HopDavid Posted July 17, 2014 Share Posted July 17, 2014 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. Your story is changing. First you were saying Sergiu clipped my mention of m(v*vb). He did not -- as shown in the screen capture upthread. Now you want to focus on the word expect. You seem to believe all expectations are accurate. Not so. For most people m(v*vb) is unexpected. They are typically surprised to learn of the Oberth benefit. You also want to argue that a period gives you license to ignore the second part of the paragraph. Comedians lampoon public figures by taking a single sentence out of context to get an obviously wrong statement. People with common sense recognize this as silliness, a comedic device. It is sad to see you using this in earnest. Ophiolite, the error was an obvious typo, acknowledged and fixed. The other errors are Swansont fictions. I have not omitted the term m(v+vb). In fact that term is the focus for much of the blog post in question. Nor have i said the relationship between velocity and energy is linear. These two bloopers come from Swansont's imagination, not me. I am justifiably annoyed. Link to comment Share on other sites More sharing options...
swansont Posted July 18, 2014 Share Posted July 18, 2014 Your story is changing. First you were saying Sergiu clipped my mention of m(v*vb). He did not -- as shown in the screen capture upthread. I think there was a misunderstanding here. When you said I clipped context, I thought you were referring to quoted material. Since the only quote in my post was of the OP, I could not have done so. Now it is clear you are referring to something else. But my position is unchanged. Your statement is that one should expect for math not to work, and I think that's silly. I explained that. You say a "naive reader" might expect that, but I stated that I was referring to people who had a basic understanding of math. Now you want to focus on the word expect. You seem to believe all expectations are accurate. Not so. For most people m(v*vb) is unexpected. They are typically surprised to learn of the Oberth benefit. No, not just now. "expect" is in my original post, which is just two sentences long. One mentions the lack of citation. The other is It's not IMO a particularly good explanation of the Oberth effect, since 1/2 mv2 + 1/2 mvb2 is not what you'd expect, assuming you had a basic understanding of kinematics and math. I've bolded part of it. Your statement about kinetic energy has no dependence on the Oberth effect, it's a simple issue of change in speed vs change in KE. The Oberth effect isn't part of that particular argument. You also want to argue that a period gives you license to ignore the second part of the paragraph. Comedians lampoon public figures by taking a single sentence out of context to get an obviously wrong statement. People with common sense recognize this as silliness, a comedic device. It is sad to see you using this in earnest. The second sentence adds no context to the first. You made a claim as to what one should expect to happen to KE when speed changes. The second sentence reinforces the fact that you are insisting that people will expect the wrong answer. My point was that people who can do math won;t expect the wrong answer. Ophiolite, the error was an obvious typo, acknowledged and fixed. The other errors are Swansont fictions. I have not omitted the term m(v+vb). In fact that term is the focus for much of the blog post in question. Nor have i said the relationship between velocity and energy is linear. These two bloopers come from Swansont's imagination, not me. I am justifiably annoyed. I never claimed you omitted anything. You are reading way more into my very short post than is actually there. I also don't give a rat's ass if you are annoyed. I stated an opinion: people who understand math will not expect the incorrect answer. Also that your explanation of the Oberth effect is poor. You are free to disregard my opinion. You are also free to ask for clarification of what I posted, if I have been unclear. But what you can't do is tell me what my intent was, because only I know that. So unless you are asking for clarification, I have no intention of continuing this discussion. Link to comment Share on other sites More sharing options...
