Does the Anti-Gravity drive as described defy the laws of physics?

[swansont]

10 hours ago, Cosmic Yoyo said:

"If the masses are equal, they are experiencing the same exact force. This result is a fairly straightforward result of kinematics."

No, gravity provides a different amount of force to objects at different altitudes.

Your example was a drop of water coming out of a faucet. The difference in the force with height is negligible - it is not responsible for the increasing separation of the drops.

If you disagree, show your work.

 

 

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"Mass and velocity does nit give you a force.

And you still have not explained what this alleged bounce is."

Mass and change in velocity do.

Show me an equation where force directly depends on mass and velocity.

 

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In my original post, I refer to the bounce force as planting a mawashi geri.

I meant in physics terms. I have not idea what "planting a mawashi geri" means. Pretty sure it's not physics terminology.

 

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"Where does this energy come from?"

In the example, Iron man and his double's legs.

Stop with your insipid examples.

We are analyzing two objects in orbit. NOTHING ELSE.

 

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"That makes no sense from a physics perspective "

I was responding to your exact words. -"(you just add in a second velocity vector)"

 

"You don’t have a “device”. You seem to have two particles in orbit. I think, because your description is far from clear."

Ironman tethered to his mech buddy is not a device?

We are analyzing two objects in orbit. NOTHING ELSE. It's not at all helpful to keep changing the subject, if your goal is to explain your idea.

 

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"You don’t really “apply” velocities. You can impart a velocity by exerting a force."

While obviously not technically correct, it's good to be able to distinguish initial velocity, applied velocity and the resultant total of the two.  I'd rather you complain about that than having too much verbosity that will make people's eyes glaze over.

I don't know what an applied velocity is, as I said. Re-using the phrase is not helpful. How does the satellite acquire this extra velocity?

 

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"Contrary motion? of what?

what is an ”internally activated” orbit? 

Again you mention a bounce without having explained what that means. "

The bounce makes contrary motion if done right.  Contrary motion means following the same shaped trajectory out of the manoeuvre that created the bounce or other interaction.  When Ironman and his double kicked the Red Dwarf, their trajectories toward and away are mirrored over their shared orbital plane through a line running from the Red Dwarf to the centre of the Earth.  They performed the bounce and made Red Dwarf elevate.

Restating a term is not an explanation of what it means. (i.e. explain contrary motion and bounce without talking about contrary motion and bounce)

And I will continue to ignore any description that includes Iron Man or anything other than two objects in orbit around a planet.

 

[Cosmic Yoyo]

1. I don't understand why you're so annoyed at me describing things in a way that makes sense to me. I didn't say anything false.

2. f = mv which is the same as f = ma if complete velocity transfer is made, say if one goes from a stop to full speed.

3. I didn't think it was a requirement on this site to label things a particular way.  I am sorry if I offend.  I really thought it would be OK to provide physical context of a kick as a force.

4,5. Insipid examples are used because some people need these examples to follow a narrative.  It's functional.  Is it breaking the rules here?

6. If I said 'velocity that arises from applied force', you'd understand me. I say 'applied velocity' to truncate that. I really can't understand why you find this so irritating.  How does which satellite acquire this velocity? See I can ask you are you referring to the Red Dwarf, or are you referring to Ironman and his double or just Ironman by himself.  See how easy that is to distinguish elements?

7. The bounce is simply a force applied to another object.  They key thing about a bounce if you like is that it's due to an ideally elastic collision.  The term contrary motion I use is derived from the playing of a musical scale in opposing directions.  It's basically a retelling of "for every action there's an equal and opposite reaction".  The bounce is simply a way to transfer momentum.

 

Edit - my presence here isn't to sell my idea, I'm asking the question does it work.

Edited December 11, 2020 by Cosmic Yoyo

[Ghideon]

1 hour ago, Cosmic Yoyo said:

1. I don't understand why you're so annoyed at me describing things in a way that makes sense to me. I didn't say anything false.

Maybe because your descriptions do not make so much sense to other members trying to understand? I would like to participate in the discussion but I can't follow the descriptions.
(This applies also to 3,4,5 and 6 above)

 

1 hour ago, Cosmic Yoyo said:

2. f = mv which is the same as f = ma if complete velocity transfer is made, say if one goes from a stop to full speed.

That does not sound correct. Mass * velocity, mv is momentum, not force. 

