Wharp drive? S.Ilinca magnetic jet engine on Oberth effect

[Ilinca Sergiu]

What do you think- my engine on the Oberth effect will work or not?

I think this is the main part explaining the Oberth effect: (citation)-
"Take 1/2 of m (v + vb)v2, and you get 1/2 mv2 as well as 1/2 mvb2, the kinetic energy you'd expect from adding these two speeds. On top of that, you also get m(v * vb). The pink rectangle above is Oberth gravy.

For example a kilogram going 10 meters/second has kinetic energy of 50 joules; a kilogram going 2 meters/second has kinetic energy of 2 joules. But a kilogram moving (10 + 2) meters/second doesn't have a kinetic energy of 52 joules, rather (50 + 2 + (10 * 2)) joules. Starting with a 10 meter/second speed and speeding up another 2 meters a second gives you a 20 joule Oberth benefit.

So accelerating a mass already moving fast gives you more kinetic energy for your buck."

 

What do you think- my engine on the Oberth effect will work or not?

I think this is the main part explaining the Oberth effect: (citation)-
"Take 1/2 of m (v + vb)v2, and you get 1/2 mv2 as well as 1/2 mvb2, the kinetic energy you'd expect from adding these two speeds. On top of that, you also get m(v * vb). The pink rectangle above is Oberth gravy.

For example a kilogram going 10 meters/second has kinetic energy of 50 joules; a kilogram going 2 meters/second has kinetic energy of 2 joules. But a kilogram moving (10 + 2) meters/second doesn't have a kinetic energy of 52 joules, rather (50 + 2 + (10 * 2)) joules. Starting with a 10 meter/second speed and speeding up another 2 meters a second gives you a 20 joule Oberth benefit.

So accelerating a mass already moving fast gives you more kinetic energy for your buck."

 

I wonder correctly I believe or not-in this case here?

 

 

 

I think in the case of accelerating, the energy of the jet stream will be:

What do you think- my engine on the Oberth effect will work or not?

I think this is the main part explaining the Oberth effect: (citation)-
"Take 1/2 of m (v + vb)v2, and you get 1/2 mv2 as well as 1/2 mvb2, the kinetic energy you'd expect from adding these two speeds. On top of that, you also get m(v * vb). The pink rectangle above is Oberth gravy.

For example a kilogram going 10 meters/second has kinetic energy of 50 joules; a kilogram going 2 meters/second has kinetic energy of 2 joules. But a kilogram moving (10 + 2) meters/second doesn't have a kinetic energy of 52 joules, rather (50 + 2 + (10 * 2)) joules. Starting with a 10 meter/second speed and speeding up another 2 meters a second gives you a 20 joule Oberth benefit.

So accelerating a mass already moving fast gives you more kinetic energy for your buck."

3magnetic jet S.Ilinca.jpgI think in the case of accelerating, the energy of the jet stream will be:

I propose to consider a model, in which supposedly thrust nozzle behaves like a mass.( resists the movement as some mass). In this case- in my opinion F=F*v=10000000kg, in Oberth opinion m=(v*vb)=10000000kg. What is it : a simple coincidence? I and Oberth are wrong? Or are both right,and we see the same process-which is outside incorrectly seen as (v+v).

Wikipedia about the Oberth effect (why the motion of a rocket increases the kinetic energy of the propellant ):
- e=av
- E=Fv
- E=Fs
From the beginning I try to show that:
-If acceleration increases the weight of the propellant in a nozzle : ( ma=F ) , F=mav,
-If to consider (F) of thrust identical to (m) mass : (F=m) , F=Fv,
-If not consider (F) of thrust identical to (m) mass : F=m(v*vb) ,that is identical to F=Fv,
-if considered covered by a nozzle distance (s) instead of speed (v) : F=Fs,

See no differences except the fact that I have the engine moves relative to the ship, Oberth relative to the selected point ( or celestial body ). and (F) instead (E) for clarity.

