magnetic propulsion system.

Re: Magnetic propulsion theory revisted.

reactor1967

Newtons Laws are in effect. For every action there is always an equal but opposite reaction. The net momentum of your device will be zero. I would suggest you try building it first. Then you can see what Newtons Laws are all about first hand.

I've built lots of mechanical devices looking for a way to circumvent Newtons Laws. Only two devices I built, showed something that suggested I should investigate further. One was a gyro drive. The other was a single loop of wire which I injected a very large jolt of current into. The loop of wire moves toward the north side of the magnetic field it produces when the current is first applied. But when the current is turned off the loop of wire moves back in the other direction, then oscillates to a stop. So the equal but opposite reaction is displaced in time. It's like dangling a carrot in front of me. But I can never quite reach the carrot. All attempts so far to amplify the propulsive effect on the loop of wire have not been successful.

But I continue to do research to this day on this simple little experiment. You can read about it in my gravity research thread.
 
Re: Magnetic propulsion theory revisted.

Newtons Laws are in effect. For every action there is always an equal but opposite reaction. The net momentum of your device will be zero. I would suggest you try building it first. Then you can see what Newtons Laws are all about first hand.

I've built lots of mechanical devices looking for a way to circumvent Newtons Laws. Only two devices I built, showed something that suggested I should investigate further. One was a gyro drive. The other was a single loop of wire which I injected a very large jolt of current into. The loop of wire moves toward the north side of the magnetic field it produces when the current is first applied. But when the current is turned off the loop of wire moves back in the other direction, then oscillates to a stop. So the equal but opposite reaction is displaced in time. It's like dangling a carrot in front of me. But I can never quite reach the carrot. All attempts so far to amplify the propulsive effect on the loop of wire have not been successful.

But I continue to do research to this day on this simple little experiment. You can read about it in my gravity research thread.

Im pretty sure your right. And, building it would not be all that bad. It would be a very simple thing to do.
 
Re: Magnetic propulsion theory revisted.

Why build it when modeling it mathematically is much easier and will tell you what you wish to know? Modeling this would actually be quite simple because all motion is, essentially, constrained to a single, linear degree of freedom. Vectors not even required. The hardest part, which is really not that hard, is to model the time-varying magnetic fields which act as the forcing functions to a system that is otherwise in mechanical equilibrium.

RMT
 
Re: Magnetic propulsion theory revisted.

You must be familiar with the so-called 'inertial propulsion' devices. I saw years ago an article in an Australian encyclopedia called "How It Works", which explained that a spinning wheel precesses without momentum. It's like the physics demonstration in which a wheel on an axle, which is too heavy for a person to raise over his head, can easily be held aloft when the wheel is spinning and allowed to precess.

Also are you familiar with Dr. Robert Beck? He's pretty old by now, but he was into the anti-gravity
and "free energy" borderland sciences.
 
Re: Magnetic propulsion theory revisted.

Packerbacker

You must be familiar with the so-called 'inertial propulsion' devices.

Yes, very familiar. Some of them produce apparently successful results. But to date nothing in the public domain has gotten off the ground. There are lots of claims. But everyone either dies with the so called secret. Or just doesn't have a working device at present (Searle).

I saw years ago an article in an Australian encyclopedia called "How It Works", which explained that a spinning wheel precesses without momentum. It's like the physics demonstration in which a wheel on an axle, which is too heavy for a person to raise over his head, can easily be held aloft when the wheel is spinning and allowed to precess.

You don't have to accept the how it works explanation. Since theory is not fact. But I believe understanding this simple little toy with alternate interpretations is the key to understanding and mastering the reality we live in.

Also are you familiar with Dr. Robert Beck? He's pretty old by now, but he was into the anti-gravity
and "free energy" borderland sciences.

No, haven't heard of him. My favorite though is Norman Dean, the inventor of the Dean machine. Actual documented claims of his device losing weight on a scale exist. Of course he died with the secret.
 
Re: Magnetic propulsion theory revisted.

Einstein:

Ah yes, Dean. That was the name I was forgetting. Bob (Beck) was telling us once that they had been playing around with a Dean drive device "down at the shop" but had not achieved any notable results.
 
Re: Magnetic propulsion theory revisted.