Edited December 11, 2020 by Ghideon
clarification

[Cosmic Yoyo]

Ghideon

Look at the page on momentum. You'll see on the right F = d/dt(mv). It's per unit time. So presumes the acceleration is one in one second, or whatever unit of time you're using for the acceleration.

https://en.wikipedia.org/wiki/Momentum

I can't describe two bodies pushing each other apart much more simply than that, but I've used Ironman and his double as the device because I don't need to explain how they do things.  I also used the example of a massive barbell form with sliding weights.  If you can think of simpler language, I'm open to suggestions.  However, right now, Ironman and his double can push each other apart, are tethered, can lock their tether and can push against other objects.  I really don't know how I can make it any simpler without having to have extraneous explanations to the point of bringing out finger puppets.  It really is almost on that level.  I can't help but think this is an effort to draw me into a rabbit hole.

If people can answer my question in the negative - fine, answer it according to the rules of this site, i.e. bring evidence.

If people can't answer, or want to hedge their bets, "I can't see why not" is a perfectly reasonable response.

Anything else from now is just nit picking.

[Phi for All]

1 hour ago, Cosmic Yoyo said:

1. I don't understand why you're so annoyed at me describing things in a way that makes sense to me. I didn't say anything false.

I said this EXACT same thing to my plumber after telling him my water-on isn't splurtling enough and it drabbles from the bendy part.

[Cosmic Yoyo]

Oh dear. Swansont made the exact same mistake I did and now you're making it too Phi!

[Ghideon]

44 minutes ago, Cosmic Yoyo said:

You'll see on the right F = d/dt(mv)

Yes. Under the assumption mass is constant then F = d/dt(mv) 

But that is not what you wrote:

2 hours ago, Cosmic Yoyo said:

f = mv which is the same as f = ma if complete velocity transfer is made, say if one goes from a stop to full speed.

mv does not have the unit newton. ma does have the unit newton. They are not the same.

 

53 minutes ago, Cosmic Yoyo said:

So presumes the acceleration is one in one second, or whatever unit of time you're using for the acceleration.

What does that mean? 

 

[Cosmic Yoyo]

f = mv is what f = ma resolves to in one second. If you accelerate to 9.8 ms^-1 in one second, you will be travelling at a velocity of 9.8 ms^-1.

Momentum and force are equivalents - if an object has a certain amount of momentum, it will require an equivalent amount of force to bring it to a stop.  They don't have to have the same units to do this.

By saying mv, you're saying full acceleration to the value of the unit time, which is just another way of saying the velocity.

How long the force transfer takes is irrelevant here except for how the units relate to each other.  When you start to talk about specifying the duration, you start to talk about impulse.

[swansont]

37 minutes ago, Cosmic Yoyo said:

f = mv is what f = ma resolves to in one second. If you accelerate to 9.8 ms^-1 in one second, you will be travelling at a velocity of 9.8 ms^-1.

But mv does not have units of force. v=at for constant acceleration, but that assumes constant acceleration, so it’s not generally true. 

9.8 ms^-1 is not an acceleration. It should be 9.8 ms^-2

These details matter.

 

37 minutes ago, Cosmic Yoyo said:

Momentum and force are equivalents - if an object has a certain amount of momentum, it will require an equivalent amount of force to bring it to a stop.  They don't have to have the same units to do this.

No they are not equivalent. F = dp/dt

You do need proper units to get valid answers

 

37 minutes ago, Cosmic Yoyo said:

By saying mv, you're saying full acceleration to the value of the unit time, which is just another way of saying the velocity.

Physics is pretty well-established. I would prefer it if you wouldn’t make up stuff as you went.

 

37 minutes ago, Cosmic Yoyo said:

How long the force transfer takes is irrelevant here except for how the units relate to each other.  When you start to talk about specifying the duration, you start to talk about impulse.

On the contrary, F = dp/dt means the duration of the force tells you the change in momentum. A constant force in effect for twice the time will impart twice the momentum. And yes, that’s known as an impulse. 

 

[Cosmic Yoyo]

1. If you accelerate *to*

2. They are equivalent numerically.

3. This is a lay translation, trying to communicate the concept. It isn't fictitious.

4. Arbitrary duration that's defined by the units, which for time is seconds. If you're talking about f = mv, it's that force for one second and at any time within that second, the instantaneous force equals the force over the duration of one second.  Force isn't time based itself, but time is used to define it as well as other units.

[swansont]

4 hours ago, Cosmic Yoyo said:

1. I don't understand why you're so annoyed at me describing things in a way that makes sense to me. I didn't say anything false.

The same reason people get annoyed by foreign visitors who butcher (or refuse to learn) the language of the place they are visiting.

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4,5. Insipid examples are used because some people need these examples to follow a narrative.  It's functional.  Is it breaking the rules here?

In this case it’s a distraction, and seems to be used in the place of rigorous analysis.

