Silvestru Posted April 3, 2018 Share Posted April 3, 2018 8 minutes ago, OlegGorokhov said: I can answer again already today (thank the admins). Please ask any questions.I will answer all the questions that arise. You can start by answering Strange's questions that you kept ignoring or straw-manning. Link to comment Share on other sites More sharing options...
DrP Posted April 3, 2018 Share Posted April 3, 2018 10 hours ago, OlegGorokhov said: Thus, for this 20kg object in order to move every second the displacement (D2), there is no need for Work (Pushing) of the gravitational Force, because the object moves by inertia, i.e. without the Work (Pushing) of the gravitational Force! Yes - but it is accelerating due to the force applied. The inertia just keeps it going until another force counters it. What am missing? Link to comment Share on other sites More sharing options...
OlegGorokhov Posted April 3, 2018 Author Share Posted April 3, 2018 6 minutes ago, Strange said: Please provide the predictions made by your model which are different from standard physics and the corresponding experimental results. Once again My data (displacements !!!) are not diffrent from standard physics. Can you get it?I explain WHY the “current” formula for Work and Energy is erroneous, since it relates the Work (W) of the gravitational Force (F) to the displacement of the pushed object (D) Why can not you understand that? As I said before the whole text is too big to be put here (this info is for the admins too, I have tried to put the main text here) Please, read the whole texts (on Medium) carefully! https://goo.gl/aETJdQ (3 min read)https://goo.gl/YffRQL (5 min read)https://goo.gl/1q9HaU (30 min read) Link to comment Share on other sites More sharing options...
Strange Posted April 3, 2018 Share Posted April 3, 2018 Just now, OlegGorokhov said: My data (displacements !!!) are not diffrent from standard physics. Can you get it? Yes, I know that. So stop talking about it. I am asking you to show the predictions of your model which are DIFFERENT from standard physics. In other words, how do we test your idea? We compare the predictions against experiment. And against current theory to see which best matches the data. So, you say that you have a different equation for energy. So please show how your calculated energy compares with measurements. Link to comment Share on other sites More sharing options...
OlegGorokhov Posted April 3, 2018 Author Share Posted April 3, 2018 (edited) 25 minutes ago, DrP said: Yes - but it is accelerating due to the force applied. The inertia just keeps it going until another force counters it. What am missing? Thank you. Did you read the Karlson's example? Focus on the 2nd second only. Importantly, as a result of Carlson’s pushing, during this 2nd second, the stone will only move the 4.9m displacement (of 14.7m), and the 9.8m displacement will not be the result of Carlson’s pushing, but only will be the result of moving by inertia, since exactly with this velocity the stone was separated from you. And on this displacement of the stone (9.8m), during the 2nd second, Carlson’s Energy will not be spent! The 2nd second of the movement of the stone (under the action of the pushing Carlson) and the 2nd second of free fall of any stone (under the action of the pushing gravitational force) are completely identical processes. These two processes are completely identical, because in both cases: 1) at the beginning of the 2nd second the stone already has the velocity of 9.8m/s to move at this velocity by inertia (during the 2nd second); 2) as a result of pushing the outer Force (Carlson in the first case and the gravitational Force in the second case), the velocity of the stone increases by 9.8m/s (during the 2nd second); 3) the total dispacement (during the 2nd second) is: 14.7m = 9.8m (inertial displacement) + 4.9m (non-inertial displacement). Now science ERRONEOUSLY THINKS that the Energy of the gravitational Force (Karlsson’s Energy) is spent on this inertial displacement of the stone (9.8m) during the 2nd second.Do you agree with the fact that Carlson’s Energy will not be spent on this displacement of the stone (9.8m), during the 2nd second? Because this displacement (9.8m, during the 2nd second) would be even without Karlson. Edited April 3, 2018 by OlegGorokhov Link to comment Share on other sites More sharing options...
Silvestru Posted April 3, 2018 Share Posted April 3, 2018 Man made the most important discovery in physics but can't decide whether to spell Carlson with a C or K. Also the fact that you need Carlson at all isn't inspiring trust in your "finding". I bet Carlson is wearing a propeller just to get away from you speculations. Link to comment Share on other sites More sharing options...