Take a look from the other side .
1 kg of burning hydrogen has approximately 120 MJ energy. This is - work in 12000000 kgm. This energy can move 12000000 kg per 1 meter . In a vacuum, this will mean that a lot had the effect of F in 12000000 kg . So when almost isochorically process 1 kg of hydrogen can give a great power - 12000000 kg . ( for more isochorically process (F) increases proportionally) .
Now look at the nozzle my motor which moves with acceleration equal to the acceleration of a jet . (Compare with my previous models).

 

What do you think- my engine on the Oberth effect will work or not?

I think this is the main part explaining the Oberth effect: (citation)-
"Take 1/2 of m (v + vb)v2, and you get 1/2 mv2 as well as 1/2 mvb2, the kinetic energy you'd expect from adding these two speeds. On top of that, you also get m(v * vb). The pink rectangle above is Oberth gravy.

For example a kilogram going 10 meters/second has kinetic energy of 50 joules; a kilogram going 2 meters/second has kinetic energy of 2 joules. But a kilogram moving (10 + 2) meters/second doesn't have a kinetic energy of 52 joules, rather (50 + 2 + (10 * 2)) joules. Starting with a 10 meter/second speed and speeding up another 2 meters a second gives you a 20 joule Oberth benefit.

So accelerating a mass already moving fast gives you more kinetic energy for your buck."

3magnetic jet S.Ilinca.jpgI think in the case of accelerating, the energy of the jet stream will be:
0magnetic jet S.Ilinca.jpg

I propose to consider a model, in which supposedly thrust nozzle behaves like a mass.( resists the movement as some mass). In this case- in my opinion F=F*v=10000000kg, in Oberth opinion m=(v*vb)=10000000kg. What is it : a simple coincidence? I and Oberth are wrong? Or are both right,and we see the same process-which is outside incorrectly seen as (v+v).

Wikipedia about the Oberth effect (why the motion of a rocket increases the kinetic energy of the propellant ):
- e=av
- E=Fv
- E=Fs
From the beginning I try to show that:
-If acceleration increases the weight of the propellant in a nozzle : ( ma=F ) , F=mav,
-If to consider (F) of thrust identical to (m) mass : (F=m) , F=Fv,
-If not consider (F) of thrust identical to (m) mass : F=m(v*vb) ,that is identical to F=Fv,
-if considered covered by a nozzle distance (s) instead of speed (v) : F=Fs,

See no differences except the fact that I have the engine moves relative to the ship, Oberth relative to the selected point ( or celestial body ). and (F) instead (E) for clarity.

magnetic jet Oberth5 S.Ilinca.jpgTake a look from the other side .
1 kg of burning hydrogen has approximately 120 MJ energy. This is - work in 12000000 kgm. This energy can move 12000000 kg per 1 meter . In a vacuum, this will mean that a lot had the effect of F in 12000000 kg . So when almost isochorically process 1 kg of hydrogen can give a great power - 12000000 kg . ( for more isochorically process (F) increases proportionally) .
Êîïèÿ magnetic jet Oberth6 S.Ilinca.jpgNow look at the nozzle my motor which moves with acceleration equal to the acceleration of a jet . (Compare with my previous models).

 

Turns out the same.
If there is a point of reference:
-If acceleration increases the weight of the propellant in a nozzle : ( ma=F ) , F=mav,
-If to consider (F) of thrust identical to (m) mass : (F=m) , F=Fv,
-If not consider (F) of thrust identical to (m) mass : F=m(v*vb) ,that is identical to F=Fv,
-if considered covered by a nozzle distance (s) instead of speed (v) : F=Fs,
and,
If no point of reference:
-transformation of Isobaric process in Isochorically and increases static component of thrust.

 

 

 

 

[swansont]

 

What do you think- my engine on the Oberth effect will work or not?

I think this is the main part explaining the Oberth effect: (citation)-

"Take 1/2 of m (v + vb)v2, and you get 1/2 mv2 as well as 1/2 mvb2, the kinetic energy you'd expect from adding these two speeds. On top of that, you also get m(v * vb). The pink rectangle above is Oberth gravy.

 

You mention a citation but don't actually give one. It's not IMO a particularly good explanation of the Oberth effect, since 1/2 mv² + 1/2 mv_b² is not what you'd expect, assuming you had a basic understanding of kinematics and math.

