of orbits. Should. Perhaps (see below) But have you shown that it actually does? It only depends on the initial velocity, v and the gravity vector, g. But you're right if the step size is big, it may not end up at the exact point. But this is not what I am referring to. In the drawing there is a clear precession effect The drawing you gave in the OP does not show a full orbit, so how do you conclude that the precession is clear? And how can you conclude that it's not an artifact of the calculation method? And I don't think it's a matter of "may not". A finite step size will give you errors in the position, The issue is how big they are, and what the effect is over a full cycle, not whether they exist. If you have a circle of radius 1, so it has a circumference of 2π and let's use a step size of 1% of that. You start with a tangent, and move .02π. The distance to the center is not 1 anymore, it's 1.002. You have introduced a .2% error in your path in just one step, and these errors tend to accumulate. You have 99 more to go to complete the circle. The assertion that this will just work does nothing to convince me that you will end up where you started. So until you have run this simulation on a circle and shown how much error is introduced, you can conclude nothing about other orbits. Link to comment Share on other sites More sharing options...
Professional Strawman Posted March 10, 2015 Author Share Posted March 10, 2015 I know that I have not shown the full orbit. But it is evident that it will precess on its own. What you call errors are not really errors, that's how vectors work. If you use smaller step sizes the orbit will be a smooth curve, instead of the bumpy orbit which my drawings shows. Are you aware of an algebraic equivalent of my summation method? Link to comment Share on other sites More sharing options...
swansont Posted March 10, 2015 Share Posted March 10, 2015 I know that I have not shown the full orbit. But it is evident that it will precess on its own. What you call errors are not really errors, that's how vectors work. If you use smaller step sizes the orbit will be a smooth curve, instead of the bumpy orbit which my drawings shows. Are you aware of an algebraic equivalent of my summation method? The solution which takes the limit of the interval going to zero is called calculus. So a solution exists, but it is not algebraic. But it removes the errors (and yes, they are errors). It's not that that's how vectors work, it's how the approximation works. Link to comment Share on other sites More sharing options...
Professional Strawman Posted March 11, 2015 Author Share Posted March 11, 2015 I know there are several techniques to smoothen my orbits/trajectories with Perturbation theory. http://en.wikipedia.org/wiki/Perturbation_theory I was asking if there was an algebraic equation that was equivalent to my summation method. For example, I start with v and g [via inverse square] obtain the resultant and then the next vector, g, should point itself to the origin again and repeat the process. I was wondering if you knew a mathematical equivalent of this process? And you are free to call them errors, I wouldn't. The parallelogram law gives us a resultant without an error. Link to comment Share on other sites More sharing options...
swansont Posted March 11, 2015 Share Posted March 11, 2015 And you are free to call them errors, I wouldn't. The parallelogram law gives us a resultant without an error. But they are errors. I just showed you that you get a different answer than what you get for a situation that has a known solution. i.e. the answer is wrong, by some quantifiable amount. It is called discretization error. http://en.wikipedia.org/wiki/Discretization_error Link to comment Share on other sites More sharing options...
Professional Strawman Posted March 11, 2015 Author Share Posted March 11, 2015 (edited) The smoothing techniques of my orbits are a different matter. If you were referring to them as something that produces an error based on step sizes, then that's a problem with that technique. And I understand that. But I was not referring to them. My orbits are vectors and there are no errors in my drawings. The precession shown in the OP is not based on errors, they are what the vectors predicted. If you dispute them, then you are free to do so. But there are no errors in that drawing. They are what the vectors predicted. Edited March 11, 2015 by Goeton Link to comment Share on other sites More sharing options...
swansont Posted March 11, 2015 Share Posted March 11, 2015 The smoothing techniques of my orbits are a different matter. If you were referring to them as something that produces an error based on step sizes, then that's a problem with that technique. And I understand that. But I was not referring to them. My orbits are vectors and there are no errors in my drawings. The precession shown in the OP is not based on errors, they are what the vectors predicted. If you dispute them, then you are free to do so. But there are no errors in that drawing. They are what the vectors predicted. So you think that a satellite actually moves in a straight line from point to point? Why don't you just go ahead and do the solution for a circular orbit and show me that I'm wrong? Link to comment Share on other sites More sharing options...