Packerbacker

Bob (Beck) was telling us once that they had been playing around with a Dean drive device "down at the shop" but had not achieved any notable results.

Yes, the device Dean took out a patent on apparently did not perform like the devices he demonstrated. At the time he wanted to sell his secret to the military for a million dollars. But the military wasn't buying. So he kept the secret and took it to his grave.

But that doesn't stop me, or others from trying to find out what that secret was. My analysis so far indicates Newton's Laws are incomplete. Forces don't come in pairs. They come in threes. All along there is a hidden third player. I think Dean put together a machine using that hidden third player to his advantage.
 
Re: Magnetic propulsion theory revisted.

Why build it when modeling it mathematically is much easier and will tell you what you wish to know? Modeling this would actually be quite simple because all motion is, essentially, constrained to a single, linear degree of freedom. Vectors not even required. The hardest part, which is really not that hard, is to model the time-varying magnetic fields which act as the forcing functions to a system that is otherwise in mechanical equilibrium

I would be glad to try if you want to help. My questions are:
What calculations do you think I need?
Do I need software? (I can write my own if needed.)
How is it set up? Basicly where would you start?
What mathematical units would I be using? Beside using gauss on the magnet I am unsure of other units.
I am willing to try. With modeling chances are I would write a program and run it and post it with code for others to try. If I could understand it well enough to do it I think it would be fun.

So lets see. The device needs a weight. How much weight we are pushing. I need to look into that.
The springs have ratings from what I remember. Don,t know what unit that rating is just as of yet. So as the spring compresses it will put pressure on the magnet in some unit. I need to figure out how much pressure to use. I need to covert from pressure to acceleration. The electro-magnet will have a gauss rating. The piece of metal attached to the spring should have a weight associated with it. The magnet attracts the metal with some force in some unit. That force causes acceleration by compressing the spring which has to be enough to compress the spring. Now that is the bottom side. The top side does the same thing counter-reacting the force pushing up. But, I want to do it so the magnet does not accelerate faster than the metal on the spring coming toward it. So far that is my estimation. Before I can calculate the top side I need to calculate the bottom side so I know the rating for the spring on the top side. I need to do some homework on my design getting the proper information before I can model it. It does not explain enough at this point. Feel free to point me in the right direction.

Units to use.

1. Gauss
2. PSI - pounds per square inch
3. On acceleration I am guessing milimeters or meters per second.
4. I need to to some electrical calculations. Being a while for me but I can handle that. I need to figure out what my power supply will be how many volts and miliamps. I am going to say lets use 12 volts and some current I need to come up with. To get gauss I need look into how big my metal is and how many windings. The more windings the more gauss.

This is what I am coming up with so far for modeling.
 
Re: Magnetic propulsion theory revisted.

Why build it when modeling it mathematically is much easier and will tell you what you wish to know? Modeling this would actually be quite simple because all motion is, essentially, constrained to a single, linear degree of freedom. Vectors not even required. The hardest part, which is really not that hard, is to model the time-varying magnetic fields which act as the forcing functions to a system that is otherwise in mechanical equilibrium.

Correct. We don't have to get into complex math to show the roblem. A simple analogy will do just fine.

We can see from the diagram that the metal plates that will be attracted by the magnet are of different masses. That is supposed to be the variation on the momentum as they move toward and away from the magnet between them.

It's a problem in vector analysis but, as I said, we'll keep it simple.

Let's start with both springs in their fully extended position - magnet off. OK. Now turn the magnet on. The plates move toward the center in the +x and -x directions (toward each other).

For the sake of argument we plug in some arbitrary numbers to represent the momentum of each plate: "a" is the large plate and "b" the smaller one.

P = momentum (mv)

P(a) = +2
P(b) = -1
P(a,b) = +1

The system has a net momentum of +1 along the x axis.

Next, we turn off the magnet and the spring returns the plates to their original positions and in so doing imparts a further momentum to the system:

P'(a) = -2
P'(b) = +1
P'(a,b) = -1

We now have a net momentum of -1 along the x axis. (We primed the P's here to differentiate them from the first half of the cycle).

