Quote from: Celebrimbor on 02/20/2013 06:49 pmQuote from: LegendCJS on 02/20/2013 05:18 pmI didn't provide this simplification, Newton did. p is momentum. F = dp/dt is the proper form of Newtons second law. F = ma is only true if m is a constant.This is a physics question not a maths one but hey: what is the experimental evidence that F = dp/dt is more correct than F = ma? Without observing mass fluctuations, how can we tell the difference?It may be that nature is somewhere in between no?Because you can apply F = dp/dt to a leaking water balloon or any other system whose mass is changing in a conventional way and get all the experimental evidence you need, silly! Its also in every physics/ mechanics textbook ever written for an audience that is familiar with calculus.
Quote from: LegendCJS on 02/20/2013 05:18 pmI didn't provide this simplification, Newton did. p is momentum. F = dp/dt is the proper form of Newtons second law. F = ma is only true if m is a constant.This is a physics question not a maths one but hey: what is the experimental evidence that F = dp/dt is more correct than F = ma? Without observing mass fluctuations, how can we tell the difference?It may be that nature is somewhere in between no?
I didn't provide this simplification, Newton did. p is momentum. F = dp/dt is the proper form of Newtons second law. F = ma is only true if m is a constant.
Quote from: LegendCJS on 02/20/2013 07:01 pmQuote from: Celebrimbor on 02/20/2013 06:49 pmQuote from: LegendCJS on 02/20/2013 05:18 pmI didn't provide this simplification, Newton did. p is momentum. F = dp/dt is the proper form of Newtons second law. F = ma is only true if m is a constant.This is a physics question not a maths one but hey: what is the experimental evidence that F = dp/dt is more correct than F = ma? Without observing mass fluctuations, how can we tell the difference?It may be that nature is somewhere in between no?Because you can apply F = dp/dt to a leaking water balloon or any other system whose mass is changing in a conventional way and get all the experimental evidence you need, silly! Its also in every physics/ mechanics textbook ever written for an audience that is familiar with calculus.Before I flood my kitchen, I checked good old Wikipedia...The second law is only valid for a fixed set of particles. If you loose some of them from your "balloon" then the law does not apply.http://en.wikipedia.org/wiki/Newtons_second_law#Newton.27s_second_law
Quote from: GeeGee on 02/20/2013 06:01 pmQuote from: LegendCJS on 02/20/2013 04:32 pmWoodwards effect or not, mass fluctuations or not, the math to show how you can not get a force out of any push heavy/pull light scheme is pretty simple, and can be worked by anyone who has seen the chain rule in intro calculus.With words:Force = change in momentummomentum = mass*velocityif both mass and velocity are functions of time then the chain rule applies.F = v*dm/dt + m*dv/dt.Now find a periodic function of mass and a periodic function of velocity of your choice that gives you a non periodic Force? YOU CAN NOT DO IT!(happy to be proven wrong.)I'm not sure, but I think you're referring to the vdm/dt term argument, which is addressed in the following papers on page 1 of this thread: "Origin of inertia JF Woodward 2004" in Appendix B , "Refutation 02 ORNL of Woodward" and "Refutation 03 ORNL Woodward of ORNL" on page 7.What can those possibly say besides "I get to ignore Newtons Laws because I'm special" anyway?*this response deliberately flippant because I want to provoke someone into doing the reading and summarizing it for me because the burdon of proof is on them/ the supporters and I do not have the time or will to dig into it myself.
Quote from: LegendCJS on 02/20/2013 04:32 pmWoodwards effect or not, mass fluctuations or not, the math to show how you can not get a force out of any push heavy/pull light scheme is pretty simple, and can be worked by anyone who has seen the chain rule in intro calculus.With words:Force = change in momentummomentum = mass*velocityif both mass and velocity are functions of time then the chain rule applies.F = v*dm/dt + m*dv/dt.Now find a periodic function of mass and a periodic function of velocity of your choice that gives you a non periodic Force? YOU CAN NOT DO IT!(happy to be proven wrong.)I'm not sure, but I think you're referring to the vdm/dt term argument, which is addressed in the following papers on page 1 of this thread: "Origin of inertia JF Woodward 2004" in Appendix B , "Refutation 02 ORNL of Woodward" and "Refutation 03 ORNL Woodward of ORNL" on page 7.
Woodwards effect or not, mass fluctuations or not, the math to show how you can not get a force out of any push heavy/pull light scheme is pretty simple, and can be worked by anyone who has seen the chain rule in intro calculus.With words:Force = change in momentummomentum = mass*velocityif both mass and velocity are functions of time then the chain rule applies.F = v*dm/dt + m*dv/dt.Now find a periodic function of mass and a periodic function of velocity of your choice that gives you a non periodic Force? YOU CAN NOT DO IT!(happy to be proven wrong.)
Quote from: LegendCJS on 02/20/2013 04:32 pmWoodwards effect or not, mass fluctuations or not, the math to show how you can not get a force out of any push heavy/pull light scheme is pretty simple, and can be worked by anyone who has seen the chain rule in intro calculus.With words:Force = change in momentummomentum = mass*velocityif both mass and velocity are functions of time then the chain rule applies.F = v*dm/dt + m*dv/dt.Now find a periodic function of mass and a periodic function of velocity of your choice that gives you a non periodic Force? YOU CAN NOT DO IT!(happy to be proven wrong.)Hi, is this the point I brought up earlier? A heavy mass going one way and a light mass going the other doesnt really define what happens. The important bit is, was it heavy or light at the precise moment that it was given a shove in the other direction? If it were both heavy and light during this shove, dependent on its current velocity (moving right = heavy, moving left = light) then everything would cancel.If however it were heavy (or light) for the entire shove it is easy to get an unbalanced force.Think of two people passing a medicine ball between them. If the ball becomes heavy before it reaches the left hand person and light before it reaches the right, then the left hand person will be pushed left faster than the right hand person is pushed right. If they are connected by a rope the total system would get a net push left.
