Professional Strawman

A nice video about centrifugal forces.

Mike, thanks for your reply. I sort of understand your idea but I did not understand the electrical connection to it all. But I have a better idea of what you're trying to do with regard to centrifugal forces. Sounds interesting.

Mike, I am not sure I understand your idea. But I have seen Prof Laithwaite's video and I understand that the item feels lighter when it spins because the centrifugal forces literally hold the flywheel in that position. This is pretty much why a top stands erect when it spins. Angular momentum is a manifestation of centrifugal forces.

Mike, the answer to your question in the OP is centrifugal force since Einstein redefined inertia/centrifugal-force as a distortion of spacetime. In your bucket example, water does not fall to the ground because of centrifugal forces which is also a distortion of spacetime. How this distortion looks or works is irrelevant since GR is not a mechanical theory, it's only a mathematical assertion, way too complicated compared to Newton's math. And Einstein's redefinition of inertia/centrifugal-force comes from his misunderstanding of Newton's Bucket experiment. He believed that centrifugal-forces are the result of an action-at-a-distance force, when it is clearly the result of a contact-force. This is why Einstein's new definition of "inertia/centrifugal-force" is listed under speculative ideas in Wikipedia.

Oh okay. Many thanks.

You are free to interpret that experiment as you see fit. I still go with Lev B, on this.

No thanks. I know you said you're a Physicist and stuff, but I will go with Lev B on this. A real test of e=m, needs an experiment that verifies an increase in the weight/gravity of a test-mass when it is heated or when it is made to spin. Do you have a link on such experiments?

fiveworlds, I tried and learned a little canvas, but I was never going to automate this. But the Excel program works exactly right. My summation method is equivalent to the Leapfrog method, so that was just the program I was looking for. So I found what I was looking for. Many thanks for trying. If you want to see other Excel programs, here they are: http://galileo.phys.virginia.edu/classes/581/

Lev's paper was in reference to your claim that mass increase has been observed in the context of e=m. There is no mention of Lense-Thirring because everybody knows that Lense-Thirring are referring to centrifugal forces. And this is so because Einstein redefined Centrifugal forces in terms of curvature of spacetime, based on a misunderstanding of the Newton's Bucket experiment. His redefinition of "inertia/centrifugal forces" listed under Speculative Ideas in Wikipedia. This paper is about free fall experiments and rotors. Here he reports a discrepancy, which I think is a manifestation of the centrifugal forces causing a drag on the projectile. He also reports other experiments that show that the weight/gravity of an object drops when it is heated, just the opposite of e=m. http://www.researchgate.net/publication/48194960_Frequency_Dependence_of_Rotor%27s_Free_Falling_Acceleration_and_Inequalityof_Inertial_and_Gravity_Masses [accessed Mar 21, 2015].

"According to the standard cliche, these experiments were done, “to test the velocity dependence of longitudinal and transverse masses” actually they tested the velocity dependence of momentum." http://www.physics.uoguelph.ca/~des/Phys2320/concept%20of%20mass.pdf The dragging due to centrifugal forces that Lense-Thirring are referring to might be a small value in the context of an earth orbit. But this dragging should show up in free fall experiments. A rotating body's free fall will be dragged compared to a non-rotating body, which falls without a drag. I see there are a few papers documenting this Newtonian effect. http://www.worldsci.org/pdf/abstracts/abstracts_6367.pdf

nec, that's just centrifugal forces[fictitious forces] holding the wheel in that position. The same effect that causes a top to stay erect [or a bicycle wheel] when it spins. [Draw an fbd to understand this yourself]. This is also related to frame dragging also known as Lense-Thirring effect, which is a rotating body's orbit will be slightly different or dragged because of centrifugal forces. Therefore, I am thinking that a rotating body might not drop at the same rate as a non-rotating body. By the way, according to e=m, a rotating body weighs heavier than a non-rotating body. A hotter body weighs heavier than a colder body. Of course there are no experiments that support this aspect of e=m, but there are many experiments that show the opposite.

Hi fiveworlds, I found an Excel program at the following link: http://www.phys.virginia.edu/classes/581/GeneralPlanet.xls The precession shown in Dan's video can be obtained by changing n to 0.5 [very strong gravitational field]. This is nearly what my drawing predicts. So many thanks for trying.

Fiveworlds is not a physicist, and even if he were, he is not doing that bad. He will eventually get around to it, and did I mention the code for friction?

Another straw man. No thank you.

fiveworlds, this video helps you see the orbital prcession of a marble. See how the marble's orbit begins to precess? It traces a flower orbit, if you notice carefully.

Watch out for the orbital precession of the marble. Go to 1:50.

fiveworlds, many thanks for your explanation. You clearly misunderstood my drawing. The diagonals are the bold arrows in the drawing. Do you notice the black bold arrows in my drawing? [this drawing is also my profile picture] http://3.bp.blogspot.com/-tFDkItTrHWY/Ug-bh0x_NRI/AAAAAAAAE6U/PKx81419uBA/s1600/Kepler%27s+Second+Law_4.bmp At 1, you see the arms of the parallelogram: v and g. Therefore the diagonal is r. I use the resultant r, at 2, and point g towards the origin and get the diagonal again and so. Sensei, I think fiveworlds misunderstood my drawing.

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.

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.

Thanks fiveworlds, for your program. I tried on it an online html platform and I did not see an orbit/graph. How do I run this program? Also, the diagonals should not be of equal lengths and they won't since Kepler's 2nd law is inherent in this Summation method. That is, at the farther point in the orbit, relative to the origin, the diagonals will get smaller and become longer at the nearest points. So you can ignore the length of the diagonals.

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.

Hi all. I am looking to plot a graph of the Parallelogram Law summations shown in the link below. I am not sure how to do this, since the vector g, needs to continuously point to the origin after each summation. Does anyone here know how to do this or on what platform to do this? Using the graph, one can determine the rate of precession per revolution. Also the vector, g, needs to incorporate the inverse square law. The initial velocity is something the user determines. The rest should be done by the software. Thanks in advance for the help. http://3.bp.blogspot.com/-tFDkItTrHWY/Ug-bh0x_NRI/AAAAAAAAE6U/PKx81419uBA/s1600/Kepler For some background on the question, please see the link below. http://www.scienceforums.net/topic/88054-parallelogram-law-summations-precession-of-orbits/ ------ The precession looks like the images in the following links. 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 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.

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?

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?