Now we add the momena together to see what we have after one complete cycle:

P(a) + P(b) + P'(a) + P'(b) = P(a,b) + P'(a,b) = +1-1 = 0

The net momentum is zero along the x axis. Repeat the cycle and what you really get is the system oscillating about the point x=0, the original position of the system.
 
Re: Magnetic propulsion theory revisted.

Correct. We don't have to get into complex math to show the roblem. A simple analogy will do just fine.

We can see from the diagram that the metal plates that will be attracted by the magnet are of different masses. That is supposed to be the variation on the momentum as they move toward and away from the magnet between them.

It's a problem in vector analysis but, as I said, we'll keep it simple.

Let's start with both springs in their fully extended position - magnet off. OK. Now turn the magnet on. The plates move toward the center in the +x and -x directions (toward each other).

For the sake of argument we plug in some arbitrary numbers to represent the momentum of each plate: "a" is the large plate and "b" the smaller one.

P = momentum (mv)

P(a) = +2
P(b) = -1
P(a,b) = +1

The system has a net momentum of +1 along the x axis.

Next, we turn off the magnet and the spring returns the plates to their original positions and in so doing imparts a further momentum to the system:

P'(a) = -2
P'(b) = +1
P'(a,b) = -1

We now have a net momentum of -1 along the x axis. (We primed the P's here to differentiate them from the first half of the cycle).

Now we add the momena together to see what we have after one complete cycle:

P(a) + P(b) + P'(a) + P'(b) = P(a,b) + P'(a,b) = +1-1 = 0

The net momentum is zero along the x axis. Repeat the cycle and what you really get is the system oscillating about the point x=0, the original position of the system.

All right net momentum of zero. Well I guess that is it. A little simplier than what I was thinking. Thanks Darby.
 
Re: Magnetic propulsion theory revisted.

Well, Darby beat me to it! But he also simplified it quite a bit more than I was going to suggest. I was thinking that since reactor enjoys coding so much, that we could use his example as a good sample problem to introduce dynamic modeling with a simple 1 Degree Of Freedom simulation. We would draw on the classical "spring-mass-damper" problem used to introduce 2nd order dynamical systems, except in this case we can omit the damper (because there is none in the setup, per se), which then keeps the whole system a 1st order with the spring constant (k).

Here are some charts that introduce the more complex spring-mass-damper system modeling theory.

http://ecow.engr.wisc.edu/cgi-bin/getbig/me/340/duffie/notes/340-09spring-mass-damper.pdf

We can generalize the problem down to a spring-mass problem and write the equilibrium 1 DOF equation of motion first, and then we can introduce a simple forcing function for the magnetic field to see how the dynamics come into play with the spring constant.

Please let me know if you wish to continue this line of modeling, reactor, as it will be a bit more involved than Darby's approach, but it can lead to a 1 DOF simulation and give you some insight into how we build 6 DOF simulations of aircraft and spacecraft. This will be a multi-post effort. For now, let us just understand the equilibrium situation and write the equilibrium equation of motion:

All springs are typically defined by their linear "spring constant" which is a measure of Force per unit Distance (F/d). By looking at reactor's diagram with the magnet turned off, we can see that the lower spring must support the total weight of the magnet, the upper spring, and the upper plate to achieve equilibrium. From Newton's Laws we can write the simple form of the equilibrium as:

Summation of all vertical forces = 0

Now we account for the forces:

1) Forces due to gravity of the upper plate, magnet, and upper spring:
F1 = (Mup + Mus + Mm) * g
F1 = Total force due to gravity of upper plate, magnet, and upper spring.
Mup = Mass of upper plate.
Mus = Mass of upper spring.
Mm = Mass of magnet
g = gravitational constant (9.8 m/sec^2 or 32.174 feet/sec^2)

2) Reaction force of the lower spring pushing up against the Mup, Mus, and Mm masses:
F2 = Reaction force of lower spring against the masses above it.
F2 = kls*xls
kls = spring constant of the lower spring (Lbs per inch or Lbs per foot) - This is a known just as we assume the masses are also known. It represents how much energy the spring can store.
xls = equlibrium distance between the lower plate and the magnet when the lower spring is compressed by the upper masses.