[Ilinca Sergiu]

 

You mention a citation but don't actually give one. It's not IMO a particularly good explanation of the Oberth effect, since 1/2 mv² + 1/2 mv_b² is not what you'd expect, assuming you had a basic understanding of kinematics and math.

Of course m(v*vb) does not have Oberth effect but this is the essence on which to base this effect.

link : http://hopsblog-hop.blogspot.com/2013/10/what-about-mr-oberth.html

 

[swansont]

Of course m(v*vb) does not have Oberth effect but this is the essence on which to base this effect.

link : http://hopsblog-hop.blogspot.com/2013/10/what-about-mr-oberth.html

 

 

Thanks for the cite. It's a bad explanation. Kinetic energy is not linear in speed, so expecting it to be so, e.g. a doubling of speed to double the energy, is a misunderstanding of physics, not some kind of extra benefit. It's not good to base a project off of an incorrect idea.

 

The issue here is that work is a function of displacement. A rocket traveling slowly doesn't move very far in a given amount of time. Thus for a constant thrust, not much work is done. But a rocket moving faster, a greater distance is traveled and more work is done. The rocket is more efficient in increasing the kinetic energy. Even though nothing about the function of the rocket itself has changed.

 

———

 

On to the next material:

 

Force is not mass. F ≠ mav, F ≠ Fv, F ≠ Fs

 

Fv is instantaneous power if they are in the same direction (there's a dot product). Fs is work (again, it's a dot product). You need to use terminology as it's already defined (and not repurpose it) to communicate effectively. There's legitimate physics here, but you have to do that physics to analyze this system.

[Ilinca Sergiu]

Thanks for the cite. It's a bad explanation. Kinetic energy is not linear in speed, so expecting it to be so, e.g. a doubling of speed to double the energy, is a misunderstanding of physics, not some kind of extra benefit. It's not good to base a project off of an incorrect idea.

 

The issue here is that work is a function of displacement. A rocket traveling slowly doesn't move very far in a given amount of time. Thus for a constant thrust, not much work is done. But a rocket moving faster, a greater distance is traveled and more work is done. The rocket is more efficient in increasing the kinetic energy. Even though nothing about the function of the rocket itself has changed.

 

———

 

On to the next material:

 

Force is not mass. F ≠ mav, F ≠ Fv, F ≠ Fs

 

Fv is instantaneous power if they are in the same direction (there's a dot product). Fs is work (again, it's a dot product). You need to use terminology as it's already defined (and not repurpose it) to communicate effectively. There's legitimate physics here, but you have to do that physics to analyze this system.

You are right.

Yes Kinetic energy is not linear in speed- but in acceleration?

But does it matter changing something in functions of rocket or not - if there is a real increase its speed ?( your explanation is not something abstract - real increase in the velocity of a rocket).

Of course it would be better to explain my engine with work and power.

Indeed (F) no (m) but offer such an experiment with a truck.

[swansont]

F ≠ mv and kg is not a unit of force. This is all gibberish until you deal in proper physics equations and units.

[Endy0816]

@OP:

 

You are mixing up mass and force.

 

Force = Mass x Acceleration

 

Similarly, if you multiply Mass and a Velocity, you end up with Momentum.

[Ilinca Sergiu]

F ≠ mv and kg is not a unit of force. This is all gibberish until you deal in proper physics equations and units.

What unit you prefer ? -

• Dyne
• Kilogram-force
• Kip (unit)
• Newton (unit)
• Planck force
• Pound (force)
• Poundal
• Sthène
• Ton-force
• ----at your discretion, include them in your calculations. My question is not about unit - about how can the force in some cases (my case) to behave as mass, what is the difference?

@OP:

 

You are mixing up mass and force.

 

Force = Mass x Acceleration

 

Similarly, if you multiply Mass and a Velocity, you end up with Momentum.

Yes and do it intentionally.

[swansont]

What unit you prefer ? -

 

 

Consistency is more important than the specific choice, but since this is physics related, Newtons will do. mv does not get you Newtons.

[Ilinca Sergiu]

 

Consistency is more important than the specific choice, but since this is physics related, Newtons will do. mv does not get you Newtons.

