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### Author Topic: re: energy producing experiments  (Read 134851 times)

#### Tarsier_79

• Full Member
• Posts: 117
##### Re: re: energy producing experiments
« Reply #315 on: February 09, 2022, 02:44:07 AM »
" The experiment shows that the disk is harder to rotate than 1/2mr² predicts. "

https://en.wikipedia.org/wiki/Moment_of_inertia
Half way down, a computer simulation clearly showing the disk is not 2x the distance of the ring, a similar result to the youtube vid.

I am no math guru, but there is usually more to the picture than a simple formula.
You have the formula for inertia of both bodies. You then need to take into account angular momentum, rolling vs accelerating around an axle and gravity at an angle.

I found this:

#### Johnsmith

• Full Member
• Posts: 211
##### Re: re: energy producing experiments
« Reply #316 on: February 15, 2022, 01:50:56 AM »
Math typically works that way.

#### Delburt Phend

• Full Member
• Posts: 222
##### Re: re: energy producing experiments
« Reply #317 on: February 23, 2022, 08:22:01 PM »
Some request models;

Gooble  “A scientific model is a physical and/or mathematical and/or conceptual representation of a system of ideas, events or processes. Scientists seek to identify and understand patterns in our world by drawing on their scientific knowledge to offer explanations that enable the patterns to be predicted.”

You say that I have not presented a model, and then you show how the
Law of Conservation of Momentum would work. Apparently, I have presented a model and you are fully aware of it.

35 kg dropped one meter can accelerate 140 kg to 2.21 m/sec. Model: Atwood’s Machine

A 35 meter high stack of evenly spaced 1 kilogram masses exert the same force as a 35 kilogram mass. Model: scale balance

After a drop of one meter the stack can be reconfigured by accelerating one kg to 26.24 m/sec, costing 26.24 units of momentum. The one kilogram moving 26.24 m/sec will rise 35 m. Model: kinematic equations.

140 kg moving 2.21 m/sec equals 310 units of momentum and you only need 26.24 units to recycle the system. Model: kinematic equations.

If the 310 units of momentum is placed in one kg the rise would be 4900 m.  Only 35 of those meters is needed to reconfigure the stack. Model: kinematic equations.

The momentum of a large mass can be given to a small mass. Model: cylinder and spheres machine where all the motion is returned to the cylinder; and the Law of Conservation of Momentum.

The stack of 35 1 kg masses could accelerate a 140 kg rim and stack to 2.21 m/sec and then the rim could throw one kg 140 times higher than the original drop. Model: cylinder and spheres machine; the Law of Conservation of Momentum: kinematic equations.

The double yo-yo despin ‘cylinder and sphere’ transfers all the motion of the “cylinder and sphere” to only the spheres. Then the spheres transfer the motion back to the cylinder and spheres combination.

The conservation of energy could not return the motion of the spheres back to the cylinder. Because as the conservation of momentum would require a 10 m/sec speed for the spheres the conservation of energy would only require a 3.16 m/sec velocity.

When masses collide, or transfer motion, smaller masses do not and can not give their energy to a larger mass. The smaller masses can and do however give all of their momentum to the larger mass.

So the double yo-yo de-spin demonstrates the transfer of momentum to the spheres and then back again to sharing the motion with a mass that is about nine times more massive.

This means that the double yo-yo de-spin: https://www.youtube.com/watch?v=YaUmzekdxTQ  has a ten fold increase in energy twice.

#### Tarsier_79

• Full Member
• Posts: 117
##### Re: re: energy producing experiments
« Reply #318 on: February 23, 2022, 08:50:34 PM »
Your oversimplified math does not prove anything. Prove it with a build and a measurable gain in Potential energy. Measurable input---measurable output. It should be simple for you, you have many models.

ADD: This is a much better experiment. You can measure the speed of the cylinder and calculate the speed of the ball bearings from where they land.