OlegGorokhov Posted April 3, 2018 Author Share Posted April 3, 2018 8 minutes ago, Strange said: Yes, I know that. So stop talking about it. I am asking you to show the predictions of your model which are DIFFERENT from standard physics. In other words, how do we test your idea? We compare the predictions against experiment. And against current theory to see which best matches the data. So, you say that you have a different equation for energy. So please show how your calculated energy compares with measurements. First of all understand that the current formula for Work and Energy is erroneous. To understand this it needs standart physics (displacements) and the logics. Read again these texts about the Karlson's example and the proof of the erroneousness of using the Joule as the unit of measure for Work and Energy and the erroneousness of the “current” formula for Work and Energy. Engage the logics.Do you agree with the fact that Carlson’s Energy will not be spent on this displacement of the stone (9.8m), during the 2nd second?Because this displacement (9.8m, during the 2nd second) would be even without Karlson. Link to comment Share on other sites More sharing options...
Strange Posted April 3, 2018 Share Posted April 3, 2018 9 minutes ago, OlegGorokhov said: Did you read the Karlson's example? Yes. Now, please show the predictions of your model and how they compare with measurement. (Stores about [CK]arlson are not science. Mathematics and data are needed.) Link to comment Share on other sites More sharing options...
DrP Posted April 3, 2018 Share Posted April 3, 2018 6 minutes ago, OlegGorokhov said: Do you agree with the fact that Carlson’s Energy will not be spent on this displacement of the stone (9.8m), during the 2nd second? Because this displacement (9.8m, during the 2nd second) would be even without Karlson. The force he applies to keep it accelerating at 9.8m/ss is the same as the gravitational force and will require energy. With gravity it is potential energy being converted to kinetic energy. If he stopped pushing, then yes, of course it will continue on at constant velocity until some external thing slowed it (via friction or plain obstruction). If he keeps the force on the object to achieve 9.8m/ss going then the stone will get faster and it will cost energy. I don't see your point. Link to comment Share on other sites More sharing options...
swansont Posted April 3, 2018 Share Posted April 3, 2018 21 minutes ago, OlegGorokhov said: As I said before the whole text is too big to be put here (this info is for the admins too, I have tried to put the main text here) Please, read the whole texts (on Medium) carefully! https://goo.gl/aETJdQ (3 min read)https://goo.gl/YffRQL (5 min read)https://goo.gl/1q9HaU (30 min read) ! Moderator Note Stop doing this Link to comment Share on other sites More sharing options...
Strange Posted April 3, 2018 Share Posted April 3, 2018 13 minutes ago, OlegGorokhov said: The 2nd second of the movement of the stone (under the action of the pushing Carlson) and the 2nd second of free fall of any stone (under the action of the pushing gravitational force) are completely identical processes. This is the equivalence principle and has been known about since at least 1915. 1 Link to comment Share on other sites More sharing options...
OlegGorokhov Posted April 3, 2018 Author Share Posted April 3, 2018 4 minutes ago, Silvestru said: Also the fact that you need Carlson at all isn't inspiring trust in your "finding". I bet Carlson is wearing a propeller just to get away from you speculations. You do not understand it even with Karlson . Then you will never understand it without Karlson. The Karlson's example is needed for such as you. Read again, again and again So do you agree with the fact that Karlson’s Energy will not be spent on this displacement of the stone (9.8m), during the 2nd second? Because this displacement (9.8m, during the 2nd second) would be even without Karlson. Link to comment Share on other sites More sharing options...
Strange Posted April 3, 2018 Share Posted April 3, 2018 Now.... about that evidence ... Link to comment Share on other sites More sharing options...
OlegGorokhov Posted April 3, 2018 Author Share Posted April 3, 2018 1 minute ago, Strange said: This is the equivalence principle and has been known about since at least 1915. So do you agree with the fact that Karlson’s Energy will not be spent on this displacement of the stone (9.8m), during the 2nd second?Because this displacement (9.8m, during the 2nd second) would be even without Karlson. Link to comment Share on other sites More sharing options...
swansont Posted April 3, 2018 Share Posted April 3, 2018 9 minutes ago, OlegGorokhov said: First of all understand that the current formula for Work and Energy is erroneous. Since basic physics works, this is obviously false. You need to do better in explaining why you think this is so. If you integrate a force through a displacement, you get the kinetic energy. That's not erroneous. It's the work-energy theorem. Solve, using your method, for the kinetic energy of an object subject to a force F through a displacement d. Link to comment Share on other sites More sharing options...
DrP Posted April 3, 2018 Share Posted April 3, 2018 Just now, OlegGorokhov said: So do you agree with the fact that Karlson’s Energy will not be spent on this displacement of the stone (9.8m), during the 2nd second?Because this displacement (9.8m, during the 2nd second) would be even without Karlson. So what? The stone has inertia. Spell it out for us because we are clearly missing your point. Link to comment Share on other sites More sharing options...