Professional Strawman Posted March 12, 2015 Author Share Posted March 12, 2015 No. I don't think that a satellite moves in a straight line. My drawings are essentially coordinates for a graph computed by Parallelogram Law summations. And I am not interested in attacking a straw man. Link to comment Share on other sites More sharing options...
swansont Posted March 12, 2015 Share Posted March 12, 2015 No. I don't think that a satellite moves in a straight line. My drawings are essentially coordinates for a graph computed by Parallelogram Law summations. And I am not interested in attacking a straw man. And yet point 1 and point 2, etc. are physically separated. I am not sure I understand what errors you're referring to. The summation method, uses the initial velocity v, and the vector g, at position 1 [shown in the drawing] which is obtained by Newton's force law. Then it's just a matter of computing the resultant and using it again at the next parallelogram, shown at 2 and so on. When I use smaller steps, I essentially get the same orbit, only it looks smoother. The only way this makes sense to me is if you are computing position with some finite amount of time between steps. Which, by definition, means that satellite is modeled as moving in a straight line during that step. Is that, or is that not, what you are doing? (A literal reading has you finding the resultant of an acceleration vector and a velocity vector, which is nonsense, so I've assumed that you aren't doing that) Link to comment Share on other sites More sharing options...
Professional Strawman Posted March 13, 2015 Author Share Posted March 13, 2015 (edited) Okay, I probably confused you when I said smaller step sizes. The smaller step sizes are one of the techniques used to draw a smooth curve [via perturbation theory] using the coordinates plotted by the Parallelogram Law summations. There are no errors in the Parallelogram Law. Perhaps I am not explaining myself correctly. The summation method forms the basis for a new theory of gravity, which works on the mechanics of a screw. I do not wish to explain this again in this forum. I have done this on other forums. If you are curious, you may take a look. If not, you are free to do whatever you wish to do with this thread. As I said before I have no interest in attacking a straw man. There are no errors in the Parallelogram Law, and therefore no errors in my summation method. PS. If you're very confident that you've discovered an error in the Parallelogram Law, then you should talk to a Physicist, perhaps publish a paper or two about your discovery. But I have no interest in your take on the Parallelogram Law. I think you are wrong when you say there are errors in my drawing. And you are free to have your opinion on my drawings. Edited March 13, 2015 by Goeton Link to comment Share on other sites More sharing options...
Professional Strawman Posted March 13, 2015 Author Share Posted March 13, 2015 Watch out for the orbital precession of the marble. Go to 1:50. Link to comment Share on other sites More sharing options...
swansont Posted March 13, 2015 Share Posted March 13, 2015 Watch out for the orbital precession of the marble. Go to 1:50. You do realize there's friction here, right? Okay, I probably confused you when I said smaller step sizes. The smaller step sizes are one of the techniques used to draw a smooth curve [via perturbation theory] using the coordinates plotted by the Parallelogram Law summations. There are no errors in the Parallelogram Law. Perhaps I am not explaining myself correctly. The summation method forms the basis for a new theory of gravity, which works on the mechanics of a screw. I do not wish to explain this again in this forum. I have done this on other forums. If you are curious, you may take a look. If not, you are free to do whatever you wish to do with this thread. As I said before I have no interest in attacking a straw man. There are no errors in the Parallelogram Law, and therefore no errors in my summation method. PS. If you're very confident that you've discovered an error in the Parallelogram Law, then you should talk to a Physicist, perhaps publish a paper or two about your discovery. But I have no interest in your take on the Parallelogram Law. I think you are wrong when you say there are errors in my drawing. And you are free to have your opinion on my drawings. I am a physicist. The error is not in the notion of vector addition, it is in how you calculate the path — these are not the same thing. Then again, you haven't explained in any sort of detail what you are actually doing. You just say "Parallelogram Law", so I don't really know if you're applying it correctly. The error I point out is not worthy of publication, as it is well-known. I even pointed you to the wikipedia article explaining it. Further, nobody is going to be interested in a method you have not independently tested. "There are no errors" is not worth the electrons used to render them on the screen. "Trust me, I'm right" is not the currency of science. Link to comment Share on other sites More sharing options...
Professional Strawman Posted March 13, 2015 Author Share Posted March 13, 2015 Another straw man. No thank you. Link to comment Share on other sites More sharing options...
swansont Posted March 13, 2015 Share Posted March 13, 2015 ! Moderator Note Seeing as how you are refusing to answer questions, this is now closed. Don't re-introduce the topic. 1 Link to comment Share on other sites More sharing options...
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