#### Delburt Phend

• Full Member
• Posts: 222
##### Re: re: energy producing experiments
« Reply #319 on: February 23, 2022, 09:43:18 PM »
You apparently are not familiar with the physics. You have no information about the mass of the spinning (space station shaped) shaft or its distribution of mass.  Without that knowledge you cannot predict motion conservation of any type.

The cylinder gives you a known mass at a known radius, for a known momentum. The disk is better than the space station, but not as easy as a cylinder.

#### Tarsier_79

• Full Member
• Posts: 117
##### Re: re: energy producing experiments
« Reply #320 on: February 24, 2022, 08:19:24 AM »
Quote
You apparently are not familiar with the physics. You have no information about the mass of the spinning (space station shaped) shaft or its distribution of mass.  Without that knowledge you cannot predict motion conservation of any type.
Wow, really?...

I was saying this is a more elegant experiment than yours, as is this one: https://www.youtube.com/watch?v=_wsDEOW9DuU&t=0s
Due to the ability to properly measure kinetic energies of the components together, then separately afterwards.

Wild claims require irrefutable proof.

#### Delburt Phend

• Full Member
• Posts: 222
##### Re: re: energy producing experiments
« Reply #321 on: February 24, 2022, 03:35:42 PM »
I don’t know why anyone would call these legitimate experiments wild claims. It is only that you (and others) do not like the data they produce.

This disk throw that you posted is a legitimate experiment; if you know the mass of the disk, and you and also know the distribution of mass, then you can evaluate the momentum of the spheres alone and the momentum of the combination of disk and spheres.

You can eliminate the bearing resistance by letting the unwind occur while the system is falling. I have used mechanical releases and video cams can separate the motion into 1/240th of a second. It is easy to tell what is happening.

#### Tarsier_79

• Full Member
• Posts: 117
##### Re: re: energy producing experiments
« Reply #322 on: February 24, 2022, 07:50:14 PM »
Quote
You can eliminate the bearing resistance by letting the unwind occur while the system is falling. I have used mechanical releases and video cams can separate the motion into 1/240th of a second. It is easy to tell what is happening.
/?
What horizontal movement did the system have? What was the final speed of the yo-yos at the point in time the cylinder stopped?

There are excellent bearings available. Your video is from 2017...If it is such a perfect experiment, why has no-one duplicated it, or from what I can tell taken much notice of it? This experiment will not produce consistently repeatable results. For me, it doesn't prove a gain in energy. If it doesn't prove it for me, it probably isn't proving it for a number of other people.

If you actually find a way to create energy and prove it, the whole world will sit up and listen.

#### Delburt Phend

• Full Member
• Posts: 222
##### Re: re: energy producing experiments
« Reply #323 on: February 24, 2022, 08:43:09 PM »
These problems are on your side of the fence Tarsier; not mine.

#### lancaIV

• elite_member
• Hero Member
• Posts: 5166
##### Re: re: energy producing experiments
« Reply #324 on: February 24, 2022, 09:25:51 PM »
#322 :  " There are excellent bearings available." price/full load rotations ratio
10 Mio full load rotations means much :  30 000 RPM transformer(m. or g./a.) x 60 min = per hour x 24 = per day

as R.O.I. calculateable friction part : exchange costs per annum,by own skills by external service ?
a good source about ultra-long life bearings for recommend ?

Sincere
OCWL

#### Tarsier_79

• Full Member
• Posts: 117
##### Re: re: energy producing experiments
« Reply #325 on: February 25, 2022, 08:35:56 AM »
I don't have any problems. I am not the one claiming to have broken energy conservation with a half baked experiment.

Quote
35 kg dropped one meter can accelerate 140 kg to 2.21 m/sec. Model: Atwood’s Machine
This isn't energy gain. I do not see a video of this. Your atwoods video is not 140 or 35kg, and does not have a scale for the distance traveled over what time.
Quote
A 35 meter high stack of evenly spaced 1 kilogram masses exert the same force as a 35 kilogram mass. Model: scale balance
Balance has very little to do with Potential or kinetic energy.
Quote
After a drop of one meter the stack can be reconfigured by accelerating one kg to 26.24 m/sec, costing 26.24 units of momentum. The one kilogram moving 26.24 m/sec will rise 35 m. Model: kinematic equations.
Sounds like energy conservation in an imaginary world with no resistance from your description.

