My partial summary of pequaide's "energy producing experiments" thread

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My partial summary of pequaide's "energy producing expe

Post by Furcurequs »

Hello,

After wading through some of pequaide's 90 page long "energy producing experiments" thread, I thought I might summarize some of what I believe he is claiming.

First of all, (somewhere in that thread) he seems to point out - and I believe rightly so - that the following three scenarios are essentially the same in regards to the acceleration of the total mass and thus all would have basically the same gain in momentum and kinetic energy.

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The first consists of two equal masses hanging from a massless Atwood machine which are accelerated along with the mass M0 - by the weight of M0.

The second is a flywheel with all the mass in the rim being accelerated together with the mass M0 - by the weight of M0 - with the accelerating force on the flywheel applied tangentially at the rim.

The third is a massive block resting on a frictionless table that is being accelerated horizontally at the same rate that the mass M0 is being accelerated downwardly - both masses together being accelerated by the weight of mass M0 - thanks to both being connected by a massless line running over a massless, frictionless pulley.

...yet then he seems to be claiming that since this:

Image

..is statically equivalent to this - as in both are in balance:

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...that this:

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Is in some ways equivalent to this dynamically:

Image


He seems to believe they are equivalent in terms of the impedance - the moment of inertia - experienced by the extra weight on the small diameter shaft which he uses to accelerate the two - saying there is no difference in how that weight descends.

...yet while he also seemingly believes that the outer mass gains many times the kinetic energy that would come from the driving weight moving downward through its vertical distance alone.

For that to be the case he doesn't seem to want to accept that the moment of inertia of a point mass - or of that (approximately) of a flywheel in which most all of the mass is at the rim - is equal to m*r^2

He seems to believe and states that it is simply m*r.

So, consider the following:

When he removes the mass at r1 and places 1/20th of that mass at r2 - which is now 20 times farther away from the center of rotation, of course.

(I only showed 10 times in my drawings due to space constraints.)

...this would indeed statically balance...

...but it would also mean that the leveraged force from the driving weight that would accelerate that mass would also now be 1/20th of what it was on the larger mass before.

So, 1/20th the mass being accelerated by 1/20th the force means it would be the same acceleration as the larger mass experienced before when it was closer in and hanging from the drive shaft.

So, in a given amount of time, the small mass would be accelerated through the same arc length. ...but that arc length on a circle with 20 times the radius of that of the drive shaft would mean only 1/20th the angle of rotation of the shaft.

Image

(Again, I used a 10x radius instead of 20 in my drawings.)

This means that the driving weight can't possibly be accelerating downward as much as it did before in the prior scenario - and so can't reach the same speed as before when moving through the same distance.

Since the speed of the rim - and the speed of the smaller mass - is directly tied to the speed of the driving "weight", and the driving weight would be moving much more slowly after the same distance of "fall", pequaide's claims just do not seem to add up.

That extra "r" is very important.

I=m*r*r=m*r^2

Pequaide seems to want to take advantage of the leveraging effect - even while ignoring it (in the dynamic case).

Ironically, the greater impedance - the greater moment of inertia - would mean that more of the energy from the "fall" of the driving weight should be stored in the motion of the smaller mass(es), but I see no reason at all to believe it would be any more than conventional science would state.

I, like Fletcher and others, though, would certainly consider looking at some actual data along with a clear description of the experimental setup.

...rather than to just continue to see more confusing and/or confused claims.

Reality, of course, could trump anything I have to say. Pequaide's reality and mine, though, don't currently seem to be the same.

Also, I believe a correct understanding of what is really going on here would be beneficial to us all.

Feel free to correct me if you think I've gotten something wrong. Also feel free to use any of these images in other threads if they could be of help.

Take care.

Dwayne
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re: My partial summary of pequaide's "energy producing

Post by pequaide »

Apparently my descriptions have been quite clear; as you have accurately drawn diagrams of what I have been claiming.

And you continue to claim things that are not supported experimentally. It may seem good to you that the acceleration of one tenth at ten times the radius should rotate slower; but it does not.

The drive force MO is seven and one half pounds.

M1 and M2 are 20.54 kilograms.

M1replacement is .900 kilograms.

I can balance these two masses with .900 kg at R2 and 20.54 kg at R1.

