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Author Topic: TD replications  (Read 55042 times)

Offline telecom

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Re: TD replications
« Reply #75 on: February 10, 2017, 01:21:16 AM »
Hi Luc,
the actual gain , I think, is 2 times higher than you calculated because
your output goes back and forth during the cycle, so the total distance
is 22 mm by 540 grams = 11880 gram x mm

Free Energy | searching for free energy and discussing free energy

Re: TD replications
« Reply #75 on: February 10, 2017, 01:21:16 AM »

Offline telecom

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Re: TD replications
« Reply #76 on: February 10, 2017, 01:23:16 AM »
Considering that the input is 3700 gram x mm per cycle,
you have a gain of 3.

Offline Floor

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Re: TD replications
« Reply #77 on: February 10, 2017, 08:10:17 AM »
@GoToLuc
       
                 very nice work
                           thanks                         
                                floor

Free Energy | searching for free energy and discussing free energy

Re: TD replications
« Reply #77 on: February 10, 2017, 08:10:17 AM »
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Offline gotoluc

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Re: TD replications
« Reply #78 on: February 10, 2017, 03:35:58 PM »
Hi Luc,
the actual gain , I think, is 2 times higher than you calculated because
your output goes back and forth during the cycle, so the total distance
is 22 mm by 540 grams = 11880 gram x mm

Hi telecom,

It would be great if there was twice the movement. However, I don't see that.
I made a video just for you to count them.

Link: https://www.youtube.com/watch?v=dpBaeJD38HI

Regards

Luc

Offline gotoluc

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Re: TD replications
« Reply #79 on: February 10, 2017, 06:03:39 PM »
What is still a question is how the 11mm push is measured over that distance, with the wheels' magnet aligned in the center, or with it passing by.

Since the device is very dynamic the only way I could measure it was to do each step at a time. So to answer your question quickly, the 11mm stroke is measured once the rotating magnet is centered with it.

Luc

Free Energy | searching for free energy and discussing free energy

Re: TD replications
« Reply #79 on: February 10, 2017, 06:03:39 PM »
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Offline dieter

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Re: TD replications
« Reply #80 on: February 10, 2017, 06:24:55 PM »
Then I would suggest to measure the push with each measurement step of the rotation, that was 5mm, maybe try to do finer steps due to the peaks.

Offline telecom

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Re: TD replications
« Reply #81 on: February 10, 2017, 07:10:45 PM »
Hi telecom,

It would be great if there was twice the movement. However, I don't see that.
I made a video just for you to count them.

Link: https://www.youtube.com/watch?v=dpBaeJD38HI

Regards

Luc

Ok, I guess it was a wishful thinking on my end!
In this case you probably need to calculate the work balance for the linear movement to come to the initial position.
Regards

Free Energy | searching for free energy and discussing free energy

Re: TD replications
« Reply #81 on: February 10, 2017, 07:10:45 PM »
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Offline conradelektro

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Re: TD replications
« Reply #82 on: February 10, 2017, 07:19:10 PM »
If you do a replication please do compare output-POWER and input-POWER (and not FORCE or WORK)

See: http://overunity.com/17097/magnet-force-shield/msg499715/#msg499715 (concerning POWER versus WORK, TIME is of the essence)

FORCE

WORK = FORCE * DISTANCE

POWER = WORK over TIME

Also see: http://overunity.com/17097/magnet-force-shield/msg499638/#msg499638 (concerning a magnetic shield, if you plan one in your replication; note, a magnet is also a "known material")

Greetings, Conrad

Offline gotoluc

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Re: TD replications
« Reply #83 on: February 10, 2017, 07:23:22 PM »
That really would complicate measurements to level that may not be possible since if the 11mm magnet is allowed to move while the rotating magnet is moving in its first  55mm of travel the rotating magnet force needed to pull in goes down as the 11mm magnet moves. And once the 11mm magnet has reached it's 5.5mm center the rotating magnet requires 0 grams to move through.

A very complex measurement setup would be needed to probably find it all comes to the same.

The way I did it (in steps) measures all maximums. By holding down the 11mm magnet while the rotating magnet moves in, the rotating magnet needs the most pull force to pull it in the first 55mm and the opposite happens once I release the 11mm magnet and hold it down now the rotating needs maximum force to be pulled out of the remaining 55mm of travel.

Hope you understand

Luc


Free Energy | searching for free energy and discussing free energy

Re: TD replications
« Reply #83 on: February 10, 2017, 07:23:22 PM »
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Offline gotoluc

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Re: TD replications
« Reply #84 on: February 10, 2017, 07:28:36 PM »
In this case you probably need to calculate the work balance for the linear movement to come to the initial position.