Quote
140 kg moving 2.21 m/sec equals 310 units of momentum and you only need 26.24 units to recycle the system. Model: kinematic equations.

If the 310 units of momentum is placed in one kg the rise would be 4900 m.  Only 35 of those meters is needed to reconfigure the stack. Model: kinematic equations.
or Naive mathematics...

Quote
The momentum of a large mass can be given to a small mass. Model: cylinder and spheres machine where all the motion is returned to the cylinder; and the Law of Conservation of Momentum.
With a claim of 10x energy and an unconvincing proof.

Quote
The stack of 35 1 kg masses could accelerate a 140 kg rim and stack to 2.21 m/sec and then the rim could throw one kg 140 times higher than the original drop. Model: cylinder and spheres machine; the Law of Conservation of Momentum: kinematic equations.
More naive maths. Show physical proof.

Quote
The double yo-yo despin ‘cylinder and sphere’ transfers all the motion of the “cylinder and sphere” to only the spheres. Then the spheres transfer the motion back to the cylinder and spheres combination.

The conservation of energy could not return the motion of the spheres back to the cylinder. Because as the conservation of momentum would require a 10 m/sec speed for the spheres the conservation of energy would only require a 3.16 m/sec velocity.
Again uninformed mathematics.

Quote
When masses collide, or transfer motion, smaller masses do not and can not give their energy to a larger mass. The smaller masses can and do however give all of their momentum to the larger mass.
Do they? Show proof and how this helps in energy creation.

#### Kator01

• Hero Member
• Posts: 894
##### Re: re: energy producing experiments
« Reply #326 on: March 20, 2022, 02:38:13 AM »
Tarsier,

concerning your formula ( relpy#315):

Have you ever made the effort of reading for what a condition this formula is true ?
It needs sharp eyes to read the bottom right sentence

My impression :  the discussion is - because of a lot of false assumptions and arguments- is rather
exausting and not constructive.

Try first to understand instead of endless arguments and build it...its not so complicated.

@Delburt:
how is this possible that the rotation of the Cylinder is CCW since the steel-ball is hanging at the right of the cylinder and it seems the steel-ball  remains at the same hight in the first moments ? Can you explain it ? Did you give it a ccw-spin before releasing the cylinder ?

Mike

I ask because I am working on a concept of repeating one of the experiments

#### lancaIV

• elite_member
• Hero Member
• Posts: 5166
##### Re: re: energy producing experiments
« Reply #327 on: March 20, 2022, 03:06:03 AM »
#325 :                        " ....  More naive maths. ... "

https://en.wikipedia.org/wiki/Mathematics

"maths" = ALL spectrum ,non naive/naif

dialectical materialism = theory in images and/or words/numbers about object :

material,non dialectical = without words/numbers/image

#315

FORMULA : as AXIOMATIK part right/wrong ?
https://de.wikipedia.org/wiki/Axiom
the basic law/Fixpoint : Anti-/ Hypo-/These
https://de.wikipedia.org/wiki/Kondition_(Mathematik)
https://de.wikipedia.org/wiki/Vorkonditionierung

: not applyable for this case,conditioning !?