Sums of periodic functions either cancel out or are periodic. Derivatives (where they exist) of periodic functions are periodic, and products of periodic functions are periodic. This is just high school math being applied to a formula from college freshmen (at an engineering school) physics.
Changing the subject somewhat: If Woodward's mass fluctuation theory fails on such an elementary analysis as yours purports to be, how come he's presenting papers still, like at the Summer 2012 (Joint Propulsion Conf. AIAA)? If his theory is so obviously wrong, how come nobody else is publicly deriding it with absolute certainty? Is there something wrong with the peer review process?
Can you put it in writing? And use the medicine ball analogy. It's a word problem, with what I think are all the necessary assumed numbers:We have person A and B to the left and right, respectively. They are separated by distance d, 1.5 m. We have the medicine ball M, with an at rest mass of 1kg. Person A throws it to person B with a velocity of .75m/sec. Person B catches it, but it has a mass of 1.1kg. Person B throws it back to A at .75 m/sec, and it has a mass of 0.9kg. Repeat. Ignore the effects of air friction and gravity.
Quote from: LegendCJS on 02/21/2013 03:04 pmSums of periodic functions either cancel out or are periodic. Derivatives (where they exist) of periodic functions are periodic, and products of periodic functions are periodic. This is just high school math being applied to a formula from college freshmen (at an engineering school) physics. I appreciate your taking the time for these explanations. Sadly, I still need tutoring.Can you put it in writing? And use the medicine ball analogy. It's a word problem, with what I think are all the necessary assumed numbers:We have person A and B to the left and right, respectively. They are separated by distance d, 1.5 m. We have the medicine ball M, with an at rest mass of 1kg. Person A throws it to person B with a velocity of .75m/sec. Person B catches it, but it has a mass of 1.1kg. Person B throws it back to A at .75 m/sec, and it has a mass of 0.9kg. Repeat. Ignore the effects of air friction and gravity.I don't see how you turn this description into a periodic function. And this just assumes that the medicine ball changes mass, which I understand you to be saying doesn't matter, even if it were to be true. Reword the description, if I haven't descibed what Kelvin suggests.What you seem to be describing is a tuning fork, which just vibrates back and forth, and does not change its momentum with respect to an external frame of reference. Ignoring the friction of the material of the tuning fork, it would vibrate forever, bu not go anywhere. If the momentum is to be changed in a preferential direction, energy should be injected into the equation somehow, presumably causing this imbalance in the mass of the medicine ball. Or an imbalance in the masses of the two forks in the tuning fork. I don't think that your equation is complete.
Quote from: LegendCJS on 02/21/2013 03:04 pmSums of periodic functions either cancel out or are periodic. Derivatives (where they exist) of periodic functions are periodic, and products of periodic functions are periodic. This is just high school math being applied to a formula from college freshmen (at an engineering school) physics. I appreciate your taking the time for these explanations. Sadly, I still need tutoring.Can you put it in writing?
and call me crazy if F turns out not to be periodic (and please share your details is it isn't, believe me I want "star trek" style space drives as much as any dreamy eyed science fiction fan)
If you want a function representing discrete shove events start with a(t) and make it a square wave, and integrate to get v(t) and position.
So because it is getting more mass it must slow down.
Quote from: LegendCJS on 02/21/2013 06:01 pmSo because it is getting more mass it must slow down.I guess you missed it the first time...Ever heard of Galilean invariance?You have to be very careful with that v*dm/dt term.Or, to put it another way - getting more mass from where?
Quote from: JohnFornaro on 02/21/2013 04:40 pmQuote from: LegendCJS on 02/21/2013 03:04 pmSums of periodic functions either cancel out or are periodic. Derivatives (where they exist) of periodic functions are periodic, and products of periodic functions are periodic. This is just high school math being applied to a formula from college freshmen (at an engineering school) physics. I appreciate your taking the time for these explanations. Sadly, I still need tutoring.Can you put it in writing? Well since this is a MATH ONLY thread, lets not necessarily appeal to intuition about physics.The claim is that "products of periodic functions are periodic". Well, lets find out (without appealing to infinite examples in Matlab or the authority of high school math):What is a periodic function? Lets say it is a continuous, smooth function of some parameter, t, with the property that,a(t) = a(t+T) (with T>0; we also assume T to be the smallest value and we call it the period).And now lets double down on the periodic functions. No need for them to have the same period, so we havea: a(t)=a(t+Ta); and b: b(t)=b(t+Tb)Now definef: f(t)=a(t)b(t).Is f(t) periodic? Lets try shifting f(t) along by a whole number multiple of Ta:f(t + nTa) = a(t+nTa)b(t+nTa) = a(t)b(t+nTa). This will only be equal to f(t) if nTa happens to be some whole number multiple of Tb.Oh... so the product of two periodic functions is periodic if and only if the ratio of the two periods is rational.
You continue to miss the point.The v in v*dm/dt is not frame invariant. This throws a huge wrench in your explanation.Take the medicine ball's frame of reference. Why should gaining more mass suddenly add momentum in a particular direction where before it had none?