That's the beauty of this design, there is no work "balance" needed for the next position (cycle)... as the fist cycle goes through the next cycle is right there and needs exactly the same as the first cycle.

Luc

Offline gotoluc

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Re: TD replications
« Reply #85 on: February 10, 2017, 07:38:16 PM »
If you do a replication please do compare output-POWER and input-POWER (and not FORCE or WORK)

See: http://overunity.com/17097/magnet-force-shield/msg499715/#msg499715 (concerning POWER versus WORK, TIME is of the essence)

FORCE

WORK = FORCE * DISTANCE

POWER = WORK over TIME

Greetings, Conrad


Dear Conrad,

Please reply to the simple question (in bold) of this post:

Dear Conrad,

Lets look at a test device which can test your distance time beliefs.

Test device parts needed:
A DC electric motor which has a flywheel attached to its shaft and use of photo switch to turn the motor on and off.

First test:
We attach a scale to the outer circumference of the flywheel and adjust the current to the motor so it can pull 34 grams.
Then we adjust the photo switch to power the motor 110 mm distance of the flywheel outer circumference.
We note of the RPM in this condition.

Second test:
We attach a scale to the outer circumference of the flywheel and adjust the current to the motor so it can pull 538 grams.
Then we adjust the photo switch to power the motor 11 mm distance of the flywheel outer circumference.
We note of the RPM in this condition.

If your belief is correct, the RPM should be greater on the first test compered to the second test, correct?

Regards

Luc

Free Energy | searching for free energy and discussing free energy

Re: TD replications
« Reply #85 on: February 10, 2017, 07:38:16 PM »
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Offline telecom

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Re: TD replications
« Reply #86 on: February 10, 2017, 08:12:30 PM »
That's the beauty of this design, there is no work "balance" needed for the next position (cycle)... as the fist cycle goes through the next cycle is right there and needs exactly the same as the first cycle.

Luc

So, to return the linear stage to the initial position will take the same work, and the linear stage will generate the same work?
Approximately 1.5 times more than input?

Offline dieter

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Re: TD replications
« Reply #87 on: February 10, 2017, 08:28:20 PM »
How about to slap a rudimentary crankshaft together, using the 11mm push over leverage to turn the wheel? Would probably be less timeconsuming than all measurements an defendings ^^

Offline gotoluc

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Re: TD replications
« Reply #88 on: February 10, 2017, 09:23:25 PM »
So, to return the linear stage to the initial position will take the same work, and the linear stage will generate the same work?
Approximately 1.5 times more than input?


You don't need to return the linear stage to the initial position. The rotor magnet alternate N-S-N-S which creates the back and forth linear action which give 11mm linear force (538 grams) in each direction. No rest needed.

Luc

Offline conradelektro

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Re: TD replications
« Reply #89 on: February 10, 2017, 09:23:48 PM »
Dear Conrad,

Please reply to the simple question (in bold) of this post:

If your belief is correct, the RPM should be greater on the first test compered to the second test, correct?[/size]

Luc,

it is not a simple question and I have no simple answer. I have not studied physics (only mathematics and law) therefore I would have to read up about flywheels in my physics text books. This would take hours which I am not prepared to put in today.

The quick answer: I do not see the connection between a flywheel and your machine. A flywheel is continuous movement. Your machine has two movements, a continuous turning movement (like a flywheel, if you want to see that probably false analogy) and an intermittent reciprocal movement.

Very important and constantly overlooked fact: your sledge also pauses during its movement (cycle, reciprocal movement). And during the pauses it does not do work.

Like a LED that is driven intermittently, whose average light output is less than a constantly driven LED (with the same Voltage and Amperage, this is what dimming is all about), your sledge outputs less energy than thought if the pauses are taken into consideration. (And exactly there I see no continuous fly wheel).

So, please forgive me that I do not want to study fly wheels at the moment. Your fly wheel example might have the answer "yes", but there is no Konnex to your machine

Just think for a moment, it is important how long work is done. Only when knowing how long a certain work is done one knows its power output (energy).

I have said more than often: POWER is WORK over TIME. If there is no time, there is no power (when the sledge pauses). A power company sells Watt-Hours and not Watt. You have to get the difference between Watt and Watt-Hours (between Work and Power or Energy). Which in your case is the difference between Gram or Gram-Hours. Or, if we would do the conversion from Gram to Joule, it is the difference between Joule and Joule-Seconds (or Joule-Hours if you want).

This is not my opinion, this is a fact. Everybody would complain if his work would not be paid by the hour. It matters very much how many hours you do your work. The average power output of your work is less (per hour) if you drink coffee for 15 minutes every hour (a pause of 15 minutes every hour). Your power output will be 25% less. All employers know that simple fact. All power companies know that it is important how long you switch on your loads.

Greetings, Conrad

 

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