This means that R2 is 22.8 times longer than R1. R1 comes off of a .75 inch shaft and R2 comes off of a 19 inch wheel. 19/.75 = 25.33 to 22.8; this small discrepancy has to do with rope diameter.

I have given you data. MO drops the same distance in the same 2 seconds if the two M1 and M2 are suspended at .75 inches or if two .900 kilograms masses are suspended at the 19 inch circumference.

I have done experiments with at least four different wheels now and I have given data for all of them. So how you can say I don't give you data is beyond my understanding.

There are many picture and a few videos. [list=][/list]
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re: My partial summary of pequaide's "energy producing

Post by pequaide »

I think you understand this; but I would like to make it clear that the drive force of 7.5 lbs remains at the .75 inch shaft.

That drive mass can rotate two 20.54 kilogram masses at the shaft just as easily as it can rotate two .900 masses at the 19 inch circumference.

After the drive mass drops the same distance in the same two seconds then the rotational velocity at the shaft, and everywhere else, is the same in both arraignments. So you either have (20.54 *2) 41.08 kilograms moving 1 m/sec or you have (.900 * 2 ) 1.800 kilograms moving 22.8 m/sec.

1/2mv²: ½ * 41.08 kg *1 m/sec *1 m/sec = 20.54 joules; or ½ *1.800 kg* 22.8 m/sec* 22.8 m/sec= 467.8 joules.

This is real data from real experiments.
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re: My partial summary of pequaide's "energy producing

Post by Wubbly »

Furcurequs,

There are several problems with pequaide's latest round of Atwoods experiments.

One main problem is that he doesn't bother to quantify the error potential in his experiments.

He uses stiff bearings that produce a counter torque that he never bothers to try to measure or quantify. What affect does this counter torque have on the time measurements and on the energy calculations? pequaide has no idea, so he just ignores it. The counter torque alone could account for the times being approximately equal.

How accurate are the mass measurements? His larger mass is measured to .01 kg yet the smaller mass is measured to .001 kg. Does this mean he only knows his large mass to the nearest .01 kg? What if the larger mass was .004 kg heavier or lighter? How would this affect the time or the energy calculations? pequaide has no idea.

How accurate is the time measurement? Sometimes he will say the accelerations of two experiments take 'about' 2 seconds, or 'about' 3 seconds. Does that mean the time is accurate only to the nearest second? This is hardly enough precision to say the times of the two systems are the same.

How accurate are the radius (distance) measurements? How does the potential error in the distance measurements affect the error in the time measurements or the energy calculations? pequaide has no idea. And is dropping an inner mass 2 or 3 centimeters really going to give you good data to claim energy creation? Hardly.

He also completely ignores the rotational energy in his flywheel as if it is of no consequence. This was quite comical when he was using a huge, massive flywheel. We have to give him credit for using a lighter pully in his later experiments, but he still ignores its energy as if it has no affect on the experiment.

In the Atwoods Analysis thread an excel spreadsheet was posted that had a tab simulating one of pequaide's experiments. The time difference required to account for the difference in accelerations should have been 0.001 seconds. And that was assuming the masses were accurate to the nearest gram, the distances were accurate to the nearest millimeter, and that the bearings were frictionless. And who has frictionless bearings to play with? pequaide's conclusion was that since his observed times were about the same, then mr² is incorrect, and he was creating energy.

The error potential in pequaide's experiments completely drowns out any of his supposed claims of energy creation.

Since he never, ever, ever backs down from his stance that he is creating energy, (despite numerous, patient people trying to explain to him what he may be doing wrong), there are some members who think he is just messing with people's minds.

.
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re: My partial summary of pequaide's "energy producing

Post by Furcurequs »

Hello pequaide,

Thanks for responding and thanks for confirming that I have both understood and presented your experimental claims accurately in my drawings. (Well, all but maybe that last one, right?)

I must confess that I haven't followed your "energy producing experiments" thread from the very beginning and maybe haven't even read it all. I also seem to have missed information in some other related threads - as I now see thanks to Wubbly.

I'll admit, then, I may not have done my due diligence in tracking down all your data. So, instead of wading into this, I may have just dived into it head first without first checking the depth. ...ouch.

So, my apologies if I have left out any relevant information.

...and thanks for giving actual numbers for the variables shown in my drawings.

