energy producing experiments

[daanopperman]

Hi pequaide ,

You asked how the drawings is made . My OS is Windows , which has a program called Paint , but not even L.d. V. could paint with it , very limited capability to draw something but helpful to explain an idea . I think they ( M.S ) put it in just to inflate a bad product .

BTW , I use XP , which is the best product of MS that I have used , from Windows 3.1 right up to present day . The only thing that I miss is the ability to change the config sys. files . The other thing I miss is loading a new HD within the same PC with the OS used , it will shut down and you have to contact MS to clear your shutdown . You cannot unplug your current HD and install a new HD or the system will shut down rendering your PC useless .

[johannesbender]

daan to configure your config.sys file (if you need to) without booting to command prompt simply in windows run sysedit.exe from your start menu run shortcut .

i use to install new hdd all the time ...

[Wubbly]

pequaide wrote:

I go on the internet and try to find real experiments; because I know that any real experiment will prove that I am correct. ... No data collection that I have found supports mrr; none.

Tarsier 79 said he had one where the time doubled; he did not even know what false data he wanted to present.

So lets do a little challenge here. Find experiments with real data collection and bring them to the attention of the group.

Good idea pequaide. Here's an experiment found by Tarsier79 done by an unbiased third party. You can download the ".swf" file and step through it, counting the number of frames for one rotation at the first radius, then for one rotation at twice the radius.
http://www.animations.physics.unsw.edu. ... ation.html

If the mrr theory is correct, if you double the radius you should see the time doubling for the same angle travelled. This is exactly what Tarsier79 should have seen for an equal drop distance of his driver mass. The time should have doubled.

pequaide, here are the equations to use:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
1) τ = I x alpha
2) a = alpha x r
3) circumference = 2 PI r.
4) d = 1/2 at²
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Lets assume current physics is corrent and the moment of inertia for point masses is: I=mr².
If you move the weights to twice the radius, calculate the time difference to complete one rotation of the wheel:

1) τ = I x alpha This is the equivalent of F=ma for the rotational world. Torque = Moment_Of_Inertia times the angular acceleration. Or to look at it another way, Rotational Force = Rotational Mass x Angular Acceleration. Since the input torque of the driver mass is the same for both radius cases, we can call that a constant. Solve the equation for angular acceleration "alpha". Since we assumed I=mr², then at twice the radius the moment of inertia is four times the original value, and equation 1 predicts the angular acceleration must be one fourth its initial value.

2) a = alpha x r This is a standard equation relating linear acceleration to angular acceleration. It's also simple geometry. The linear acceleration is the angular acceleration multiplied by the radius. When the radius is doubled and the angular acceleration is one fourth its initial value, then equation 2 says the linear acceleration is equal to one half its initial value.

3) circumference = 2 PI r. This is a simple geometry equation. If you double the radius, you double the circumference (you double the distance the balls travel).

4) d = 1/2 at² This is a basic physics equation. Solve the equation for time "t". At twice the radius, the balls travel twice the distance (equation 3) with half the linear acceleration (equation 2). Plug in twice the distance and half the linear acceleration into equation 4, and it predicts the time to make one complete revolution at twice the radius should take twice as long.

Tarsier 79 said he had one where the time doubled; he did not even know what false data he wanted to present.

If I=mr², at twice the radius, that's exactly what he should have seen if his experiment was set up correctly.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Getting back to the experiment in the .swf file ...
This experiment showed 10.5 frames for one complete revolution of the rod with no balls attached.
It showed 33 frames for one complete revolution of the rod with the balls attached at a given radius.
It showed 59 frames for one complete revolution of the rod with the balls attached at twice the radius.

You are not going to see a perfect doubling of the times because you are also accelerating the rod (which is not massless). This will cause the times to be closer to each other, so you see 33 and 59 instead of 33 and 66. If the rod was massless, both frame counts would be smaller and would be closer to a perfect doubling. Peq attaches a massive flywheel to the system and throws the numbers way off and measures the acceleration of the flywheel instead of the acceleration of the attached masses, or he uses stiff bearings to throw the numbers way off and comes to incorrect conclusions.