OlegGorokhov Posted April 3, 2018 Author Share Posted April 3, 2018 4 minutes ago, DrP said: The force he applies to keep it accelerating at 9.8m/ss is the same as the gravitational force and will require energy. With gravity it is potential energy being converted to kinetic energy. If he stopped pushing, then yes, of course it will continue on at constant velocity until some external thing slowed it (via friction or plain obstruction). If he keeps the force on the object to achieve 9.8m/ss going then the stone will get faster and it will cost energy. I don't see your point. Thank you Keep in your mind the fact that if there is the pushing Force that provides the 9.8 (m/s^2) acceleration, then ALWAYS the displacement from this pushing is 4.9m (ONLY.) Just think about the 1st second of fall where we also have the pushing Force that provides the 9.8 (m/s^2) acceleration BUT the displacement is 4.9m. 4.9m ! Understand it? 8 minutes ago, DrP said: So what? The stone has inertia. Spell it out for us because we are clearly missing your point. Thank you. First of all, it is important to understand the meaning of the word “inertia”.“Inertia” means “inaction” (or “inactivity”). When the stone is moving through this displacement (9.8m, during 2nd second) by inertia, then no Energy (of Karlson) is being spent on this displacement. The stone moves this displacement (9.8m, during 2nd second) with total inactivity of all Forces -- nothing pushes the stone to move this displacement (9.8m, during 2nd second) . Link to comment Share on other sites More sharing options...
swansont Posted April 3, 2018 Share Posted April 3, 2018 11 minutes ago, OlegGorokhov said: Thank you Keep in your mind the fact that if there is the pushing Force that provides the 9.8 (m/s^2) acceleration, then ALWAYS the displacement from this pushing is 4.9m (ONLY.) Just think about the 1st second of fall where we also have the pushing Force that provides the 9.8 (m/s^2) acceleration BUT the displacement is 4.9m. 4.9m ! Understand it? How does that help me solve a problem? How about solving: How far does an object drop under gravity in 2 seconds? Link to comment Share on other sites More sharing options...
OlegGorokhov Posted April 3, 2018 Author Share Posted April 3, 2018 17 minutes ago, swansont said: Since basic physics works, this is obviously false. You need to do better in explaining why you think this is so. If you integrate a force through a displacement, you get the kinetic energy. That's not erroneous. It's the work-energy theorem. Solve, using your method, for the kinetic energy of an object subject to a force F through a displacement d. Please read again the 2 texts (about Karlson and the proof of the erroneousness of using the Joule as the unit of measure for Work and Energy and the erroneousness of the “current” formula for Work and Energy. ) Only repeated reading can help. The question isSo do you agree with the fact that Karlson’s Energy will not be spent on this displacement of the stone (9.8m), during the 2nd second?Because this displacement (9.8m, during the 2nd second) would be even without Karlson. Keep in your mind the fact that if there is the pushing Force that provides the 9.8 (m/s^2) acceleration, then ALWAYS the displacement from this pushing is 4.9m (ONLY.) Just think about the 1st second of fall where we also have the pushing Force that provides the 9.8 (m/s^2) acceleration BUT the displacement is 4.9m. 4.9m !Understand it? First of all, it is important to understand the meaning of the word “inertia”.“Inertia” means “inaction” (or “inactivity”). When the stone is moving through this displacement (9.8m, during 2nd second) by inertia, then no Energy (of Karlson) is being spent on this displacement. The stone moves this displacement (9.8m, during 2nd second) with total inactivity of all Forces -- nothing pushes the stone to move this displacement (9.8m, during 2nd second) . Link to comment Share on other sites More sharing options...
DrP Posted April 3, 2018 Share Posted April 3, 2018 14 minutes ago, OlegGorokhov said: Thank you Keep in your mind the fact that if there is the pushing Force that provides the 9.8 (m/s^2) acceleration, then ALWAYS the displacement from this pushing is 4.9m (ONLY.) Just think about the 1st second of fall where we also have the pushing Force that provides the 9.8 (m/s^2) acceleration BUT the displacement is 4.9m. 4.9m ! Understand it? OK - but the force will have to increase each step to maintain the 9.8m/ss yea? With gravity it is because the objects are closer - with Karlson it must be because he pushes harder each step. Inverse square yea? So more work is done each step to obtain that constant 4.9m Sorry if I am misunderstanding your point - I'm trying to see where our confusion lies. Link to comment Share on other sites More sharing options...