I did look at your picture "album" on this site, however, and didn't see there a clear representation of this experiment, though I may have overlooked it. If you are going to use just photographs it might help to have some things labeled - at least for us (sort of) newbies.

I didn't bother to sign in to the yahoo group to see the files there, either.

I did understand from your descriptions that the drive mass remained suspended from the inner shaft after you removed the other masses and replaced them with smaller masses suspended from the larger diameter wheel.

I had drawn this but did not upload it before due to the apparent 6 attachment limit per post of the forum software:

Image

With all that said, it appears that I'm not the only one with some misgivings about your claims, though, apparently including some people who have been following them longer than I have.

You do claim the creation of excess energy in this experiment, however, which of course does not fit with accepted science.

A way to gain excess energy is what most of us here are trying to accomplish, but I just don't see it happening in this particular experiment.

I have formal engineering training. My degree was in electrical engineering rather than mechanical, but I do remember those college physics classes and my statics and dynamics classes - which if I remember correctly I also did well in.

So, your claims don't agree with what I have been taught, either, nor with my own understanding as to what should be happening in this specific experiment.

I was never the sort to just accept an equation, plug numbers into it and crank out an answer, either. I like to first try to understand things on a fundamental level so that I can derive the equations myself.

It has been over two decades since I worked in engineering, though, so I'm sure I could be a bit rusty when it comes to some of that.

I have not done this specific experiment, either, but since I do like to see things with my own eyes, maybe at some point I will.

It is understood that in a flywheel the motion of any bit of mass is constrained to a given circular path. That is why we speak of angular momentum of the mass in a flywheel about its center. The linear momentum of the moving mass in a stationary spinning flywheel sums to zero, though.

In the case of the Atwood with suspended moving balanced masses the linear momentums, of course, also sum to zero. The angular momentums don't, though.

The total angular momentum of the two vertically moving masses around the center of the Atwood, I believe, would be the same as if their total mass was distributed around the (moving) rim of the Atwood.

The equations, then, that describe your experiment seem to be the same as those governing the storage of energy in a flywheel - and that seems to have been thoroughly researched and understood for hundreds of years.

In other words, there has been more experimental data that would seemingly deny your claim than back it up - just not from you.

I'll probably just continue to scratch my head about your claims as I get back to my own experiments and builds.

Oh, if you tried to post a list of drawings and videos, it doesn't appear to have shown up. I'll try to look at more of your stuff as time goes by and maybe make more comments later depending upon what I see.

I will add that maybe some of your other experiments that I've not yet addressed in this thread, I might find more interesting - not that I necessarily believe your claims about them, either, but that they may at least show an efficient way of impedance matching to help in transferring energy from one moving mass to another.

Wubbly,

Thanks for your response, too, and thanks for the link to the other thread.

I see that you and others have already been addressing some of my concerns there - and for some while.

Until I've made more of an effort to look at pequaide's videos and other data, though, I guess I better not be too critical regarding specific issues, but in what I've seen so far, I have had concerns similar to yours as to the rigorousness of his experiments. I think I, too, may have even already seen an "about" (or two) here or there.

I also see that you and others have already made some pretty nice diagrams and so I didn't really need to add my drawings.

(I am also trying out some vector graphics programs on my Linux machine, though. I once had a really nice freeware Windows program that had both vector graphics and a scientific spreadsheet, but I can't remember the name of it. I need to find my old hard drive and dig it out and see if it will run in WINE on Linux.)

I guess I'll now go back to scratching my head and catching up on my forum reading.

Thanks again to both of you.

Dwayne
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re: My partial summary of pequaide's "energy producing

Post by pequaide »

I believe you are predicting that the one kilogram at a radius of 20 accelerates around the arc of the circle at the same rate as 20 kilograms at a radius of 1. This is not the same rotational rate; this is the distance around the arc, and the same distance around the arc in the larger circle would be only 1/20 the angular displacement of that displacement occurring in the smaller circle at the shaft.

From the experimental data: the acceleration of 41.08 kg (2*20.54 kilograms) at the shaft results in a half rotation in two second. This is a distance of about 3.325 cm for an acceleration of .016625 m/sec/sec.

If the 1.800 kg (2 * .900 kilograms) has the same rate of acceleration (.016625 m/sec/sec) then it will only travel .03325 m in two second. But it has to travel .758 meters for the stopwatch to be triggered the second time.