Based on peq's current math skills, he should be able to follow these simple equations and simple logic and understand why doubling the time was not false data.

[pequaide]

Stay off the thread W

I do the experiments myself. I know how they work.

[pequaide]

Focus on the center of mass for a physical pendulum and forget W. Remember you can start the motion and then flip the hammer so to speak; the center of mass rules.

[Wubbly]

We can also start with the assumption that peq's I=mr hypothesis is correct, and calculate the time difference the experiment should show for the driver mass to drop any given distance when you double the radius.

Let's assume peq is correct and I=mr. If you double the radius, then the Moment of Inertia "I" would double.

1) τ = I x alpha .:The input torque "τ" of the driver mass is still constant whether or not you double the radius. Since the moment of inertia doubles from the initial assumption I=mr, then at twice the radius, equation 1 yields an angular acceleration "alpha" of half its initial value.

2) a = alpha x r .:If equation 1 predicts an angular acceleration of half its initial value at double the radius, then equation 2 predicts the linear acceleration will be equal in both radius cases.

3) circumference = 2 PI r. .:At twice the radius, the distance travelled doubles.

4) d = 1/2 at² .:If equation 2 predicts the same linear acceleration for both radius cases, you can solve equation 4 for time, and plug in twice the distance with the same linear acceleration. You find that at twice the radius, the time for twice the distance should be longer by a factor of the square root of two. It should be 1.414 times longer. So if one revolution took 33 frames at the first radius, then at double the radius the time predicted should be (33 frames x 1.414) = 47 frames. But since you are also accelerating the rod, this would push these two number closer together and the experiment would show twice the radius coming in at less than 47 frames, which is not even in the ballpark of 59 frames that the experiment showed. The results of this experiment are much closer to the I=mrr theory.

There is nothing in the peq hypothesis explaining why the experiment showed a time at twice the radius as being much longer than his "mr" hypothesis predicts.

I go on the internet and try to find real experiments; because I know that any real experiment will prove that I am correct. ... No data collection that I have found supports mrr; none.

This experiment supports the mrr theory. The frame count in this experiment did not increase by a factor of the square root of two as predicted by the pequaide hypothesis. This experiment proved the pequaide "mr" hypothesis is incorrect.

[Furcurequs]

pequaide wrote:

I go on the internet and try to find real experiments; because I know that any real experiment will prove that I am correct. ... No data collection that I have found supports mrr; none.

Tarsier 79 said he had one where the time doubled; he did not even know what false data he wanted to present.

So lets do a little challenge here. Find experiments with real data collection and bring them to the attention of the group.

Good idea pequaide. Here's an experiment found by Tarsier79 done by an unbiased third party. You can download the ".swf" file and step through it, counting the number of frames for one rotation at the first radius, then for one rotation at twice the radius.
http://www.animations.physics.unsw.edu. ... ation.html

If the mrr theory is correct, if you double the radius you should see the time doubling for the same angle travelled. This is exactly what Tarsier79 should have seen for an equal drop distance of his driver mass. The time should have doubled.

pequaide, here are the equations to use:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
1) τ = I x alpha
2) a = alpha x r
3) circumference = 2 PI r.
4) d = 1/2 at²
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Lets assume current physics is corrent and the moment of inertia for point masses is: I=mr².
If you move the weights to twice the radius, calculate the time difference to complete one rotation of the wheel:

1) τ = I x alpha This is the equivalent of F=ma for the rotational world. Torque = Moment_Of_Inertia times the angular acceleration. Or to look at it another way, Rotational Force = Rotational Mass x Angular Acceleration. Since the input torque of the driver mass is the same for both radius cases, we can call that a constant. Solve the equation for angular acceleration "alpha". Since we assumed I=mr², then at twice the radius the moment of inertia is four times the original value, and equation 1 predicts the angular acceleration must be one fourth its initial value.

2) a = alpha x r This is a standard equation relating linear acceleration to angular acceleration. It's also simple geometry. The linear acceleration is the angular acceleration multiplied by the radius. When the radius is doubled and the angular acceleration is one fourth its initial value, then equation 2 says the linear acceleration is equal to one half its initial value.

