TD replications

[gotoluc]

Luc,

Thanks for sharing your latest test results, it takes great courage to do that.
Even though the test results are not what we have hope it would be, it is still an advancement to our knowledge.

Regards,
Alex

Thanks Alex, I'm happy to help by sharing what I find, even if the results are not favorable.
It's nothing out of the ordinary for me!...  I've been doing the same for the past 10 years.
What fuels me is hope that one day we find an energy solution for those in need.
Not for fame or fortune.

Thanks for your willingness to help

Luc

[dieter]

When I first read "Twist Drive" I thought it would utilize sheering force, rather than attraction / repulsion.


Maybe it does?


There is a force, turning a parallel 2nd magnet. When stopped at 90deg, it can be removed fro the 1st magnet without force, eg. by gravity. Then again by gravity it can brought in parallel position. The torque of the sheering is significantly higher than the gravity force alone.


Maybe that is also a TD.


I have made a little Toy to demonstrate it, maybe I'll post a picture later.

[Floor]

@Gotoluc

I'm not even close to being done with the PMs yet. 

Your last design was,   I guess,    near unity,  don't really know though?.  My own examinations
of interactions similar to that design left me with no understanding of why that design should  have been more than unity.  Although I did let myself get a bit carried away with your initial report.

From 60 % plus to 10% plus is a major oversight.  Can you give us some details
of that over sight ?
also
Your energy, enthusiasm and many hours of work in the shop are much appreciated.

      regards
            floor

[Floor]

@Gotoluc

Notes..

  The integration of the work done in the inputs via the rotating bicycle rim experiment....

1. position magnet by rotation above the sliding magnet
2. remove magnet by rotation from near the sliding magnet

may together (attractions and repulsions) come to a net work of less than either
1. or 2.  alone (just above)......

except that their peak forces were not matched / canceling one the other out.

Other wise your complete set might have shown some OU ?

                            floor

[gotoluc]

@Gotoluc

I'm not even close to being done with the PMs yet. 

Great to hear!... please make a video demo once you have found something so I can evaluate it as well

From 60 % plus to 10% plus is a major oversight.  Can you give us some details
of that over sight ?

This link to the below quote explained the oversight: http://overunity.com/16987/td-replications/msg501737/#msg501737

this morning I decided to re-measure the first device with the most care to details.
Now the first device is showing a 10% gain which could be caused by accumulative errors from the 5 gram resolution scale.
What I found could of cause the 60% gain error is by using a different input rotor magnet then the one used for the 11mm output.
I check the rotor magnets and found they have different magnetization force. So most likely that's what happened plus the scale resolution problem.

So to revise what may of caused the 60% gain error.

1. I must of use a different rotor magnets to measure the input force then the output force when I first measured the v.1 device.
    Seems this alone can cause a 30% difference. I was surprised of how much each rotor magnet vary in force.

2. The scale I use are 20kg max luggage scale. It has a 5 gram resolution.
     The rotor input force of the first device range from 1g to 85g max. However, the scale only starts to display at 15g and above.
     So I figure it's unsuitable for accuracy when measuring below 100 grams.
     I would estimate the math averaging on the input of the first device could be off by 10 to 25% based on this resolution issue alone.

-----------------------------------------------------------------------------------------------------------------------------------------------------------------------


The below are the measurements done on the v.2 device

The input rotor force measured between 110g to 2.4kg, so very good data was obtained as far as the scales resolution ability.
Half of the input rotor distance (to make one output stroke) is 16 1/8 inches of circumference.
Samples were taken at every 1/8 inch distance, so 129 input distance samples were recorded in total!... giving a very good input average calculated to be 1.1kg over the 16 1/8 inch half rotor circumference.

The output force was adjusted to slide 5 inches of distance. The gram pull force measurements varied between 2.3Kg to 14.5Kg.
40 samples were taken at every 1/8 inch making an average of 6.46Kg over the 5 inch output stroke.

I've just realized I made an error (a few days back) on my final math!!!... I had the calculations of the rotor input engaging and disengaging averages calculated separately and added them together 1.24Kg + 0.957Kg = 2.2Kg but the error is, the 2.2Kg should then be divided by 2 = 1.1Kg to get the correct input rotor average over the 16.125" for half of the rotor circumference as I correctly did above.

So if we take the 16.125" rotor input travel distance and divide it by the 5" output slider distance = 3.23 times more distance the input rotor needs to travels at 1.1Kg average compared to the 5"output distance at 6.46Kg average.
So if we multiply the input average 1.1Kg x 3.23 times =  3.55Kg of comparable input force to distance needed compared to the output.
Now if we subtract this 3.55Kg of comparable input force to the 6.46Kg output force = 2.91Kg left over which is a 82% Gain over the Input.

Your energy, enthusiasm and many hours of work in the shop are much appreciated.

      regards
            floor

Thanks

Can someone look over my calculations to see if the reasoning looks to be correct.

Regards

Luc