A half rotation of the wheel has a circumference distance of .758 m and the stopwatch is only stopped after this half rotation. For the .900 kilograms to travel .758 m, when it has a (predicted) acceleration of .016625 m/sec/sec, it will have to accelerate for 9.54 second.

The experiment shows that with 20.54 kilograms on both sides at the shaft the 7.5 pound force accelerates the wheel through one half rotation in about 2 seconds.

The experiment shows that with .900 kilograms on both sides of the circumference of the wheel, the 7.5 pound force at the shaft causes the wheel to accelerate through one half rotation in about 2 second. So: is 2 seconds closer to 2 seconds or is 2 seconds closer to 9.54 seconds?

Your prediction for acceleration seems logical: that; one twentieth the force is working on 1/20th the mass which should give the same acceleration. But the experiment proves that your prediction is not correct.

Another way to predict the motion would be this: that; one kilogram of mass at 20 inches acts like 20 kilograms at one inch. And the driving force will rotationally accelerate the wheel as if the one at twenty were 20 at 1. And; in this arraignment the one will have twenty times the velocity of the driving force. And the experiment proves that this prediction is correct.

Apparently the prediction must be made from the reference point of the applied force, and not from the position of the resistance.

You are correct that many have followed this tread for a long time; But most fail to do experiments. I think I have four wheels that demonstrate the Principal above. I know what these wheels do. I am fully convinced that free energy can be made from gravity.

For something this large to be missed this long there have to be several erroneous concepts afloat in the world of physics. One is that an Atwood's, or a spinning rim, has zero momentum. Another is that angular momentum is conserved in the lab (with moment of inertia).

Kepler's angular momentum conservation works in space where there is a huge difference in linear velocity between apogee and perigee. This linear velocity difference is caused by gravity, and there is no such effect in the lab. You simply can not use the same formula on events that are with and without this gravitation acceleration. This has been discussed in depth; and I will try and direct you to some of the files as I get time.

The Atwood's is the machine that proves F = ma; which is of course a Newtonian formula. Newton's Three Laws of Motion are referring to momentum conservation not energy conservation. The a in F = ma is v/t; or Ft = mv. So how can the force in the Atwood's produce zero momentum? The up and down vectors in the Atwood's do not cancel; they enhance each other. And the same is true of the spinning rim; all points of motion enhance each other, they do not cancel. This is also discussed in those 90 pages.

Thanks for the drawings; I will refer to them.

Sorry about the sign in thing on yahoo. I don't think the pictures should be too far back in the 90 pages: I will go look.
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re: My partial summary of pequaide's "energy producing

Post by pequaide »

Page 89: The weights and the wheel are in different pictures. The big one inch bolts (.900 kilograms) are suspended from the circumference of the plywood wheel. The big 45 pound bar bell masses are suspended from the white rope on the shaft. This rope increases the diameter of the radius of suspension. The 7.5 pounds of force are with the bar bell masses on one side.
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re: My partial summary of pequaide's "energy producing

Post by Wubbly »

Dwayne,

Try not to get dragged too far down pequaide's rabbit hole.

People have been trying to talk to him for years with no success.

A quick internet search can find examples of experiments that easily show conservation of angular momentum, despite pequaide's belief that is does not exist in the lab: http://www.youtube.com/watch?v=9MGQJar8dNg
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re: My partial summary of pequaide's "energy producing

Post by pequaide »

Galileo's pendulum proves that the halving of the radius does not change the linear velocity of the mass.

I don't know if the student is adding motion when he PULLS it in or if he just thinks it is rotating four times as fast.

The student assumes that he is adding no force and motion to the system. Yet he says the linear motion double, ( if rotation quads then linear motion doubles).

Well if you can double linear motion without the use of force then you already have a energy producing devise. Because the batteries at double the linear motion will rise four times as high, and they will have 4 times the energy without the application of an outside force. This is not possible according to Newton.

It is quite possible that the pull is like gravity and that this force accelerates the batteries. But it is not possible that you have a change in motion without the application of force.

Remember: you can always release the batteries from their circular path and their motion become linear. And direction is easily changed without a change in motion.