To be in equilibrium, these two forces must counter each other in the vertical direction. Hence:

F1 = F2

What does this give us? Well, if we know the masses of the magnet, upper plate, and upper spring, and we are given the spring constant of the lower spring, then this will allow us to solve for the equilibrium distance (xls) of the lower spring. This would be considered an "initial condition" for the simulation program which we could write from this exercise.

<font color="red">Problem for the student:[/COLOR] Assuming we have solved for the equilbrium distance for the lower spring (xls), use the same analytical technique to solve for the equilibrium distance of the upper spring (xus). This will be another "initial condition" for the dynamic simulation we can eventually build.

RMT
 
Re: Magnetic propulsion theory revisted.

Allow me to explain where I am going with this (I am sure Darby can already see it, but for the edification of others...)

(BTW, please let me know if I should continue. If you are not interested in building this model, I can avoid wasting my time...thanks)

Where we are headed with this modeling effort is to come up with equations that solve for the linear (1-DOF) accelerations of the magnet itself and the upper plate as a function of the force exerted by the magnet on both plates at any given time. Once we have equations for their accelerations, we can then implement them in a time domain simulation which will numerically integrate those accelerations over time to yield their velocities, and then further numerically integrate those velocities over time to yield their positions (heights above the reference table top upon which the whole thing sits). The fact that we are using integrals is why the initial lengths of the springs are required... they become the initial conditions to the velocity integrators. The initial conditions for the acceleration integrators are assumed to be zero because the magnet and upper plate are at equilibrium before we turn the magnet on.

This approach is on par with how we build numerical simulations of aircraft and spacecraft in 6 Degrees Of Freedom. When we can develop the aerodynamic and thrust force and moment equations for pitch, roll, yaw rotations and longitudinal, lateral, and vertical translations, we can write these equations so as to solve for the linear accelerations and rotational accelerations (all 6 of them). We then apply the numerical integration technique described above for 1-DOF to all 6 of these accelerations and that is how we simulate the acceleration, velocity, and position of the vehicle changing with respect to time.

As I said before, this is a very good exercise to help people understand how we model dynamical systems in the real world of aerospace engineering. It is a powerful tool that, once learned, can be extended to equations that describe any type of dynamical system.

RMT
 
Re: Magnetic propulsion theory revisted.

I need to spend some more time reading your last two post. No, I would not want to wast your time. That would be rude of me. Right off the bat I am boggled already. I want to look thru what you said and see if I think my theory still holds.
 
Re: Magnetic propulsion theory revisted.

Reactor,

I need to spend some more time reading your last two post. No, I would not want to wast your time. That would be rude of me. Right off the bat I am boggled already. I want to look thru what you said and see if I think my theory still holds.

Please do not hesitate to ask questions for those things you do not understand. I will be happy to explain them in more detail with simpler language, if necessary. While I think L. Ron Hubbard's Scientology is a load of BS, he did make a very important point in some of his writings: People most often get confused when they read a word, do not understand it, but continue to read on thinking they will "make up" for the misunderstanding later. I am a firm believer that this is a problem many people create for themselves, and thus I want to help you understand anything you are missing.

One question (well, really two) for you that I would appreciate a straight-forward answer:

Q: Do you fully understand the concept of the integral &amp; derivative in calculus? And do you understand how velocity is the derivative with respect to time of position, and that the integral of velocity with respect to time will yield position?

This is a fundamental understanding of calculus that is necessary before much of what I wrote above will make sense. So let me know if I need to review derivatives and integrals first, which I will gladly do. That might clear some things up.

Thanks,
RMT
 
Re: Magnetic propulsion theory revisted.

One question (well, really two) for you that I would appreciate a straight-forward answer:

Q: Do you fully understand the concept of the integral &amp; derivative in calculus? And do you understand how velocity is the derivative with respect to time of position, and that the integral of velocity with respect to time will yield position?

This is a fundamental understanding of calculus that is necessary before much of what I wrote above will make sense. So let me know if I need to review derivatives and integrals first, which I will gladly do. That might clear some things up.

