Author Topic: Woodward's Effect - MATH ONLY  (Read 33394 times)

Offline Celebrimbor

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Re: Woodward's Effect - MATH ONLY
« Reply #40 on: 02/20/2013 09:02 pm »

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.


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

Offline LegendCJS

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Re: Woodward's Effect - MATH ONLY
« Reply #41 on: 02/20/2013 09:14 pm »

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.


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

Indeed, start loosing particles (i.e. mass) and F=ma goes out the window.  F = dp/dt still applies.  No disagreement from me.
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Online Chris Bergin

Re: Woodward's Effect - MATH ONLY
« Reply #42 on: 02/20/2013 09:43 pm »
This thread's a bit over my head, but I want everyone to remember not to shout other people's opinions down, or be distruptive. It's a big forum, always fine to have other threads on your own opinions.
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Offline GeeGee

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Re: Woodward's Effect - MATH ONLY
« Reply #43 on: 02/20/2013 10:56 pm »
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 momentum

momentum = mass*velocity

if 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.

Uh, what? I just told you where your complaint is addressed. The OP said "In here, math, discussion of math and preferably some links to other people explaining the math." So do you really want to be proven wrong, or are you just here to sling mud?

Offline KelvinZero

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Re: Woodward's Effect - MATH ONLY
« Reply #44 on: 02/20/2013 11:04 pm »
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 momentum

momentum = mass*velocity

if 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.

Offline LegendCJS

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Re: Woodward's Effect - MATH ONLY
« Reply #45 on: 02/21/2013 03:04 pm »
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 momentum

momentum = mass*velocity

if 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.

That doesn't change the outcome.  Please take the time to do the numerical experiment if you don't trust me.  Its only 3 steps.

Take this medicine ball being passed back and forth.  Open excel or matlab or whatever program you like.  Pick a periodic function for its position.  Any function you like, but if you don't have a favorite a good old sinusoid will do.  Now differentiate it to get velocity v(t), and again to get acceleration a(t).  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.

You are done with step 1 of 3.

Woodward and you are saying that the mass alternates periodically. I.e. m(t) is periodic.  Pick any function you want for m(t) as long as it is periodic.  Make sure it is heavy when you want it to be and light when you want it to be.  Please do it.  Make it satisfy your condition: "was it heavy or light at the precise moment that it was given a shove."  You know when it was given a shove because you can refer to v(t) and a(t).  Have you picked an m(t) that does this yet? yes? good. Step 2 of 3 complete.

Now plug these functions into this formula to finish step 3:

F = v*dm/dt + m*dv/dt

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) 

But there is no use arguing with wordy posts and analogies about something that is this easily verifiable or falsifiable.

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.
« Last Edit: 02/21/2013 03:14 pm by LegendCJS »
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Offline JohnFornaro

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Re: Woodward's Effect - MATH ONLY
« Reply #46 on: 02/21/2013 04:40 pm »
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.

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.

**************************************

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?
Sometimes I just flat out don't get it.

Offline GeeGee

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Re: Woodward's Effect - MATH ONLY
« Reply #47 on: 02/21/2013 05:29 pm »


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?

If you look back to the bottom of page 2, you'll see that I specified the papers in which the vdm/dt term argument (which LegendJCS is appealing to) is addressed.

Offline MikeAtkinson

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Re: Woodward's Effect - MATH ONLY
« Reply #48 on: 02/21/2013 05:53 pm »

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.

When person A throws the ball he is accelerating it (with lets assume constant) acceleration a1 (for a time t1), lets also assume that when he catches the ball he decelerates it with a constant deceleration which is also a1. Lets assume person B catches and throws the ball with an acceleration a2 (for a time t2), which is in the opposite direction to a1. So the acceleration graph will have a period of positive acceleration a1, followed by zero acceleration followed by a period of negative acceleration a2.

To impart the same velocity change to the ball we need a1 * t1 = - a2 * t2 = 1.5 m/s.

In the medicine ball example the ball does not change mass during the catch and throw, so we can use F = ma. Integrating we get a net sum of zero assuming m1 == m2.

In the medicine ball example the mass change occurs between the throws and catches. This is where Woodward comes in, you have a momentum change (a change of mass, but not velocity).  Using F = v*dm/dt + m*dv/dt. V is constant, so m*dv/dt = 0, using classical dynamics the forces from the mass changes will exactly cancel out the missmatch in forces on the throwers A and B.

Woodward somehow thinks that he v*dm/dv term does not exist and so the missmatch in forces on A and B are not cancelled.

To be fair to Woodward it seems he has both the mass and velocity varying at the same time, rather than separated as in the medicine ball example, I'm still trying to follow his assumptions and math, they are rather complicated, so they obscure the situation.


Offline LegendCJS

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Re: Woodward's Effect - MATH ONLY
« Reply #49 on: 02/21/2013 06:01 pm »
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.

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.

Going with the medicine ball analogy for now.

You left out a vital point of the description.  When person A has the ball the ball is 0.9kg.  He throws the ball at 0.75 m/s.  Somewhere in the middle the ball changes mass to 1.1kg.  Let us say this change takes place at the midpoint of the distance between person A and B. When this change happens the ball must conserve kenetic energy.  So because it is getting more mass it must slow down. The ball that was 0.9kg and moving at 0.75m/s is now a ball at 1.1kg and moving at 0.6136 m/s when person B catches it, not 0.75 m/s.

Person B tosses the 1.1kg ball back to person A at 0.75 m/s.  When the ball reaches half way it sheds mass to become 0.9kg again, but it has to speed up, so it is now moving at 0.9167 m/s.

The cycle repeats.