Yes maybe on this, many people express thrust of jet engine through (kgf),(Tf).

1kgf=9.8N

[swansont]

Yes maybe on this, many people express thrust of jet engine through (kgf),(Tf).

1kgf=9.8N

A kgf is a shortcut, wherein 1 kg is multiplied by g, which is an acceleration. You still can't get there from mv. Force is not momentum, nor is it work, nor power. These things are not interchangeable.

[Ilinca Sergiu]

A kgf is a shortcut, wherein 1 kg is multiplied by g, which is an acceleration. You still can't get there from mv. Force is not momentum, nor is it work, nor power. These things are not interchangeable.

Not interchangeable formally. What essentially differ these two examples?

[swansont]

There's no way to analyze the lower drawings, as the information is incomplete. You need the mass rate of the exhaust as well as its speed to find a thrust (and the effects of the pressure drop in the nozzle). As the mass of the rocket would be decreasing with time, this will give a non-constant acceleration.

[Ilinca Sergiu]

There's no way to analyze the lower drawings, as the information is incomplete. You need the mass rate of the exhaust as well as its speed to find a thrust (and the effects of the pressure drop in the nozzle). As the mass of the rocket would be decreasing with time, this will give a non-constant acceleration.

Maybe these calculations to leave more experienced readers ?

I think F(thrust) = m(mass) can check with a simple experiment(with isochoric process more difficult).

[swansont]

I think the more experienced one is, the easier it is to recognize that equating terms that have different units is wrong.

[Ilinca Sergiu]

Many perceive Oberth effect only with move the jet engine (rockets) in a circular orbit. So I suggest the model of my engine with circular motion of jet engine. Maybe it will be more clear what I mean.
The only difference here is what I have instead of the gravity - acceleration (but according to Einstein they are identical).

[HopDavid]

 

You mention a citation but don't actually give one. It's not IMO a particularly good explanation of the Oberth effect, since 1/2 mv² + 1/2 mv_b² is not what you'd expect, assuming you had a basic understanding of kinematics and math.

 

1/2 (v + v_b)² = 1/2 v² + v*v_b + 1/2 v_b² just as I said in my explanation. If you don't think this is true, I would suggest you lack a basic understanding of math. This is high school math.

 

Thanks for the cite. It's a bad explanation. Kinetic energy is not linear in speed, so expecting it to be so, e.g. a doubling of speed to double the energy, is a misunderstanding of physics,

 

My explanation does not suggest doubling speed would double the energy. Kinetic energy is 1/2 m v². Doubling the speed would quadruple the kinetic energy. Tripling speed would increase kinetic energy nine fold.

 

My explanation is not bad. Rather your reading comprehension is bad.

[swansont]

 

1/2 (v + v_b)² = 1/2 v² + v*v_b + 1/2 v_b² just as I said in my explanation. If you don't think this is true, I would suggest you lack a basic understanding of math. This is high school math.

 

My explanation does not suggest doubling speed would double the energy. Kinetic energy is 1/2 m v². Doubling the speed would quadruple the kinetic energy. Tripling speed would increase kinetic energy nine fold.

 

My explanation is not bad. Rather your reading comprehension is bad.

 

 

"Take 1/2 of m (v + v_b)v², and you get 1/2 mv² as well as 1/2 mv_b², the kinetic energy you'd expect from adding these two speeds" is a blatantly incorrect statement, and you seem to agree, as you've admitted that 1/2 (v + v_b)² = 1/2 v² + v*v_b + 1/2 v_b²

 

^{But by all means, put the blame on me. The customer is always wrong.}

[HopDavid]

 

 

"Take 1/2 of m (v + v_b)v², and you get 1/2 mv² as well as 1/2 mv_b², the kinetic energy you'd expect from adding these two speeds" is a blatantly incorrect statement, and you seem to agree, as you've admitted that 1/2 (v + v_b)² = 1/2 v² + v*v_b + 1/2 v_b²

 

^{But by all means, put the blame on me. The customer is always wrong.}

 

You've brought to my attention a typo. I had an extra v in the equation. I've corrected that. That section now looks like this:

 

Take 1/2 of m (v + v_b)², and you get 1/2 mv² as well as 1/2 mv_b², the kinetic energy you might expect

from adding these two speeds. On top of that, you also get m(v * v_b).