3) circumference = 2 PI r. This is a simple geometry equation. If you double the radius, you double the circumference (you double the distance the balls travel).

4) d = 1/2 at² This is a basic physics equation. Solve the equation for time "t". At twice the radius, the balls travel twice the distance (equation 3) with half the linear acceleration (equation 2). Plug in twice the distance and half the linear acceleration into equation 4, and it predicts the time to make one complete revolution at twice the radius should take twice as long.

Tarsier 79 said he had one where the time doubled; he did not even know what false data he wanted to present.

If I=mr², at twice the radius, that's exactly what he should have seen if his experiment was set up correctly.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Getting back to the experiment in the .swf file ...
This experiment showed 10.5 frames for one complete revolution of the rod with no balls attached.
It showed 33 frames for one complete revolution of the rod with the balls attached at a given radius.
It showed 59 frames for one complete revolution of the rod with the balls attached at twice the radius.

You are not going to see a perfect doubling of the times because you are also accelerating the rod (which is not massless). This will cause the times to be closer to each other, so you see 33 and 59 instead of 33 and 66. If the rod was massless, both frame counts would be smaller and would be closer to a perfect doubling. Peq attaches a massive flywheel to the system and throws the numbers way off and measures the acceleration of the flywheel instead of the acceleration of the attached masses, or he uses stiff bearings to throw the numbers way off and comes to incorrect conclusions.

Based on peq's current math skills, he should be able to follow these simple equations and simple logic and understand why doubling the time was not false data.

QFT

Dwayne

[Fletcher]

QFT

Quadratic Frobenis Test
Quantum Field Theory
Quantitative Feedback Theory
Quoted For Truth

[Furcurequs]

Yeah, I had to look it up, too, when I first saw it. As a matter of fact I looked it up again just before I used it. ...lol

I, of course, meant the latter - "Quoted For Truth".

Dwayne

ETA: Wait! I'm making an assumption there. Maybe you already knew all those things it could mean!

[pequaide]

On a beam like the aluminum tube experiment F is accepted as an Fd or Fr application of force. Starting with the mass and the force in the same spot the application of force upon the mass is one to one. If you double the d of the force you double the force working on the mass. But if you had halved the d of the mass you would also double the force applied to the mass. The mass is md also.

But W, F, F, and T can not be told simple truths; they insult or physiologically evaluate their disdained opponent and some times even lie. They are never swayed from their errant reasoning. They are here to trash others.

You have your partial evaluation; drop your trash there.

I am here because this is being done; I have dozens of functioning energy producing experiments. You, WFFT, have all the rest of the site; let there be one positive thread. Go away.

[daxwc]

“A positive attitude may not solve all your problems, but it will annoy enough people to make it worth the effort.� - Herm Albright (1876 - 1944)

But Then again so does trying to hoodwink people.


.

[pequaide]

http://lecturedemo.ph.unimelb.edu.au/Me ... of-Inertia

The r distance is about six to one; so is it moving 6 time faster or 36 times faster?


No comments from WFFT necessary; thanks.

[pequaide]

Here is functionally what I did; in different arangements.

First: I measured the spin with photo gates to eliminate the guess work of; is that 72° or 91°. With gates you get a number .0623 sec or such.

Then I reduce the mass on one side to one sixth and moved it to the end. And I got that same number; .0623 plus or minus .0002.

Then I reduced the mass of the other side to one sixth and placed that mass on the other end. And I again got the same number.

In this way you move the same bearing at the same speed to get the same friction. And there is nothing to guess at.

[pequaide]

Did you notice that you can hit the pause button on the left and then hit the pause arrow forward or the pause arrow backward on the right? You can do a frame by frame and count the fastest frames for a 180° for the short radius and then the fasted frames for the long radius and then compare.

[pequaide]

At six times the rotational velocity the sliding masses, in both arrangements, have the same energy; because the sliding masses have the same linear velocity.

They can be released at any time.

Energy increases will occur as you reduce the mass at the end point location. As you reduce the mass the linear velocity builds. You trade mass for velocity; but energy is the square of velocity, and the energy quickly begins to dramatically increase.