Ilinca Sergiu Posted August 3, 2014 Author Share Posted August 3, 2014 To understand my question needed to simplify these physical model to one very simple. Probably easier is impossible. Far away in space is the space station. At the station there is a Elevator without ceiling and the Elevator shaft is also open to space. On the floor of the Elevator has the ball weighing 1kg . In this Elevator is an astronaut he takes the ball and throws it up with the speed 10 m/s. The ball flies out of the station with the speed 10 m/s. The thrust obtained station from throwing the ball is Ftrast=m*v=10kgf. This Elevator is also the ability to move up and down with very large accelerations (10000g). The movement of the Elevator up and down with any accelerations and velocities has no impact on the total momentum of the station. He will always be ( 0 ). In the absence of Fthrust the station will make only oscillatory movements. If the Elevator will be acceleration in 10000g the ball on the floor will have a weight of 10,000kg or 10T. The astronaut will not be able to pick up the ball and throw up. And the astronaut will not withstand this acceleration. Now to throw the ball during acceleration of the lift we need Superman. I.e. this person should be 10,000 times stronger than ordinary cosmonaut. During acceleration Elevator Superman takes the ball with weight 10T and throws it up with the speed in 10 m/s. Now what may be the pressure on the floor of the Elevator from this shot and accordingly Fthrust ? Obviously it will be much greater than in the case without acceleration. And it is : the weight of the ball M=10000 kg, speed 10m/s,.. F thrust = M*v=100000kgf (I know weight not mass, but how differently? ). Do not forget that the total momentum of the system without Fthrust is ( 0 ) and these 100,000kgf will be increase the total momentum of the system? I think this is Oberth effect. But today it is used only reverse view .When the spacecraft passes periapsis gravity celestial body brake the spacecraft ( the bigger the ship's speed and gravity the more will be this brake and the more engine thrust). I think Oberth effect on the orbit can be compared with swings. Appears in periapsis when appears the weight of burning fuel. Oberth effect on the multi-stage rocket (change of acceleration). The passage periapsis of the rocket can be compared with the passage obstacles of bullet. The increase of kinetic energy in Obert effect does not arise from any relative velocities but from the increased potential energy of the fuel in nozzle. Oberth effect are the reflection of the law of energy conservation. Link to comment Share on other sites More sharing options...
swansont Posted August 3, 2014 Share Posted August 3, 2014 The Oberth effect is fairly simple: a given amount of thrust will be more efficient at increasing KE if the rocket is moving. That's because a set amount of fuel will do more work, since the thrust is exerted over a greater distance. i.e. it takes a certain amount of energy to get to a speed v (1/2mv2). It takes four times as much to get to 2v, so to get from v to 2v you need an additional 3/2mv2 of energy, i.e. you need to add three times as much energy. But you don't need three times the fuel to do it, because you do more work for a given thrust as you move faster. So when you ask "100,000kgf will be increase the total momentum of the system?" the answer is no. I don't see the connection of your idea to the Oberth effect. I also think you will get nowhere if your units don't match. kgf is a unit of force (and a non-SI unit at that) and not momentum. Making examples more complicated is often not the best approach to solving a problem. Success is usually found in making the example as simple as you can, without losing important detail. Most basic analysis is done without numbers. Link to comment Share on other sites More sharing options...
Ilinca Sergiu Posted August 5, 2014 Author Share Posted August 5, 2014 (edited) Sorry in the third picture I made a mistake. The picture must be ... The Oberth effect is fairly simple: a given amount of thrust will be more efficient at increasing KE if the rocket is moving. That's because a set amount of fuel will do more work, since the thrust is exerted over a greater distance. i.e. it takes a certain amount of energy to get to a speed v (1/2mv2). It takes four times as much to get to 2v, so to get from v to 2v you need an additional 3/2mv2 of energy, i.e. you need to add three times as much energy. But you don't need three times the fuel to do it, because you do more work for a given thrust as you move faster. So when you ask "100,000kgf will be increase the total momentum of the system?" the answer is no. I don't see the connection of your idea to the Oberth effect. I also think you will get nowhere if your units don't match. kgf is a unit of force (and a non-SI unit at that) and not momentum. Making examples more complicated is often not the best approach to solving a problem. Success is usually found in making the example as simple as you can, without losing important detail. Most basic analysis is done without numbers. Here's a more simple. Edited August 5, 2014 by Ilinca Sergiu Link to comment Share on other sites More sharing options...
swansont Posted August 5, 2014 Share Posted August 5, 2014 The impulse will be found by conservation of momentum. The momentum of the container and the ejected ball will be equal in magnitude and opposite in direction if you are in the center-of-momentum frame. Thrust would be found if this were a continuous action, and F = dP/dt Link to comment Share on other sites More sharing options...
Ilinca Sergiu Posted August 12, 2014 Author Share Posted August 12, 2014 The impulse will be found by conservation of momentum. The momentum of the container and the ejected ball will be equal in magnitude and opposite in direction if you are in the center-of-momentum frame. Thrust would be found if this were a continuous action, and F = dP/dt You calculate the momentum for resting or uniformly moving system ( which is the same) of the box-ball. And saying that it is right for accelerating system box-ball. So you deny the influence of the "gravity-acceleration" inside of the box on the momentum. I can't agree with ,gravity affects the momentum. Link to comment Share on other sites More sharing options...
swansont Posted August 12, 2014 Share Posted August 12, 2014 Over arbitrarily short time intervals, such as the instant before and after a particle is ejected, the effect is negligible. Link to comment Share on other sites More sharing options...