I appreciate the help.
Do I understand the intergral and derivative in calculus? I don,t remember about the intergral but I remember the derivative is a rate of change. I also remember being shown the hard way to compute a derivative then the easy way.
Volocity is a derivative with respect to time of position. It seems to me volocity is also a rate of change per time and position. At t1 the volocity is # and position is #. At t2 the volocity is and position is #. The volicity itself can speed up or slow day per rate of change which is a derivative. Does it go something like this?
I saved what you wrote above to review later. I will need to spend some time with it to understand it. I have been looking into software for reviewing and then studying up on calculas. I have not found much yet though. I figure I will come across something.
 
Re: Magnetic propulsion theory revisted.

The other was a single loop of wire which I injected a very large jolt of current into. The loop of wire moves toward the north side of the magnetic field it produces when the current is first applied. But when the current is turned off the loop of wire moves back in the other direction, then oscillates to a stop. So the equal but opposite reaction is displaced in time. It's like dangling a carrot in front of me. But I can never quite reach the carrot. All attempts so far to amplify the propulsive effect on the loop of wire have not been successful.

On youtube.com I saw a video where a guy was trying to explain how a ufo works. He had a coil of wire that he plugged into an outlet. The coil jumped into its own magnetic field and levitated. He could not leave it up long because the coil was becoming dangerously hot. I have been trying to find that video again. It was on a video where there was a bright yellow light at the bottom of the craft. I figured that someone could use what he did but control the current in sections of the coil so that the front of the coil could be lowered and the back raised to make it move forward or backwards and so forth. Also this was a AC system not DC.

Links to youtube videos of levitation coils.

http://www.youtube.com/watch?v=Pi-22r0OJJA

http://www.youtube.com/watch?v=WX1fkfJPWpY&amp;feature=related

And the video I saw.
http://www.youtube.com/watch?v=mP5JgG1-0jg&amp;NR=1
 
Re: Magnetic propulsion theory revisted.

I started a download of math tutor dvd for calculus 1 &amp; 2. That is a big download so it will take a while. I went back and read your post but I am having trouble finding the equations you mentioned. I was getting journals which probably was not a bad thing but I could not immediately find the equations. I wanted to take a look at them. Would you please post them for me. I am talking about the.
"modeling effort is to come up with equations that solve for the linear (1-DOF)" and I think you said this could goto.
"modeling effort is to come up with equations that solve for the linear (6-DOF)"
I am going to have to learn more about integrals. Derivatives I might be able to muster through though a review would not hurt me. It has been a while for me on calculus. I did better in college at calculus than I did at algebra. For some reason it was easier for me.

On the equations please let me know. Thanks.
 
Re: Magnetic propulsion theory revisted.

Reactor,

I appreciate the help.

You are to be commended for deciding to study this, and learn (or re-learn as the case may be) the powerful tool that is calculus. It can help you not only in your magnetic ideas but also in your AI experiments.

Do I understand the intergral and derivative in calculus? I don,t remember about the intergral but I remember the derivative is a rate of change. I also remember being shown the hard way to compute a derivative then the easy way.

1) The integral is merely the mathematical function that "undoes" or reverses the derivative. It is most instructive to view and describe these functions of calculus within the domain of analytic geometry, which includes simple X-Y graphs...which we will do below.
2) I have a feeling (not sure) that when you say the "easy way" to compute a derivative you might be referring to the finite difference method, which is actually only an approximation to the derivative. And in some cases it can be a very good approximation, especially if the quantity you wish to take the derivative of is linear in nature. But let's not get ahead of ourselves...

Volocity is a derivative with respect to time of position. It seems to me volocity is also a rate of change per time and position.

Let's make sure we are clear here. The first statement is definitely true. I am not sure exactly what you are saying in the second (i.e. an equation might clear it up). But let's write the finite-difference approximation for velocity in terms of position and time:

Velocity = (x2 - x1)/(t2 - t1) = Delta(X)/Delta(t) = Difference in Positions/Difference in Time

This is an approximation because depending upon how large the granulaity is for the time difference (t2-t1), the numerical value of the velocity could change within that time interval and this equation would not represent that change. The way we go from the finite difference approximation to the instantaneous derivative is through the application of the calculus principle of "allow the difference in time to approach zero". Symbolically, this is represented as:

(t2-t1)-&gt;0

Where we represent the finite-difference approximation of a derivative as Delta(x)/Delta(t), when we wish to represent the precise, calculus version of the derivative we write it as:

velocity = dx/dt

This notation probably refers to what you have termed the "hard way" to compute the derivative. I agree it may be hard, but it is hard because it is computing the precise derivative, rather than just the approximation. The classic example is if we assume position can be represented by the function "x squared" (x^2):

Position = x^2

The "hard" way to take the derivative would be to move the exponent down to be a multiplier of "x" and then reduce the exponent by 1. Hence:

dPosition/dt = 2*x (we brought the exponent "2" down as a multipler on "x" and reduced the power of "x").