For person A they gain the momentum of mass time change in velocity for their catch and throw: 0.9*(0.75+0.9167) = 1.5

Person B does the same: 1.1*(0.75+0.6136) = 1.5. in the other direction.

Net Force = 0.

Guess what 1kg*(0.75 + 0.75)  equals....
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Offline MikeAtkinson

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Re: Woodward's Effect - MATH ONLY
« Reply #50 on: 02/21/2013 06:10 pm »
Yes, I used conservation of momentum, you used conservation of energy, in both cases Woodard says they are broken locally, with the excess momentum and energy being transferred the rest of the universe.


Offline MikeAtkinson

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Re: Woodward's Effect - MATH ONLY
« Reply #51 on: 02/21/2013 06:14 pm »
That is where Woodward's analogy of sand falling off a train breaks down. The mass of the system (train + falling sand) does not change, so there is no v*dm/dv term, until the sand reaches the ground - then the system either has to include the earth or the force decelerating the sand needs to be taken into account.

Offline Celebrimbor

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Re: Woodward's Effect - MATH ONLY
« Reply #52 on: 02/21/2013 06:50 pm »
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.

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 have
a: a(t)=a(t+Ta); and b: b(t)=b(t+Tb)

Now define

f: 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. 

Offline Celebrimbor

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Re: Woodward's Effect - MATH ONLY
« Reply #53 on: 02/21/2013 07:16 pm »

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) 


Now that we know that F can be aperiodic.  Why should this allow space drives?

The thing that we need to know is whether the average net force has positive magnitude when averaged over many cycles.  A lack of periodicity necessary but not sufficient for this to be true.

Offline MP99

Re: Woodward's Effect - MATH ONLY
« Reply #54 on: 02/21/2013 07:21 pm »
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.

ISTM that "square wave" *requires* your input periodic functions to be symmetrical around zero? (Please correct me if not.) If so, are you sure that doesn't guarantee the result you state?

Take the attached image (the headline illustration at http://en.wikipedia.org/wiki/Periodic_function) as being my a(t) function, with zero at the "x" line. I don't need to do anything more than integrate that to end up with a position which trends upwards. It's completely unphysical for the system we're considering, but is more like the pulsed engine on a V-1 rocket.

ISTM a constant mass with an a(t) *rectangular* function with positive & negative segments of equal duration but peaks of +2 and -1 will get a net force over the cycle. (Vary mass from 0.999 to 1.001 if you like.) Same for a rectangular wave from +1 to -1 where the positive part of the curve is twice the duration of the negative part.

Again, I'm not saying this represents the system, so can you just list all the requirements for the mass & position curves to represent the system as described by it's advocates?

Also, I'm not clear quite what the position curve is supposed to represent? It's clearly not the absolute position of an unconstrained propulsive element, because a requirement that it be a periodic system would *require* it to have no net movement. So, I'm thinking that it would represent an absolute position in a system which is bolted to an "infinite" mass like the Earth. Then the net of F over a cycle is the thrust an untethered system would experience.

cheers, Martin
« Last Edit: 02/21/2013 07:25 pm by MP99 »

Offline 93143

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Re: Woodward's Effect - MATH ONLY
« Reply #55 on: 02/21/2013 07:36 pm »
So 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?
« Last Edit: 02/21/2013 07:48 pm by 93143 »

Offline LegendCJS

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Re: Woodward's Effect - MATH ONLY
« Reply #56 on: 02/21/2013 07:45 pm »
So 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?
My point isn't to argue that mass fluctuations are true or not.  It is to say that it doesn't matter, you can't use them to make a propulsive force.  Let a magic fairy give the medicine ball more mass or take it away for all I care.
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Offline 93143

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Re: Woodward's Effect - MATH ONLY
« Reply #57 on: 02/21/2013 07:49 pm »
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?
« Last Edit: 02/21/2013 07:50 pm by 93143 »

Offline LegendCJS

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Re: Woodward's Effect - MATH ONLY
« Reply #58 on: 02/21/2013 07:58 pm »
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.

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 have
a: a(t)=a(t+Ta); and b: b(t)=b(t+Tb)

Now define

f: 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. 

Remember the whole framing is "push heavy, pull light" thrusting.  The periods of the pushing and pulling oscillation and the periods of the mass fluctuations had better be the same or at least some integer ratio or after a little time you will be pushing light and pulling heavy and there would be no hope of getting anywhere.

You might end up displacing your thruster some distance form its starting position (in a frictionless world) but the instant you turned it off and all the moving parts came to rest the thruster would be back where it started.

The product of two periodic function, if they have different frequencies (inverse periods) will be a function composed of a component at the sum of the two frequencies and a component at the difference of the two frequencies.  If the ratio of periods isn't rational the resulting function's period will be infinite.
« Last Edit: 02/21/2013 08:17 pm by LegendCJS »
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Offline LegendCJS

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Re: Woodward's Effect - MATH ONLY
« Reply #59 on: 02/21/2013 08:05 pm »
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?

I never said it does.  I agree. It can not.  We are on the same page.  Preaching to the choir. We are citing the very same thing to show why thrusting form any Woodwards effect will not work. Check my numbers from my example.  Or if you are not replying to me don't make your posts ambiguous about who you are replying too.

I'll repeat:

A 0.9kg ball moving at 0.75 m/s has momentum of: m*v = 0.6750.

All of the sudden it gains mass (not that I believe this can happen either) of 0.2kg to become 1.1kg.  But it didn't/can't gain any momentum in any direction.  So momentum has to be the same as the 0.9kg ball.  That is why the speed decreased to  0.6136 m/s, so the p = m*v product doesn't change.
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