Ilinca Sergiu Posted August 16, 2014 Author Share Posted August 16, 2014 Over arbitrarily short time intervals, such as the instant before and after a particle is ejected, the effect is negligible. Yes,but we are interested in short time but a certain interval of time. What's not essential to the acceleration-gravity 10m/s2 and particles with 400000m/s2 , can be significant for return values or equal - 400,000 and 400,000 m/s2 values. If the relative speed is close to zero - i=W*v Link to comment Share on other sites More sharing options...
swansont Posted August 16, 2014 Share Posted August 16, 2014 Yes,but we are interested in short time but a certain interval of time. Then calculate the impulse imparted by gravity. That's the amount of overall change in momentum of the system. Link to comment Share on other sites More sharing options...
Ilinca Sergiu Posted August 16, 2014 Author Share Posted August 16, 2014 Then calculate the impulse imparted by gravity. That's the amount of overall change in momentum of the system. No need to imparted by gravity here. Link to comment Share on other sites More sharing options...
Ilinca Sergiu Posted August 20, 2014 Author Share Posted August 20, 2014 As will be burning fuel when nozzle acceleratingin in 30000g or 300000m/s2 (acceleration of normal railgun) ?If the fuel acceleration is approximately equal to the acceleration of nozzle-I think that will be a nuclear fusion reaction similar to the reaction of the white dwarf. I think if make this engine with the nuclear fusion reaction - will look something like this.If without nuclear fusion reaction-- will look too, something like this but without nozzle coil. Link to comment Share on other sites More sharing options...
Ilinca Sergiu Posted August 28, 2014 Author Share Posted August 28, 2014 Perhaps more relevant is if consider a similar system from a position of (N) power. (With position (N) and (A) starts and the explanation of Oberth effect). Link to comment Share on other sites More sharing options...
Ilinca Sergiu Posted October 28, 2014 Author Share Posted October 28, 2014 (edited) Or easier. Imagine himself instead of the person in the picture and try to understand what he would push these carts (in what conditions it will be equally hard to push two carts?). Edited October 28, 2014 by Ilinca Sergiu Link to comment Share on other sites More sharing options...
Ilinca Sergiu Posted November 15, 2014 Author Share Posted November 15, 2014 I'll try to explain in more detail. The man pushes the nozzle against the force of thrust. Physically, it is difficult for him to do. If he pushes the nozzle certain distance within a certain time, he spends some power. The sooner he does it, the more he gets tired (consumes more power).This power is transferred to his feet on the ground surface or to another mass (for example a moving platform). The same thing happens if someone pushes some mass which applies a very small acceleration. If this acceleration not exist ,that people will push any weight without power consumption. In this case, the equivalence between (F) and (m) will not be ,the nozzle will resist movement and the mass will not. I.e. this equivalence will shift giving speed only the mass but not the nozzle. But this is a perfect case because in reality, it is necessary to give first some acceleration to get the velocity (which occurs when nozzle is in acceleration I described at the beginning). If such an ideal case was that ,the system could unbalance even more than the equivalence of (F) and (m). Link to comment Share on other sites More sharing options...
Ilinca Sergiu Posted December 18, 2014 Author Share Posted December 18, 2014 Sublight spaceship with the engine on Oberth effect,will look like this- Link to comment Share on other sites More sharing options...
Ilinca Sergiu Posted December 24, 2014 Author Share Posted December 24, 2014 It would be very interesting to know: - that there is speculation? My physical model based on Oberth effect, a brilliant mathematical expression from HopDavid of this effect, or the Oberth effect itself ? Link to comment Share on other sites More sharing options...
imatfaal Posted December 24, 2014 Share Posted December 24, 2014 You have a diagram of a man-made object with a tag saying a=300000 m/s^2 - about 3000 times higher than the human body can withstand, twice as high as the electronic circuitry in artillery shells need, and about the average acceleration of a bullet from a luger whilst within the barrel* - and you ask where the speculation is? * gotta love wikipedia Link to comment Share on other sites More sharing options...
Ilinca Sergiu Posted December 27, 2014 Author Share Posted December 27, 2014 In my drawings the only thing which is moved with high acceleration ( including 300000 m/s2)- the nozzle of the jet engine(similar object to the nozzle). Where you saw people under acceleration 300000 m/s2 ?. Maybe it horror picture belongs to you personally? 150000 m/s2 acceleration limit for electrical circuit- these data are taken from the archives of what century? Link to comment Share on other sites More sharing options...
Ilinca Sergiu Posted December 30, 2014 Author Share Posted December 30, 2014 I think this engine will look something like this. Link to comment Share on other sites More sharing options...
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