The pink rectangle above is Oberth gravy.

 

---------------------

 

A naive reader often does expect v_b to boost kinetic energy by only 1/2 mv_b². People are often surprised that they also get m(v * v_b). But some of my readers may have more accurate expectations so I inserted the word "might".

 

The context you clipped is also there: On top of that, you also get m(v * v_b). That part I didn't change. The illustrations as well as the text are now quite clear. 1/2 (v + v_b)² = 1/2 v² + v*v_b + 1/2 v_b²

[swansont]

The context you clipped is also there: On top of that, you also get m(v * v_b). That part I didn't change. The illustrations as well as the text are now quite clear. 1/2 (v + v_b)² = 1/2 v² + v*v_b + 1/2 v_b²

I didn't clip any context, Ilinca Sergiu did. In my original objection I was quoting the OP, not your website. I wasn't even aware of your website at that point, since the link wasn't given until afterwards. (at my prompting, I will add)

[HopDavid]

I didn't clip any context, Ilinca Sergiu did.

 

False. Sergiu included the phrase in his quotes. It is you who clipped context.

[swansont]

 

False. Sergiu included the phrase in his quotes. It is you who clipped context.

 

 

My quote in post 2 is verbatim from the OP. I clipped his example, but nothing from his quote.

 

And I can't believe that this discussion is happening.

[HopDavid]

 

 

My quote in post 2 is verbatim from the OP. I clipped his example, but nothing from his quote.

 

And I can't believe that this discussion is happening.

 

Believe me, under normal circumstances I wouldn't be here. I have zero interest in "Wharp Drives".

 

It's forgivable if you give Sergiu's scheme only a cursory reading. For your benefit I did a screen capture:

 

 

As you can see, Sergiu did not clip context. See the portion with the red underline.

 

The clipping was done by you. Yes, if you clip the latter part, the statement's wrong. Thank you for stating the excruciatingly obvious. Kudos for ignoring context.

 

I make mistakes on my blog. I welcome criticism and correction. But misinterpretation coming from a clipping of context is a waste of my time. If you have a valid criticism, I'd like to hear it. If not, please admit as much and I will be done with this silly thread.

Edited July 16, 2014 by HopDavid

[Ophiolite]

 

If you have a valid criticism, I'd like to hear it.

Launching into an aggressive attack from the outset and failing to apologise for an acknowledged error, does not promote productive discussion. Courtesy does not incur any increase in your personal taxes.

[swansont]

 

Believe me, under normal circumstances I wouldn't be here. I have zero interest in "Wharp Drives".

 

It's forgivable if you give Sergiu's scheme only a cursory reading. For your benefit I did a screen capture:

 

 

As you can see, Sergiu did not clip context. See the portion with the red underline.

 

The clipping was done by you. Yes, if you clip the latter part, the statement's wrong. Thank you for stating the excruciatingly obvious. Kudos for ignoring context.

 

I make mistakes on my blog. I welcome criticism and correction. But misinterpretation coming from a clipping of context is a waste of my time. If you have a valid criticism, I'd like to hear it. If not, please admit as much and I will be done with this silly thread.

 

The statement is wrong once you hit the period at the end of the sentence.

 

"Take 1/2 of m (v + vb)v2, and you get 1/2 mv2 as well as 1/2 mvb2, the kinetic energy you'd expect from adding these two speeds." That's a full stop. And that's not what you'd expect, if you understood math. "On top of that, you also get m(v * vb)." is a true statement, but my objections is the claim of what you'd expect (and you can tell, because I explicitly use that phrase) and that last term was not included in the description of "what you'd expect" — you say "on top of that". The clear implication, to me, is that the additional term is unexpected. I can't find a way to read this another way.

 

Your little cartoon conveniently omits the what you'd expect phrase, which is the focus off my criticism. i.e. you have (oh, the irony) clipped the context from it.

 

My opinion of the description is unchanged.