Furcurequs: you are meeting the negatives, you will have to do you own thinking.
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re: My partial summary of pequaide's "energy producing

Post by Wubbly »

Wow. Another experiment completely lost on pequaide. No wonder you don't believe angular momentum applies in the lab, if you can't even understand an experiment that simple.
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re: My partial summary of pequaide's "energy producing

Post by pequaide »

Why doesn’t the experimenter let the batteries wind up on an immovable center post? That way the experiment would have no outside force applied. But his friends would not have been happy with the experimenter. Unfortunately most experimenters do there work to get accolades from their friends; not to find the truth. I would like to see the experiment that he mentions did not work, or did it?

Frankly Wubbly you should be embarrassed to have presented such an obviously fallacious experiment.

I have done the immovable center post experiment on a frictionless plane and there is no linear velocity change and no angular momentum conservation. It is just the difference between knowing the truth or impressing your friends.
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re: My partial summary of pequaide's "energy producing

Post by eccentrically1 »

Here is an example from a physics textbook showing the conservation of AM between the earth and the moon; the tidal bulge makes it easier to visualize, hopefully.

Image

"A view of the earth-moon system from above the north pole. All distances have been highly distorted for legibility. The earth's rotation is counterclockwise from this point of view (arrow). The moon's gravity creates a bulge on the side near it, because its gravitational pull is stronger there, and an “anti-bulge� on the far side, since its gravity there is weaker. For simplicity, let's focus on the tidal bulge closer to the moon. Its frictional force is trying to slow down the earth's rotation, so its force on the earth's solid crust is toward the bottom of the figure. By Newton's third law, the crust must thus make a force on the bulge which is toward the top of the figure. This causes the bulge to be pulled forward at a slight angle, and the bulge's gravity therefore pulls the moon forward, accelerating its orbital motion about the earth and flinging it outward."

In a rotating object, nothing is required to keep it rotating once it is set in motion. This follows from Newton's first law. The earth, the moon, the planets, the solar system, etc., will rotate, conserve their angular momentum between them, with negligible change due to a near frictionless environment. The moon doesn't get closer and closer to the earth; because of conservation of AM. It gets very fractionally farther away (in a slower orbit) because of the tidal friction slowly overcoming the gravitational attraction.

So the batteries in the utube video change "orbit" distance and speed because the guy pulled on the fishing line. Gravity was already pulling on the batteries. If he had hung a weight on the fishing line, the same thing would have occurred, only the weight would have taken the place of his arm force.
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re: My partial summary of pequaide's "energy producing

Post by Fletcher »

He could indeed have hung a weight beneath the center of the hanger & let that overcome Cf's eccentrically1.

However perhaps a more objective way would be to make a more sophisticated rotating platform with a small battery powered motor at the center, this has a cam arm that pulls the radial weights inwards against Cf's, sliding along a track for example, to predetermined stops positions.

It would mean that the rotational speed would be given to the setup & known - then the motor could be switched on & off by a RC unit etc.

The weights would slide back to original start positions by Cf's & again it's rotational velocity measured.

N.B. the motor would consume a small amount of electric energy converted to mechanical energy & since the whole assembly rotates & the motor battery isolated it would have no baring on the experimental data gathered.

That would conclusively prove whom is correct, what is the truth & who wants to impress - it could be made from wood, mechano or knex for example, quite easily, in probably a few hours.
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Post by eccentrically1 »

Hi Fletcher
That would conclusively prove it. Maybe pequaide would like to try that experiment for his own assurance.
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re: My partial summary of pequaide's "energy producing

Post by pequaide »

There is a much simpler method. Tie a thirty cm string onto an air table puck. Launch the puck so that it rotates nicely on the end of the string. Arrange for the string to be interrupted by a pin that extends from the table. Let this pin be 15 cm from the center of mass of the puck.

In this way the puck starts in a 30 cm circle and then it will drop down to a 15 cm circle. The puck's rate of rotation around the second smaller circle will double when the string strikes the pin. The distance traveled around the arc of the circle will remain constant.

For clarity lets say that the puck is moving 10 cm per second. In one second it will travel 19° around the arc of the large circle and it will cover a distance of 10 cm. After the string strikes the pin the puck will travel 38° around the smaller circle in one second and it will cover a distance of 10 cm.

You can video tape the motion to confirm the velocities.

Plug these numbers into the angular momentum formula and it will not be conserved.

I have an air table; and I have done these experiments, so I already have my assurance.

Note that; once the motion of the puck is started, there is no application of an outside force.
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