OlegGorokhov Posted April 3, 2018 Author Share Posted April 3, 2018 (edited) 4 minutes ago, DrP said: OK - but the force will have to increase each step to maintain the 9.8m/ss yea? With gravity it is because the objects are closer - with Karlson it must be because he pushes harder each step. NOOOOOOOO the force does not to increase each step !!!!! Any second of fall the force (that is the size of the pushing per any second) is constant!!!!! Edited April 3, 2018 by OlegGorokhov Link to comment Share on other sites More sharing options...
Strange Posted April 3, 2018 Share Posted April 3, 2018 4 minutes ago, OlegGorokhov said: Please read again the 2 texts (about Karlson and the proof of the erroneousness of using the Joule as the unit of measure for Work and Energy and the erroneousness of the “current” formula for Work and Energy. ) Why? If your model is different you should be able to show that mathematically. Link to comment Share on other sites More sharing options...
swansont Posted April 3, 2018 Share Posted April 3, 2018 Just now, OlegGorokhov said: Please read again the 2 texts (about Karlson and the proof of the erroneousness of using the Joule as the unit of measure for Work and Energy and the erroneousness of the “current” formula for Work and Energy. ) No. That's not how this works. You answer my questions. You defend your claims. You explain the situations where you think standard physics fails, and/or how your solution is required to come up with the correct answer. Just now, OlegGorokhov said: Only repeated reading can help. The question isSo do you agree with the fact that Karlson’s Energy will not be spent on this displacement of the stone (9.8m), during the 2nd second?Because this displacement (9.8m, during the 2nd second) would be even without Karlson. I don't care about Karlson's energy. I just want to solve the problem. Solve the problem with your method. Explain why standard physics doesn't give you the right answer. Just now, OlegGorokhov said: Keep in your mind the fact that if there is the pushing Force that provides the 9.8 (m/s^2) acceleration, then ALWAYS the displacement from this pushing is 4.9m (ONLY.) Just think about the 1st second of fall where we also have the pushing Force that provides the 9.8 (m/s^2) acceleration BUT the displacement is 4.9m. 4.9m !Understand it? Answer the questions, as you promised to do. Don't just repeat yourself. Link to comment Share on other sites More sharing options...
DrP Posted April 3, 2018 Share Posted April 3, 2018 (edited) 9 minutes ago, OlegGorokhov said: NOOOOOOOO the force does not to increase each step !!!!! Any second of fall the force (that is the size of the pushing) is constant!!!!! The force from gravity increases the closer the objects get via an inverse square law. The KE of the object is 4 times the amount it had when it had half the velocity. So it requires more energy the faster it goes... thus more force to keep the same rate of acceleration. KE = 1/2 MV^2 Please hit me with -ve if I am wrong here - I am getting this from my head rather than looking it up. The force between 2 objects falls off via an inverse square law with distance. F = G.M1M2 / Cr^2 where C here is something like 4pi x Epsilon(0) or something like that. G is gravitational constant. Edited April 3, 2018 by DrP G - Gravitational constant Link to comment Share on other sites More sharing options...
OlegGorokhov Posted April 3, 2018 Author Share Posted April 3, 2018 (edited) 31 minutes ago, swansont said: No. That's not how this works. You answer my questions. You defend your claims. You explain the situations where you think standard physics fails, and/or how your solution is required to come up with the correct answer. I don't care about Karlson's energy. I just want to solve the problem. Solve the problem with your method. Explain why standard physics doesn't give you the right answer. Answer the questions, as you promised to do. Don't just repeat yourself. The Karlson's example is needed to understand this point. It's important to understand thatThe 2nd second of the movement of the stone (under the action of the pushing Carlson) and the 2nd second of free fall of any stone (under the action of the pushing gravitational force) are completely identical processes. These two processes are completely identical, because in both cases: 1) at the beginning of the 2nd second the stone already has the velocity of 9.8m/s to move at this velocity by inertia (during the 2nd second); 2) as a result of pushing the outer Force (Carlson in the first case and the gravitational Force in the second case), the velocity of the stone increases by 9.8m/s (during the 2nd second); 3) the total dispacement (during the 2nd second) is: 14.7m = 9.8m (inertial displacement) + 4.9m (non-inertial displacement). Now science ERRONEOUSLY THINKS that the Energy of the gravitational Force (Karlsson’s Energy) is spent on this inertial displacement of the stone (9.8m) during the 2nd second. So do you agree with the fact that Karlson’s Energy will not be spent on this displacement of the stone (9.8m), during the 2nd second?Because this displacement (9.8m, during the 2nd second!) would be even without Karlson. Edited April 3, 2018 by OlegGorokhov Link to comment Share on other sites More sharing options...
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