 

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6. If I said 'velocity that arises from applied force', you'd understand me. I say 'applied velocity' to truncate that. I really can't understand why you find this so irritating.

Yes, and I’d ask what was supplying the force. 

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 How does which satellite acquire this velocity? 

Whichever one you referred to in your example. You just called it “particle”

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7. The bounce is simply a force applied to another object.

Applied by what source, and to what object?

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 They key thing about a bounce if you like is that it's due to an ideally elastic collision.

Collision between what objects? Why do you keep changing the conditions? There was no collision described in your paper.

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 The term contrary motion I use is derived from the playing of a musical scale in opposing directions.  It's basically a retelling of "for every action there's an equal and opposite reaction".  The bounce is simply a way to transfer momentum.

That’s Newton’s 3rd law. Action-reaction. There is existing terminology. Making up new vocabulary is one of the things that annoys me.

As mentioned in our guidelines for these discussions 

https://www.scienceforums.net/topic/86720-guidelines-for-participating-in-speculations-discussions/

you need to be familiar with the area of science into which your idea would fit, or the material you are critiquing. You must also know the terminology. You can't effectively communicate if you are using different definitions than everyone else, or making up nomenclature for things where it already exists. 

(the value of posting an abstract is also mentioned)

 

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Edit - my presence here isn't to sell my idea, I'm asking the question does it work.

You have to clearly explain your idea in order to answer that.

 

2 minutes ago, Cosmic Yoyo said:

1. If you accelerate *to*

But then you don’t know what the force is. F=mv doesn’t tell you

2 minutes ago, Cosmic Yoyo said:

2. They are equivalent numerically.

3. This is a lay translation, trying to communicate the concept. It isn't fictitious.

4. Arbitrary duration that's defined by the units, which for time is seconds. If you're talking about f = mv, it's that force for one second and at any time within that second, the instantaneous force equals the force over the duration of one second.  Force isn't time based itself, but time is used to define it as well as other units.

Numerical equivalence means nothing in physics. Units matter.

I assure you, forces can be a function of time

[Janus]

44 minutes ago, Cosmic Yoyo said:

f = mv is what f = ma resolves to in one second. If you accelerate to 9.8 ms^-1 in one second, you will be travelling at a velocity of 9.8 ms^-1.

Momentum and force are equivalents - if an object has a certain amount of momentum, it will require an equivalent amount of force to bring it to a stop.  They don't have to have the same units to do this.

By saying mv, you're saying full acceleration to the value of the unit time, which is just another way of saying the velocity.

How long the force transfer takes is irrelevant here except for how the units relate to each other.  When you start to talk about specifying the duration, you start to talk about impulse.

f=ma does not resolve to f=mv in one sec.   Neither are force and momentum equivalent.  Reducing to base units of KGS:

f(Newtons) = kg-m/s^2, and momentum = kg-m/s.  Even over one sec these these are not equivalent.

1kg-1m/1s^2 ≠ 1kg-1m/1s,

because 1s^2 = (1s)^2 = 1s x 1s

For the two to be equivalent, everything in one equation as to cancel out everything in the other, but this doesn't

1kg-1m/(1s x 1s) ≠ 1kg-1m/1s,   or 1= 1/s

one of the s's on the left side doesn't cancel out.

What you are trying to do is like trying to argue that 1 linear foot is equivalent to one square foot,  when one is a linear measure and the other a measure of area.

 

 

 

 

[Cosmic Yoyo]

1² ≠ 1^1?

^{You sure bout that?}

[studiot]

22 minutes ago, Cosmic Yoyo said:

1² ≠ 1^1?

^{You sure bout that?}

No more than 1 elephant = 1 glass of whisky.

[Janus]

25 minutes ago, Cosmic Yoyo said:

1² ≠ 1^1?

^{You sure bout that?}

That's not what I said.   1² and 1¹ are dimensionless numbers.   1s and 1s²    involve units of dimension and are not equivalent, in the same way that 1ft and 1ft² are not equivalent.

1s = 1¹s¹

1s² =1²s²

While the 1¹ and1² parts of these equations are equal to each other, the s¹ and s²  are not.

Edited December 11, 2020 by Janus

[Ghideon]

3 hours ago, Cosmic Yoyo said:

Momentum and force are equivalents - if an object has a certain amount of momentum, it will require an equivalent amount of force to bring it to a stop.  They don't have to have the same units to do this.

Thanks for your attempt to explain. I believe I have basic knowledge about Newtonian physics and for any improvements needed I'll use other sources than your posts. 

As for the errors in your physics other members have already provided you with corrections; no need for me to repeat.