I know that this seems complicated. Let's leave that behind for now and address a few other points you made:

At t1 the volocity is # and position is #. At t2 the volocity is <font color="red">#[/COLOR] and position is #.

I added the "#" which I think you omitted. But yes, this is correct. This represents the finite-difference approximation of the derivative that I explained above. Very good!

The volicity itself can speed up or slow day per rate of change which is a derivative. Does it go something like this?

Yes. Again, this is very good. And when the velocity itself goes up or down, this is what we call acceleration. Acceleration (linear) is nothing more than the derivative of velocity with respect to time. And since velocity is the derivative of position with respect to time, we could also desribe acceleration as "the second derivative of position with respect to time". In other words: If we apply the calculus derivative function with respect to time TWICE on the measurements on position, then we arrive at the measure of the body's acceleration.

I saved what you wrote above to review later. I will need to spend some time with it to understand it. I have been looking into software for reviewing and then studying up on calculas. I have not found much yet though. I figure I will come across something.

Very good. Now let me give you an example that uses the domain of analytic geometry I mentioned above. Taking the derivative of a quantity is nothing more than figuring out the instantaneous slope of a line (or curve). Here is a graphical example I use in my ARO 101 course to teach students one of the most important stability derivatives used in aerodynamic analysis of an airfoil or wing... We call this stability derivative the Lift-Curve Slope and this diagram may help you understand it a bit better. (but maybe not!) /ttiforum/images/graemlins/smile.gif

Clalpha.jpg


In this graphic we are using the finite-difference approximation to calculate the slope of the lift coefficient curve. As the angle of attack of an airfoil increases, the total lift increases... up to the point where the airfoil stalls. So the lift-curve slope is a derivative of the lift coefficient with respect to angle of attack in much the same way that velocity is a derivative of position with respect to time. You could also draw an "x-y" plot of position (y values) vs. time (x-values). The slope of that line at any point represents the velocity.

We will deal with the integral in a later post, if you wish. But as I said above, the integral is just the "undoing" of the derivative. So if the derivative is equivalent to finding the slope of a curve anywhere along that curve, we will come to find that the integral seeks to find the area underneath a curve.

Hope this helps more than it confuses...
RMT
 
Re: Magnetic propulsion theory revisted.

Hope this helps more than it confuses...

No actually this clears up some of my misconceptions. For one I never new the way I thought was easy was an approximation. When I first learned to do a derivative we used a whole sheet of paper to work it. Then on the chalk board the next day he showed us a quick way. Now I know. It never occurred to me that a derivative could be applied to a graph though I am pretty sure down the line somewhere someone told me. The velocity equation explains a little more than d = r * t. Thank you for explaining this to me. It has been a while for me. The software I am downloading will take me around 16 hours or more to work through once I get it installed. Math Tutor DVD has everything from grade school on up to calculus then they have physics. But, each course is sold seperately. So, if it works out I may get some more of their stuff. I would like to catch up on my math then get their physics CD. Will now see if I can sit down and work a derivative.

The integral seeks to find the area underneath a curve.

1 question. How much area?

-------------------------------------------------------------------------------
Here I found a very good example of using slops to compute a derivative.

http://www.worsleyschool.net/science/files/curve/slope.html

I think I like this way. I can relate to it. And, I understand the approximation.
We have to have two points to compute a slope. So we draw another point on the line a ways down from where we want our derivative and connect the dots and compute the slope. Then we get closer to our point and pick another point and connect the dots and compute our slope. Then we do this again and again each time getting closer to where we want our derivative. When we are done we have a bunch of derivatives. Now we can see in what direction our derivative is going then make an approximation as to what our derivative will be. Slope is.

Deleted naughty boy tirade.
 
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