Application and physical result ,between

Hub-Kraft

AUFTRIEBSKRAFT,LASTSCHWERKRAFT,"GRAVITY"

https://www.mathematik.uni-muenchen.de/~diening/ws11/numerik/numscript.pdf

and ,point of view :

https://de.wikipedia.org/wiki/Normalkraft

Zug-Kraft
ANTRIEBSKRAFT

https://studyflix.de/ingenieurwissenschaften/zugkraftdiagramm-1171

Force/s their scalar,vectorial,tensorial orientation from source to target

Pressure : https://de.wikipedia.org/wiki/Druck_(Physik)                                                            https://www.linguee.com/german-english/translation/spezifischer+druck.html

Force,specific :

https://www.linguee.com/german-english/translation/spezifische+kraft.html

The      FORMAT        https://de.wikipedia.org/wiki/Form_(Philosophie)

LOGOS         https://en.wikipedia.org/wiki/Tractatus_Logico-Philosophicus

https://de.wikipedia.org/wiki/Logos
ALGORYTHM  https://de.wikipedia.org/wiki/Algorithmus

wmbr
OCWL
« Last Edit: March 20, 2022, 05:42:23 AM by lancaIV »

#### Tarsier_79

• Full Member
• Posts: 117
##### Re: re: energy producing experiments
« Reply #328 on: March 20, 2022, 03:32:29 AM »
Quote
@Delburt:
how is this possible that the rotation of the Cylinder is CCW since the steel-ball is hanging at the right of the cylinder and it seems the steel-ball  remains at the same hight in the first moments ? Can you explain it ? Did you give it a ccw-spin before releasing the cylinder ?

This is a much better experiment IMO. Input can be measured, but it is still difficult to measure the output.

I suspect the Ball hangs in the same spot because the tube is wound up on a piece of fishing line/string, its drop causes the unwind and spin. In the initial moments, the ball is being wound up at the same rate the tube unwinds as it is accelerated by gravity.

Quote
concerning your formula ( relpy#315):

http://hyperphysics.phy-astr.gsu.edu/hbase/hoocyl.html
The formula on the top left is a cylinder rolling without slipping, the formula top right is a hoop rolling without slipping.
The formula at the bottom is either slipping without rolling, the text on the bottom right refers to that bottom formula.

It was in response to:
Quote
Quote
1/2mr² is a claim that the disk has ½ as much resistance to motion as a hoop; and should therefore accelerate down the ramp twice as fast. If the disk is rotating down the ramp twice as fast, then it should cover twice the distance in the same period of time. The disk should reach the end of the ramp when the hoop is only halfway down the ramp.

The experiment shows that the disk is harder to rotate than 1/2mr² predicts.

This is the computer calculated race under perfect conditions in a simulation:
https://en.wikipedia.org/wiki/Moment_of_inertia#/media/File:Rolling_Racers_-_Moment_of_inertia.gif
It closely resembles the outcome of the physical test.
Delberts assumption that the hoop should only be half way down the ramp based on 1/2mrr shows a naive view of the math involved.

Quote
Try first to understand instead of endless arguments and build it...its not so complicated.
1. I am not convinced there is enough evidence to justify a build.
2. My build time is currently allocated elsewhere.
3. If I were to build it, I would be complicating it enough to measure input vs output.

#### Tarsier_79

• Full Member
• Posts: 117
##### Re: re: energy producing experiments
« Reply #329 on: March 20, 2022, 05:37:43 AM »
IancaIV, what point are you trying to make? Let us look at context:

Quote
140 kg moving 2.21 m/sec equals 310 units of momentum and you only need 26.24 units to recycle the system. Model: kinematic equations.
Using online calculators:
https://www.omnicalculator.com/physics/free-fall
https://www.calculatorsoup.com/calculators/physics/gravitational-potential.php
https://www.calculatorsoup.com/calculators/physics/kinetic.php

The 140KG falls from 0.25m to achieve this speed. (343 J)
1kg can rise to 35m at 26.24 kgm/s, which also costs 344 J
There is a difference in momentum, bu where is the energy gain?

Quote
If the 310 units of momentum is placed in one kg the rise would be 4900 m.  Only 35 of those meters is needed to reconfigure the stack. Model: kinematic equations.
So unless I am missing something, it takes 51,200 J to "place" 310 units of momentum on 1kg. How did we magically transfer 310 units of momentum from a 140kg at 2.21m/s mass to a 1kg mass at 310m/s?